Question #183631

4. There are supplied 320 KJ/cycle of Diesel engine operating on 250 g of air P 1 =97.91 KPa, T 1 =48.9C.

At the end of compression, P 2 =3930KPa. Assume that air and the product within the cycle have air

properties. Determine compresion ratio, cut-off ratio, Wnet, thermal efficiency, and mean effective

pressure.


1
Expert's answer
2021-04-22T07:23:08-0400

Compression ratio

r=v1v2=(p2p1)1/γ=(393097.91)1/1.4=13.97r=\frac{v_1}{v_2}=(\frac{p_2}{p_1})^{1/\gamma}=(\frac{3930}{97.91})^{1/1.4}=13.97

Cut-off ratio

rc=v3v2=(T3T1)r_c=\frac{v_3}{v_2}=(\frac{T_3}{T_1})

But T2T1=(v2v1)γ1    T2=322.05(13.97)1.41=924.7K\frac{T_2}{T_1}=(\frac{v_2}{v_1})^{\gamma -1} \implies T_2=322.05(13.97)^{1.4-1}=924.7 K

320=0.25×1.005×(T3924.7)    T3=2198.33184K320=0.25 \times1.005 \times(T_3-924.7) \implies T_3=2198.33184 K

So, rc=(2198.33924.7)=2.4r_c=(\frac{2198.33}{924.7})=2.4


Thermal efficiency

11rγ1[1γ×rcγ1rc1]1-\frac{1}{r^{\gamma -1}}[\frac{1}{\gamma} \times \frac{r_c^{\gamma}-1}{r_c-1}]

1113.971.41[11.4×2.41.412.41]=0.572401-\frac{1}{13.97^{1.4 -1}}[\frac{1}{1.4} \times \frac{2.4^{1.4}-1}{2.4-1}]=0.57240

Wnet

0.57240×320=183.168kJ/cycle0.57240 \times320=183.168 kJ/cycle

Mean effective pressure.

Pm=WnetV=0.5724×1.4×3930×(2.41)(13.971)(1.41)=849.86174kPaP_m=\frac{W_{net}}{V}=\frac{0.5724 \times 1.4 \times 3930 \times(2.4-1)}{(13.97-1)(1.4-1)}=849.86174 kPa


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