Chapter 11
Hydrogen Storage
Problem Solutions
Fund. of Renewable Energy Processes
Prob. Sol. 11.1
Page 1 of 3
395
Prob 11.1 An alloy, Mg2 Ni, interacts with hydrogen so that
the following reversible reaction takes place:
Mg2 Ni + 2H2 ↔ Mg2 NiH4 .
The thermodynamic data for the absorption reaction are (per
kilomole of H2 ):
∆Hf 0 = −64.4 MJ,
∆Sf 0 = −122.3 kJ/K.
The relevant atomic masses (in daltons) are
H
1
Mg
24.3
Ni
58.7
In the questions below, all pressures are in the plateau region.
a. What is the dissociation pressure at 300 K?
.....................................................................................................................
∆H
−122.2 × 103
−64.4 × 106
∆S
+
=−
+
ln p = −
R
RT
8314
8314 × 300
= 14.7 − 25.82 = −11.12
(1)
p = 14.8 × 10−6 atmos.
(2)
Owing to the great stability of the hydride (high energy
of formation), the hydrogen equilibrium
pressure at 300 K is surprisingly low:
15 millionths of an atmosphere.
b. At what temperature is the equilibrium pressure 1 MPa?
.....................................................................................................................
Inverting Equation 2,
T =
T =
∆H
−64.4 × 106
=
R ln p + ∆S
8314 × 10 + 122.2 × 103
−64.4 × 106
∆H
=
= 625
R ln p + ∆S
−122.2 × 103 + 8314 ln 10
(3)
K.
(4)
The equilibrium pressure reaches 1 atmosphere at 625 K.
Solution of Problem
11.1
090723
396
Page 2 of 3
Prob. Sol. 11.1
Fund. of Renewable Energy Processes
c. What are the proportions, by mass, of the elements in Mg2 Ni?
.....................................................................................................................
The molecular mass of the alloy is 2 × 24.3 + 58.7 = 107.3 daltons.
Hence, magnesium represents 2 × 24.3/107.3 = 0.453 or 45.3%, while Ni
represents 58.7/107.3 = 0.547 or 54.7%.
The composition of the alloy (by weight) is
45.3% magnesium and 54.7% nickel.
d. A vessel with an internal volume of 1000 cm3 contains 1.56
kg of Mg2 NiH4 . The density of this material is 2600 kg/m3 .
The saturated hydride is placed in the evacuated vessel. The
temperature is raised to 400 C, but no hydrogen is allowed to
escape. How many kg of hydrogen are desorbed? How many
remain in the hydride?
.....................................................................................................................
1.56 kg of alloy occupy a volume of 1.56/2600 = 600 × 10−6 m3 or 600
3
cm . Thus, the dead space inside the vessel is 400 cm3 .
The pressure at 400 C (673 K) can be calculated the same way as in
Question 1:
ln p = −
∆H
−122.2 × 103
−64.4 × 106
∆S
+
=−
+
= 3.19
R
RT
8314
8314 × 673
p = 24.2 atmos
or
p ≈ 2.4 × 106 Pa.
From the perfect gas law, the amount of H2 in gas form is,
µ=
2.4 × 106 × 400 × 10−6
pV
=
= 170 × 10−6 kmole.
RT
8314 × 673
(5)
(6)
(7)
(8)
This corresponds to 340 mg of H2 . The molecular mass of the hydride is
107.3+4 = 111.3 The hydrogen represents 4/111.3=0.036 of the total mass.
While cold, the amount, MH , of hydrogen in the hydride was 0.036×1.56 =
0.056, or 56,000 mg. Thus, some 0.6% of the hydrogen was desorbed. The
amount remaining is 56,000 − 340 = 55,660 mg.
340 mg of hydrogen were desorbed,
55,700 mg remain in the hydride.
e. How much heat must be added to desorb the rest of the hydrogen? Assume that desorption stops when the empirical
Solution of Problem
090723
11.1
Fund. of Renewable Energy Processes
Prob. Sol. 11.1
Page 3 of 3
397
formula of the material reaches the Mg2 NiH0.4 composition.
The hydrogen desorbed is removed from the vessel so that
the pressure remains constant.
.....................................................................................................................
The amount of hydrogen in the material after desorption represents
10% of the amount in the saturated hydride, or 5600 mg. Thus, the amount
desorbed is 55500 − 5600 = 49900 mg or 0.0499 kg or 0.025 kmoles. The
energy necessary to desorb this hydrogen is
W = 0.025 × 64.4 = 1.6 MJ.
(9)
The energy necessary for “total” desorption is 1.6 MJ.
f. If a hydrogen compressor were built using the vessel above,
and if the hydrogen were fed into the system at 105 Pa and
85 C, what would the theoretical efficiency of the compressor
be? The output gas is at 85 C and 5 MPa. Neglect all losses.
.....................................................................................................................
The efficiency would be
η=
RT ln r
,
∆H
where r is the compression ratio, 50, in the present case.
η=
8314 × (273 + 85) × ln 50
= 0.18.
64.4 × 106
The efficiency is 18%.
Observe, however, that when all losses are included the efficiency may
be as low as some 5%.
Solution of Problem
11.1
090723
398
Page 1 of 2
Prob. Sol. 11.2
Fund. of Renewable Energy Processes
Prob 11.2 Figure 11.8 (text) shows the pressure-composition isotherms for the reaction of hydrogen with LaNi5 . Assume
that, in the plateau region, the isotherms are horizontal—that is,
that the pressure does not depend on the composition. Assume
further that the pressure at 40 C is 0.3 MPa and at 140 C is 5
MPa.
One kg of LaNi5 H5 is placed inside a pressure vessel and
heated to 140 C. Hydrogen is slowly withdrawn until the hydride
is left with the empirical formula LaNi5 H2 .
Molecular masses: La, 138.9; Ni, 58.7.
The ∆H for absorption for this alloy is -32.6 MJ per kilomole
of H2 , and the ∆S is -107.7 kilojoules per kelvin per kilomole.
a. To maintain the temperature at 140 C during the desorption,
must heat be supplied to or withdrawn from the hydride?
.....................................................................................................................
Desorption is endothermic. If heat is not supplied, temperature falls.
To maintain the temperature at a constant 140 C
during desorption, heat must be added to the hydride.
b. How many joules of heat must be exchanged with the hydride
to perform the desorption?
.....................................................................................................................
We must first calculate the density of LaNi5 H5 :
The molecular mass of LaNi5 H5 is 1 × 138.9 + 5 × 58.7 + 5 × 1 =
437.4 daltons. This means that 1 kilomole of the alloy masses 437.4 kg.
Consequently, 1 kg corresponds to 2.29 × 10−3 kilomoles.
When the hydride changes from LaNi5 H5 to LaNi5 H2 , 1.5 kilomoles of
H2 (3 kilomoles of H) are desorbed for each kilomole of hydride. Thus, the
hydrogen desorbed is 2.29 × 10−3 × 1.5 = 3.43 × 10−3 kilomoles of H2 .
The energy required for the desorption is 32.6 × 106 × 3.43 × 10−3 =
112, 000 J.
112 kJ of heat are required to desorb the hydrogen.
c. What is the change in entropy of the system during desorption? Assume that ∆S is temperature independent. Does the
entropy increase or decrease during the desorption?
.....................................................................................................................
During desorption, the hydrogen goes from a highly ordered form in
the crystal to a highly disordered for in the gas—the entropy increases.
The entropy increases is 107.7 × 103 × 3.43 × 10−3 = 369 J K−1 .
Solution of Problem
090723
11.2
Fund. of Renewable Energy Processes
Prob. Sol. 11.2
Page 2 of 2
399
Entropy change due to desorption is 369 J K−1 .
d. How many kilograms of hydrogen were withdrawn?
.....................................................................................................................
From Answer 2, we see that 3.43 × 10−3 kilomoles of H2 or 6.86 × 10−3
kg of H2 were desorbed.
6.86 grams of hydrogen were withdrawn.
Solution of Problem
11.2
090723
400
Page 1 of 1
Prob. Sol. 11.3
Fund. of Renewable Energy Processes
Prob 11.3 Hydrogen is to be transported. Two solutions
must be considered:
1. Liquefy it.
2. Convert it to ammonia. Later, when hydrogen is to be used,
the ammonia can be catalytically cracked with 85% recovery
of hydrogen. 46.2 MJ per kg of ammonia are required to
dissociate the gas.
Purely from the energy point of view, which is the more economical solution?
From the Textbook (Section 11.2), it can be seen that liquefying hydrogen costs about 40 MJ/kg or 80 MJ/kmole. To obtain liquid hydrogen
we invest 286 MJ/kmole to obtain the gas and 80 MJ/kmole to liquefy it.
We recover only 286 MJ/kmole when using the liquid. Thus, the process
efficiency is
286
= 0.78.
ηcryo =
286 + 80
In the ammonia method, we produce 2/3 kilomole of NH3 for each
kilomole of H2 . The reaction that produces ammonia is exothermic but the
heat produced cannot be stored.
To recover the hydrogen, we must dissociate the ammonia into its
elements and this takes 46.2 MJ per kilomole of ammonia or 2/3 × 46.2 =
30.8 MJ per kilomole of hydrogen produced. Of this, only 85% of the gas
is saved. The efficiency of the process is
ηammonia =
286 × 0.85
= 0.77.
286 + 30.8
From the energy point of view, the two processes are about equal.
Solution of Problem
090723
11.3
Fund. of Renewable Energy Processes
Prob. Sol. 11.4
Page 1 of 1
401
Prob 11.4 Calorimeter measurements show that when a
compound, AB, reacts with H2 forming a hydride ABH, 18.7
MJ of heat are released for each kilomole of H2 absorbed. The
manufacturer of this hydride consults you about the advisability of shipping it (saturated with hydrogen) inside a normal gas
pressure cylinder (which is rated at 200 atmospheres). During
shipment the cylinder may be exposed to the sun. Without having any additional data on the hydride, what would you advise
the manufacturer? Could you tell him the maximum temperature
that the hydride can safely reach?
.....................................................................................................................
You must be cautious in your advise to the manufacturer. Since you do
not know the entropy change owing to hydration, you could make a guess
at ∆S = −100 kJ per kelvin per kilomole of hydrogen. Then, the hydride
would reach the limiting pressure of 200 atmospheres at a temperature that
satisfies
RT ln p = ∆H − T ∆S
from which
T =
∆H
−18.7 × 106
=
= 334
R ln p + ∆S
8314 ln 200 − 100, 000
K.
This temperature (61 C) is uncomfortably low. Moreover, the estimate
of the entropy change may be wrong: if the instead of -100 kJ K−1 kmole−1 ,
the value is -110 K−1 kmole−1 , then the maximum permissible temperature
is a cool 284 K (about 10 C). Recommendation: do not transport hydrogen
this way. You can use a more stable hydride, one with higher ∆H (in
absolute value).
Solution of Problem
11.4
090723
402
Page 1 of 5
Prob. Sol. 11.5
Fund. of Renewable Energy Processes
Prob 11.5
a. Estimate the enthalpy change for the hydrogen absorption reaction of an alloy that has a plateau pressure of 1 atmosphere
when T = 0 C.
.....................................................................................................................
For all alloys, the temperature at which the plateau pressure is 1 atmosphere is given (roughly) by
T = 10∆H,
where ∆H is expressed in megajoules. Consequently,
∆H = T /10.
For T = 0 C or 273 K, ∆H ≈ −27.3 MJ/kmole of H2 . The negative
sign results from the absorption being exothermic.
By rough estimate, the enthalpy change owing to the absorption
of hydrogen in Alloy A is about -27.3 MJ/kmole of hydrogen.
b. Estimate the enthalpy change for the hydrogen absorption reaction of an alloy that has a plateau pressure of 1 atmosphere
when T = 30 C.
.....................................................................................................................
For T = 30 C, i.e., T = 303 K, ∆H ≈ −30.3 MJ/kmole of H2 .
By rough estimate, the enthalpy change owing to the absorption
of hydrogen in Alloy B is about -30.3 MJ/kmole of hydrogen.
Before proceeding with the rest of the solution, it is useful to tabulate
the different plateau pressures that will occur.
The plateau pressure of either alloy is
∆S
∆H
peq = exp −
atmos.
+
R
RT
Using the two values of ∆H calculated above and our guess of −100 kJ
K−1 kmoles−1 for the entropy change during absorption, we can construct
the table below:
Alloy
A
A
B
B
TEMP.
TEMP.
(C)
(K)
PLATEAU.
PRESS.
(atmos.)
0
28
28
100
273
301
301
373
1.0
3.1
0.9
9.6
Solution of Problem
090723
11.5
Fund. of Renewable Energy Processes
Prob. Sol. 11.5
Page 2 of 5
403
Assume that Alloy A is in an environment that causes its
temperature to be 0 C (perhaps the outside of a house) and that
Alloy B is at 28 C (say, inside a room). The alloy containers are
interconnected by means of a pipe so that the hydrogen pressure
is the same in both. The amount of hydrogen in the system is
such that, in any phase of the cycle, when one alloy is saturated,
the other is depleted.
c. What is the system pressure?
.....................................................................................................................
Hydride B
Hydride A
9.6 atmos
10
100 C
5
3.1 atmos
A2
B2
28 C
2
1
A1
1.0 atmos
0C
0.9 atmos
28 C
B1
0.5
0.2
0.1
At this phase of the cycle, the conditions are as follows:
TA = 273 K,
pA = 1.0 atmos,
TB = 301 K,
pB = 0.9 atmos.
Assume that initially the system contains no hydrogen and that a
third container is used to load the gas. The pressure rises until it reaches
0.9 atmos where it stabilizes as Alloy B starts absorbing the gas. When
saturated, the pressure once again starts up. However, the problem specifies
that when one alloy is saturated the other must remain depleted. Thus,
loading must be stopped an the gas inside the system will be at a pressure
larger than 0.9 atmos (to saturate B) but less than 1.0 atmosphere (to keep
A depleted). The alloys will be at points A1 and B1.
The system pressure will be between 0.9 and 1.0 atmos.
Solution of Problem
11.5
090723
404
Page 3 of 5
Prob. Sol. 11.5
Fund. of Renewable Energy Processes
d. Which alloy is depleted and which contains most of the hydrogen?
.....................................................................................................................
Alloy A is depleted while Alloy B is saturated.
Raise the temperature of Alloy A to 28 C.
e. What is the plateau pressure? What is the actual hydrogen
pressure?
.....................................................................................................................
The plateau pressure of Alloy A is now raised to 3.1 atmos.
However, the alloy was depleted, so, raising the plateau pressure only
accentuates this condition. Actually, the pressure might rise slightly because the hydrogen is warmed up a bit.
The hydrogen pressure remains essentially unchanged.
Circulate hot water (from a hot source) through the heat
exchanger of Alloy B and heat it up to 100 C (373 K). Maintain
this temperature.
f. What is the pressure of the system and what happens to the
two alloys?
.....................................................................................................................
At this new phase of the cycle, the conditions are now:
TA = 301 K,
pA = 3.1 atmos,
TB = 373 K,
pB = 9.6 atmos.
Hydrogen is desorbed from B and absorbed by A. Gas flows from B
to A. When the process is finished, the gas will be at a pressure somewhat
above 3.1 atmos and well below 9.6 atmos. The exact pressure depends on
how much gas is in the system.
The system pressure will be between 3.1 and 9.6 atmos.
Alloy A will be saturated and Alloy B will be depleted.
Return the temperature of Alloy B to 28 C, and that of Alloy
A to 0 C (by putting it in contact with the exterior).
g. What happens?
.....................................................................................................................
Solution of Problem
090723
11.5
Fund. of Renewable Energy Processes
Prob. Sol. 11.5
Page 4 of 5
405
This re-establishes the initial conditions in which Alloy A was depleted
and B was saturated. Thus desorption occurs in A and absorption, in B.
Hydrogen flows from A to B.
Alloy A will desorb, becoming depleted.
Alloy B will absorb becoming saturated.
h. Tabulate all the sorption heat inputs and outputs of the system and the corresponding temperatures. Ignore all other
heat inputs and outputs. For example, ignore the heat necessary to change the temperature of the system.
.....................................................................................................................
All heat quantities are per kilomole of hydrogen.
In Step 6, Alloy B is heated to 373 K. As soon as the rising temperature
exceeds 334 K (the temperature that causes its plateau pressure to exceed
the 3.1 atmospheres of Alloy A), Alloy B starts desorbing and draws 30.3
MJ from the hot source. This energy drives the heat pump. The hydrogen
is absorbed by Alloy A (at 301 K) and 27.3 MJ are rejected into the room.
In Step 7, Alloy B is returned to 301 K by stopping the hot water
circulation. The temperature falls toward 301 K while that of Alloy A falls
toward 273 K. As soon as the plateau pressure of Alloy A exceeds that of
Alloy B, the latter starts absorbing hydrogen and delivers 30.3 MJ of heat
to the room. Alloy A, at 273 K, desorbs hydrogen and draws 27.3 MJ from
the environment outside the house.
i. How much heat is delivered (per kmole of hydrogen) to the
environment at 28 C in a complete cycle as described above.?
.....................................................................................................................
The room receives 30.3 MJ from Alloy B and 27.3 MJ from Alloy A,
a total of 57.6 MJ (all per kilomole of H2 ).
The room receives 57.6 MJ of heat for
each kilomole of hydrogen handled.
j. Of this amount of heat, how much must come from the hot
water source (and must be paid for)?
.....................................................................................................................
The hot water source has to furnish 30.3 MJ for
each kilomole of hydrogen handled.
Solution of Problem
11.5
090723
406
Page 5 of 5
Prob. Sol. 11.5
Fund. of Renewable Energy Processes
k. What is the coefficient of performance of this heat pump—
that is, what is the ratio of the heat delivered to the environment at 28 C to the heat required from the hot water?
.....................................................................................................................
57.6
COP =
= 1.90.
30.3
The coefficient of performance is 1.90.
Solution of Problem
090723
11.5
Fund. of Renewable Energy Processes
Prob. Sol. 11.6
Page 1 of 4
407
Prob 11.6 A recently developed binary compound is to be
used as a hydrogen storage medium because it forms a reversible
(ternary) monohydride.
The data of the system include:
Molecular mass of the compound (no hydrogen): 88 daltons.
Density of the compound (hydrided or depleted): 8900 kg
m−3 .
Enthalpy of hydriding: −26.9 MJ per kilomole of H2 .
Change of entropy owing to absorption of hydrogen: −100 kJ
K−1 per kilomole of H2 .
Type of isotherm: single plateau.
Heat capacity of the compound: 200 J K−1 kg−1 .
Heat capacity of the container and of the hydrogen gas: negligible.
Heat capacity of water: 4180 J K −1 kg−1 .
Density of hydrogen: 0.089 kg m−3 .
2.5 kg of the compound (activated and in a fine powder at
a temperature of 0 C) are placed inside a container measuring
internally 10 by 10 by 10 cm.
We want to adjust the system so that the alloy is saturated
and the gas pressure is exactly the plateau pressure. To find
this point we will have to observe the behavior of the pressure
in the alloy canister as hydrogen is added. The pressure initially
rises, then as the plateau is reached, it stabilizes until saturation is
achieved. We will first overshoot this point, then, having removed
the hydrogen source, we will purge some of the gas, observing
the pressure and stopping the purge just when the pressure falls
to the previously observed plateau pressure. During this whole
operation the temperature is carefully maintained at 0 C.
A 5-atmosphere hydrogen source is used to fill the container.
The gas is applied for a time long enough to allow equilibrium to
be established. Any heat absorbed or released by this operation
is removed or added to the system so that at the end of the
operation, the pressure is still 5 atmospheres and the temperature
is still 0 C. Next the hydrogen source is removed. The pressure
inside the container remains at 5 atmospheres. A valve is cracked
open and hydrogen leaks out while the pressure is monitored. As
soon as the pressure stabilizes, the valve is closed.
a. How many grams of hydrogen were lost in the bleeding above?
.....................................................................................................................
Solution of Problem
11.6
090723
408
Page 2 of 4
Prob. Sol. 11.6
Fund. of Renewable Energy Processes
The volume occupied by the compound is
Vcompound =
2.5
= 2.8 × 10−4
8900
m3 .
The volume of the container is 0.001 m3 . Thus the “dead space” is
720 × 10−6 m3 . This is the volume occupied by the gaseous hydrogen.
The equilibrium pressure is
∆H
∆S
+
p = exp −
R
RT
−100 × 103
−26.9 × 106
= exp −
+
= 1.21
8314
8314 × 273.2
atmos.
Hydrogen will be bled until its pressure falls from 5 to 1.21 atmospheres
(5.06 × 105 to 1.23 × 105 pascals). At the higher pressure the hydrogen in
gaseous for corresponded to
µ=
5.06 × 105 × 720 × 10−6
pV
=
= 160 × 10−6
RT
8314 × 273.2
kilomoles
At the lower pressure, the hydrogen in the dead space is 39 × 10−6
kilomoles.
Thus, 160 × 10−6 − 39 × 10−6 = 121 × 10−6 kilomoles of H2 (242 × 10−6
kg) were bled off.
242 milligrams of hydrogen were lost in the bleeding.
b. How much hydrogen remains in storage? Express the loss as
a percentage of the stored gas.
.....................................................................................................................
Assuming that the alloy is fully hydrided, there is one kilomole of H
for each kilomole of alloy.
One kilomole of the alloy masses 88 kg. Hence, 2.5 kg of alloy corresponds to 2.5/88 = 0.0284 kilomoles of alloy and, consequently, 0.0284
kilomoles of H (28.4 × 10−3 kg of H2 ). This is the amount of hydrogen in
the storage after the bleeding. The amount of hydrogen lost in bleeding is
0.242 g.
28.4 grams of hydrogen remain in storage after the bleeding.
The bleeding caused a loss of 0.85% of the stored gas.
Solution of Problem
090723
11.6
Fund. of Renewable Energy Processes
Prob. Sol. 11.6
Page 3 of 4
409
c. What is the pressure of the stored gas?
.....................................................................................................................
The gas remaining in storage it at a pressure of 1.21 atmospheres.
The container is now immersed in a tank of water at 40 C.
This tank contains 0.3 liters of water and is entirely adiabatic.
Its walls have negligible heat storage capacity.
d. What is the temperature of the system (water tank plus alloy
container) after equilibrium is reached?
.....................................................................................................................
Let T be the equilibrium temperature.
As the new temperature, T , is established, the water cooled down by
40 + 273.2 − T = 313.2 − T K. This means that an amount, Qwater =
4180 × 0.3 × (313.2 − T ) = 392, 800 − 1, 254T joules, of heat energy was
transferred to the hydride.
The heat removed from the water heats up the hydride (Qhydride ) and
causes some hydrogen to be desorbed (Qdesorb ):
Qhydride = chydride Mhydride ∆T
J,
where
chydride = heat capacity of the hydride = 200 J/K per kilogram, and
Mhydride = 2.5 kg.
Thus, Qhydride = 500(T − 273.2) = 500T − 136, 600 J.
Qdesorb = µdesorb × 26.9 × 106
J.
The heat balance requires
392, 800 − 1, 254T = 500T − 136, 600) + µdesorb × 26.9 × 106 ,
1754T − 529, 400 + µdesorb × 26.9 × 106 = 0.
The amount of hydrogen in gaseous form is the sum of the amount
that was present at 0 C plus the amount desorbed:
µ = 39 × 10−6 + µdesorbed =
pV
.
RT
Hence,
pV
− 39 × 10−6 ,
RT
pV
6
−6
1754T − 529, 400 + 26.9 × 10
= 0,
− 39 × 10
RT
µdesorb =
Solution of Problem
11.6
090723
410
Page 4 of 4
Prob. Sol. 11.6
Fund. of Renewable Energy Processes
pV
= 0,
RT
p
1754T − 530, 449 + 2.33
= 0.
T
At T , the equilibrium pressure of the hydrogen is
1754T − 530, 449 + 26.9 × 106
∆H
∆S
+
p = exp −
R
RT
−100 × 103
3236
−26.9 × 106
= exp −
= exp 12.03 −
+
8314
8314 × T
T
3236
3236
= 167, 700 exp −
atmos or 17 × 109 exp −
T
T
Pa.
3236
1754T 2 − 530, 450T + 39.6 × 109 exp −
= 0.
T
A numerical solution leads to T = 300.8 K.
The equilibrium temperature is 300.8 K.
Solution of Problem
090723
11.6
Fund. of Renewable Energy Processes
Prob. Sol. 11.7
Page 1 of 3
411
Hydride
B
Hydride
A
Prob 11.7 An inventor proposes the following device to cool
drinks at a picnic. It consists of two sturdy containers (something like small portable oxygen bottles) one of which (Container
A) can be placed inside a styrofoam box in which there are 12
beer cans. The other (Container B) is outside the box and can
be placed over a fire. A pipe interconnects the two containers.
See the figure. The containers are filled with different alloys capable of absorbing hydrogen. Alloy A is TiFe. Alloy B has to be
described.
When both containers are at the same temperature, 298 K,
Alloy A is depleted and B is saturated.
a. For this to happen, what are the required characteristics of
Alloy B? Establish a relationship between the thermodynamic
parameters of the two alloys.
.....................................................................................................................
If, when both alloys are at the same temperature, alloy A is depleted
and B is saturated, then the plateau pressure of the former must be higher
than that of the latter.
ln pA > ln pB ,
−
∆HA
∆SB
∆HB
∆SA
+
>−
+
,
R
RT
R
RT
∆HB − T ∆SB < ∆HA − T ∆SA
∆GB < ∆GA
The free energy of desorption of Alloy B must
be smaller than that of Alloy A.
The various ∆H and ∆S, above are for absorption. This means that
both ∆H and ∆S are smaller than zero.
b. Which of the alloys from the table below can be used as Alloy
B in the device under consideration?
Solution of Problem
11.7
090723
412
Page 2 of 3
Prob. Sol. 11.7
Hydride
Fund. of Renewable Energy Processes
∆H
MJ/kmole
∆S
kJ/K
per kmole
AB
-21.0
-96.5
CD
-26.1
-99.4
EF
-27.9
-106.8
GH
-32.1
-101.8
IJ
-32.6
-110.5
KL
-33.4
-98.3
.....................................................................................................................
To answer this question, one must calculate the plateau pressure of the
different hydrides at T = 298. This is done by applying the formula
∆S
∆H
p = exp −
.
+
R
RT
The result is
Hydride
MJ/kmole
AB
CD
EF
GH
IJ
KL
∆H
kJ/K per kmole
-21
-26.1
-27.9
-32.1
-32.6
-33.4
∆S
atmos.
-96.5
-99.4
-106.8
-101.8
-110.5
-98.3
p
22.9
4.1
4.9
0.5
1.1
0.2
Alloy EF is the TiFe of Container A. The only alloy, other than EF,
that cannot be used is AB whose plateau pressure is higher than that of
EF.
.....................................................................................................................
To operate the system, Container A is placed in a bucket
of water and kept at 25 C. This requires refreshing the water
occasionally. Container B is placed over the picnic fire so that
hydrogen is transferred to Container A whose hydride becomes
saturated.
Next, container A is placed inside the styrofoam box, in contact with the 355-ml beer cans which, for good thermal contact,
are immersed in 4.5 liters of water.
Container B is now cooled to 298 K returning the system to
its original state.
The heat required to desorb the hydrogen from A, will cool
the 12 cans of beer from 25 C to 10 C.
Assume that beer behaves as water, at least as far its thermal capacity is concerned. The styrofoam box is, essentially, adi-
Solution of Problem
090723
11.7
Fund. of Renewable Energy Processes
Prob. Sol. 11.7
Page 3 of 3
413
abatic. Assume that during the cycle the composition of the
hydride in A varies from TiFeH0.95 to TiFeH0.4
The atomic mass of Ti is 47.9 daltons and that of Fe is 55.8
daltons.
c. Estimate the minimum mass of TiFe required.
.....................................................................................................................
Inside the styrofoam box, the volume of liquid to be cooled is 4.5 +
12 × 0.355 = 8.76 liters or 0.00876 m3 . The heat capacity of this liquid
(essentially, all water) is 4.2 MJ/K per cubic meter. Thus the heat to be
removed is
Wcooling = 4.2 × 106 × 0.00876 × (25 − 10) = 550, 000
J.
When 1 kmole of H2 is desorbed, it removes 27.9 MJ of heat. To
remove 550,000 J one needs 0.0197 kmoles of hydrogen.
Each kilomole of alloy will desorb (0.95 − 0.4)/2 = 0.275 kilomoles of
H2 . For the amount of hydrogen to be handled, we need 0.0197/0.275 =
0.0716 kilomoles of alloy.
The atomic mass of the alloy is 47.9 + 55.8 = 103.7 daltons. 0.0716
kilomoles of alloy mass 0.0716 × 103.7 = 7.42 kg.
7.4 kg of TiFe are needed.
Considering that the mass of the alloy in Canister B must be of the
same order as that in Canister A and considering also the mass of canisters
and pipe, the whole system is a bit too heavy to become a popular pick
nick beer cooler.
Solution of Problem
11.7
090723
414
Page 1 of 1
Prob. Sol. 11.8
Fund. of Renewable Energy Processes
Prob 11.8 Two canisters are interconnected by a pipe.
Canister “A” contains TiFe and is at 300 K, while Canister “B”
contains CaNi5 and is at 350 K. The system is filled with hydrogen
at a pressure of 4 atmospheres.
In which canister is the bulk of the hydrogen? No guesses,
please! Use the thermodynamic data of Table 11.4 of the Text.
.....................................................................................................................
The plateau pressure of hydrogen in contact with a hydride is
ln p = −
∆S
∆H
+
.
R
RT
For Canister “A”,
ln pA = −
−106, 100 −28.0 × 106
+
= 12.76 − 11.23 = 1.534,
8314
8314 × 300
pA = 4.64
atmospheres.
Thus the hydrogen above the TiFe is a lower pressure than the plateau
pressure. Consequently the alloy is depleted.
For Canister “B”,
ln pB = −
−101, 200 −31.8 × 106
+
= 12.17 − 10.93 = 1.240,
8314
8314 × 350
pB = 3.46
atmospheres.
Thus the hydrogen above the CaNi5 is at higher pressure than the
equilibrium pressure. Consequently the hydride is saturated.
In all probability, most of the hydrogen is in Canister ˝B˝
Solution of Problem
090723
11.8
Fund. of Renewable Energy Processes
Prob. Sol. 11.9
Page 1 of 2
415
Prob 11.9 WARNING: This problem contains units, such as
grams, centimeters, and so on, that are not of the SI.
ln p
x = 0.1
0
x = 0.9
Stoichiometric index, x
1
Consider a vessel containing a hydrogen-absorbing alloy, AB.
Vessel:
Volume: 200 cm3 ,
Thermal insulation: adiabatic,
Heat capacity: negligible.
Alloy:
Formula mass: 120 daltons,
Amount: 200 g,
Heat capacity: 1700 J/K per kg of alloy (same for hydrided
alloy).
Hydride:
Heat of formation of ABH (absorption): −30.0 MJ per kmole
of H2 ,
Entropy change owing to absorption: −110 kJ/K per kmole
of H2 ,
Density (of ABH0.9 ): 1600 kg/m3 .
System:
Initial temperature: 300 K,
Initial hydrogen pressure: equilibrium.
Hydrogen is forced into the vessel until the average alloy composition is ABH0.9 What is the minimum pressure required to
force all the needed hydrogen into the vessel? How much hydrogen is forced in?
.....................................................................................................................
Let us start by calculating the plateau pressure, p300 .
p300
∆S
∆H
= exp −
+
R
RT
−110, 000
−30 × 106
= exp −
= 3.33
+
8314
8314 × 300
Solution of Problem
atmos.
11.9
090723
416
Page 2 of 2
Prob. Sol. 11.9
Fund. of Renewable Energy Processes
Next, let us determine the number, N , of kmoles of alloy in the vessel.
N = 0.2 kg ×
1 kmole
= 0.00167
120 kg
kmole.
The amount of hydrogen (H) absorbed is 0.00167∆x, where ∆x is the
change in the stoichiometric index, In this case, the change is 0.8. Thus,
1336 × 10−6 kilomoles of H or 668 × 10−6 kilomoles of H2 are absorbed.
This causes a release of Q = 668 × 10−6 × 30 = 0.02 MJ of heat in the
vessel.
Since the specific heat capacity of the alloy is 1700 J/K per kilogram,
its total heat capacity is c = 0.2 × 1700 = 340 J/K (for the amount of alloy
in the vessel).
This causes a temperature rise of
∆T =
Q
0.02 × 106
=
= 59
c
340
K.
The temperature reaches 300 + 59 = 359 K. The pressure is now
−110, 000
−30 × 106
= 27.1
p418 = exp −
+
8314
8314 × 359
atmos.
The minimum hydrogen feed pressure is 27.1 atmos.
The volume occupied by the alloy is
0.2 kg ×
1 m3
= 125 × 10−6
1600 kg
m3 .
This is 125 cm3 . Since the vessel has a capacity of 200 cm3 , there is
an “dead” space of 75 cm3 .
At 27.1 atmospheres (2.75 MPa), the amount of gas in this space is
µ=
pV
2.75 × 106 × 75 × 10−6
=
= 69 × 10−6
RT
8314 × 359
kmoles.
The amount of hydrogen pumped in must be 668 × 10−6 kmoles absorbed plus 69 × 10−6 kilomoles of compressed gas, a total of 737 × 10−6
kilomoles.
The amount of hydrogen pumped into the vessel is 0.737 mole.
Solution of Problem
090723
11.9
Fund. of Renewable Energy Processes
Prob. Sol. 11.10
Page 1 of 2
417
Prob 11.10 A perfectly adiabatic (heat insulated) vessel has
an internal volume of 100 cm3 and contains 240 g of an alloy powder, AB, that forms a monohydride, ABH. The thermodynamic
data for absorption are:
∆H = −28 MJ per kilomole of H2 ,
∆S = −100 kJ/K per kilomole of H2 ,
Heat capacity, cv = 400 J/K per kg of alloy.
Additional data include
Formula mass of the alloy = 150 daltons,
Density of the alloy = 8000 kg/m3 ,
Bulk density of the alloy powder = 4000 kg/m3 .
The vessel has been charged with hydrogen so that the pressure is 10 atmospheres. The system is at 30 C.
a. Hydrogen is withdrawn. How many milligrams of the gas can
be removed without causing a change in the temperature of
the hydride? Remember that the container is adiabatic. The
heat owing to the work that the withdrawn hydrogen may
exert is exchanged with the gas outside the hydride container
and does not influence the temperature of the latter.
.....................................................................................................................
If the system is at 30 C (303 K), then the plateau pressure is
ln p = −
∆H
100 × 103
28 × 106
∆S
+
=
−
= 0.913.
R
RT
8314
8314 × 303
(1)
This means that the plateau pressure is 2.49 atmospheres, much lower
than the actual gas pressure. We cab release hydrogen until the pressure
reaches the plateau value without causing any appreciable desorption.
1 m3
−6
The volume occupied by the alloy itself is 0.24 kg × 8000
kg = 30×10
3
3
3
m or 30 cm . Hence, the dead volume inside the vessel is 70 cm .
The amount of hydrogen in this dead volume is
µ=
106 × 70 × 10−6
pV
=
= 27.8 × 10−6
RT
8314 × 303
kmole.
(2)
When the pressure is down to 2.49 atmospheres (about 2.49 × 105
pascals), the amount of hydrogen left in the container is
µ=
pV
2.49 × 105 × 70 × 10−6
=
= 6.92 × 10−6
RT
8314 × 303
kmole.
(3)
Thus, 20.9 × 10−6 kilomoles of hydrogen or 41.7 mg can be removed.
Solution of Problem
11.10
090723
418
Page 2 of 2
Prob. Sol. 11.10
Fund. of Renewable Energy Processes
A total of 41.7 mg of hydrogen can be removed
without changing the temperature of the hydride.
b. If more hydrogen is released, it will cause the cooling of the
hydride. Assume that the vessel has no heat capacity. How
much hydrogen is released if the pressure falls to 1 atmosphere?
.....................................................................................................................
The plateau pressure of 1 atmosphere corresponds to a temperature,
T1 , of
100 × 103
28 × 106
ln 1 =
= 0.
(4)
−
8314
8314T1
From this, T1 = 280 K.
Thus hydrogen will be released until the hydride reaches this temperature. Since the only source of heat is that stored in the hydride itself, the
latter must loose an amount of heat
∆H ′ = 400
J
× 0.240 kg × (303 − 280) K = 2210
K kg
J.
(5)
So, the amount of heat available for desorption is 2210 J. This is enough
to desorb 2210/28 × 106 = 79 × 10−6 kilomoles of H2 or 158 mg of the gas.
158 mg of hydrogen were released.
c. What is the value of x in the empirical formula ABHx after
the above desorption?
.....................................................................................................................
The amount of alloy in the container is
µ′ =
0.240 kg
= 0.0016
150 kg per kmole
kmole.
(6)
For each kilomole of the alloy AB, there is half a kilomole of H2 at
the beginning of the desorption. Thus the amount of H2 in the hydride
was 800 × 10−6 kmole. At the end of the desorption, it was 800 × 10−6 −
79 × 10−6 = 721 × 10−6 kilomoles. Thus x = 721/800 = 0.90. Only a small
fraction of the stored hydrogen has escaped.
After the release, x =0.90.
Solution of Problem
090723
11.10
Fund. of Renewable Energy Processes
Prob. Sol. 11.11
Page 1 of 2
419
Prob 11.11 The Pons and Fleishman cold fusion experiment
employs an electrolytic cell consisting of a palladium negative
electrode and a platinum positive electrode. The electrolyte is a
concentrated solution of LiOH in D2 0. The palladium electrode
is a cylindrical rod 10 cm long and 1.2 cm in diameter. Just prior
to the experiment, the rod is completely degassed by heating it
up in a vacuum.
When a 0.5 A current is forced through the cell, nothing
unusual happens for a long time. To be sure, normal electrolysis
occurs with D2 evolving at the palladium and 02 at the oxygen
electrodes. The D2 O used up is continually replenished.
In some rare instances, it is claimed, after the electrolysis has
proceeded for a long time, heat suddenly begins to be produced
in substantial amounts—73 W, in this case. This heat production
rate is sustained for 120 hours after which the cell is disconnected.
If you look up the enthalpies of formation of all palladium
compounds, you will find that the largest value is associated with
the formation of palladium hydroxide: 706 MJ/kmole.
You will also find that the atomic mass of palladium is 106
daltons and that the density of the metal is 12 g cm−3 .
Can you prove that the energy generated is not chemical in
nature?
To explain the delay, assume that D2 -D2 fusion occurs at a
rapid rate only if the deuterium is packed with sufficient density and that this will happen only when the palladium is completely saturated with deuterium and the formation of palladium
di-deuteride begins. How long would you expect the cell to operate before it heats up?
.....................................................................................................................
The volume of the palladium electrode is
V =
πd2
π × 1.22
×L=
× 10 = 11.3
4
4
cm3 ,
where d is the diameter and L is the length of the electrode.
The mass of the electrode is
M = V δ = 11.3 × 12 = 136
g.
Here, δ is the density in g/cm3 .
106 kg of palladium correspond to 1 kilomole, hence, the number of
kilomoles of palladium in the electrode is
N=
0.136
= 1.28 × 10−3
106
Solution of Problem
kmoles (Pd).
11.11
090723
420
Page 2 of 2
Prob. Sol. 11.11
Fund. of Renewable Energy Processes
If all the palladium in the electrode reacted chemically to produce
palladium hydroxide (releasing 706 MJ/kmole), the released energy would
ne
Wchem = 706 × 106 × 1.28 × 10−3 = 906, 000 J.
This is about 1 MJ.
However, the claim is that a total of 73 W were released over 120 hours.
This corresponds to an energy
J
s
W experiment = 73 × 120 hr × 3600 = 31.5 × 106
s
hr
J.
Not only is the claimed energy release much larger than that you could
expect from a chemical reaction, but, also, it was stated that there was no
obvious corrosion of the electrode, in other words, the palladium did not
react chemically.
A 0.5 A current corresponds to a flow of
0.5
electrons
electrons
coulombs
×
= 3.1 × 1018
,
s
1.60 × 10−19 coulombs
s
Each electron corresponds to 1 deuterium atom, hence, the electrolysis
liberated 3.1 × 1018 deuterium atoms.
The total number of palladium atoms in the electrode is 1.28 × 10−3 ×
6.02 × 10−26 = 770 × 1021
Assuming that all the liberated D2 was absorbed by the palladium, it
would take 770×1021/3.1×1018 = 247, 000 seconds (68 hours) to transform
all the Pd into PdH.
If the fusion reaction would be initiated by the formation of PdH2 ,
then you would expect a delay of some 68 hours between the beginning of
the experiment and the beginning of massive heat production.
Solution of Problem
090723
11.11
Fund. of Renewable Energy Processes
Prob. Sol. 11.12
Page 1 of 6
421
Prob 11.12 A canister contains a mixture of two alloys (Alloy 1 and Alloy 2). A hydrogen source equipped with a valve is
connected to this canister. A measured amount of the gas can be
delivered to it.
Describe the behavior of psystem (the hydrogen pressure, in
pascals, read by a manometer connected to the canister) versus
the amount, µH2 of H2 (in moles) introduced into the system.
Sketch a rough psystem vs µH2 graph.
Estimate all the break points in the above sketch—that is, all
the values of µH2 at which the curve changes abruptly its character.
Do the above for T=400 K, following the detailed instructions
in Items 1 through 5.
Here are some data:
Internal volume of the canister: 1 liter.
The idealized p vs x characteristics of the alloys are as follows:
Region 1 – The equilibrium pressure is proportional to the
stoichiometric coefficient, x, for 0 < p < pplateau . When x = xcritℓ , p
reaches pplateau .
Region 2 – The plateau pressure is perfectly constant until x
reaches a critical value, xcritu .
Region 3 – For x > xcritu , the pressure rises linearly with x
with the same slope as that of Region 1.
Alloy 1: 0.8 kg of alloy AB having a density of 2750 kg/m3 ,
∆S = −110 kJ K−1 kmole−1 , ∆H = −35 MJ kmole−1 .
At 400 K, xcritℓ1 (the minimum value of x in the plateau region) is 0.3, and xcritu1 (the maximum value of x in the plateau
region) is 3.55.
Alloy 2: 1.0 kg of alloy CD having a density of 2750 kg/m3 ,
∆S = −90 kJ K−1 kmole−1 , ∆H = −28 MJ kmole−1 .
At 400 K, xcritℓ2 (the minimum value of x in the plateau region) is 0.3, and xcritu2 (the maximum value of x in the plateau
region) is 4.85.
The formula masses are
A – 48 daltons,
B – 59 daltons,
C – 139 daltons,
D – 300 daltons.
The density of the above alloys is (unrealistically) the same
whether hydrided or not.
Define “gas-space” as the space inside the canister not occupied by the alloys.
Solution of Problem
11.12
090723
422
Page 2 of 6
Prob. Sol. 11.12
Fund. of Renewable Energy Processes
a. Calculate the volume, Vgas−space of the gas space.
.....................................................................................................................
Vcanister = 0.001 m3 .
Valloy
1
=
0.8 kg
= 0.000291 m3 .
2750 kg per m3
Valloy
2
=
1.0 kg
= 0.000364 m3 .
2750 kg per m3
Vgas−space = 0.001 − 0.000291 − 0.000364 = 0.000345 m3 .
The volume of the gas-space inside the canister is 345 ml.
b. What is the number of kilomoles of each alloy contained in
the canister?
.....................................................................................................................
The molecular mass of Alloy 1 is 48 + 59 = 107 daltons; that of Alloy
2 is 139 + 300 = 439 daltons.
For Alloy 1, with a molecular mass of 107 daltons, the total number of
kilomoles is
0.8 kg
= 0.0075 kmoles.
N1 =
107 kg/kmole
For Alloy 2, with a molecular mass of 439 daltons, the total number of
kilomoles is
1.0 kg
= 0.0023 kmoles.
N2 =
439 kg/kmole
The canister contains 7.5 moles of Alloy 1 and
2.3 moles of Alloy 2.
c. Tabulate the plateau pressures of the two alloys for 300 and
400 K.
.....................................................................................................................
pplateau
∆H
∆S
+
= exp −
R
RT
Solution of Problem
090723
11.12
Fund. of Renewable Energy Processes
Prob. Sol. 11.12
Page 3 of 6
423
For T = 300 K:
pplateau1
−110 × 103
−35 × 106
= exp −
+
8314
8314 × 300
= 0.448
atmos.
pplateau2
−28 × 106
−90 × 103
= 0.670
+
= exp −
8314
8314 × 300
atmos.
For T = 400 K:
−110 × 103
−35 × 106
pplateau1 = exp −
= 14.97
+
8314
8314 × 400
atmos.
−28 × 106
−90 × 103
= 11.08
+
= exp −
8314
8314 × 400
atmos.
pplateau2
Alloy
300 K
400 K
1
2
0.448
0.670
14.97
11.08
The figure below shows the 400 K isotherms for the two alloys. Observe
that in this figure, the ordinate is linear with pressure and not with the
natural log of pressure, as usual.
Pressure (atmos)
BP3
x = 0.3
14.97 atmos.
Alloy 1 (7.5 moles)
BP1
x = 0.3
11.08 atmos.
Alloy 2 (2.3 moles)
BP1
BP2
BP3
BP4
x = 3.55
BP4
x = 4.85
BP2
T = 400 K
0
1
2
3
4
Stoichiometric index, x
5
d. Describe, in words, the manner in which the system pressure,
psystem , varies as hydrogen is gradually introduced in the system. Use µH2 as the measure of the number of moles of H2
introduced. Sketch a psystem vs µH2 graph.
.....................................................................................................................
Solution of Problem
11.12
090723
424
Page 4 of 6
Prob. Sol. 11.12
Fund. of Renewable Energy Processes
At low values of µH2 , both alloys are depleted and the psystem grows
linearly with µH2 until it reaches 11.08 atmospheres (the plateau pressure
of Alloy 2). Increasing µH2 will not alter psystem until Alloy 2 is saturated
which occurs when x2 = 4.85. From here on, psystem grows again linearly
with µH2 until it reaches 14.97 atmospheres (the plateau pressure of Alloy
1). It then remains steady until Alloy 1 becomes saturated (x1 = 3.55).
Finally, the pressure grows again with µH2 .
e. Calculate the values of µH2 that mark the break-points (points
of abrupt change in the psystem behavior—that is, points where
the pressure changes from growing to steady, or vice-versa).
.....................................................................................................................
Under all circumstances the amount of H2 in the system is the sum of
1. The hydrogen, µgas−space in gaseous form stored in the gasspace.
2. The hydrogen, µ1 stored in Alloy 1.
3. The hydrogen, µ2 stored in Alloy 2.
µH2 = µ1 + µ2 + µgas−space .
.......................................... The 1st Break-Point
The first break-point occurs when the pressure (that was growing linearly with µH2 ) reaches the plateau value of 11.08 atmospheres (1.12 MPa)
of Alloy 2. At this point, xcritℓ 2 = 0.3. This means that for each kilomole
of alloy, there are 0.3 kilomoles of H or 0.15 kilomoles of H2 .
Since there are 0.0023 kmoles of Alloy 2 in the canister, the amount of
H2 in this alloy is
µ2 = 0.15 × 0.0023 = 345 × 10−6 kmoles.
When the hydrogen pressure is 11.08 atmospheres, Alloy 1 is depleted
(Region 1) because its plateau pressure is 14.97 atmospheres. In this region
p = αx,
where α is a proportionality constant that has to be determined. Region 1
ends when x = xcritℓ 1 = 0.3 and p=14.97 atmosphere, hence
α=
At 11.08 atmospheres,
14.97
= 49.9.
0.3
11.08
= 0.222.
49.9
This means that for each kilomole of alloy, there are 0.222 kilomoles
of H or 0.111 kilomoles of H2 .
Since there are 0.0075 kmoles of Alloy 1 in the canister, the amount of
H2 in this alloy is
x=
µ1 = 0.111 × 0.0075 = 832 × 10−6
kmoles.
Solution of Problem
090723
11.12
Fund. of Renewable Energy Processes
Prob. Sol. 11.12
Page 5 of 6
425
When the hydrogen pressure is 11.08 atmospheres (1.12 MPa), the
gas-space contains
µgas−space =
pV
1.12 × 106 × 0.000345
=
= 116 × 10−6
RT
8314 × 400
kmoles.
The total amount of H2 at the first break-point is µ1 +µ2 +µgas−space =
832 × 10−6 + 345 × 10−6 + 116 × 10−6 = 1.293 × 10−3 kilomoles of H2 .
.......................................... The 2nd Break-Point
The next break-point occurs when Alloy 2 saturates at x = 4.85 and
the gas pressure is still 11.08 atmospheres.
The amount of hydrogen stored in Alloy 2 is
µ2 =
4.85
× 0.0023 = 5, 580 × 10−6
2
kmoles.
Since the pressure is still 10.08 atmosphere, the amounts of hydrogen
in Alloy 1 and in the gas-space are the same as before. Thus, the total
amount of H2 at the second break-point is µ1 + µ2 + µgas−space = 832 ×
10−6 + 5, 580 × 10−6 + 116 × 10−6 = 6.53 × 10−3 kilomoles of H2 .
.......................................... The 3rd Break-Point
The third break-point occurs when the pressure, having risen from
11.08 atmospheres, reaches 14.97 atmospheres (1.52 MPa), the plateau pressure of Alloy 1.
0.3
× 0.0075 = 1, 120 × 10−6 kmoles.
2
The amount of hydrogen in Alloy 2 is the same as in the previous
break-point (µ2 = 5, 580 × 10−6 kmoles.), plus the small amount of gas
dissolved in the saturated alloy.
The slope of the p-x line is
µ1 =
α=
11.08
= 36.9,
0.3
hence, the amount dissolved is
xdissolved =
14.97 − 11.08
= 0.10.
36.9
Thus the value of x at this break-point is x = 4.85 + 0.10 = 4.95, and
the amount of hydrogen in Alloy 2 is
µ2 =
4, 95
× 0.0023 = 5, 690 × 10−6
2
Solution of Problem
kmoles.
11.12
090723
426
Page 6 of 6
µgas−space =
Prob. Sol. 11.12
Fund. of Renewable Energy Processes
pV
1.52 × 106 × 0.000345
=
= 158 × 10−6
RT
8314 × 400
kmoles.
Thus, the total amount of H2 at the third break-point is µ1 + µ2 +
µgas−space = 1, 120 × 10−6 + 5, 690 × 10−6 + 158 × 10−6 = 6, 970 × 10−6
kmoles of H2 .
.......................................... The 4th Break-Point
The amounts of gas in the gas-space and in Alloy 2 are the same as in
the previous step because the gas pressure is the same.
µgas−space = 158 × 10−6
µ1 =
3.55
× 0.0075 = 13, 300 × 10−6
2
µ2 = 11, 400 × 10−6
Pressure (atmos.)
kmoles.
Breakpoint
Pres.
(atmos.)
µ1
1
2
3
4
11.08
11.08
14.97
14.97
0.83
0.83
1.12
13.30
BP3
kmoles.
kmoles.
µ2
µgs
(moles)
0.34
5.58
5.69
5.69
0.12
0.12
0.16
0.16
µ
1.29
6.53
6.97
19.15
14.97 atmos
BP4
BP2
11.08 atmos
BP1
T = 400 K
0
5
10
15
20
Hydrogen in the system (moles)
Solution of Problem
090723
25
11.12
Fund. of Renewable Energy Processes
Prob. Sol. 11.13
Page 1 of 5
427
Prob 11.13 A container able to withstand high pressures has
an internal capacity of 0.1 m3 . It contains 490 kg of an alloy, AB,
used to store hydrogen.
The properties of this alloy are:
Atomic mass of A
60 daltons
Atomic mass of B
70 daltons
Density of AB
8900 kg/m3
Heat capacity
1 kJ K−1 kg−1
∆H
-25 MJ
/kmole of H2
∆S
-105 kJ K−1
/kmole of H2
Depletion end
of plateau
Saturation end
x=0.01
of plateau
x=1
x is the stoichiometric coefficient of hydrogen in ABHx . The
values correspond to 300 K. The plateau pressure is essentially independent of x in the above interval. Above x = 1, the equilibrium
pressure rises very rapidly with x so that x does not appreciably
depend on the hydrogen pressure.
a. What volume inside the container can be occupied by gas?
.....................................................................................................................
490 kg of AB occupy 490/8900 = 0.055 m3 so that the space for the
gas inside the container is 0.100 − 0.055 = 0.045 m3 .
In the container, the space that can be filled by gas is 45 liters.
b. How many kilomoles of alloy are inside the container?
.....................................................................................................................
The molecular mass of AB is 130 daltons or 130 kg/kilomole. Thus,
490 kg of alloy correspond to 490/130 = 3.77 kilomoles.
In the container, there are 3.77 kilomoles of the Alloy AB.
c. What is the plateau pressure of the hydrogen in the alloy at
300 k?
.....................................................................................................................
ln p = −
∆S
∆H
−105 × 103
−25 × 106
+
=−
+
= 2.606,
R
RT
8314
8314 × 300
Solution of Problem
(1)
11.13
090723
428
Page 2 of 5
Prob. Sol. 11.13
p = 13.5
Fund. of Renewable Energy Processes
atmos.
(2)
The plateau pressure is 13.5 atmospheres.
d. Introduce 10 g of H2 . Give an upper bound for the pressure
of the gas in the container.
.....................................................................................................................
10 g of H2 is the same as 0.01 kilomoles of H. To reach the plateau
(which starts when x=0.01), the amount of hydrogen required is 0.01 ×
3.77 = 0.0377 kilomoles of H. The amount introduced is insufficient to
reach the plateau. A small amount of gas will dissolve in the alloy, the rest
will remain in gaseous form. An upper bound for the pressure can be found
by assuming that no gas is in the alloy, all of it being in gaseous form.
p=
µRT
0.005 × 8314 × 300
=
= 277
V
0.045
kPa.
(3)
Notice that in the above equation, the number of kilomoles of hydrogen
was taken as 0.005, not 0.01, because the hydrogen gas is in the H2 , not H,
form.
The gas pressure will be 2.77 atmospheres or less.
e. Now introduce additional hydrogen so that the total amount
introduced (Steps 4 and 5) is 100 g. The temperature is kept
at 300 K. What is the pressure of the gas?
.....................................................................................................................
100 g of hydrogen correspond to 0.10 kilomoles of H or 0.05 kilomoles
of H2 .
In the plateau region, the alloy must have more than 0.01 × 3.77 =
0.0377 and less than 3.77 kmoles of H. The amount introduced seems to be
sufficient to cause the alloy to operate in the plateau, hence the gas pressure
is probably the plateau pressure (13.5 atmospheres). We must, however,
check how much hydrogen is gaseous form.
At 13.5 atmospheres (≈ 1.35 MPa), the amount of gas in the container
is
pV
1.35 × 106 × 0.045
µ=
=
= 0.024 kmoles of H2
(4)
RT
8314 × 300
The rest of the hydrogen (0.05 − 0.024 = 0.026 kilomoles of H2 or 0.052
kilomoles of H) is in the hydride. Thus, there is sufficient hydrogen to cause
the alloy to reach its plateau. The gas pressure is 13.5 atmospheres.
The gas pressure is 13.5 atmospheres.
Solution of Problem
090723
11.13
Fund. of Renewable Energy Processes
Prob. Sol. 11.13
Page 3 of 5
429
f. What is the stoichiometric index, x, of the hydride, ABHx ?
.....................................................................................................................
From the previous question, the amount of hydrogen absorbed is 0.052
kilomoles of H. Since there are 3.77 kilomoles of alloy, the stoichiometric
index is 0.052/3.77 = 0.014.
The empirical formula for the hydride is ABH0.014 .
g. Finally, introduce sufficient hydrogen so that the total
amounts to 4 kg. The temperature remains at 300 K. What
is the pressure of the gas?
.....................................................................................................................
4 kg of hydrogen correspond to 4 kilomole of H or 2 kilomoles of H2 .
Since the alloy, at the saturated end of the plateau has an empirical formula,
ABH1 , the 3.77 kilomoles of alloy can hold, at this point, 3.77 kilomoles of
H. If the amount of H is increased further, no more H is absorbed because
of the very steep p vs x characteristic. All the excess (4 − 3.77 = 0.23
kilomoles of H or 0.115 kilomoles of H2 ) hydrogen must be in gaseous form.
Its pressure is
p=
µRT
0.115 × 8314 × 300
=
= 6.37 × 106
V
0.045
Pa.
(5)
The gas pressure is 63.7 atmospheres.
h. Assume that the container has negligible heat capacity. Withdraw adiabatically (that is, without adding any heat to the
system) 1 kg of hydrogen. What will be the pressure of the
gas inside the container after the 1 kg of hydrogen have been
removed?
.....................................................................................................................
A certain amount of hydrogen, µdes , will be desorbed owing to the
removal of 1 kg of gas from the container. Qdes joules of heat are required
for this desorption:
Qdes = µdes ∆H = 25 × 106 µdes
joules.
Notice that µdes is the number of kilomoles of H2 , not of H.
Solution of Problem
11.13
090723
430
Page 4 of 5
Prob. Sol. 11.13
Fund. of Renewable Energy Processes
Since the processes is adiabatic and assuming that the hydrogen gas
has negligible heat capacity, the heat necessary for desorption must come
from the alloy itself.
Qdes = 25 × 106 µdes = cM ∆T = 1000 × 490(300 − T2 )
Where T2 is the temperature after desorption, c is the heat capacity of the
alloy and M is the mass of the alloy.
µdes = 5.88 − 19.6 × 10−3 T2
kmoles of H2 .
Before the removal of the hydrogen, the amount of gas in the container
was 0.115 kmoles of H2 . After the removal, the temperature falls to T2 and
the plateau pressure is
3007
.
p2 = 105 exp 12.63 −
T2
The 105 factor, above, converted the pressure from atmospheres to
pascals.
At that pressure, the amount of gas in the container is
3007
3007
5
×
0.0.45
10
exp
12.63
−
exp
12.63
−
T2
T2
p2 V
µ2 =
=
= 0.541
RT2
8314T2
T2
Hence, the amount of hydrogen in gaseous form was reduced by
exp 12.63 − 3007
T2
kmoles of H2 .
∆µ = 0.115 − 0.541
T2
3 kg of hydrogen removed corresponds to 1.5 kilomoles of H2 . Consequently, the amount desorbed is
exp 12.63 − 3007
T2
µdes = 1.5 − ∆µ = 1.385 + 0.541
kmoles of H2 .
T2
But we had already determined that
µdes = 5.88 − 19.6 × 10−3 T2
kmoles of H2
hence
5.88 − 19.6 × 10−3 T2 = 1.385 + 0.541
exp 12.63 −
3007
T2
T2
Solution of Problem
090723
11.13
Fund. of Renewable Energy Processes
Prob. Sol. 11.13
Page 5 of 5
431
A numerical solution yields, T2 = 229 K.
The pressure of the gas is the plateau pressure at 229 K:
3007
= 61.4
p2 = 105 exp 12.63 −
229
kPa.
The gas pressure is 61.4 kPa or 0.606 atmos.
Solution of Problem
11.13
090723
432
Page 1 of 1
Prob 11.14
properties:
Prob. Sol. 11.14
Fund. of Renewable Energy Processes
Two hydrogen storing alloys have the following
Hydride
∆H
MJ kmole−1
∆S
kJ K−1 kmole−1
A
B
-28
-20
-100
?
a – (10 pts) –What must the value of ∆SB be to make the
plateau pressure of the two hydrides be the same when T = 400
K?
∆SB is, of course, the ∆S of Alloy B.
.....................................................................................................................
The plateau pressure is given by
exp p = −
∆S
∆H
+
,
R
RT
hence, the two hydrides have the same plateau pressure when
−
∆SA
∆HA
∆SB
∆HB
+
=−
+
,
R
RT
R
RT
1
1
(∆HA − ∆HB ) − ∆SA =
(−28 + 20) × 106 + 100 × 103
T
400
= −80 × 103 kJ K−1 kmole−1
−∆SB =
The ∆S of Hydride B must be 80 kJ−1 kmole−1 .
b – (5 pts) – How do you rate your chances of finding an alloy
with the properties of Alloy B? Explain.
.....................................................................................................................
The ∆S of alloy B seems to be substantially lower than that of any
alloy listed.
Chances of finding an alloy with such a low ∆S are poor.
Solution of Problem
090723
11.14
Fund. of Renewable Energy Processes
Prob. Sol. 11.15
Page 1 of 1
433
Prob 11.15 TiFe is sold by Energics, Inc under the label
HYSTORE 101. Pertinent data are found in Table 11.4 through
11.6 in the Text. Treat this alloy as “ideal” (no hysteresis).
The atomic mass of iron is 55.8 daltons and that of titanium
is 47.9 daltons. The saturated alloy (TiFeH0.95 ) is at 350 K and
is in a perfectly adiabatic container with negligible heat capacity.
A valve is opened and hydrogen is allowed to leak out until
the pressure reaches 2 atmospheres. What is the composition of
the hydride at the end of the experiment—that is, what is the
value of x in TiFeHx ?
To simplify the problem, assume that there is no“gas space”
in the container (patently impossible). Also, neglect the heat capacity of the hydrogen gas and any Joule-Thomson heating owing
to the escaping gas.
.....................................................................................................................
The data for HYSTORE 101 are
∆S = −106.1 kJ K-1 kmole−1 .
∆H = −28 MJ kmole−1 .
Heat capacity of the alloy, c = 540 J K−1 kg−1 .
Since the formula mass of the alloy is 103.7 daltons, c = 540 × 103.7 =
56 kJ K−1 kmole−1 .
The plateau pressure is
1
1
106.1 × 103 28.0 × 106
= exp 12.762 − 3367.8
.
p = exp
−
×
8314
8314
T
T
For T = 350 K, p = 23.1 atmos.
A 2 atmos pressure corresponds to a temperature of 279.0 K.
Hydrogen will be desorbed cooling the alloy until it reaches 279.0 K,
i.e, until the temperature falls by 350 − 279 = 71 K. Since the container has
negligible heat capacity, all the heat must come from the alloy. The heat
removed per kilomole of alloy is
Q = c∆T = 56, 000 × 71 = 3.98 × 106
joules/kmole of alloy.
To remove this much heat the amount, N, of hydrogen that must be
desorbed is
3.98 × 106
N=
= 0.14 kmoles (H2 )/kmole alloy.
28 × 106
or
0.28 kmoles (H)/kmole alloy
Since initially the hydride had the empirical formula, TiFeH0.95 it will
have a formula TiFeH0.67 after the release of the gas.
The stoichiometric index of H in the alloy is 0.67.
Solution of Problem
11.15
090723
434
Page 1 of 2
Prob. Sol. 11.16
Fund. of Renewable Energy Processes
Prob 11.16 A canister contains an alloy, AB, that forms a
monohydride, ABH. The canister is perfectly heat-insulated—
that is, adiabatic and contains 0.01 kmoles of the alloy and a free
space of 600 ml which is, initially, totally empty (a vacuum).
The molecular mass of the alloy is 100 daltons and its thermodynamic characteristics for absorption are ∆H = −25 MJ and
∆S = −100 kJ/K all per kilomole of H2 . For simplicity, make the
unrealistic assumption that the plateau extents from x = 1 all the
way to x = 0. x is the stoichiometric coefficient in ABHx . Assume also that the plateau is perfectly horizontal and that there
is no hysteresis. The heat capacity of the alloy is 500 J kg−1 K−1 .
Again, for simplicity, assume that neither the canister itself nor
the hydrogen gas has significant heat capacity. Finally, assume
that the volume of the alloy is independent of x.
Introduce 0.002763 kmoles of H2 into the canister. Both canister and hydrogen are at 300 K. What will the gas pressure be
inside the canister?
.....................................................................................................................
Inside the canister, part of the gas will occupy the free space (designate
this quantity as µF ree ) and part, µHydride , will be absorbed causing the
average composition of the alloy to become ABHx . These two parts are
µF ree =
pVF ree
,
RT
where VF ree is the volume of the free space, and T is the final temperature
of the system, and
µHydride = 12 xµAlloy ,
where µAlloy is the number of kilomoles of the alloy inside the canister.
When the hydrogen is introduced and µHydride kilomoles of hydrogen
is absorbed, an amount of heat
Q = 12 xµAlloy |∆H|
is released. This causes a temperature rise of
∆T =
Q
,
cM
where c is the heat capacity of the alloy and M is its mass in kg. Since
each kilomole of AB masses 100 kg,
M = 100µAlloy = 100 × 0.01 = 1
kg.
Solution of Problem
090723
11.16
Fund. of Renewable Energy Processes
Prob. Sol. 11.16
Page 2 of 2
435
The heat capacity of the alloy is c = 500 J kg−1 K−1 so that
∆T =
Q
=
cM
1
2 xµAlloy |∆H|
500
,
and the final temperature of the alloy will be
T = 300 +
Q
=
cM
1
2 xµAlloy |∆H|
500
= 300 +
12.5 × 104
x = 300 + 250x.
500
The plateau pressure will be
∆H
−25 × 106
∆S
100 × 103
p = exp
= exp
−
−
RT
R
8314 × (300 + 250x)
8314
−3007
+ 12.03
= exp
(300 + 250x)
The amount of free hydrogen will be
−3007
+ 12.03 × 6 × 10−4
105 × exp (300+250x)
pVF ree
µF ree =
=
RT
8314 × (300 + 250x)
−3007
exp (300+250x) + 12.03 × 7.22 × 10−3
=
(300 + 250x)
The amount of H2 in the hydride is
µHydride = 12 µAlloy x = 0.005x.
Finally, the total amount of hydrogen introduced into the canister is
µT otal = 0.002763 = µF ree + µHydride
−3007
+ 12.03 × 7.22 × 10−3
exp (300+250x)
+ 0.005x.
=
(300 + 250x)
A numerical solution of the above equation leads to x = 0.32. This
allows the calculation of the pressure,
−3007
p = exp
+ 12.03 = 61.4 atmos.
(300 + 250x)
The temperature is
T = 300 + 250x = 380
K.
The final hydrogen pressure is 61.4 atmospheres.
Solution of Problem
11.16
090723
436
Page 1 of 4
Prob. Sol. 11.17
Fund. of Renewable Energy Processes
Prob 11.17 HELIOS is an electric airplane developed by
AeroVironment to serve as a radio relay platform. It is supposed
to climb to fairly high altitudes (some 30 km) and orbit for a
prolonged time (months) over a given population center fulfilling
the role usually performed by satellites. Although its geographic
coverage is much smaller, HELIOS promises to be substantially
more economical.
The plane is propelled by 14 electric motors of 1.5 kW each.
Power is derived from photovoltaic cells that cover much of the
wing surface. In order to stay aloft for many days, the plane
must store energy obtained during the day to provide power for
nighttime operation. The solution to this problem is to use a water electrolyzer that converts the excess energy, provided during
daytime hours by the photovoltaics, into hydrogen and oxygen.
The gases are then stored and, during darkness feed a fuel cell
that provides the power required by the airplane.
Although the specifications of the HELIOS are not know, let
us take a stab into providing the outline of a possible energy
storage system. To that end, we will have to make a number of
assumptions that may depart substantially from the real solution
being created by AeroViroment.
a. Calculate the amount of hydrogen and of oxygen that must
be stored. Assume that during take off and climbing to cruise
altitude, the full 1.5 kW per motor is required, but, for orbiting at altitude only half the above power is required. Assume
also that the power needed for the operation of the plane
(other than propulsion, but including the energy for the radio equipment) is 3 kW. Assume also that the longest period
of darkness last 12 hours. The fuel cell has an efficiency of
80%.
.....................................................................................................................
14 motors operating at half power require 14× 21 1.5 = 10.5 kW. Adding
to this the 3 kW for house keeping, leads to a power of 13.5 kW required
for orbiting. This calls for a 15 kW fuel cell that, in an emergency can
supply much more power for a short time. The total energy spent during
a 12 hour night would be 13.5 × 12 × 3600 = 583, 000 kJ or 583 MJ.
Since the fuel cell has an 80% efficiency, the required fuel input is
583/.8 = 729 MJ. The free energy of 1 kilomole of hydrogen combining with
half a kilomole of oxygen is 228 MJ. Thus, 729/228 = 3.2 kmol of hydrogen
(6.4 kg). Correspondingly, 1.6 kmol of oxygen (51.2 kg) are needed.
Solution of Problem
090723
11.17
Fund. of Renewable Energy Processes
Prob. Sol. 11.17
Page 2 of 4
437
Fuel needed:
6.4 kg of hydrogen,
51.2 kg of oxygen.
b. The amount of fuel calculated in Item 1 must be stored. Assuming STP conditions, what is the volume required?
.....................................................................................................................
The density of hydrogen at STP is
ρH2 =
2 kg/kmol
= 0.0893
22.4 km3 /kmol
kg/m3 ,
and that of oxygen is
ρO2 =
32 kg/kmol
= 1.43
22.4 km3 /kmol
3
kg/m .
The volume of hydrogen is
VH2 =
6.4
= 71.7
0.0893
m3 ,
and that of oxygen is
VO2 =
51.2
= 35.8
1.43
m3 .
Of course, the volume of hydrogen is twice that of oxygen.
The volumes are
Hydrogen: 71.7 m3
Oxygen: 35.8 m3 .
c. Clearly, the volumes calculated in Item 2 are too large to
fit into HELIOS. The fuel cell busses being operated experimentally in Chicago, have hydrogen tanks that operate at
500 atmospheres and that allow a gravimetric concentration
of 6.7% when storing hydrogen. If such tanks were adopted
for the HELIOS, what would be the mass of the fully charged
fuel storage system?
.....................................................................................................................
At 500 atmosphere the volume of hydrogen is reduced to 71.7/500 =
0.143 m3 .. With a gravimetric concentration of 6.7%, the tank would mass
6.4/0.067 = 95.5 kg and the total (tank plus gas) would be 102 kg. The
mass-to-volume ratio of the tank is 95.5/0.143 = 668 kg/m3 .
Solution of Problem
11.17
090723
438
Page 3 of 4
Prob. Sol. 11.17
Fund. of Renewable Energy Processes
The 51.2 kg of oxygen would occupy 35.8/500 = 0.07 m3 . The mass of
the tank is 668 × 0.07 = 46.8 kg.
The total mass of the fully charged fuel storage system (not counting
compressors) is 95.5 + 6.4 + 46.8 + 51.2 = 149 kg.
The mass of the fully charge fuel storage system is 149 kg.
d. Assume that the efficiency of a mechanical hydrogen compressor is 60% and that of an oxygen compressor is 80%. How
much energy do these compressors require to compress the
gases isothermally to their 500 atmosphere operating temperature. Assume, for simplicity that the electrolyzer that
produces these gases is pressurized to 5 atmospheres.
.....................................................................................................................
The energy to compress a gas isothermally is
W = µRT ln
p1
.
p0
The compression ratio is, for both hydrogen and oxygen, 500:5, thus,
ln 100 = 4.61. The RT product is 8314 × 273 = 2.27 × 106 .
For hydrogen,
WH2 =
1
× 3.2 × 2.27 × 106 × 4.61 = 55.8 × 106
0.6
J.
1
× 1.6 × 2.27 × 106 × 4.61 = 20.9 × 106
0.8
J.
For oxygen,
WO2 =
The total compression energy is 76.7 MJ
e. While orbiting, during daylight, what is the total energy that
the photovoltaic collectors have to deliver? The electrolyzer
is 80% efficient.
.....................................................................................................................
Solution of Problem
090723
11.17
Fund. of Renewable Energy Processes
Prob. Sol. 11.17
Page 4 of 4
439
The total amount of hydrogen that has to be produced is 3.2 kilomoles.
An ideal electrolyzer produces 1 kilomole of hydrogen for each 237.2 MJ of
electric energy used. An electrolyzer with 80% efficiency needs 237.2/0.8 =
297 MJ per kilomole or a total of 950 MJ.
The photovoltaic collectors have to supply:
950 MJ for propulsion and housekeeping.
76.7 MJ for the gas compressors,
or 1.03 GJ total. which are expended during the 12 daylight hours,
and correspond to some 1.8 kW.
The photovoltaic system has to generate a total of 1.03 GJ daily.
Solution of Problem
11.17
090723
440
Page 1 of 4
Prob. Sol. 11.18
Fund. of Renewable Energy Processes
Prob 11.18 Two 100-liter canisters are interconnected by a
pipe (with negligible internal volume). Canister “A” contains
37.2 kg of FeTi and Canister “B”, 37.8 kg of Fe0.8 Ni0.2 Ti.
Although the gas can freely move from one canister to the
other, there is negligible heat transfer between them. Thus the
gas can be at different temperatures in the two canisters. The
gas always assumes the temperature of the alloy it is in contact
with.
The pertinent data are summarized in the two boxes below:
Element
Ti
Fe
Ni
Atomic
Mass
(daltons)
Density
(kg/m− 3)
47.90
55.85
58.71
4540
7870
8900
Alloy
∆H
∆S
(MJ
(kJ
kmol−1 ) K−1 kmol−1 )
FeTi
Fe0.8 Ni0.2 Ti
-28.0
-41.0
-106.1
-118.8
ln p
Plateau
x
0
1
To simplify the solution, assume that the ln p vs x characteristics of the alloys consist of a perfectly horizontal plateau followed
by a vertical line in the saturated region. In other words, the
characteristics look as sketched in the figure. The alloys are depleted when x = 0 and are saturated when x = 1. Also, neglect all
the hydrogen dissolved in the saturated alloy.
Initially, Canisters A is at 300 K and Canister B, at 400 K.
They have been carefully evacuated (the gas pressure is zero).
Assume that the density of the alloys is the average of that
of the component elements.
Enough hydrogen is introduced into the system so that one
of the alloys becomes saturated. This will cause the temperature
of the alloys to change.
To simplify things, assume that the alloys and the canisters
themselves have negligible heat capacity.
a. Does the temperature increase or decrease?
.....................................................................................................................
Solution of Problem
090723
11.18
Fund. of Renewable Energy Processes
Prob. Sol. 11.18
Page 2 of 4
441
Hydrogen will be absorbed by the alloy. This is an exothermic process.
The temperature will rise.
The temperature will rise.
b. The temperature is now adjusted to the values of 300 K and
400 K, as before. How many kg of hydrogen had to be introduced into the system to make the gas pressure, at this stage,
10% higher than the plateau pressure of the saturated alloy
while leaving the other alloy depleted? Please be accurate to
the gram.
.....................................................................................................................
The density of FeTi is
δF eT i =
7870 + 4540
= 6200
2
3
kg/m .
The density of Fe0.8 Ni0.2 Ti is
δF e0.8 N i0.2 T i =
0.8 × 7870 + 0.2 × 8900 + 4540
= 6310
2
kg/m3 .
The volume of 37.2 kg of FeTi is
VF eT i =
37.2
= 6 × 10−3
6200
m3
or 6 liters.
The volume of 37.8 kg of Fe0.8 Ni0.2 Ti is
VF e0.8 N i0.2 T i =
37.8
= 6 × 10−3
6200
m3
or 6 liters.
Thus, the “dead space” in each canister is 94 liters.
The plateau pressure is
∆S
∆H
pP lateau = exp −
.
+
R
RT
For the two alloys,
pP lateauA
pP lateauB
3368
,
= exp 12.76 −
T
4931
= exp 14.29 −
,
T
Solution of Problem
11.18
090723
442
Page 3 of 4
Prob. Sol. 11.18
Fund. of Renewable Energy Processes
This leads to a plateau pressure for Alloy A of 4.63 atmos at 300 K and,
for Alloy B, at 400 K, 7.11 atmos. We have to introduce sufficient hydrogen
so that the pressure of the system reaches 1.1 × 4.63 = 5.09 atmospheres.
Since each canister has a 94 liter dead space, when the gas pressure
is 5.09 atmospheres (509 kPa), the amount of hydrogen in gas form in the
Canisters A is
µ=
pV
509, 000 × 0.094
=
= 0.0192
RT
8314 × 300
kmoles of H2 .
and in Canister B is
µ=
509, 000 × 0.094
pV
=
= 0.0144
RT
8314 × 400
kmoles of H2 .
Alloy A is completely saturated. Its stoichiometric index is x = 1. The
mass of one kilomole of FeTi is
mF eT i = 55.85 + 47.90 = 103.8
kg/kmole,
and of the Fe0.8 Ni0.2 Ti alloy is
mF e0.8 N i0.2 T i = 0.8 × 55.85 + 0.2 × 58.71 + 47.90 = 104.3
kg/kmole.
37.2 kg of FeTi correspond to 37.2/103.8 = 0.358 kmoles, while 37.8
kg of Fe0.8 Ni0.2 Ti correspond to 37.8/104.3 = 0.362 kmoles of H.
Thus in the just saturated FeTi, we have 0.358 kmoles of H absorbed
or, 0.179 kmoles of H2 .
Alloy B is depleted and holds no hydrogen.
The total amount of hydrogen in the system is 0.0192+0.0144+0.179 =
0.213 kmoles of H2 .
The system must be filled with 0.425 kg of H2 .
c. Now raise the temperature of Alloy A to 400 K. Describe
what happens:
3.1 what is the final gas pressure,
3.2 what is the stoichiometric value of H in each alloy,
3.3 how many joules of heat had to be added,
3.4 how many joules of heat had to be removed.
To simplify things, assume that the alloys and the canisters
themselves have negligible heat capacity.
.....................................................................................................................
When the temperature of Alloy A is raised to 400 K, its plateau pressure becomes 76.9 atmos. Hydrogen desorbs and is absorbed by Alloy B
Solution of Problem
090723
11.18
Fund. of Renewable Energy Processes
Prob. Sol. 11.18
Page 4 of 4
443
which clamps the pressure at 7.1 atmospheres while it absorbs the gas. Is
there enough hydrogen to saturate this alloy?
At a gas pressure of 7.1 atmosphere and 400 K (in both canisters),
the 94 liters of dead space in each canister will hold 0.020 kmoles of H2 ,
a total of 0.04 kmoles of hydrogen. Since the total hydrogen introduced
was 0.213 kmoles, that leaves 0.213 − 0.040 = 0.173 kmoles of H2 (0.346
kmoles of H) to be absorbed by Alloy B. There are 0.362 kmoles of this
alloy and, thus, when saturated (x = 1) it can hold 0.362 kmoles of H.
Consequently, there is not enough hydrogen in the system to saturate this
alloy. Its stoichiometric index will be
x=
0.346
= 0.956.
0.362
The stoichiometric index in Alloy A is zero (totally depleted).
The stoichiometric index in Alloy B is 0.956 (not quite saturated).
Since there is not enough hydrogen to saturate Alloy B, the pressure
of the gas will be equal to the plateau pressure of this alloy.
The gas pressure will be 7.1 atmospheres.
Alloy A was saturated before its temperature was raised. It then became completely depleted. All hydrogen (0.358 kmol of H) was desorbed.
This corresponds to 0.178 kmoles of H2 . It take 28 MJ to desorb 1 kmole
of H2 , thus the amount of heat that had to be introduced was
QA = 0.178 × 28 = 4.98
MJ.
The amount of heat to desorb the gas from Alloy A is about 5 MJ.
Alloy B has to absorb 0.346 kmoles of H (0.173 kmoles of H2 ). The
heat that has to be removed is
Q = 0.173 × 41 = 7.09
MJ.
The amount of heat that must be removed from Alloy B is 7.1 MJ.
Solution of Problem
11.18
090723
444
Page 1 of 2
Prob. Sol. 11.19
Fund. of Renewable Energy Processes
Prob 11.19 A canister with 1 m3 capacity contains 3000 kg
LaNi5. Although initially the system was evacuated, an amount,
µ1 , of hydrogen is introduced so that the pressure (when the
canister and the alloy are at 298 K) is exactly 2 atmospheres.
Assume that once the alloy is saturated, it cannot dissolve any
more hydrogen, i.e., assume that the p-x characteristic just after
the beta phase is vertical.
The density of lanthanum (atomic mass 138.90 daltons) is
6145 kg/m3 and that of nickel (atomic mass 58.71 daltons) is
8902 kg/m3 . Assume that the density of the alloy is equal to that
of a mixture of 1 kilomole of lanthanum with 5 kilomoles of nickel.
a. What is the value of µ1 ?
.....................................................................................................................
One kilomole of LaNi masses 138.90 + 5 × 58.71 = 432.5 kg.
5
One kg of La occupies 1/6145 m3 , hence 1 kilomole (138.90 kg) occupies
0.0226 m3 . One kg of Ni occupies 1/8902 m3 , hence 5 kilomoles (293.6 kg)
occupie 0.0330 m3 . The density of LaNi5 is 432.5/(0.0226+0.0330)=7780
kg/m3 .
3000 kg of LaNi5 correspond to 3000/432.5 = 6.94 kilomoles and occupy 3000/7780 = 0.386 m3 . This means that in the canister there is an
empty space of 0.614 m3 .
From a table in the Text, one obtains for the alloy used, ∆Hf = −31.0
MJ/kilomole (H2) and ∆Sf = −107.7 kJ/K−1 kmole−1 (H2). This allows
us to calculate the plateau pressure at 298 K.
p = exp
∆H
∆S
−
RT
R
−31.0 × 106
107.7 × 103
= exp
=1.556 atmos.
−
8314 × 298
8314
The hydride formed when LaNi5 fully combines with hydrogen is
LaNi5H5. Since the gas pressure inside the canister exceeds the plateau
pressure, the alloy must be saturated an must hold 5 kilomoles of H for
each kilomole of alloy. Thus the total amount of monoatomic hydrogen
absorbed by the alloy is 6.94 × 5 = 34.68 kilomoles of H or 17.34 kilomoles
of H2.
The rest of the gas fills the dead space in the canister which has a
volume. Vdead = 0.614 m3 . At a pressure of 2 atmospheres, the amount of
hydrogen in this space is
µdead =
pV
2 × 105 × 0.614
=
= 49.56 × 10−3
RT
8314 × 298
kmoles.
(1)
The total amount of hydrogen introduced into the canister was 17.34 +
0.05 = 17.39 kmoles.
Solution of Problem
090723
11.19
Fund. of Renewable Energy Processes
Prob. Sol. 11.19
Page 2 of 2
445
The amount of gas introduced was 17.39 kmoles of H2
b. 100 MJ of heat are introduced into the canister whose walls
are adiabatic and have no heat capacity. Ignore also the heat
capacity of the free hydrogen gas.
Calculate the pressure of the free hydrogen gas.
.....................................................................................................................
After the heat injection and after a new equilibrium has been established, an amount, µ, of hydrogen has been desorbed. The pressure of the
free gas must equal the plateau pressure of the alloy:
5
1.0133 × 10 exp
−31 × 106
1
107700
× +
8314
T
8314
=
µ + 0.05
× 8314T. (2)
0.614
This simplifies to
exp
−3728.7
+ 12.954 = 0.13363(µ + 0.05)T
T
(3)
The 100 MJ of heat injected into the alloy will be used up in providing
the desorption energy and in heating up the alloy:
108 = 31 × 106 µ + 420 × 3000(T − 298).
(4)
Solving this equation for µ,
µ = 15.338 − 0.040645T.
(5)
And introducing this value of µ into Equation 3,
−3728.7
+ 12.954 = 2.0563T − 0.0054314T 2,
exp
T
(4)
whose numerical solution leads to
T = 369.8 K and µ = 0.307 kilomoles.
Using this temperature, the plateau pressure equation yields
p = exp
−3728.7
+ 12.954 = 17.654
369.8
atmos.
(5)
The gas pressure in the canister is 17.7 atmospheres.
Solution of Problem
11.19
090723
446
Page 1 of 3
Prob. Sol. 11.20
Fund. of Renewable Energy Processes
Prob 11.20
Please use R = 8314.
Here is the setup:
A hydrogen source is connected to a canister containing an
alloy, A, that can be fully hydrided to AH.
The hydrogen source is a high pressure container with an internal capacity of Vs = 1 liter. It is charged with enough hydrogen
to have the gas at p0 = 500 atmospheres when the temperature is
T0 = 300 K. The container is in thermal contact with the hydride
canister. In steady state, the container, the hydrogen and the
canister are all at the same temperature. The heat capacity of
the container is 300 J/K.
The whole system— container and canister—is completely
adiabatic: no heat is exchanged with the environment.
There is a pipe connecting the hydrogen source to the hydride
canister. A valve controls the hydrogen flow. According to the
Joule-Thomson law, the gas escaping from the source and flowing
into the canister will warm up. However, in this problem, assume
that there is no Joule-Thomson effect.
Initially, while the hydrogen delivery valve is still shut off,
there is no gas pressure inside the canister and canister and contents are at 300 K.
Canister:
Volume, V = 1 liter.
Heat capacity, ccan = 700 J/K.
The alloy has the following characteristics:
Amount of hydride, m = 5.4 kg.
Density of hydride, δ = 9000 kg/m3 .
Molecular mass of alloy, A, 100 daltons.
Heat of absorption: -28 MJ/kmoles.
Entropy change of absorption: -110 kJ K−1 kmole−1 .
Heat capacity, chyd = 500 J kg−1 K−1 ,
The ln p versus x characteristics of the alloy are a perfectly
horizontal plateau bound by vertical lines corresponding to
the depleted and the saturated regions.
The valve is opened and hydrogen flows into the canister. If
you wait long enough for the transients to settle down, what is
the pressure of the gas in the hydrogen source? Is the hydride in
the plateau region?
.....................................................................................................................
Solution of Problem
090723
11.20
Fund. of Renewable Energy Processes
Prob. Sol. 11.20
Page 2 of 3
447
The alloy occupies a volume of
Valloy =
m
5.4
=
= 0.0006
δ
9000
m3 ,
(1)
hence the volume of the dead space is
Vds = 0.001 − 0.0006 = 0.0004
m3 .
(2)
The total volume available for has is Vgas = 0.0014 m3 (the sum of the
dead space in the alloy canister plus the space in the hydrogen container).
The initial amount of hydrogen in the source is
µ0 =
p0 Vs
500 × 1.0133 × 105 × 0.001
=
= 20.3 × 10−3
RT0
8314 × 300
kmoles (H2 ).
(3)
Hydrogen from the source will flow into the canister filling the dead
space and hydriding the alloy.
There are 5.4 kg of alloy (molecular mass, 100 daltons) in the canister.
This amounts to 5.4/100 = 0.054 kilomoles of alloy. Since when fully
hydrided, the hydride has an empirical formula, AH, the alloy can, at a
maximum, absorb 0.054 kilomoles of H or 0.027 kmoles of H2 . This is more
than the available hydrogen (0.0203 kilomoles of H2 ). Thus the system
cannot move into the saturated region—it must be in the plateau region.
This means that, after all settles down, the gas pressure will be equal to the
plateau pressure, p. Owing to the hydrogen absorption, the temperature
of the system (hydrogen container, alloy canister, and alloy) will rise from
the original 300 K to a final temperature, T .
∆H
∆S
+
p = 1.013 × 10 exp −
R
RT
−1.1
×
105
−28 × 106
= 1.013 × 105 exp −
+
8314
8314T
3367.8
= 1.013 × 105 exp 13.2307 −
Pa.
T
5
(4)
The amount of hydrogen in gas form is
µgas =
0.0014p
p
pVgas
=
= 168.4 × 10−9 .
RT
8314T
T
(5)
Since no gas was lost,
µhy = µ0 − µgas = 20.3 × 10−3 − 168.4 × 10−9
Solution of Problem
p
T
(6)
11.20
090723
448
Page 3 of 3
Prob. Sol. 11.20
Fund. of Renewable Energy Processes
Here, µhy is the amount of hydrogen (H2 ) stored in the hydride.
We need more independent information relating µhy to T , so as to
eliminate µhy from the above formula.
The heat source is the absorption of hydrogen by the alloy. This
amounts to
Qabs = ∆Hµhy = 28 × 106 µhy .
(7)
This amount of heat must go to raising the temperature of the source
container, the canister, the alloy and the gas in the dead space,
Qin = (T − 300) (Csource + Ccan + mChy + µds CH )
= (T − 300) 300 + 700 + 500 × 5.4 + 29 × 103 µds
p
= (T − 300) 3700 + 29 × 103 × 48.1 × 10−9
T
−3 p
= (T − 300) 3700 + 1.395 × 10
.
T
(8)
Since, in steady state, Qabs = Qin ,
p
28 × 106 µhy = (T − 300) 3700 + 1.395 × 10−3
T
µhy =
T − 300 −3 p
.
3700
+
1.395
×
10
28 × 106
T
(9)
(10)
Equation 6 becomes,
p
20.3 × 10−3 − 168.4 × 10−9
T
p T − 300 =0
(11)
3700 + 1.395 × 10−3
−
6
28 × 10
T
p
p 1 − 8.296 × 10−6 − (T − 300) 0.006509 + 2.454 × 10−9
= 0 (12)
T
T
By combining Equations 4 and 12, we can solve numerically for T (by
trial and error). The result is (to an unwarranted precision): T = 407.1 K.
A this temperature, the plateau pressure is 142.2 atmosphers.
The final hydrogen pressure is 142 atmospheres
and the alloy is in the plateau region.
Solution of Problem
090723
11.20
Fund. of Renewable Energy Processes
Prob. Sol. 11.21
Page 1 of 7
449
Prob 11.21 A hydrogen source consists of a 5-liter container with hydrogen at an initial pressure of 200 atmospheres.
Throughout the experiment the hydrogen in this container remains at a constant temperature of 300 K.
a - How many kilomoles, µ0 , of hydrogen are initially in the container?
.....................................................................................................................
µ0 =
pV
200 × 1.013 × 105 × 0.005
=
= 0.0406 kilomoles.
RT
8314 × 300
(1)
The source contains 0.0406 kilomoles of hydrogen.
A separate 2-liter canister (“alloy canister”) contains 3.5 kg
of a metallic alloy, AB, with the following properties:
Density
Heat of absorption
Entropy change (absorp.)
Formula mass
Composition (fully hydrided)
δ
∆H
∆S
3.5
-30
-100
100
kg/liter
MJ/kmole (H2)
kJ/(kmole K)
daltons
ABH
Assume an extremely simplified ln(p) vs x characteristic:
- Horizontal plateau.
- No depletion region.
- Vertical saturation region.
ln p
Plateau
x
0
1
The following heat capacities are relevant:
Canister
Alloy
Hydrogen (H2 )
100
J/K
440
J/K per kg
20,800 J/K per kmole
b - If the alloy were fully hydrided, how many kilomoles of hydrogen would be absorbed by the 3.5 kg of the alloy in the canister?
.....................................................................................................................
Solution of Problem
11.21
090723
450
Page 2 of 7
Prob. Sol. 11.21
Fund. of Renewable Energy Processes
When fully hydrided, there is one kilomole of hydrogen per kilomole of
AB. Since there are 3.5 kg of the latter in the canister and since its formula
mass is 100 daltons, then the canister contains 0.035 kmoles of AB or 0.035
kmoles of H or 0.0175 kilomoles of H2 .
If fully hydrided, the alloy would hold 0.0175 kmoles of hydrogen.
c - The alloy does not completely fill the canister. A “dead
space” is left which is initially a vacuum but will later contain
some hydrogen. What is the volume of this dead space?
.....................................................................................................................
The density of the alloy is 3,500 kg/m3 , hence 3.5 kg occupy 0.001 m3 .
The canister has a volume of 0.002 m3 , hence the dead space has a volume
of 0.001 m3 .
The volume of the dead space is 0.001 m3 .
Before the beginning of the experiment, the alloy is completely degased—there is hydrogen neither in the dead space nor
absorbed by the alloy. Both canister and the alloy are at 300 K.
The experiment consists of opening the valve and waiting
until the system settles into steady state.
As hydrogen flows from source to alloy canister, the pressure
in the former drops (but the temperature stays put) while the
pressure in the dead space rises (and so does the temperature).
d - When steady state is reached, there is an unique pressure
in the dead space. The calculation of this pressure is laborious.
Assume that it is 10 atmospheres. Calculate the number of kilomoles, µds of H2 in the dead space and the number of kilomoles,
µabs , of H2 absorbed by the alloy. Prove that the assumed pressure cannot correspond to a steady state situation.
.....................................................................................................................
[ Assume p = 10 atmos.]
Note that, unless the alloy is saturated, the gas pressure must equal
the plateau pressure, which is
∆S
∆H
5
p = 1.013 × 10 exp −
+
R
RT
100,000 −30 × 106
5
= 1.013 × 10 exp
+
8314
8314T
3608
5
Pa.
(1)
= 1.013 × 10 exp 12.03 −
T
Solution of Problem
090723
11.21
Fund. of Renewable Energy Processes
From this,
T =
Prob. Sol. 11.21
Page 3 of 7
451
3608
12.03 − ln
.
p
5
1.013 × 10
(2)
[T = 370.9 K.]
The amount of gas in the dead space must be:
µds =
pVds
10 × 1.013 × 105 × 0.001
=
= 3.285 × 10−4
RT
8314 × 370, 9
(3)
[ µds = 0.0003285 kmoles of H2 .]
The introduction of hydrogen into the alloy canister resulted in a temperature rise of
∆T = T − 300 = 370.9 − 300 = 70.9 K.
(4)
[ ∆T = 70.9 K.]
Since the alloy canister is adiabatic, this temperature rise must be the
result of the release of an amount Q of absorption heat, assuming that C
is the total heat capacity of the system.
∆T =
Q
.
C
(5)
Here, C = Calloy +Chyd +Ccan is the total head capacity of the system.
The 3.5 kg of alloy contribute
Calloy = 440 × 3.5 = 1540 J/K.
(6)
The µds kmoles of hydrogen in the dead space contribute
Chyd = 20,800µds ,
J/K.
(7)
and the rest of the canister contributes Ccan = 100 J/K. Consequently,
C = 1640 + 20,800µds.
(8)
Since we have a values for µds , we can calculate C.
[ C=1646.8. J/K]
Solution of Problem
11.21
090723
452
Page 4 of 7
Prob. Sol. 11.21
Fund. of Renewable Energy Processes
Q = C∆T = (1640 + 20,800 × 0.0003285) × 70.9 = 1.1676 × 105
J. (9)
[ Q = 116,760 J.]
Each kmole of H2 absorbed will release ∆H joules of heat, thus
µabs =
Q
116,760
= 0.003892 kmoles of H2 .
=
0.0175∆H
×30 × 106
(10)
[ µabs = 0.003892 kmoles of H2 .]
The total amount of hydrogen removed from the hydrogen source is
µtrans = µds + µabs = 0.0003285 + 0.003892
= 4.22 × 10−3 kmoles of H2 .
(11)
[ µtrans = 0.00422 kmoles of H2 .]
The original amount of gas in the source as calculated in part a of this
problem is µ0 = 0.0406 kmoles of H2 .
The amount, µ1 , of hydrogen left in the source is
µ1 = µ0 − µ = 0.0406 − 0.00422 = 0.03638 kmoles of H2 .
(12)
[ µ1 = 0.03638 kmoles of H2 .]
Consequently, the pressure becomes
p=
0.03638 × 8314 × 300
µ1 RTsource
=
= 1.815 × 107 Pa or 179.1 atmos.
Vsource
0.005
(13)
Here we have a contradiction: the pressure of the gas in the dead space
is 10 atmos while that of the hydrogen source is 179 atmos, yet the two
containers are interconnected. A stream of hydrogen must be flowing from
source to alloy canister—it cannot be a steady state situation.
e - Calculate the correct steady state pressure in the dead space.
Make sure you check the correctness of any assumption you may
have made
.....................................................................................................................
Solution of Problem
090723
11.21
Fund. of Renewable Energy Processes
Prob. Sol. 11.21
Page 5 of 7
453
Initially, the source canister contains µ0 kmoles of hydrogen, under a
pressure, pinit , and a temperature of 300 K.
µ0 =
pinit V
0.005
=
pinit = 2.00 × 10−9 pinit .
RT
8314 × 300
(14)
When Valve “a” is opened, an amount, µtrans , of hydrogen flows into
the alloy canister. Of this, an amount, µds will fill the dead space in that
canister and an amount, µabs , will be absorbed by the alloy.
µtrans = µds + µabs .
(15)
After the transfer, a total of µ0 − µtrans is left in the source.
At the end of the experiment, the pressure in the source is
pf =
8314 × 300
(µ0 − µtrans ) = 4.988 × 108 (µ0 − µtrans ).
0.005
(16)
This pressure is the same as that in the alloy canister because the
source canister and the alloy canisters are interconnected. Since the dead
space has a volume, Vds = 0.001 m3 , it now contains
µds =
pf Vds
4.988 × 105 (µ0 − µtrans )
60(µ0 − µtrans )
=
=
RT
8314T
T
kmoles (H2 ).
(17)
Assuming that the alloy is not saturated, the pressure of the gas, both
in the plateau and in the dead space, is
∆H
∆S
+
pf = 1.013 × 10 exp −
R
RT
100,000 −30 × 106
5
= 1.013 × 10 exp
+
8314
8314T
3608
5
Pa.
= 1.013 × 10 exp 12.03 −
T
5
(18)
Observe that the pressure is in pascals and that the pressure in the
alloy canister is the same as that in the source, that is, Equations 16 and
18, represent the same pressure:
3608
4.988 × 108 (µ0 − µtrans ) = 1.013 × 105 × exp 12.03 −
T
µ0 − µtrans = 2.031 × 10
Solution of Problem
−4
3608
× exp 12.03 −
T
(19)
(20)
11.21
090723
454
Page 6 of 7
Prob. Sol. 11.21
T =
Fund. of Renewable Energy Processes
3608
µ0 −µtrans
12.03 − ln 2.031×10
−4
(21)
From the above equation, one can find the temperature of the alloy
canister for any chosen value of µtrans . Of course, only a single temperature
(and a single value of µtrans ) is correct, because the temperature has also
to satisfy
Q
∆Hµabs
T = 300 +
,
(22)
= 300 +
C
Calloy + Chyd + Ccan
where Q is the heat of absorption of µabs kilomoles of H2 , and C = Calloy +
Chyd + Ccan , is the total head capacity of the system.
The 3.5 kg of alloy contribute
Calloy = 440 × 3.5 = 1540 J/K.
(23)
The µds kmoles of hydrogen in the dead space contribute
Chyd = 20,800µds ,
J/K.
(24)
and the rest of the canister contributes Ccan = 100 J/K. Consequently,
C = 1640 + 20,800µds = 1640 + 20,800
60(µ0 − µtrans )
,
T
(25)
and
T = 300 +
= 300 +
30 × 106 µabs
trans )
1640 + 20,800 60(µ0 −µ
T
trans )
30 × 106 µtrans − 60(µ0 −µ
T
trans )
1640 + 20,800 60(µ0 −µ
T
(26)
The temperature we are seeking must satisfy simultaneously Equations
21 and 26. One easy way to find T is to use a numerical method:
Use µtrans as a parameter. Start by assigning to it an arbitrary value
(which must be smaller than µ0 ). Using Equation 21, calculate T . Introduce this value of T into the right hand side of Equation 21.This will yild
a value of T that, in all probability, is different from the one found from
Equation 21. Change the value of µtrans until the T from the two equations
is the same.
This leads to the following results (when the initial pressure in the
source is 200 atmos):
Solution of Problem
090723
11.21
Fund. of Renewable Energy Processes
Prob. Sol. 11.21
Page 7 of 7
455
H2 transferred to alloy canister: µtrans = 0.01460 kmoles of H2 .
H2 in dead space: µds = 0.003097 kmoles of H2 .
H2 absorbed by the alloy: µabs = 0.01120 kmoles of H2 .
Temperature in the alloy canister: T = 502.5 kelvins.
System pressure: pf = 127.6 atmos.
Note that the amount of H2 in the alloy is less that necessary to saturate it (0.0175 kmoles).
Solution of Problem
11.21
090723
456
Page 1 of 3
Prob. Sol. 11.22
Fund. of Renewable Energy Processes
Prob 11.22 The (very idealized) data of the TiFe alloy are:
∆H= -28 MJ kmoles−1, for absorption.
∆S= -106.1 KJ K−1 kmoles−1 , for absorption.
Specific heat capacity, c = 540 J kg−1 K−1
Density = 6200 kg/m3 .
The plateau pressure extends all the way from x = 0 to x = 1.
A this latter point, the pressure rises independently of x.
100 kg of the above alloy are placed inside a 160-liter canister
which is then carefully pumped out so that the internal pressure
is, essentially, zero. The container is perfectly heat insulated from
the environment, and is (as well as the alloy inside) at 298.0 K.
A large, 1000-liter, separate container filled with hydrogen
(pressure = 500.00 atmospheres) is kept at a constant 298.0 K
throughout the whole experiment.
A pipe, equipped with a shut-off valve interconnects the two
containers above. The valve is, initially, closed.
The experiment begins with the momentary opening of the
valve, allowing some hydrogen to enter the alloy-containing canister. This process lasts long enough for the hydrogen source pressure to fall to 475.54 atmospheres, at which moment the valve
is shut. Disregard any Joule-Thomson effect on the temperature
of the gas entering the alloy container, i. e., assume that the
hydrogen enters this container at 298.0 K.
How much hydrogen was allowed into the alloy-containing
canister?
.....................................................................................................................
The volume of the hydrogen source is Vsource = 1 m3 .
The temperature of the gas in the hydrogen source is Tsource = 298 K
(always).
The initial pressure of the gas in the hydrogen source is psource0 =
500 × 1.013 × 105 = 50.65 MPa.
As a consequence, the initial amount of hydrogen in the source is
µsource0 =
50.65 × 106 × 1
= 20.443
8314 × 298
kmole (H2 ).
(1)
When the pressure falls to psource1 = 475.54 × 1.013 × 105 = 48.172
MPa, the amount of gas is
µsource1 =
48.172 × 106 × 1
= 19.443
8314 × 298
kmole (H2 ).
(2)
Exactly 1 kmole of H2 has been withdrawn and introduced in the alloycontaining canister.
Solution of Problem
090723
11.22
Fund. of Renewable Energy Processes
Prob. Sol. 11.22
Page 2 of 3
457
1.00 kmoles of H2 was allowed into the alloy-containing canister.
What is the final pressure of the hydrogen in this alloycontain-ing canister after steady state has been reached?
Ignore the heat capacity of the H2 . The heat capacity of the
canister is 20,000 J K−1 .
.....................................................................................................................
Since the alloy has no α-phase, the hydride will either be in the plateau
or in the saturated region. Let us assume the former (if it proves wrong,
we will assume saturation).
The amount of hydrogen, µintro , introduced will be divided into two
parts:
– one part, µalloy , will be absorbed by the alloy, forming a hydride,
and
– another part, µds , will fill the dead space whose volume is
Vds = 0.160 − Valloy .
(3)
The volume of the alloy, Valloy is
Valloy =
100
Malloy
=
= 0.01613
δalloy
6200
m3 .
(4)
hence
Vds = 0.160 − 0.01613 = 0.14387
m3 .
(5)
The pressure of the gas in he dead space must be equal to the plateau
pressure (if the hydride is in the β-phase):
3367.8
∆H
∆S
= exp 12.762 −
.
(6)
+
p = exp −
R
(298 + ∆T )R
298 + ∆T
The amount of hydrogen in the dead space is
3367.8
5
exp
12.762
−
298+∆T × 1.013 × 10 × 0.14387
pVds
=
.
µds =
8314(298 + ∆T )
8314(298 + ∆T )
(7)
In the last two equations, we represented the temperature of the contents of the canister as the sum of the initial temperature, 298 K, and the
temperature elevation, ∆T , owing to the absorption process.
The amount of gas absorbed by he alloy is
3367.8
exp 12.762 − 298+∆T
1.753
µalloy = µintro − µds = 1.0 −
.
(8)
298 + ∆T
Solution of Problem
11.22
090723
458
Page 3 of 3
Prob. Sol. 11.22
Fund. of Renewable Energy Processes
The absorption of µalloy kmoles of H2 causes the release of an amount,
Qrelsd , of heat

Qrelsd = 28 × 106 1.0 −
exp 12.762 −
3367.8
298+∆T
298 + ∆T
1.753

.
(9)
This heat causes temperature of both the alloy and the canister itself
to rise by an amount, ∆T .
Ignoring the heat capacity of the gas, the total heat capacity, Csys , of
the system is the sum of the heat capacity, Ccan , of the canister itself and
the heat capacity, Calloy , of the alloy which is
Malloy × Calloy = 100 × 540 = 54, 000
J/K. (10)
Thus,
Csys = 20, 000 + 54, 000 = 74, 000
J/K,
(11)
Intermediate Results
∆T
139.23 K
µds 0.6316 kmole
µalloy 0.3683 kmole
Qrelsd 10.31 MJ
and,
Qreleased = 74, 000∆T,


3367.8
exp 12.762 − 298+∆T
1.753
,
74, 000∆T = 28 × 106 1.0 −
298 + ∆T

∆T = 378.38 1.0 −
exp 12.762 −
3367.8
298+∆T
298 + ∆T
1.753

.
(12)
(13)
(14)
The solution of the above equation is ∆T = 139.25 which leads to
T = 437.27 K.
Introducing the correct value for ∆T into Equation 6, one finds that
the gas pressure is 157.57 atmospheres,
The pressure of the hydrogen in the dead space is 158 atmospheres.
Solution of Problem
090723
11.22
Fund. of Renewable Energy Processes
Prob. Sol. 11.23
Page 1 of 1
459
Prob 11.23 Solving this problem “exactly” is somewhat laborious.
What we want is simply an estimate of the final temperature. You must
make some simplifying assumptions. If you do this the hard way, you will
lose some points (even if you get the right answer) because you will be
wasting time. Reasonable simplifying assumptions will lead to estimates
with less than 2% error which is good enough for government work.
A small metal canister with a volume of 2 liters contains 3
kg of TiFeH0.95 (HY-STOR alloy 101) at 300 K. Alloy density is
6200 kg/m3 . The atomic mass of titanium is 47.90 daltons and
that of iron is 55.85 daltons.The canister has no heat capacity
and is completely heat-insolated from the environment. Estimate
the temperature of the gas in the cylinder after 2 g of H2 are
withdrawn.
.....................................................................................................................
The atomic mass of TiFeH0.95 is 104.3 daltons. The number of kilomoles of alloy in the canister is 3/104.3 = 28.8 × 10−3 kilomoles and the
amount of hydrogen in the alloy is 12 28.8 × 10−3 × 0.95 = 13.66 × 10−3
kmoles of H2 .
Volume of the “dead space” is
3
Vds = 0.002 −
= 0.001516 m3 .
(1)
6200
Clearly, if there are 0.95 kmoles of H in each kilomole (formula) of
alloy, the latter must be in the plateau region, i.e., the gas must be at a
pressure of
3368
−106,100 −28 × 106
= exp 12.76 −
.
(2)
+
p = exp −
8314
8314T
T
For T = 300 K, p=4.64 atmos.
Assume that the 2 g (0.001 kilomoles) of hydrogen withdrawn all come
from the hydrogen in the hydride. This ignores the fact that there is a
change in the amount of hydrogen in gas form, which, in this problem will
be considered negligible. Under this assumption, the energy absorbed by
the alloy to permit the desorption is Qdes = 28 × 106 µrem , where µrem =
0.001 kmoles (H2 ), is the number of kilomoles desorbed. Consequently,
Qdes = 28,000 J. This amount of energy must come from the cooling of the
alloy whose heat capacity is 3 × 540 = 1620 J/K,
1620 × (300 − Tf ) = 28,000.
(3)
where Tf is the final temperature of the alloy and is, in this case, 282.7 K
or, since we are only estimating this temperature, Tf = 283 K.
The gas temperature in the cylinder will be 283 K.
Solution of Problem
11.23
090723
460
Page 1 of 2
Prob. Sol. 11.24
Fund. of Renewable Energy Processes
Prob 11.24 An adiabatic canister having a volume of 0.05
m3 , contains 30 kg of alloy, HY-STOR 205 at 300 K. However
there is no gas of any kind in this canister.
The heat capacity of the canister, by itself, is negligible.
Admit µin = 0.1 kilomoles of hydrogen into the canister. After
a while, things will settle to a new equilibrium. What is the
pressure of the gas?
The density of HY-STOR 205(totallydepleted) is 8400 kg/m3 .
.....................................................................................................................
The volume of the depleted alloy is
Valloy =
30
= 0.00357 m3 .
8400
(1)
The dead space has a volume, Vds , of 0.05 − 0.00357 = 0.0464 m3 .
A fraction, α, of the hydrogen will be absorbed by the alloy, and a
fraction, 1 − α, will occupy the dead space.
The absorption will cause the release of heat from he alloy:
Q = µin α|∆H| = 0.1α × 31 × 106 = 3.1 × 106 α J.
(2)
This heat will raise the temperature of the alloy and of the free gas.
They have a heat capacity of 30 × 420 = 12,600 J/K for the alloy and
29,000 × 0.1 × (1 − α) = 2,900(1 − α) J/K for the gas. This totals 12,600 +
2,900(1 − α) J/K.
Consequently, a temperature raise of
∆T =
3.1 × 106 α
12,600 + 2,900(1 − α)
(3)
results leading to a final temperature of,
T = 300 + ∆T = 300 +
3.1 × 106 α
12,600 + 2,900(1 − α)
(4)
At this temperature, the plateau pressure will be,
3729
∆H
∆S
5
= 1.013 × 10 × exp 12.95 −
+
p = 1.013 × 10 × exp −
R
RT
T
(5)
The gas in the dead space will be at the same temperature as the alloy.
From the perfect gas law,
5
µds = µin (1 − α) =
Vds p
.
RT
Solution of Problem
090723
(6)
11.24
Fund. of Renewable Energy Processes
Prob. Sol. 11.24
Page 2 of 2

(1 − α) =


3729

6

3.1 × 10 α
300 +
12,600 + 2,900(1 − α)
3.1 × 106 α
8314 × 300 +
12,600 + 2,900(1 − α)
(7)



0.0464×1.013×105×exp
12.95 −
0.1(1−α) =
461

3729

6

3.1 × 10 α
300 +
12,600 + 2,900(1 − α)
3.1 × 106 α
300 +
12,600 + 2,900(1 − α)

5.654 × exp 
12.95 −
(8)
Although Equation 8 looks complicated, it can be seen that the only
unknown is α, and that it can be solved numerically. The result is α =
0.423.
We can now go to Equation 4 and find out T which turns out to be
T = 403.9 K. This value can, in turn, be applied to Equation 5. . This
yields p = 4,166,000 pascals or 41.1 atmospheres.
The hydrogen pressure is 41.1 atmospheres.
Solution of Problem
11.24
090723
462
Page 1 of 4
Prob. Sol. 11.25
Fund. of Renewable Energy Processes
Prob 11.25 The hydrogen distribution system of a given city
delivers the gas at a pressure of 20 atmos. An automobile refueling station must increase the pressure to 400 atmospheres to fill
the pressurized gas containers in the cars which, typically, has a
storage capacity of 6 kg of the gas.
Design a hydride hydrogen compressor for this application.
The compressor should have the ability of delivering 6 kg of compressed hydrogen (or somewhat less) in one single “stroke”, i.e.,
one single compression cycle. To make sure that there is an adequate gas flow from the city pipeline to the input of the compressor, the plateau pressure of the hydride used, at 25 C, should be
a bit lower than the 20 atmosphere pipe line pressure. Assume,
however that the intake plateau pressure is 20 atmos.
Normally, you should have complete freedom to select the
alloy to be used, however, to avoid disparate solutions to this
problem, we will impose some constraint in this choice. The alloy
must have:
∆Sabsorption = −106.8 kJ K−1 kmole−1 ,
A beginning of the plateau (transition between depletion and
plateau) when the stoichiometric coefficient, xbeg = 0.15.
An end of the plateau (beyond which the alloy saturates)
when xend = 1.05 − 0.00033T where T is the temperature in kelvins.
Use idealized characteristics of an alloy to build a hydrogen compressor capable of raising the pressure from 0.5 atmospheres to 50 atmospheres. Idealize characteristics have horizontal plateaus. The alloy has a heat capacity of 540 J kg−1 K−1 .
The hydrogen compressor has the following phases:
– Intake. Start from point “A” of the characteristics and move
to Point “B”, at constant pressure.
– Compression. Begins at point “B and goes to Point “C. Temperature and pressure increase.
– Exhaust. Begins at Point “C” and goes to Point “D”. Gas is
delivered at constant pressure.
– Reset. Goes from Point “D” to Point “A” completing the
cycle. Both pressure and temperature are reduced.
During the operation of the compressor, you must always
remain in the plateau region. You are not allowed to go into
either the depletion or the saturation region. It makes sense to
design the system so that the end of the exhaust phase (Point D)
is exactly at the beginning of the plateau (this allows the delivery
of the largest amount of hydrogen). Since Point A must have a
stoichiometric index larger than that of Point D, Point A cannot
be at the beginning of the the plateau. Use a stoichiometric index
Solution of Problem
090723
11.25
Fund. of Renewable Energy Processes
Prob. Sol. 11.25
Page 2 of 4
463
at A, xa = 0.36.
The alloy you are going to employ has the formula AB and
forms an hydride ABH. “A” has an atomic mass of 48 and “B”, of
56 daltons. The alloy has a density of 7000 kg/m3 . The internal,
empty, volume of the compressor is Vempty . Owing to the granular
nature of the alloy, only 60% of this volume is actually occupied
by the alloy, the rest is “dead space”, usually occupied by H2 gas.
The granules exactly fill Vempty , hence the “dead space” is only
the intergranular space.
a – What is the ∆H (absorption) that the alloy must have.
.....................................................................................................................
The requirement that at 25 C (298 K) the plateau pressure must be 20
atmos translates into
∆S
−106,800
∆H
∆H
= exp −
pin = exp −
+
+
R
RT
8314
8314 × 298
∆H
= exp 12.84 +
= 20.
(1)
2.478 × 106
12.84 +
∆H
= ln 20 = 3.0,
2.478 × 106
(2)
from which, ∆H = −24.4 MJ/kmole.
The alloy must have an absorption enthalpy of -24.4 MJ/kmole
b – Determine to what temperature must the alloy be heated
to achieve the desired compression.
.....................................................................................................................
The requirement that the output pressure be 400 atmos, leads to
ln pout = 12.84 +
−24.4 × 106
= ln 400 = 6.0
8314Tout
(3)
from which, Tout = 429 K or 156 C.
To effect the desired compression, the alloy must be heated to 156 C.
c – Assume that in the intake phase (A→B) the compressor
takes in 6 kg of hydrogen. Calculate how many kilograms of the
alloy are required.
.....................................................................................................................
To maximize the amount of hydrogen handled in the intake phase, the end
of that phase, Point B, must be at the edge of the plateau, i.e., xB = 1.05 −
Solution of Problem
11.25
090723
464
Page 3 of 4
Prob. Sol. 11.25
Fund. of Renewable Energy Processes
0.00033 × 298 = 0.952. Between Point A and Point B, the stoichiometric
index changed by ∆x = 0.952 − 0.36 = 0.591 and the amount of H-gas
taken in is 0.591 × µAlloy = 6 kmoles of H (6 kg of hydrogen as specified
by the problem statement. Hence, µAlloy = 6/0.591 = 10.1 kilomoles of
AB. Each kmole of alloy masses 48 + 56 = 104 kg. The total mass of alloy
needed is 10.1 × 104 = 1050 kg.
1050 kg of alloy are needed.
d – The above mass of alloy granules, exactly fills the internal
volume of the compressor. This leaves, as dead space, the intergranular volume. Calculate the internal volume of the compressor
and the volume of the dead space.
.....................................................................................................................
The density of the alloy is 7000 kg/m3 , but owing to the granularity of
the product, the alloy has an effective density of 0.6 × 7000 = 4200 kg/m3 .
Hence the granules will occupy a volume of 1050/4200= 0.250 m3 . The
dead space volume is 0.250 × 0.4 = 0.100 m3 .
The compressor has a volume of 0.250 m3 .
The dead space has a volume of 0.100 m3 .
e – How much H2 is desorbed while going from Point B to
Point C?
.....................................................................................................................
The amount of gas in the dead space at Point B is
µdsB =
20 × 105 × 0.1
pB Vds
=
= 0.081 kmole (H2 ).
RTB
8314 × 298
(4)
Repeating for Point C, we get µdsC = 1.125 kmole (H2 ).
The amount desorbed is 1.125 − 0.081 = 1.04 kmole (H2 ).
1.04 kmoles of hydrogen (H2 ) gas (2.08 kg) are desorbed.
f – Considering only the energy for desorption form Point
B to Point C and that from Point C to Point D plus the energy
necessary to heat up the alloy from the intake temperature to the
exhaust temperature, estimate the efficiency of the compressor.
The compressor itself has zero heat capacity; ignore it.
.....................................................................................................................
We have, from the previous item, knowledge of the amount of gas
desorbed in Step B→C: 1.04 kmoles of H2 . The amount desorbed in Step
C→D ought to be equal to the 6 kg (6 kmoles of H or 3 kmoles of H2 ),
Solution of Problem
090723
11.25
Fund. of Renewable Energy Processes
Prob. Sol. 11.25
Page 4 of 4
465
absorbed in Step A→B†. Thus the total desorbed from B→D is 4.04 kmoles
of H2 . This takes an energy of
Wdesor = 24.4 × 106 × 4.04 = 98.6 × 106
.
(5)
In addition, we need energy to raise the temperature of 1050 kg of alloy
from 298 K to 429 K (Step B→C), a ∆T of 131 K:
Wtemp = 1050 × 540 × 131 = 74.3 × 106 .
(6)
The total energy spent is 98.6 × 106 + 74.3 × 106 = 173 × 106 J.
The useful output is the energy, Wout , to isothermally compress the
gas from 20 to 400 atmos:
Wout = 3 × 8314 × 298 × ln(400/20) = 22.3 × 106
The efficiency is
η=
22.3 × 106
= 0.129.
172 × 106
J.
(7)
(8)
The efficiency of the compressor is 12.9%.
† If you calculate exactly the amount of hydrogen pumped out by the compressor, you will find that it is slightly larger than the amount taken in. In other
words, the compressor is not in exact steady state. The reason for this minor
discrepancy is that the initial value for xA = 0.36 that I suggested you use is not
the precise value for steady operation—the correct value is 0.3563.
Solution of Problem
11.25
090723
466
Page 1 of 5
Prob. Sol. 11.26
Fund. of Renewable Energy Processes
*************************** **************************** Prob
11.26 A system consists of two perfectly adiabatic 1-liter canisters interconnected by a pipe. A closed valve impedes the transfer of gas from one canister to another. There is no heat transfer
from one canister to the other even when the valve is opened.
Both canisters are at 300 K.
Canister A contains 5 kg of TiFe alloy and a total of 0.04 kg of
hydrogen (part in the form of gas and part chemically combined
with the alloy.) Canister B contains 5 kg of MgNi5 alloy and a
total of 0.02 kg of hydrogen.
Canisters and hydrogen have negligible heat capacity.
If the valve is opened, what will be the temperature of the
alloy in A and that of the alloy in B after equilibrium has been
established?
Data
ln p
xmin
xmax
x
Material
FeTi
MgNi5
∆Sf
kJ K−1
kmole−1
∆Hf
MJ
kmole−1
xmin
xmax
Heat
J K−1
kg−1
Density
kg
m−3
Molecular
mass
daltons
-106.1
-107.7
-28.0
-31.0
0.1
0.6
1.0
5.0
540
420
7,800
6,600
103.7
317.8
.....................................................................................................................
In solving this problem it is a good idea to make an educated guess of
what is going to happen. A plausible guess is that both alloys are in the
plateau region. To make sure, determine the initial conditions (before the
valve interconnecting the two canisters is opened). This is ease to do.
Next, consider that the plateau pressure of the two alloys—in this case,
the hydrogen pressure in the dead space of the two canisters—are going to
be different. When the valve is opened, hydrogen will be desorbed from one
alloy and absorbed by the other. Consequently, one alloy will cool down (its
temperature will be below 300 K) and the other will warm up (its temperature will be above 300 K). You can’t expect that the temperature will change
very much because there is only a modest amount of hydrogen in the system.
A preliminary wild guess would suggest something like 250 K for the colder
Solution of Problem
090723
11.26
Fund. of Renewable Energy Processes
Prob. Sol. 11.26
Page 2 of 5
467
alloy and 350 for the warmer.
To proceed with the calculations we assume that the amount of hydrogen
in the free space is small compared with that in the alloy (check later to see
if this is true). Write down the obvious condition that when the valve is
opened, the hydrogen pressure must be the same in the two canisters. This
establishes a relation between the two final temperatures, TA and TB .
Now we need another piece of information. The amount of hydrogen
desorbed by one alloy must equal (to a first approximation) the amount
absorbed by the other. This will give you the second relationship between
TA and TB .
Initial conditions
Both canisters are at 300 K. The plateau pressure is given by
∆H
∆S
+
p = exp −
R
RT
(1)
For Alloy A this leads to pplateauA0 = 4.64 atmos. and for Canister B,
pplateauB0 = 1.69 atmos.
The volume of the 5 kg TiFe (5/103.7=0.0482 kmoles) in Canister A
is 5/7800 = 6.41 × 10−4 m3 or 0.641 liters. That leaves 0.359 liters of dead
space (3.59 × 10−4 m3 ).
The 0.04 kg of hydrogen are partially in the dead space as normal
hydrogen gas at a pressure of 4.64 atmos or 4.70 × 105 Pa. From the
perfect gas law:
µdead
spaceA0
=
4.7 × 105 × 3.59 × 10−4
= 6.76 × 10−5
8314 × 300
kilomoles of H2
(2)
This leaves 0.02-0.00007 or 0.0199 kmoles of H2 or 0.0399 kilomoles of
H as part of the hydride. The stoichiometry is
xA0 =
0.0399
= 0.83.
0.0482
(3)
So, the alloy is not quite saturated.
The volume of the 5 kg MgNi5 (5/317.8=0.0157 kmoles) in Canister A
is 5/6600 = 7.58 × 10−4 m3 or 0.758 liters. That leaves 0.242 liters of dead
space (2.42 × 10−4 m3 ).
The 0.02 kg of hydrogen are partially in the dead space as normal
hydrogen gas at a pressure of 1.69 atmos or 1.71 × 105 Pa. From the
perfect gas law:
µdead
spaceB0
=
1.71 × 105 × 2.42 × 10−4
= 1.66 × 10−5
8314 × 300
Solution of Problem
kilomoles of H2
(4)
11.26
090723
468
Page 3 of 5
Prob. Sol. 11.26
Fund. of Renewable Energy Processes
This leaves 0.01-0.00002 or 0.01 kmoles of H2 or 0.02 kilomoles of H as
part of the hydride. The stoichiometry is
xA0 =
0.02
= 1.27.
0.0157
(5)
Since the alloy allows x to be as larger as 5, the alloy is quite far
from saturation and can absorb a large amount of hydrogen. It can hold
5 × 0.0157 = 0.0785 kmoles of H. It has capacity to hold an additional
0.0785 − 0.02 = 0.0585 kmoles of H.
The initial conditions are
Canister Plateau Alloy
Dead
Alloy
H2
H2
H
x
Press. volume space
Total
Total
D.S.
Alloy
(atmos) (liters) (liters) (moles) (moles) (moles) (moles)
A
4.64
0.641
0.359
48.22
20
0.0677
39.9
0.83
B
1.68
0.758
0.242
15.73
10
0.0166
20.0
1.27
We see that the hydride in Canister A, which is at 4.64 atmos, has the
empirical formula TiFeH0.83 and is nearly saturated but not quite, while in
Canister B, which is at 1.69 atmos, the hydride is MgNi5 H1.27 and, although
on the plateau, it is very far from saturation.
This suggests that when interconnected, hydrogen will flow from A to
B, and, plausibly, both alloys may wind up in the plateau region. This
simplifies calculations, but requires confirmation.
When the valve is opened, hydrogen at higher pressure in Canister
A will flow to Canister B. Temperature of A falls, reducing the plateau
pressure of A, while the temperature of B rises (owing to hydrogen absorption) and so will the plateau pressure. Equilibrium is re-established when
pplateauA = pplateauB . This occurs at temperature TE .
∆SB
∆HA
∆HB
∆SA
= exp −
(6)
+
+
exp −
R
RTA
R
RTB
∆HA
∆HB
= −∆SB +
TA
TB
(7)
31 × 106
28 × 106
= 107,700 −
TA
TB
(8)
−∆SA +
106,100 −
From this,
TB = 31
TA
.
28 + 0.0016TA
(9)
We need another independent piece of information relating TA to TB :
TA = 300 −
µA × ∆HA
= 300 − 10,370 µA .
2700
Solution of Problem
090723
(10)
11.26
Fund. of Renewable Energy Processes
Prob. Sol. 11.26
Page 4 of 5
469
In the above, 2700 is the heat capacity of the 5 kg of TiFe, and µA is
the number of kilomoles of H2 desorbed.
Similarly,
TB = 300 +
µB × ∆HB
= 300 + 14762 µB .
2100
(11)
The amount of hydrogen in the dead spaces is small compared with
that in the hydride, therefore on can take
µA ≈ µB
(12)
This allows eliminating µ from Equations 10 and 11,thus establishing
another relationship between TB and TA :
TB = 727.06 − 1.4235TA.
(13)
Equating Equation 9 to Equation 13, and thus eliminating TB , we get
the quadratic,
(14)
TA2 + 30,599.9TA − 8,938, 042 = 0,
whose solution is
TA = 289.4 K.
(15)
TB = 315.1 K.
(15)
Using Equation 13,
Alloy A cools down to 289.4 K, while Alloy B warms up to 315.1 K.
These are the solutions sought, however, it remains to prove that the
premise that both alloys are still in the plateau region is correct.
Using either Equation 10 or Equation 11, we can find the amount, µ,
of hydrogen desorbed from A or absorbed by B. It works out to
µ = 0.00102 kmole of H2 ,
(12)
which corresponds to 0.00204 kmoles of H.
Hydride A was TiFeH0.83 . It contained 0.0399 kmoles of H. After the
valve was open this amount fell to 0.0379 kilomoles of H and the hydride
composition became TiFeH0.78 . Alloy A is still in the plateau region.
Hydride B was MgNi5 H1.27 . It contained 0.020 kmoles of H. After the
valve was open this amount grew to 0.022 kilomoles of H and the hydride
composition became MgNi5 H1.40 . Alloy B is still in the plateau region.
Conclusion: Both alloys are still in the plateau region, as assumed.
Actually, the relative amount of hydrogen transfered from Canister A to
Canister B, is quite small (see figure below).
Solution of Problem
11.26
090723
470
Page 5 of 5
Prob. Sol. 11.26
Fund. of Renewable Energy Processes
TiFe
300 K
x = 0.83
ln (p)
x = 0.37
x = 1.40
TiFe (289.4 K)
and
MgNi5 (315.1 K)
MgNi5 (300 K)
x = 1.27
0
2
4
Stoichiometric index, x
6
Solution of Problem
090723
11.26
Fund. of Renewable Energy Processes
Prob. Sol. 11.27
Page 1 of 4
471
Prob 11.27 1 kg of the alloy AB is used to store hydrogen.
Its characteristics for adsorption of hydrogen are:
∆S = −110 kJ per K per kilomole.
∆H = −30MJ per kilomole.
Heat capacity, c = 500 J per kg.
Both A and B have atomic masses of 50 daltons.
The alloy is charged with hydrogen so that its average composition is ABH0.9 . It is at 400 K and is inside an adiabatic
container with negligible heat capacity. Neglect the heat capacity of the gas and assume that the alloy fills the container leaving
no dead space.
How much hydrogen is released if a valve is cracked open until
the pressure falls to 1.5 atmosphere?
.....................................................................................................................
The plateau pressure is
p = exp
30 × 106
110 × 103
−
8314
8314T
1
= exp 13.231 − 3608
T
(0)
For T = 400 K, p = 67.4 atmos. 2 atmos correspond to T = 288 K.
The desorption of hydrogen will proceed until the alloy cools to 288 K,
a temperature drop of 400 − 288 = 112 K.
All the heat must come from the alloy:
Q = m¸∆T = 1 × 500 × 112 = 56 × 1036
J
(0)
This amount of heat must come from the desorption of
N=
56103
= 0.00187
30
kmoles or 3.7 grams of hydrogen.
(1)
3.7 grams of hydrogen are released.
Solution of Problem
11.27
090723
472
Prob. Sol. Chapter 11
Fund. of Renewable Energy Processes