Subtopic: I.C. engine systems – fuel, cooling, lubrication, ignition, electrical, intake and exhaust
Question Statement:
Ambient air at $P_1 = 100\text{ kPa}$ and $T_1 = 300\text{ K}$ enters a tractor turbocharger compressor having a pressure ratio of $r_p = 1.80$. The isentropic efficiency of the compressor is $\eta_c = 75\%$ ($0.75$). Taking $\gamma = 1.40$, the isentropic discharge temperature is $T_{2s} = T_1 (r_p)^{(\gamma-1)/\gamma} = 300 \times (1.80)^{0.2857} \approx 355.0\text{ K}$. The actual air temperature leaving the compressor ($T_2 = T_1 + \frac{T_{2s}-T_1}{\eta_c}$) in Kelvin is ________ (round off to 1 decimal place).
Answer Key:
373.3
Step-by-Step Solution & Explanation
The actual compressor exit temperature $T_2$ accounting for isentropic efficiency $\eta_c$ is:
$T_2 = T_1 + \frac{T_{2s} - T_1}{\eta_c}$
Given $T_1 = 300\text{ K}$, $T_{2s} = 355.0\text{ K}$, and $\eta_c = 0.75$:
$T_2 = 300 + \frac{355.0 - 300}{0.75} = 300 + \frac{55.0}{0.75} = 300 + 73.33 = 373.33\text{ K} \approx 373.3\text{ K}$