Respuesta :
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, [tex]c_{p} = 1.005 kJ/kg -K[/tex]
Specific heat ratio for air, [tex]\gamma = 1.4[/tex]
The mass flow rate of air, [tex]\dot{m} = 2.5 kg/s[/tex]
P₁ = 2.5 bar, T₁ = 400 K
P₂ = 2.5 bar, T₂ = 300 K
Using the steady flow energy equation:
[tex]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[/tex]
Therefore, the magnitude of the rate of heat transfer from the air to the ambient, in kw, [tex]Q_{1-2} = 251.25 kW[/tex]
a) For the isentropic process:
Power developed by the turbine is given by the relation [tex]\dot{W} = \dot{M} c_{p} (T_{2} - T_{3})[/tex]
Isentropic efficiency, [tex]\eta_{t} = 80%[/tex]
P₂ = 2.5 bar, T₂ = 300 K
P₃ = 1 bar, [tex]T_{3s}[/tex] = ? where [tex]T_{3s}[/tex] is the isentropic temperature at 100% efficiency
The isentropic relation is given by:
[tex]\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 }[/tex]
[tex]T_{3s} = 230.9 K[/tex]
To get the temperature at 80% efficiency, we will use the relation:
[tex]\eta_{t} = \frac{T_{2} - T_{3} }{T_{2} - T_{3s} } \\0.8= \frac{300 - T_{3} }{300 - 230.9 }[/tex]
T₃ = 244.72 K
Power developed by the turbine is given by the relation:
[tex]\dot{W} = \dot{M} c_{p} (T_{2} - T_{3})\\ \dot{W} = 2.5 * 1.005* (300-244.72)\\ \dot{W} = 138.89 kW[/tex]
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
