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ohaa [14]
3 years ago
10

The conditions at the beginning of compression in an Otto engine operating on hot-air standard with k=1.35 and 101.325 kPa, 0.05

m3 and 32C. The clearance is 8% and 15kJ are added per cycle. Determine the mean effective pressure.
Engineering
1 answer:
Katyanochek1 [597]3 years ago
6 0

Answer:

P_m=181.42 KPa

Explanation:

P_1=101.325 KPa,V_1=0.05m^3,T_1=32C,K=1.35

Clearance is 8%.

Heat added=15 KJ

We know that compression ratio r=1+\dfrac{1}{C}

r=1+\dfrac{1}{0.08}  

r=13.5

r=\dfrac{V_1}{V_2}

13.5=\dfrac{0.05}{V_2}

V_2=3.7\times 10^{-3}

We know that efficiency of otto cycle

\eta =1-\dfrac{1}{r^{k-1}}

\eta =1-\dfrac{1}{13.5^{1.35-1}}

\eta =0.56

\eta =\dfrac{W}{Q}

W is the work out put and Q is the heat addition.

0.56 =\dfrac{W}{15}

W=8.4 KJ

We know that Work =Mean effective pressure x swept volume.

Here swept volume V_s=V_1-V_2

V_s=0.05-3.7\times 10^{-3}

V_s=0.0463 m^3

Noe by putting the values

Work =Mean effective pressure x swept volume.

8.4=P_m\times 0.0463

P_m=181.42 KPa

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(h6 - 932.93) = ( 955.38 - 932.93) ((79.33 - 75.29) / ( 82.05 - 75.29 )

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(105 x 10³) kW = m(440.732 kJ/kg)

m = (105 x 10³) / 440.732

m = 238.2 kg/s

therefore the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s

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