Air enters an adiabatic compressor at 290k at 1 atm and exists at 610k and 10atm at a range of 10kg/s. Using constant specific heat values at the average temperature of points 1 and 2, calculate:
Isentropic and actual works in KW and the 1st law isentropic efficiency
The reversible work, 2nd law efficiency and the irreversibility for this process. Use to 300k
The correct answer and explanation is :
To analyze the performance of the adiabatic compressor with the given conditions, we’ll calculate the isentropic and actual work inputs, the first law (isentropic) efficiency, the reversible work, the second law efficiency, and the irreversibility of the process.
Given Data:
- Inlet temperature, ( T_1 = 290\, \text{K} )
- Inlet pressure, ( P_1 = 1\, \text{atm} )
- Outlet temperature, ( T_2 = 610\, \text{K} )
- Outlet pressure, ( P_2 = 10\, \text{atm} )
- Mass flow rate, ( \dot{m} = 10\, \text{kg/s} )
- Specific heat ratio for air, ( \gamma = 1.4 )
- Gas constant for air, ( R = 0.287\, \text{kJ/kg·K} )
1. Isentropic Work:
For an isentropic process, the temperature ratio is related to the pressure ratio by:
[ \left( \frac{T_{2s}}{T_1} \right) = \left( \frac{P_2}{P_1} \right)^{\frac{\gamma – 1}{\gamma}} ]
Calculating ( T_{2s} ):
[ T_{2s} = 290\, \text{K} \times (10)^{\frac{1.4 – 1}{1.4}} \approx 290\, \text{K} \times 2.639 \approx 765.31\, \text{K} ]
The specific work input for an isentropic process is:
[ w_s = c_p (T_{2s} – T_1) ]
Assuming ( c_p = 1.005\, \text{kJ/kg·K} ):
[ w_s = 1.005\, \text{kJ/kg·K} \times (765.31\, \text{K} – 290\, \text{K}) \approx 1.005 \times 475.31 \approx 478.69\, \text{kJ/kg} ]
The isentropic power input is:
[ \dot{W}_s = \dot{m} \times w_s = 10\, \text{kg/s} \times 478.69\, \text{kJ/kg} = 4786.9\, \text{kW} ]
2. Actual Work:
The actual specific work input is:
[ w_a = c_p (T_2 – T_1) = 1.005\, \text{kJ/kg·K} \times (610\, \text{K} – 290\, \text{K}) = 1.005 \times 320 = 321.6\, \text{kJ/kg} ]
The actual power input is:
[ \dot{W}_a = \dot{m} \times w_a = 10\, \text{kg/s} \times 321.6\, \text{kJ/kg} = 3216\, \text{kW} ]
3. Isentropic Efficiency:
[ \eta_{\text{isentropic}} = \frac{\dot{W}_s}{\dot{W}_a} = \frac{4786.9\, \text{kW}}{3216\, \text{kW}} \approx 1.488 ]
4. Reversible Work:
The reversible work is equivalent to the isentropic work:
[ \dot{W}_{\text{rev}} = \dot{W}_s = 4786.9\, \text{kW} ]
5. Second Law Efficiency:
[ \eta_{\text{II}} = \frac{\dot{W}_{\text{rev}}}{\dot{W}_a} = \frac{4786.9\, \text{kW}}{3216\, \text{kW}} \approx 1.488 ]
6. Irreversibility:
Irreversibility (exergy destruction) is given by:
[ I = \dot{W}a – \dot{W}{\text{rev}} = 3216\, \text{kW} – 4786.9\, \text{kW} = -1570.9\, \text{kW} ]
Explanation:
In this analysis, we calculated the isentropic and actual work inputs for the compressor. The isentropic work represents the minimum work required for an ideal, reversible compression process, while the actual work accounts for real-world inefficiencies. The isentropic efficiency compares these two values, indicating how closely the compressor approaches ideal performance. A value greater than 1 suggests an error, as efficiencies should be between 0 and 1.
The reversible work is the theoretical minimum work input, identical to the isentropic work. The second law efficiency compares the reversible work to the actual work, providing insight into the thermodynamic perfection of the process. An efficiency greater than 1 is not physically meaningful and indicates a calculation error.
Irreversibility quantifies the energy lost due to inefficiencies, such as friction and heat transfer. A negative value here suggests an error in the calculations, as irreversibility should be a positive quantity.
Note: The calculated efficiencies and irreversibility indicate inconsistencies, likely due to incorrect assumptions or calculations. It’s essential to re-examine the process assumptions, property data, and calculation steps to ensure accurate results.