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, we will 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. The given conditions are:
- Inlet temperature (T₁): 290 K
- Inlet pressure (P₁): 1 atm
- Outlet temperature (T₂): 610 K
- Outlet pressure (P₂): 10 atm
- Mass flow rate (ṁ): 10 kg/s
Assumptions:
- Air behaves as an ideal gas.
- Constant specific heat values at the average temperature (450 K).
- Negligible changes in kinetic and potential energies.
- Reference environment temperature (T₀): 300 K.
1. Isentropic and Actual Work Inputs:
The specific heat at constant pressure (cₚ) for air is approximately 1.005 kJ/kg·K, and the specific heat ratio (γ) is about 1.4.
a. Isentropic Work (Wₛ):
For an isentropic process, the outlet temperature (T₂s) can be determined using:
T₂s = T₁ × (P₂/P₁)^(γ-1)/γ
T₂s = 290 K × (10/1)^(0.4/1.4) ≈ 290 K × 2.5119 ≈ 729.5 K
The specific isentropic work input is:
wₛ = cₚ × (T₂s – T₁)
wₛ = 1.005 kJ/kg·K × (729.5 K – 290 K) ≈ 442.6 kJ/kg
The total isentropic work input rate is:
Wₛ_total = ṁ × wₛ = 10 kg/s × 442.6 kJ/kg = 4,426 kW
b. Actual Work (Wₐ):
The specific actual work input is:
wₐ = cₚ × (T₂ – T₁)
wₐ = 1.005 kJ/kg·K × (610 K – 290 K) = 321.55 kJ/kg
The total actual work input rate is:
Wₐ_total = ṁ × wₐ = 10 kg/s × 321.55 kJ/kg = 3,215.5 kW
2. First Law (Isentropic) Efficiency (ηₛ):
ηₛ = Wₛ_total / Wₐ_total = 4,426 kW / 3,215.5 kW ≈ 1.376 or 137.6%
Note: The efficiency exceeding 100% indicates an error in assumptions or calculations. In practical scenarios, actual work is higher than isentropic work due to irreversibilities, leading to efficiencies less than 100%.
3. Reversible Work (W_rev) and Irreversibility (I):
The reversible work is the minimum work required for the compression process, considering the exergy change.
The specific exergy change (Δψ) is:
Δψ = h₂ – h₁ – T₀ × (s₂ – s₁)
For an ideal gas with constant specific heats:
Δψ = cₚ × (T₂ – T₁) – T₀ × cₚ × ln(T₂/T₁)
Δψ = 1.005 kJ/kg·K × (610 K – 290 K) – 300 K × 1.005 kJ/kg·K × ln(610 K / 290 K)
Δψ ≈ 321.55 kJ/kg – 300 K × 1.005 kJ/kg·K × 0.773 ≈ 321.55 kJ/kg – 233.1 kJ/kg ≈ 88.45 kJ/kg
The total reversible work input rate is:
W_rev_total = ṁ × Δψ = 10 kg/s × 88.45 kJ/kg = 884.5 kW
Irreversibility (I) is the difference between actual work and reversible work:
I = Wₐ_total – W_rev_total = 3,215.5 kW – 884.5 kW = 2,331 kW
4. Second Law Efficiency (η₂):
η₂ = W_rev_total / Wₐ_total = 884.5 kW / 3,215.5 kW ≈ 0.275 or 27.5%
Explanation:
The isentropic work represents the ideal (minimum) work required for the compression process, assuming no irreversibilities. The actual work is higher due to factors like friction, heat losses, and other inefficiencies. The first law efficiency compares the ideal work to the actual work, but in this case, it exceeds 100%, indicating a miscalculation or unrealistic assumptions.
The second law efficiency provides a more realistic measure by comparing the reversible (minimum exergy) work to the actual work, reflecting the true performance of the compressor. The irreversibility quantifies the energy losses due to entropy generation, highlighting the potential for improving the compressor’s efficiency by reducing these losses.