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 solve the problem, let’s start by breaking down the process step by step.
Given Information:
- 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} )
- Average temperature ( T_{avg} = \frac{T_1 + T_2}{2} = \frac{290 + 610}{2} = 450 \, \text{K} )
We will assume that air behaves as an ideal gas with constant specific heat values.
Step 1: Specific Heat Calculation
For air at around 300 K, the specific heat at constant pressure ( C_p ) and constant volume ( C_v ) are approximately:
- ( C_p \approx 1005 \, \text{J/kg·K} )
- ( C_v \approx 718 \, \text{J/kg·K} )
Step 2: Isentropic Work
The isentropic process refers to an idealized process where there is no entropy generation. For an adiabatic process (no heat exchange), the work done by the compressor is related to the change in enthalpy:
[
W_{\text{isentropic}} = \dot{m} \cdot \Delta h
]
The enthalpy change ( \Delta h ) for an ideal gas is:
[
\Delta h = C_p \cdot (T_2 – T_1)
]
Substituting the values:
[
\Delta h = 1005 \, \text{J/kg·K} \cdot (610 – 290) \, \text{K} = 1005 \cdot 320 = 321,600 \, \text{J/kg}
]
So, the isentropic work:
[
W_{\text{isentropic}} = 10 \, \text{kg/s} \cdot 321,600 \, \text{J/kg} = 3,216,000 \, \text{J/s} = 3,216 \, \text{kW}
]
Step 3: Actual Work
Now, we need to calculate the actual work. The actual enthalpy change is the same as the isentropic, but due to the irreversibility of the real process, the work required will be higher. The actual work is calculated using the same formula as the isentropic work.
However, for this problem, we do not have detailed information to find the actual temperature and pressure at the outlet for the real process, so we can assume that the work required in the actual process is higher than in the isentropic case.
Step 4: 1st Law Isentropic Efficiency
The isentropic efficiency for this process is calculated as:
[
\eta_{\text{1st law}} = \frac{W_{\text{isentropic}}}{W_{\text{actual}}}
]
Step 5: Reversible Work and 2nd Law Efficiency
The reversible work refers to the minimum work required for a process. The second law efficiency is the ratio of the actual work to the reversible work. Without detailed data for the actual entropy change, we can’t calculate these precisely without making further assumptions.
Step 6: Irreversibility
The irreversibility of the process is related to the entropy generation. The larger the difference between the actual and reversible work, the higher the irreversibility.
Final Summary:
- Isentropic Work: 3,216 kW
- Actual Work: Higher than 3,216 kW (not precisely given without more data).
- Isentropic Efficiency: ( \eta_{\text{1st law}} ) is the ratio of actual work to isentropic work.
- Reversible Work and 2nd Law Efficiency cannot be fully determined without more information. However, the irreversibility increases with the difference between actual and isentropic work.
This analysis provides insight into the energy required and losses in the compressor system, where irreversibilities and inefficiencies are always present in real processes.