Answer to Question #124787 in Mechanical Engineering for hasham

Question

A fluid at 0.7 bar occuping 0.09m^3 is compressed reversibly to a pressure of 3.5 bar according to the law pv^n=constant.The fluid is then heated reversibly at constant volume until the pressure is 4bar;the specific volume is then 0.5 m^3/kg.A reversible expansion according to a law pv^2=constant restore the fluid in its initial state.Sketch the cycle on ap-v diagram and calculte (i)the mass of fluid present (ii)the value of n in the first process (iii)the net work of the cycle

Answer 1

$P_1=0.7 bar ,V_1=0.09 m^3,P_2=3.5bar$ $,V_2=V_3$ $,P_3=4bar$ ,

specific volume= $0.5 \frac{kg}{m^3}$ then reversible expansion takes place to reverse the process

$PV^2=C$

For process 3-1

$P_3V_3^2=P_1V_1^2$

here, $V_3=V_2$

$\frac{V_3}{V_1}=(\frac{P_1}{P_3})^{0.5}$

${V_3}=(V_1)(\frac{P_1}{P_3})^{0.5}$

${V_3}=(0.09)(\frac{0.7}{4})^{0.5}=0.03765 m^3$

(i) mass of fluid=$\frac{V_3}{v}$ ,v= specific volume

mass of fluid= $\frac{0.03765}{0.05}=0.753$ kg

(ii) $P_1V_1^n=P_2V_2^n$

$\frac{P_1}{P_2}=(\frac{V_2}{V_1})^n$

$\frac{0.7}{3.5}=(\frac{0.0.03765}{0.09})^n$

Taking log both sides

$log(0.2)=nlog(0.4183)$

$n=\frac{-0.69897}{-0.3784}=1.847$ n=\frac{-0.69897}{0.0774}

(iii) Area under graph = $W=\int Pdv=8777 J$