Tutorial PTT 108
Material and Energy Balance
1. Liquid methanol is pumped from a storage through a 1-in.ID pipe at a rate of 3.00
gal/min. At what rate is the kinetic energy being transported by the methanol in the
pipe. (Answer should be in SI unit).
Answer:
Check conversion unit table (all units should be converted to SI unit):
1 m = 39.37 in.
1 m3 = 264.17 gal
1 min = 60 s
ID = Inner diameter
= 1-in.
Volume flowrate
= 3.00 gal/min
+
A =πr2
Given that, velocity (m/s) = volume flowrate (m3/s) / cross-sectional area (m2)
[
]
[
]
Given that, mass flowrate (kg/s) = [volume flowrate (m3/s)] x [density (kg/m3)]
First determine the density of methanol: (Confirm the substance that you deal with in
the problem, don’t just simply take the density of water!)
Given that from Table B.1, the specific gravity (SG) for methanol is 0.792, thus the
density of methanol is 0.792 g/cm3.
̇
[
]
[
]
̇
Given that, rate of kinetic energy (J/s) = ½ x [mass flowrate (kg/s)] x [velocity (m/s)]2
̇
[
]
[
]
̇
= 0.0105 J/s
2. Use the psychrometric chart (given in the Appendix) to estimate (at temperature 30°C
and 20% relative humidity (refer Example 8.4-5 as a guideline):
i)
ii)
iii)
iv)
v)
The absolute humidity
The wet-temperature
Humid volume
The dew point
Specific enthalpy
Answer: (Refer to the attached chart)
i)
ii)
iii)
iv)
v)
The absolute humidity =0.0053 kg H2O/kg dry air
The wet-temperature = 15.6°C
Humid volume = 0.8665 m3/kg dry air
The dew point = 4.4°C
Specific enthalpy = (44 – 0.39) = 43.6 kJ/ kg dry air
3. Superheated steam at 10 bar absolute and 400°C is feed to a condenser at a flowrate
of 100 mole/h. The steam is cooled to 150 °C at 1 bar. Calculate the heat (Q) that
must be transferred to or from the condenser in kW
Q
Inlet stream = 100 mol/h
Outlet stream = 100 mol/h
Condenser
Water (400°C, 10 bar, v)
Water (150°C, 1 bar, v)
Using Steam Tables (Table B.7 in Appendix)
Answer:
Be careful with the unit,
it is in kJ/kg not kJ/mol!
At P = 10 bar, T = 400°C, Ĥ = 3264 kJ/kg
At P = 1 bar, T =150°C, Ĥ = 2776 kJ/kg
ṅin (mol/h)
100
Substance
Water (v)
Given that ̇
Thus,
Ĥin(kJ/kg)
3264
ṅout (mol/h)
100
Ĥout (kJ/kg)
2776
̇
̇
̇
The rate change of the enthalpy:
̇
∑ ̇ ̂
∑ ̇ ̂
̇
̇
Solve the energy balance equation (open system): Neglecting kinetic and potential
energy, no shaft work, thus the required heat input:
0
0
∆Ḣ + ∆Ek +∆Ep = Q -Ws
̇
̇
0
Using heat capacity data (Table B.2 in Appendix)
Answer:
∆ĤH2O
Water (400°C, 10 bar, v)
∆Ĥ1
Water (150°C, 1 bar, v)
∆Ĥ2
Water (400°C, 1 bar, v)
∆ĤH2O = ∆Ĥ1 + ∆Ĥ2
Refer to your hypothetical process paths for water.
∆Ĥ1 : Change in P from 10 bar to 1 bar at constant T. Assuming ideal gas, thus ∆Ĥ1
=0 kJ/mol
∆Ĥ2: Change in T from 400°C to 150°C at constant P.
Thus,
∆Ĥ2 ∫
From Table B.2 use equation Form 1 to determine the Cp:
Cp = a + bT + cT2 + dT3
H2O (v): Cp =33.46 x 10-3 + 0.6880 x 10-5T + 0.7604 x 10 -8T2 + (-3.593 x 10-12)T3
So do integrate,
=∫
) dT
[ ]
=
= -8.9695 kJ/mol
Thus,
̇
̇̂
̇
̇
Solve the energy balance equation (open system): Neglecting kinetic and potential
energy, no shaft work, thus the required heat input:
0
0
∆Ḣ + ∆Ek +∆Ep = Q -Ws
̇
0
̇
Using specific enthalpy data (Table B.8 in Appendix)
Answer:
Assuming ideal gas (H2O):
At T = 400°C, Ĥ = 13.23 kJ/mol
At T =150°C, Ĥ = ?
Do interpolation, since the T = 150°C falls between 100°C and 200°C:
At 100°C, Ĥ = 2.54 kJ/mol
At 200°C, Ĥ = 6.01 kJ/mol
Thus,
̂
̂
ṅin (mol/h)
100
Substance
Water (v)
Ĥin(kJ/mol)
13.23
ṅout (mol/h)
100
Ĥout (kJ/mol)
4.275
The rate change of the enthalpy:
̇
∑ ̇ ̂
∑ ̇ ̂
̇
̇
Solve the energy balance equation (open system): Neglecting kinetic and potential
energy, no shaft work, thus the required heat input:
0
0
∆Ḣ + ∆Ek +∆Ep = Q -Ws
̇
̇
0