Question

Air (ideal gas) enters a perfect insulated compressor of a temperature of 310 K. The pressure ratio of compressor is 6, Specific heat at constant pressure for air 1005 J/kgK and ratio of specific heats at constant pressure and constant volume is 1.4. Assume that specific heat of air is constant. If the isentropic efficiency of the compressor is 85 percent, the difference in enthalpies of air between the exit and inlet of the compressor is _______

• A. 148.2 kJ/kg
• B. 245.2 kJ/kg
• C. 238.2 kJ/kg
• D. 198.3 kJ/kg

Hint:

Compressor isentropic efficiency is always less than $100\%$, meaning the actual work input (and thus the actual enthalpy rise) is always greater than the ideal isentropic work.
Thus, $\Delta h_{\text{actual}} = \frac{\Delta h_s}{\eta_c}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This thermodynamics question requires us to calculate the actual enthalpy change across a gas compressor taking into account its isentropic efficiency.

Step 2: Key Formula or Approach:
The isentropic outlet temperature ($T_{2s}$) for an ideal gas compressor is given by:
\[ T_{2s} = T_1 \cdot \left(r_p\right)^{\frac{\gamma-1}{\gamma}} \]
The isentropic enthalpy change is:
\[ \Delta h_s = C_p (T_{2s} - T_1) \]
The actual enthalpy change ($\Delta h_{\text{actual}}$) is computed using the isentropic efficiency ($\eta_c$):
\[ \eta_c = \frac{\Delta h_s}{\Delta h_{\text{actual}}} \quad \Rightarrow \quad \Delta h_{\text{actual}} = \frac{\Delta h_s}{\eta_c} \]

Step 3: Detailed Explanation:

• Identify the given values:
- Inlet temperature, $T_1 = 310 \text{ K}$
- Pressure ratio, $r_p = 6$
- Specific heat, $C_p = 1005 \text{ J/(kg K)} = 1.005 \text{ kJ/(kg K)}$
- Ratio of specific heats, $\gamma = 1.4$
- Isentropic efficiency, $\eta_c = 0.85$

• Calculate the isentropic outlet temperature ($T_{2s}$):
\[ T_{2s} = 310 \times (6)^{\frac{1.4-1}{1.4}} = 310 \times (6)^{0.2857} \]
\[ T_{2s} \approx 310 \times 1.6685 = 517.24 \text{ K} \]

• Calculate the isentropic enthalpy rise ($\Delta h_s$):
\[ \Delta h_s = 1.005 \text{ kJ/(kg K)} \times (517.24 - 310) \text{ K} \]
\[ \Delta h_s = 1.005 \times 207.24 \approx 208.28 \text{ kJ/kg} \]

• Calculate the actual enthalpy rise ($\Delta h_{\text{actual}}$):
\[ \Delta h_{\text{actual}} = \frac{\Delta h_s}{\eta_c} = \frac{208.28}{0.85} \approx 245.03 \text{ kJ/kg} \]

Step 4: Final Answer:
The actual enthalpy difference of air across the compressor is approximately $245.2 \text{ kJ/kg}$.