Question

Air at 2.5 bar, 400 K is extracted from a main jet engine compressor for cabin cooling. The extracted air enters a heat exchanger where it is cooled at constant pressure to 300 K through heat transfer with the ambient. It then expands adiabatically to 1.0 bar through a turbine and is discharged into the cabin. The turbine has an isentropic efficiency of 80%.
If the mass flow rate of the air is 2.5 kg/s, determine:

(a) the power developed by the turbine, in kw.
(b) the magnitude of the rate of heat transfer from the air to the ambient, in kw

Answer 1

Answer:

a) 132.89 kW

b) 251.25 kW

Explanation:

a) For the isentropic process:

Power developed by the turbine is given by the relation \dot{W} = \dot{M} c_{p} (T_{2} - T_{3})

Isentropic efficiency, \eta_{t} = 80%

P₂ = 2.5 bar, T₂ = 300 K

P₃ = 1 bar, T_{3s} = ? where T_{3s} is the isentropic temperature at 100% efficiency

The isentropic relation is given by:

\frac{T_{3s} }{T_{2} } = (\frac{P_{3} }{P_{2} }) ^{\frac{\gamma - 1}{\gamma} } \\\frac{T_{3s} }{300 } = (\frac{1 }{2.5 }) ^{\frac{1.4 - 1}{1.4 }

T_{3s} = 230.9 K

To evaluate the temperature at 80% efficiency, we will use the following method:

\eta_{t} = \frac{T_{2} - T_{3} }{T_{2} - T_{3s} } \\0.8= \frac{300 - T_{3} }{300 - 230.9 }

T₃ = 244.72 K

The power developed by the turbine is given by the relation:

\dot{W} = \dot{M} c_{p} (T_{2} - T_{3})\\ \dot{W} = 2.5 * 1.005* (300-244.72)\\ \dot{W} = 138.89 kW

b)

The constant pressure specific heat of air, c_{p} = 1.005 kJ/kg -K

Specific heat ratio for air, \gamma = 1.4

The mass flow rate of air, \dot{m} = 2.5 kg/s

P₁ = 2.5 bar, T₁ = 400 K

P₂ = 2.5 bar, T₂ = 300 K

We going to the steady flow energy equation using this equation:

Q_{1-2} = \dot{m} c_{p} (T_{2} - T_{1} \\Q_{1-2} = 2.5 * 1.005 * (300 - 400)\\Q_{1-2} = -251.25 kW

Hence, the magnitude of the rate of heat transfer from the air to the ambient, in is going to be kw, Q_{1-2} = 251.25 kW

Answer 2

Answer:

a) Power developed by the turbine = 132.89 kW

b) magnitude of the rate of heat transfer from the air to the ambient, in kw = 251.25 kW

Explanation:

b) The process is a constant pressure process (Isobaric process)

The constant pressure specific heat of air, $c_{p} = 1.005 kJ/kg -K$

Specific heat ratio for air, $\gamma = 1.4$

The mass flow rate of air, $\dot{m} = 2.5 kg/s$

P₁ = 2.5 bar, T₁ = 400 K

P₂ = 2.5 bar, T₂ = 300 K

Using the steady flow energy equation:

$Q_{1-2} = \dot{m} c_{p} (T_{2} - T_{1} \\Q_{1-2} = 2.5 * 1.005 * (300 - 400)\\Q_{1-2} = -251.25 kW$

Therefore, the magnitude of the rate of heat transfer from the air to the ambient, in kw, $Q_{1-2} = 251.25 kW$

a) For the isentropic process:

Power developed by the turbine is given by the relation $\dot{W} = \dot{M} c_{p} (T_{2} - T_{3})$

Isentropic efficiency, $\eta_{t} = 80%$

P₂ = 2.5 bar, T₂ = 300 K

P₃ = 1 bar, $T_{3s}$ = ? where $T_{3s}$ is the isentropic temperature at 100% efficiency

The isentropic relation is given by:

$\frac{T_{3s} }{T_{2} } = (\frac{P_{3} }{P_{2} }) ^{\frac{\gamma - 1}{\gamma} } \\\frac{T_{3s} }{300 } = (\frac{1 }{2.5 }) ^{\frac{1.4 - 1}{1.4 }$

$T_{3s} = 230.9 K$

To get the temperature at 80% efficiency, we will use the relation:

$\eta_{t} = \frac{T_{2} - T_{3} }{T_{2} - T_{3s} } \\0.8= \frac{300 - T_{3} }{300 - 230.9 }$

T₃ = 244.72 K

Power developed by the turbine is given by the relation:

$\dot{W} = \dot{M} c_{p} (T_{2} - T_{3})\\ \dot{W} = 2.5 * 1.005* (300-244.72)\\ \dot{W} = 138.89 kW$