Answer:
[tex]47.97\:kW[/tex]
Step-by-step explanation:
To determine net specific work output, we use the following equation:
[tex]\omega=\eta q_{in}\\\\=(1-\frac{1}{k}\frac{r_c^k-1}{r^{k-1}(r_c-1)} )c_p(T_3-T_2)\\\\=(1-\frac{1}{k}\frac{r_c^k-1}{r^{k-1}(r_c-1)} )c_pT_1r^{k-1}(r_c-1)\\\\=(1-\frac{1}{1.4}\frac{1.8^{1.4}-1}{22^{1.4-1}(1.8-1)} )*1.005*343*22^{1.8-1}(r_c-1)\:\frac{kJ}{kg} \\\\=635 \:\frac{kJ}{kg}[/tex]
To determine net power output, we use the following equation:
[tex]\dot W=\dot m \omega\\\\=\dot N\frac{P_1V}{RT_1}\omega\\ \\=\frac{2300}{60} \frac{97*2.0*10^{-3}}{0.287*343} *635\:kW\\\\=47.97\:kW[/tex]
Answer:
635kj/kg
47.9kW
Step-by-step explanation:
A four-cylinder two-stroke 2.0-L diesel engine that operates on an ideal Diesel cycle has a compression ratio of 22 and a cutoff ratio of 1.8. Air is at 70 °C and 97 kPa at the beginning of the compression process. Using the cold-air-standard assumptions, determine how much power the engine will deliver at 2300 rpm.
To get the specific work output
w=nq
[tex](1-\frac{r^k-1}{kr^(k-1)(rc-1)} )cp(T_{3} -T_{2} )[/tex]
[tex](1-\frac{r^k-1}{kr^(k-1)(rc-1)} )cp(T_{3} -T_{2} )[/tex][tex](1-\frac{1.8^1.4-1}{1.4*22^(1.4-1)(1.8-1)} )1.005*343*22^.4(1.8-1 )Kj/kg[/tex]
635kj/kg
net power is gotten from rate of operation and mass .
from ideal gas law, we say that
W=mw
[tex]N\frac{PV}{RT} w[/tex]
[tex]\frac{2300*2*10^-3}{60*0.287*343} 635[/tex]*97
47.9kW
