TUTORIAL-5 (22/02/2017)
Thermodynamics for Aerospace Engineers (AS1300)
First Law applied to Control volume
1. Air enters compressor operating at steady state at a pressure of 1 bar, a
temperature of 290K, and a velocity of 6 m/s through an inlet with an area of 0.1 m2.
At exit, the pressure is 7 bar, the temperature is 450 K and the velocity is 2 m/s.
Heat transfer from the compressor to the surroundings occurs at the rate of 180
kJ/min. Employing the ideal gas model, calculate the power input to the
compressor. Take cp= 1.005 kJ/kgK
1
Wcv= ?
2
Solution:
The governing equation for a control volume according to the first law of Thermodynamics
can be written as follows:
2
2
𝑑𝐸
𝑢
𝑢𝑖𝑛
𝑄̇ − 𝑊̇ = 𝑑𝑡𝐶𝑉 + 𝑚̇𝑜𝑢𝑡 (ℎ𝑜𝑢𝑡 + 𝑜𝑢𝑡
+
𝑔𝑧
)
−
𝑚
̇
(ℎ
+
+ 𝑔𝑧𝑖𝑛 )
𝑜𝑢𝑡
𝑖𝑛
𝑖𝑛
2
2
[1.1]
Where 𝑚̇𝑖𝑛 & 𝑚̇𝑜𝑢𝑡 are the rates at which mass is flowing in and out of the control volume
respectively.
The continuity equation can be applied to the control volume as follows:
𝑑𝑚𝐶𝑉 ⁄𝑑𝑡 = 𝑚̇𝑖𝑛 − 𝑚̇𝑜𝑢𝑡
[1.2]
For a steady process the mass inside the control volume remains constant. So, according
to the equation 𝑑𝑚𝐶𝑉 ⁄𝑑𝑡 = 𝑚̇𝑖𝑛 − 𝑚̇𝑜𝑢𝑡 = 0 => 𝑚̇𝑖𝑛 = 𝑚̇𝑜𝑢𝑡
So the overall mas flow rate across the boundaries of the control volume can be obtained
as follows:
𝑚̇𝑖𝑛 = 𝜌𝑖𝑛 𝐴𝑖𝑛 𝑉𝑖𝑛 = 𝜌𝑜𝑢𝑡 𝐴𝑜𝑢𝑡 𝑉𝑜𝑢𝑡 = 𝑚̇𝑜𝑢𝑡
[1.3]
The velocity and area information for the inlet is mentioned in the question. Applying the
ideal gas law equation on pressure and temperature the density value at the inlet of the
control volume can be found out.
1
𝑣𝑖𝑛
= 𝜌𝑖𝑛 = 𝑅
𝑃𝑖𝑛
𝑎𝑖𝑟 ∗𝑇𝑖𝑛
105
= 287∗290 = 1.201 𝐾𝑔/𝑚3
[1.4]
Here, Rair is the specific gas constant which is 287 J/Kg K
So, the mass flow rate can be calculated from this information:
𝑚̇ = 𝜌𝑖𝑛 𝐴𝑖𝑛 𝑉𝑖𝑛 = 1.201 ∗ 0.1 ∗ 6 = 0.721 𝐾𝑔/𝑠
[1.5]
In the equation 1.1 steady state process can be applied to obtain the final expression as
follows:
2
2
𝑢
𝑢
𝑄̇ − 𝑊̇ = 𝑚̇𝑜𝑢𝑡 (ℎ𝑜𝑢𝑡 + 𝑜𝑢𝑡
) − 𝑚̇𝑖𝑛 (ℎ𝑖𝑛 + 𝑖𝑛
)
2
2
[1.6]
𝑑𝐸
Here 𝑑𝑡𝐶𝑉 = 0 , as the energy of the control volume doesn’t change in a steady process
similar to the mass of the control volume. Along with this the potential energy is also
neglected.
So the final calculation for the compressor work is as follows:
2
2
𝑢
𝑢𝑖𝑛
𝑊̇ = 𝑄̇ − 𝑚̇𝑜𝑢𝑡 (ℎ𝑜𝑢𝑡 + 𝑜𝑢𝑡
)
+
𝑚
̇
(ℎ
+
)
𝑖𝑛
𝑖𝑛
2
2
𝑄̇ = −180
𝑘𝐽
𝑚𝑖𝑛
= −3
𝑘𝐽
𝑠
[1.7]
= −3 𝑘𝑊
[1.8]
ℎ𝑖𝑛 − ℎ𝑜𝑢𝑡 = 𝐶𝑃 𝑎𝑖𝑟 ∗ (𝑇𝑖𝑛 − 𝑇𝑜𝑢𝑡 ) = 1005 ∗ (290 − 450) = −160800 𝐽/𝑘𝑔
𝑢𝑜𝑢𝑡 = 2
𝑚
𝑠
, 𝑢𝑖𝑛 = 6
𝑚
𝑠
[1.9]
𝑘𝑔
̇
& 𝑚̇ = 0.721 𝑠
So the compressor work is:
2
2
6 −2
𝐽
𝑊̇ = −3000 + 0.721 ∗ (−160800) + 0.721 ∗ ( 2 ) = −118925.26 𝑠 = −𝟏𝟏𝟖. 𝟗𝟐 𝒌𝑾 Ans
The negative sign shows that, the work is done on the system or energy is given to the
compressor.
2. In a gas turbine the gas enters at the rate of 5kg/s with a velocity of 50 m/s and
enthalpy of 900 kJ/kg leaves the turbine with a velocity of 150 m/s and enthalpy of
400 kJ/kg. The loss of heat from the gases to the surroundings is 25 kJ/kg. Assume
for gas R= 0.285 kJ/kgK and cp = 1.004 kJ/kgK and inlet conditions to be at 100 kPa
and 27°C. Determine the power output of the turbine.
1
Wcv= ?
Q
2
Solution:
The governing equation for a control volume according to the first law of Thermodynamics
can be written as follows:
2
2
𝑑𝐸
𝑢
𝑢
𝑄̇ − 𝑊̇ = 𝐶𝑉 + 𝑚̇𝑜𝑢𝑡 (ℎ𝑜𝑢𝑡 + 𝑜𝑢𝑡 + 𝑔𝑧𝑜𝑢𝑡 ) − 𝑚̇𝑖𝑛 (ℎ𝑖𝑛 + 𝑖𝑛 + 𝑔𝑧𝑖𝑛 )
𝑑𝑡
2
2
[2.1]
Where 𝑚̇𝑖𝑛 & 𝑚̇𝑜𝑢𝑡 are the rates at which mass is flowing in and out of the control volume
respectively.
The continuity equation can be applied to the control volume as follows:
𝑑𝑚𝐶𝑉 ⁄𝑑𝑡 = 𝑚̇𝑖𝑛 − 𝑚̇𝑜𝑢𝑡
[2.2]
For a steady process the mass inside the control volume remains constant. So, according
to the equation 𝑑𝑚𝐶𝑉 ⁄𝑑𝑡 = 𝑚̇𝑖𝑛 − 𝑚̇𝑜𝑢𝑡 = 0 => 𝑚̇𝑖𝑛 = 𝑚̇𝑜𝑢𝑡
The overall mas flow rate across the boundaries of the control volume is given in the
question.
𝑚̇𝑖𝑛 = 𝜌𝑖𝑛 𝐴𝑖𝑛 𝑉𝑖𝑛 = 𝜌𝑜𝑢𝑡 𝐴𝑜𝑢𝑡 𝑉𝑜𝑢𝑡 = 𝑚̇𝑜𝑢𝑡 = 5 𝐾𝑔/𝑠
[2.3]
In equation 2.1 steady state process can be applied to obtain the final expression as
follows:
2
2
𝑢
𝑢
𝑄̇ − 𝑊̇ = 𝑚̇𝑜𝑢𝑡 (ℎ𝑜𝑢𝑡 + 𝑜𝑢𝑡
) − 𝑚̇𝑖𝑛 (ℎ𝑖𝑛 + 𝑖𝑛
)
2
2
[2.4]
𝑑𝐸
Here 𝑑𝑡𝐶𝑉 = 0 , as the energy of the control volume doesn’t change in a steady process
just like mass of the control volume. Along with this the potential energy is also neglected.
So the final calculation for the turbine work is as follows:
2
2
𝑢
𝑢
𝑊̇ = 𝑄̇ + 𝑚̇(ℎ𝑖𝑛 − ℎ𝑜𝑢𝑡 ) + 𝑚̇ ( 𝑖𝑛
− 𝑜𝑢𝑡
)
2
2
[2.5]
𝑘𝐽
𝑄̇ = −25 𝐾𝑔 ∗ 𝑚̇ = −25 ∗ 5 = −125 𝑘𝑊
[2.6]
𝑚 ̇ ∗ (ℎ𝑖𝑛 − ℎ𝑜𝑢𝑡 ) = 5 ∗ (900 − 400) = 5000 𝑘𝑊
[2.7]
𝑢𝑜𝑢𝑡 = 150
𝑚
𝑠
, 𝑢𝑖𝑛 = 50
𝑚
𝑠
So the compressor work is:
2 −1502
50
𝑊̇ = −125 + 5000 + 5 ∗ (
2
) ∗ 10−3 = −125 + 5000 − 50 = 𝟐𝟑𝟐𝟓 𝒌𝑾
Ans
The positive sign shows that, the work is done by the system or energy is produced by the
turbine.
3. The air speed of a turbojet engine in flight is 270 m/s. Ambient air temperature is 15°C. Gas temperature of outlet of nozzle is 600°C. Corresponding enthalpy values
for air and gas are respectively 260 and 912kJ/kg. Fuel-air ratio is 0.0190. Chemical
energy of the fuel is 44.5 MJ/kg. Owing to incomplete combustion 5% of the
chemical energy is not released in the reaction. Heat loss from the engine is 21
kJ/kg of air. Calculate the velocity of the exhaust jet.
Solution:
The governing equation for a control volume according to the first law of Thermodynamics
can be written as follows:
2
2
𝑑𝐸
𝑢
𝑢𝑖𝑛
𝑄̇ − 𝑊̇ = 𝑑𝑡𝐶𝑉 + 𝑚̇𝑜𝑢𝑡 (ℎ𝑜𝑢𝑡 + 𝑜𝑢𝑡
+
𝑔𝑧
)
−
𝑚
̇
(ℎ
+
+ 𝑔𝑧𝑖𝑛 )
𝑜𝑢𝑡
𝑖𝑛
𝑖𝑛
2
2
[3.1]
Where 𝑚̇𝑖𝑛 & 𝑚̇𝑜𝑢𝑡 are the rates at which mass is flowing in and out of the control volume
respectively.
The continuity equation can be applied to the control volume as follows:
𝑑𝑚𝐶𝑉 ⁄𝑑𝑡 = 𝑚̇𝑖𝑛 − 𝑚̇𝑜𝑢𝑡
[3.2]
For a steady process the mass inside the control volume remains constant. So, according
to the equation 𝑑𝑚𝐶𝑉 ⁄𝑑𝑡 = 𝑚̇𝑖𝑛 − 𝑚̇𝑜𝑢𝑡 = 0 => 𝑚̇𝑖𝑛 = 𝑚̇𝑜𝑢𝑡
In this question there are two inlets and one outlet. The air and the fuel comes into the
turbojet engine through separate inlet and the burned mixture leaves the turbojet at the
exit. So, this is a multiple inlet problem. But as at the fuel inlet liquid fuel comes into the
control volume we neglect the enthalpy of the fuel. The kinetic and potential energy of the
fuel is also neglected. Instead, we can add a term to the heat side of the equation 3.1 for
the heat produced during the combustion of fuel-air mixture.
Remember, in the class we considered the enthalpy of the fuel to be 44.5 MJ/Kg and didn’t
add any extra term to the heat production term in the equation 3.1. Both of these
approaches will give correct answer but it is recommended to follow the first approach
because 44.5 MJ/Kg is the heat of the reaction i.e. heat produced per Kg of fuel when it
reacts with air.
The work done by the control volume (turbojet) is zero. This is because in a turbojet, the
turbine provides the work that is needed to run the compressor. So the total work done is
zero.
So, the final equation can be written as follows:
2
2
𝑢𝑔𝑎𝑠
𝑢
𝑄̇ = 𝑚̇𝑔𝑎𝑠 (ℎ𝑔𝑎𝑠 + 2 ) − 𝑚̇𝑎𝑖𝑟 (ℎ𝑎𝑖𝑟 + 𝑎𝑖𝑟
)
2
[3.3]
The potential energy and the contribution from fuel inlet is neglected.
Let us assume the mass flow rate of the air to be 𝑚̇𝑎𝑖𝑟 . You will note that the left and right
side of the equation 3.3 can be written as a function of 𝑚̇𝑎𝑖𝑟 and it can be cancelled from
the both side. This will give us the expression with only one unknown i.e. the exit speed
(𝑢𝑔𝑎𝑠 ) of the combusted mixture (here we call it gas).
The mass flow rate of the fuel is : 𝑚̇𝑓𝑢𝑒𝑙 = 0.019 ∗ 𝑚̇𝑎𝑖𝑟
[3.4]
The mass flow rate of the fuel-air combusted mixture (gas) from continuity equation is :
𝑚̇𝑜𝑢𝑡 = 𝑚̇𝑔𝑎𝑠 = 𝑚̇𝑖𝑛 = 𝑚̇𝑔𝑎𝑠 + 𝑚̇𝑓𝑢𝑒𝑙 = 1.019 ∗ 𝑚̇𝑎𝑖𝑟
[3.5]
The heat production term can be written as:
𝑄̇ = −𝑚̇𝑎𝑖𝑟 ∗ 21 ∗ 103 + 0.019 ∗ 𝑚̇𝑎𝑖𝑟 ∗ 44.5 ∗ 106 ∗ 0.95
[3.6]
Here 0.95 is multiplied to the second term because it is mentioned in the question that due
to incomplete combustion 5% energy is not released from the reaction. So, only 95 % of
the heat released by the complete combustion of fuel and air is taken while doing the
calculation.
So, substituting equation 3.4, 3.5 and 3.6 in equation 3.4, we can obtain the value of the
velocity of the exhaust jet (𝑢𝑔𝑎𝑠 )
2
2
𝑢𝑔𝑎𝑠
𝑢
𝑄̇ = 𝑚̇𝑔𝑎𝑠 (ℎ𝑔𝑎𝑠 + 2 ) − 𝑚̇𝑎𝑖𝑟 (ℎ𝑎𝑖𝑟 + 𝑎𝑖𝑟
)
2
2
𝑢𝑔𝑎𝑠 = √2 ∗ (
= √2 ∗ (
𝑢
𝑄̇ −𝑚̇𝑔𝑎𝑠 ℎ𝑔𝑎𝑠 +𝑚̇𝑎𝑖𝑟 (ℎ𝑎𝑖𝑟 + 𝑎𝑖𝑟 )
2
𝑚̇𝑔𝑎𝑠
)
−𝑚̇𝑎𝑖𝑟 ∗21∗103 +0.019∗𝑚̇𝑎𝑖𝑟 ∗44.5∗106 ∗0.95−1.019∗𝑚̇𝑎𝑖𝑟 ∗912∗103 +𝑚̇𝑎𝑖𝑟 ∗(260∗103 +
1.019∗𝑚̇𝑎𝑖𝑟
2702
)
2
) = 𝟓𝟒𝟏. 𝟒𝟏m/s
Ans
4. A tank of volume 0.3 m3 is initially filled with air at a pressure and temperature of
3.5 MPa and 400°C. The air is now allowed to discharge slowly through a turbine
into the atmosphere until the pressure in the tank falls to the atmospheric pressure
of 0.1 MPa. Determine the work developed by the turbine. Neglect friction, heat
loss, KE & PE changes.
Solution:
The air at an initial temperature and pressure of 4000 C and 3.5 MPa occupies a tank of
volume of 𝑉 = 0.3 m3. This air is then discharged through a turbine into the atmosphere
until the pressure inside the tank falls to the atmospheric pressure (0.1 MPa). Consider the
control volume comprising of the tank and the turbine as shown in Fig. 4. Although there is
an outflow of mass from the control volume, there is no influx of mass into the control
volume. Therefore, the mass inside the control volume continuously decreases during the
process until the final equilibrium state is reached. At this point, we must realize that we
are dealing with an unsteady problem.
Fig. 4
The continuity equation is given by 𝑑𝑚𝐶𝑉 ⁄𝑑𝑡 = 𝑚̇𝑖𝑛 − 𝑚̇𝑜𝑢𝑡
Since, 𝑚̇𝑖𝑛 = 0 and 𝑚̇𝑜𝑢𝑡 = 𝑚̇2 , the above equation reduces to 𝑑𝑚𝐶𝑉 ⁄𝑑𝑡 =
−𝑚̇2
(4.1)
Apply first law for the control volume given by
𝑄̇ − 𝑊̇ =
𝑑𝐸𝐶𝑉
𝑢𝑜𝑢𝑡 2
𝑢𝑖𝑛 2
+ 𝑚̇𝑜𝑢𝑡 (ℎ𝑜𝑢𝑡 +
+ 𝑔𝑧𝑜𝑢𝑡 ) − 𝑚̇𝑖𝑛 (ℎ𝑖𝑛 +
+ 𝑔𝑧𝑖𝑛 )
𝑑𝑡
2
2
Neglecting friction, heat loss, KE and PE changes in the problem, the first law reduces to
−𝑊̇ =
𝑑𝐸𝐶𝑉
+ 𝑚̇2 (ℎ2 )
𝑑𝑡
(4.2)
Note that ℎ𝑜𝑢𝑡 = ℎ2 . Also, 𝐸𝐶𝑉 = 𝑚𝐶𝑉 𝑒𝐶𝑉
Substitute equation (4.1) in (4.2) to obtain
−𝑊̇ =
𝑑(𝑚𝐶𝑉 𝑒𝐶𝑉 ) 𝑑𝑚𝐶𝑉
(ℎ2 )
−
𝑑𝑡
𝑑𝑡
Integrate the above equation from the initial to the final state to obtain the total work done.
−𝑊 = (𝑚𝑓 𝑒𝑓 − 𝑚𝑖 𝑒𝑖 ) − (𝑚𝑓 − 𝑚𝑖 )ℎ2
(4.3)
where the subscripts 𝑖 and 𝑗 denote the initial and final states of air in the control volume.
The next step is to find the final state of the air.
The expansion through the turbine is isentropic (adiabatic and frictionless).
Therefore, pressure and temperature inside the control volume and outside are related as:
𝛾
𝑃𝐶𝑉 ⁄𝑃2 = (𝑇𝐶𝑉 ⁄𝑇2 )𝛾−1
(4.4)
Whatever be the pressure inside the control volume, the transient process will be along
the same adiabatic curve. The exit pressure 𝑃2 , initial temperature (𝑇𝐶𝑉,𝑖 ) and pressure
(𝑃𝐶𝑉,𝑖 ) in the control volume are given. Using these information along with equation (4.4),
we can obtain the exit temperature 𝑇2 = 243.7 K, for 𝛾 = 1.4.
The exit temperature (𝑇2 ) and pressure (𝑃2 ) are always a constant. The air outflow stops
when 𝑃𝐶𝑉,𝑓 = 𝑃2 , and hence 𝑇𝐶𝑉,𝑓 = 𝑇2 .
Once we know 𝑇𝐶𝑉,𝑖 and 𝑃𝐶𝑉,𝑖 , we can calculate the initial mass inside the control volume
using the perfect gas law 𝑚𝑖 = 𝑃𝐶𝑉,𝑖 𝑉 ⁄𝑅𝑇𝐶𝑉,𝑖 = 5.436 kg, where 𝑅 = 287 J/kg K.
Similarly, using 𝑃𝐶𝑉,𝑓 and 𝑇𝐶𝑉,𝑓 we can calculate 𝑚𝑓 = 0.429 kg.
We know that 𝑒𝑓 = 𝐶𝑉 𝑇𝐶𝑉,𝑓 and 𝑒𝑖 = 𝐶𝑉 𝑇𝐶𝑉,𝑖 , where 𝐶𝑉 = 𝑅 ⁄(𝛾 − 1) = 717.5 J/kg K.
Also, ℎ2 = 𝐶𝑃 𝑇2 = 𝐶𝑃 𝑇𝐶𝑉,𝑓 , where 𝐶𝑃 = 𝛾𝐶𝑉 = 1004.5 J/kg K.
Substitute the above quantities in equation (4.3) to obtain 𝑊 = 𝟏𝟑𝟐𝟒. 𝟐𝟏 kJ.
Ans
5. An aircraft gas turbine engine has air mass flow rate of 100 kg/s. The air enters
compressor at inlet temperature of 300K and 1atm absolute pressure. The
compressed air at 15 atm from compressor is heated (Q) in a burner to 1500K and
expanded in a turbine. The shaft work output (Ws) from turbine is used for driving
the compressor.
a) Calculate the gas temperature (T4) at the exit of the turbine.
b) If the nozzle‐exit temperature (T5) is 600K. What is the exhaust velocity (V5) of
gases coming out of the nozzle?
In each of the above parts clearly indicate the appropriate control volume.
Solution:
Given mass flow rate 𝑚̇ = 100 kg/s, inlet temperature 𝑇1 = 300 K, 𝑃1 = 1 atm, 𝑃2 = 15
atm, 𝑇3 = 1500 K and 𝑇5 = 600 K.
Figure 5.1
Throughout the engine, the change in PE is small and hence neglected. The processes in
turbine and compressor are isentropic expansion and compression respectively. Also, it is
given that the work extracted by the turbine is used to run the compressor. The entire
process is steady (𝑑𝑚𝐶𝑉 ⁄𝑑𝑡 = 0 and 𝑑𝐸𝐶𝑉 ⁄𝑑𝑡 = 0). Let, 𝑚̇𝑒 = 𝑚̇𝑖 = 𝑚̇.
APPROACH 1:
(a) Consider a control volume comprising of the compressor, burner and turbine as shown
in Figure 5.2. Work is done on the gas inside the control volume by the compressor. The
same work is extracted from the gas by the turbine. Thus, the net work done on the gas
inside the control volume is zero (𝑊̇ = 0). There is a heat release inside the control
volume from the burner. The change in KE is assumed small within this control volume.
Figure 5.2
Apply first law to this control volume to get
𝑄̇ = 𝑚̇(ℎ4 − ℎ1 ) = 𝑚̇𝐶𝑃 (𝑇4 − 𝑇1 )
(5.1)
Now, in order to relate 𝑄̇ to 𝑇3 , consider a control volume around the burner. In this control
volume, the net work done is zero. The first law becomes 𝑄̇ = 𝑚̇(ℎ3 − ℎ2 ) = 𝑚̇𝐶𝑃 (𝑇3 − 𝑇2 ).
Substitute this expression in equation (5.1) to obtain (assuming 𝐶𝑃 = constant)
𝑇3 − 𝑇2 = 𝑇4 − 𝑇1
(5.2)
Apply isentropic relation across the compressor to get, 𝑇2 = 𝑇1 (𝑃2 ⁄𝑃1 )
𝛾−1
𝛾
= 650.35 K.
Now we can solve for 𝑇4 in equation (5.2). Thus,
𝑇4 ≈ 1150 K.
(b) Given 𝑇5 = 600 K. Consider a control volume consisting of only the nozzle. The gas
undergoes an isentropic expansion inside this nozzle with no heat transfer or external
work mechanism. The first law applied to this control volume becomes
𝑚̇ (ℎ5 +
𝑢5 2
𝑢4 2
) − 𝑚̇ (ℎ4 +
)=0
2
2
Here, 𝑢4 2 ⁄2 ≪ ℎ4 . Therefore neglect the term 𝑢4 2 ⁄2 in comparison to the specific
enthalpy.
Thus, 𝑢5 = √2𝐶𝑃 (𝑇5 − 𝑇4 ) ≈ 1050 m/s.
APPROACH 2:
We can use a second approach to solve part (a) of the question. In this approach,
consider a control volume around the compressor. Temperature 𝑇2 is calculated using the
isentropic relation as in the first approach. Then, apply the first law to calculate the work
done on the gas by the compressor. Next, consider a second control volume around the
turbine. Apply first law for this control volume and using the condition: 𝑊𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑜𝑟 =
−𝑊𝑡𝑢𝑟𝑏𝑖𝑛𝑒 , evaluate 𝑇4 .