Question

A fluid at a pressure of 3 bar and with specific volume of 0.18m3/kg contains in a cylinder behind a piston expand reversively to a pressure of 0.6 bar according to a law p=c/v2.where c is a constant calculate the work done by the fluid on the system.​

Answer 1

Explanation:

ans: 29840Nm/kg

this is from equation PV^2 =C

wd= integration of (pdv)

substitute p =c/v^2 and get the results

Answer 2

Given:

Pressure $P_1=3bar=3\times10^5\;N/m^2$

Pressure$P_2=0.6bar$

Volume $v_1=0.18m^{3/}kg$

To find:

Work done

Step by step explanation:

       Work done ,

                                     $W=\int_1^2\frac C{v^2}dv$

                                      $W=C\int_1^2\frac{dv}{v^2}$

                                      $W=C\left|\frac{v^{-2+1}}{-2+1}\right|_{v_1}^{v_2}$

                                      $W=C\left[-v^{-1}\right]_{v_1}^{v_2}$

                                      $W=C\left[-\frac1v\right]_{v_1}^{v_2}$

                                      $W=C\left[\frac1{v_1}-\frac1{v_2}\right]\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots(1)$

also,

                    $C=Pv^2\\=P_1v_1^2\\=3\times0.18^2$

                    $=0.0972 bar(m^3/kg)^2$

and,

              $v_2=\sqrt{\frac C{P_2}}$

                  $=\sqrt{\frac{0.0972}{0.6}}$

                  $=0.402\;m^3/kg$

Substituting the values of $C$, $v_1$ and $v_2$ value in equation $(1)$, we get

work done,

             $W=0.0972\times10^5\left[\frac1{0.18}-\frac1{0.402}\right]Nm/kg$

            $W=0.0972\times105\lbrack0.181 -0.4021\rbrack Nm/kg\;$

            $W=29840\;Nm/kg$