Question #199954
  1. An engine working on otto cycle has a volume of 0.5 m^3  pressure 1 bar and temperature 27C at the commencement of compression stroke.At the end of compression stoke,the pressure is 10 bar. Heat added during the constant volume process is 200 kJ.Determine
  2. Percentage clearance
  3. Air standard efficiency
  4. Mean effectine pressure
  5. Ideal power developed by the engine if the engine runs at 400 r.p.m.so that there are 200 cycles per minute.
1
Expert's answer
2021-05-31T06:40:15-0400

(a) percentage decrease

V2V1×100\frac{V_2}{V_1}\times100

V1=0.5m3V_1=0.5m^3

p1p2=(V2V1)1.4=110=(V20.50)1.4=V2=0.09653\frac{p_1}{p_2}=(\frac{V_2}{V_1})^{1.4}=\frac{1}{10}=(\frac{V_2}{0.50})^{1.4}=V_2=0.09653

=0.096530.5×100=19.306%=\frac{0.09653}{0.5}\times100=19.306\%

(b) Air standard efficiency

η=1T1T2=1\eta=1-\frac{T_1}{T_2}=1

T2=T1(p2p1)0.41.4=(27+273)(10)0.41.4=579.2KT_2=T_1(\frac{p_2}{p_1})^{\frac{0.4}{1.4}}=(27+273)(10)^{\frac{0.4}{1.4}}=579.2K

η=1300579.2=48.2%\eta=1-\frac{300}{579.2}=48.2\%


(c)mean effective pressure

Pmean=WV3P_{mean}=\frac{W}{V_3}

W=ηQa=48.2100×200kJ=94.47kJ=\eta Q_a=\frac{48.2}{100}\times 200kJ=94.47kJ

Vs=V1V2=0.50.09653=0.40347m3V_s=V_1-V_2=0.5-0.09653=0.40347m^3

Pmean=96.410.40347=238.95kPaP_{mean} =\frac{96.41}{0.40347}=238.95kPa


(d) Ideal power

P=w×N=96.41×1200=19282KWP=w\times N=96.41\times1200=19282 KW


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