Air (ideal gas) enters a perfectly insulated compressor at a temperature of 310 K. The pressure ratio of the compressor is 6. Specific heat at constant pressure for air is 1005 J/kg.K and ratio of specific heats at constant pressure and constant volume is 1.4. Assume that specific heats of air are constant. If the isentropic efficiency of the compressor is 85 percent, the difference in enthalpies of air between the exit and the inlet of the compressor is ________ kJ/kg (round off to nearest integer).

The correct answer is 245.

Analysis of the Given Diagram and Data:
Based on the provided schematic and the temperature-entropy (T-s) diagram:
- State 1 represents the inlet to the compressor.
- State 2 represents the ideal (isentropic) exit state.
- State 2' represents the actual exit state due to irreversibilities in the compressor.
- The pressure lines are represented by $P_1$ and $P_2$ on the T-s diagram.

We are given the following parameter values:
- Inlet temperature:
$T_1=310\text{ K}$
- Pressure ratio:
$r_p=\frac{P_2}{P_1}=6$
- Specific heat at constant pressure:
$C_p=1005\text{ J/(kg·K)}=1.005\text{ kJ/(kg·K)}$
- Ratio of specific heats:
$\gamma =1.4$
- Isentropic efficiency of the compressor:
${\eta }_s=85\%=0.85$

Step 1: Calculate the ideal (isentropic) exit temperature ($T_2$)
For an isentropic process of an ideal gas between the inlet and outlet, the temperature and pressure are related by:
$\frac{T_2}{T_1}={\left(\frac{P_2}{P_1}\right)}^{\frac{\gamma -1}{\gamma }}$
Substituting the given values:
$\frac{T_2}{310}={\left(6\right)}^{\frac{1.4-1}{1.4}}$
$\frac{T_2}{310}=6^{0.2857}$
$T_2=310\times 1.6685\approx 517.225\text{ K}$

Step 2: Relate Isentropic Efficiency to Enthalpy Difference
The isentropic efficiency (${\eta }_s$) of a compressor is the ratio of the isentropic work input to the actual work input:
${\eta }_s=\frac{W_{\text{isen}}}{W_{\text{actual}}}=\frac{h_2-h_1}{h_{2\text{'}}-h_1}$
Since specific heats are constant, the enthalpy difference is directly proportional to the temperature difference:
${\eta }_s=\frac{C_p\left(T_2-T_1\right)}{h_{2\text{'}}-h_1}$

Step 3: Calculate the actual enthalpy difference ($h_{2\text{'}}-h_1$)
Rearranging the equation to solve for the actual change in enthalpy:
$h_{2\text{'}}-h_1=\frac{C_p\left(T_2-T_1\right)}{{\eta }_s}$
Substituting the calculated and given values:
$h_{2\text{'}}-h_1=\frac{1.005\times \left(517.225-310\right)}{0.85}$
$h_{2\text{'}}-h_1=\frac{1.005\times 207.225}{0.85}$
$h_{2\text{'}}-h_1=\frac{208.261}{0.85}\approx 245.01\text{ kJ/kg}$

Rounding to the nearest integer gives:
$h_{2\text{'}}-h_1=245\text{ kJ/kg}$