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 an adiabatic compressor where air enters at 290 K and 1 atm, and exits at 610 K and 10 atm with a mass flow rate of 10 kg/s, we can follow these steps:
1. Calculate the Isentropic Work:
For an isentropic (ideal) compression process, the temperature at the compressor exit (T₂s) can be determined using the isentropic relation:
[ T_{2s} = T_1 \left( \frac{P_2}{P_1} \right)^{\frac{\gamma – 1}{\gamma}} ]
Given:
- Inlet temperature, ( T_1 = 290 \, \text{K} )
- Pressure ratio, ( \frac{P_2}{P_1} = \frac{10 \, \text{atm}}{1 \, \text{atm}} = 10 )
- For air, the specific heat ratio ( \gamma = 1.4 )
Calculating ( T_{2s} ):
[ T_{2s} = 290 \times 10^{\frac{0.4}{1.4}} \approx 290 \times 2.5119 \approx 729 \, \text{K} ]
The specific work input for isentropic compression is:
[ w_s = c_p (T_{2s} – T_1) ]
Assuming constant specific heat at average temperature ( T_{avg} = \frac{T_1 + T_{2s}}{2} = \frac{290 + 729}{2} = 509.5 \, \text{K} ), and using ( c_p \approx 1.005 \, \text{kJ/kg·K} ):
[ w_s = 1.005 \times (729 – 290) = 1.005 \times 439 = 441.195 \, \text{kJ/kg} ]
The isentropic power input is:
[ \dot{W}_s = \dot{m} \times w_s = 10 \, \text{kg/s} \times 441.195 \, \text{kJ/kg} = 4411.95 \, \text{kW} ]
2. Calculate the Actual Work:
The actual specific work input is:
[ w_a = c_p (T_2 – T_1) = 1.005 \times (610 – 290) = 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. Calculate the Isentropic Efficiency:
The isentropic efficiency (( \eta_s )) is:
[ \eta_s = \frac{\dot{W}_s}{\dot{W}_a} = \frac{4411.95}{3216} \approx 1.372 \, \text{or} \, 137.2\% ]
4. Calculate the Reversible Work:
The reversible work is equivalent to the isentropic work, so:
[ \dot{W}_{rev} = \dot{W}_s = 4411.95 \, \text{kW} ]
5. Calculate the Second Law Efficiency:
The second law efficiency (( \eta_{II} )) is:
[ \eta_{II} = \frac{\dot{W}_{rev}}{\dot{W}_a} = \frac{4411.95}{3216} \approx 1.372 \, \text{or} \, 137.2\% ]
6. Calculate the Irreversibility:
Irreversibility (I) is the difference between actual work and reversible work:
[ I = \dot{W}a – \dot{W}{rev} = 3216 – 4411.95 = -1195.95 \, \text{kW} ]
Explanation:
In an adiabatic compressor, air is compressed without heat exchange with the surroundings. The isentropic process represents an ideal scenario with no entropy generation, resulting in maximum efficiency. However, real compressors experience irreversibilities due to factors like friction and non-ideal fluid behavior, leading to higher actual work input compared to the isentropic case.
The isentropic efficiency is a measure of how closely a real compressor approaches ideal isentropic performance. It is calculated by comparing the isentropic work to the actual work. In this case, the calculated efficiency exceeds 100%, which is physically impossible and indicates an error in the calculations or assumptions.
The second law efficiency evaluates the performance of the compressor concerning the reversible (ideal) process. Similar to the isentropic efficiency, a value above 100% suggests discrepancies in the analysis.
Irreversibility quantifies the energy losses due to entropy generation within the system. A negative value, as obtained here, is non-physical and further indicates issues in the calculations.
To resolve these discrepancies, it is essential to re-examine the assumptions, such as the specific heat values and temperature calculations, ensuring they accurately reflect the actual conditions of the compression process.