Question

Q.5 In an engine working on diesel cycle inlet pressure and temperature are 1bar and 17 degree C. Pressure at the end of adiabatic compression is 35 bar. The ratio of expansion i.e. after constant pressure heat addition is 5. Calculate the work done of the cycle. Assume gamma = 1.4 , Cp = 1.005KJ/kgK, Cv = 0.717 KJ/kgK

Answer 1

Answer:

2.112

12.674

894.58kJ/kg

-348kJ/kg

57%

Explanation:

Answer 2

Answer:

The work done in the cycle is 211.66kJ.

Explanation:

Given:

$P_{1} = 1 bar$, $T_{1} = 17 C= 290K$, $P_{2}=35bar$, γ = 1.4, $C_{p}=1.005kJ/kgK$, $C_{v}=0.717kJ/kgK$

To find:

Work done (W)

Formula:

$\frac{P_{2} }{P_{1} } = \frac{T_{2} }{T_{1} } ^{\frac{1.4}{1.4-1} }$

$Q=mC_{v}dT$

$W = Q_{in}-Q_{out}$

Step 1:

$\frac{P_{2} }{P_{1} } = \frac{T_{2} }{T_{1} } ^{\frac{1.4}{1.4-1} }$

$T_{2} = T_{1}(\frac{P_{2} }{P_{1} } )^{\frac{1.4-1}{1.4} }$

Substituting the given values, we get

$T_{2} = 290 * (\frac{35}{1} )^{0.2857 }$

$T_{2}=800.83K$

The ratio of expansion is given as 5. Therefore,

$P_{4}=\frac{P_{2} }{5}$

$P_{4}=\frac{35 }{5}$

$P_{4}=7bar$

$T_{4} = T_{1}(\frac{P_{4} }{P_{1} } )^{\frac{1.4-1}{1.4} }$

$T_{4} = 290*(\frac{7 }{1 } )^{0.2857}$

$T_{4}=505.64K$

Step 2:

Heat addition:

$Q_{in}=mC_{v}dT$

$Q_{in}=mC_{v}(T_{2}-T_{1})$

$Q_{in}=1*0.717*(800.83-290)$

$Q_{in}=1*0.717*510.83$

$Q_{in}=366.27 kJ/K$

Heat rejection:

$Q_{out}=mC_{v}dT$

$Q_{out}=mC_{v}(T_{4}-T_{1})$

$Q_{out}=1*0.717*(505.64-290)$

$Q_{out}=1*0.717*215.64$

$Q_{out}=154.61kJ/K$

Step 3:

Work done = heat added - heat rejected

$W = Q_{in}-Q_{out}$

$W = 366.27-154.61$

W = 211.66kJ

Therefore, the work done is 211.66kJ

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