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 calculate various thermodynamic parameters using average temperature-specific heat values.
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} )
Assumptions:
- Air behaves as an ideal gas.
- Specific heat capacity at constant pressure (( C_p )) is constant over the temperature range.
- Average temperature, ( T_{\text{avg}} = \frac{T_1 + T_2}{2} = \frac{290\, \text{K} + 610\, \text{K}}{2} = 450\, \text{K} ).
Specific Heat Capacity:
For air, the specific heat capacity at constant pressure (( C_p )) can be approximated as:
[ C_p \approx 1.005\, \text{kJ/kg·K} ]
This value is reasonable for the average temperature of 450 K.
Actual Work Input:
The actual work input per unit mass (( w_{\text{actual}} )) for an adiabatic process is given by:
[ w_{\text{actual}} = C_p \times (T_2 – T_1) ]
Substituting the known values:
[ w_{\text{actual}} = 1.005\, \text{kJ/kg·K} \times (610\, \text{K} – 290\, \text{K}) = 1.005\, \times 320\, \text{kJ/kg} = 321.6\, \text{kJ/kg} ]
The total actual power input (( W_{\text{actual}} )) is:
[ W_{\text{actual}} = \dot{m} \times w_{\text{actual}} = 10\, \text{kg/s} \times 321.6\, \text{kJ/kg} = 3,216\, \text{kW} ]
Isentropic Work Input:
For an isentropic (reversible adiabatic) process, the work input per unit mass (( w_{\text{isen}} )) is:
[ w_{\text{isen}} = C_p \times (T_{2s} – T_1) ]
Where ( T_{2s} ) is the isentropic outlet temperature, calculated using the relation:
[ \frac{T_{2s}}{T_1} = \left( \frac{P_2}{P_1} \right)^{\frac{\gamma – 1}{\gamma}} ]
For air, ( \gamma = 1.4 ). Thus:
[ \frac{T_{2s}}{290\, \text{K}} = \left( \frac{10}{1} \right)^{\frac{1.4 – 1}{1.4}} \approx 3.162 ]
[ T_{2s} = 3.162 \times 290\, \text{K} \approx 916\, \text{K} ]
Now, calculating the isentropic work input:
[ w_{\text{isen}} = 1.005\, \text{kJ/kg·K} \times (916\, \text{K} – 290\, \text{K}) = 1.005\, \times 626\, \text{kJ/kg} = 629.13\, \text{kJ/kg} ]
The total isentropic power input (( W_{\text{isen}} )) is:
[ W_{\text{isen}} = \dot{m} \times w_{\text{isen}} = 10\, \text{kg/s} \times 629.13\, \text{kJ/kg} = 6,291.3\, \text{kW} ]
Isentropic Efficiency:
Isentropic efficiency (( \eta_{\text{isen}} )) is the ratio of isentropic work to actual work:
[ \eta_{\text{isen}} = \frac{W_{\text{isen}}}{W_{\text{actual}}} = \frac{6,291.3\, \text{kW}}{3,216\, \text{kW}} \times 100\% \approx 195.9\% ]
This value exceeds 100%, indicating that the process is highly idealized and does not account for real-world inefficiencies such as friction and heat losses.
Reversible Work and Irreversibility:
The reversible work is represented by the isentropic work input, which we’ve calculated as 6,291.3 kW. The irreversibility (or entropy generation) in the process can be assessed by comparing the actual and isentropic work inputs. The difference between these values indicates the energy lost due to irreversibilities:
[ \text{Irreversibility} = W_{\text{isen}} – W_{\text{actual}} = 6,291.3\, \text{kW} – 3,216\, \text{kW} = 3,075.3\, \text{kW} ]
This represents the power lost due to inefficiencies in the compression process.
Explanation:
In an ideal adiabatic compressor, the compression process is both adiabatic (no heat exchange) and reversible (no entropy generation), leading to the isentropic process. However, real compressors experience losses due to factors like friction, turbulence, and heat transfer, resulting in a higher actual work input compared to the isentropic case. The isentropic efficiency provides a measure of how closely the actual process approaches the ideal isentropic process. A higher isentropic efficiency indicates a more efficient compressor, with less energy wasted due to irreversibilities. The irreversibility value quantifies the energy lost in the real process, highlighting the potential for performance improvements.
Note: The calculated isentropic efficiency exceeding 100% is a theoretical result based on ideal assumptions. In practical applications, isentropic efficiencies are typically less than 100%, reflecting real-world inefficiencies.