Question

A 280 mm diameter cylinder fitted with a frictionless leakproof piston contains 0.02 kg of
steam at a pressure of 0.6 MPa and a temperature of 200 0C. As the piston moves slowly
outwards through a distance of 305 mm, the steam undergoes a fully-resisted expansion
during which the steam pressure p and the steam volume V are related by pVn = Constant,
where n is a constant. The final pressure of the steam is 0.12 MPa. Determine :
a) The Value of n
b) The work done by the steam
c) The magnitude and sign of heat transfer

Answer 1

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Answer 2

Answer:

Explanation:

One of the fundamental three-dimensional shapes in geometry is the cylinder, which has two distant, parallel circular bases. At a predetermined distance from the centre, a curved surface connects the two circular bases. The axis of the cylinder is the line segment connecting the centres of two circular bases.

The stem is initially in a superheated state with

                             $p1 = 0.6 MPa$

                             $T1 = 200 degree centigrade$

from the extremely hot stem table then we will write the property

                            $v1 = 0.35212 m^{3} / kg$

                            $u1 = 2639.4 kt / kg$

                            $h1 = 2850.6 kt / kg$

                           $m = 0.2 kg of the given stem$

Specific volume $v1 = v1 / m$

                           $v1 = 0.02 * 0.35212$

                           $v1 = 7.0424 * 10^{-3} m^{3}$

Area of the piston $Ap = \pi / 4 \alpha ^{2} = 3.14 / 4 * 0.28^{2}$

                               $Ap = 0.06154 m^{2}$

During the procedure, the piston's swept volume

                                Δ$v = Ap * L$

                                Δ$v = 0.06154 * 0.305$

                                Δ$v = 0.018771 m^{3}$

                             $V2 - V1 = 0.018771$

                                   $V2 = 0.0258 m^{3}$

              and given  $P2 = 0.12MPa$

Now , The process is given $PV^{n} = c$

                                           $P1 V1^{n} = P2 V2^{n}$

                                           $(V2 / V1)^{n} = P1 /P2$

                                          $n = ln ( P1 / P2) / ln (V2 / V1)$

                                         $n = ln (0.6 / 0.12) / ln (0.0258 / 0.00704) = 1.239$

Work done by the stem  $\alpha cut$$w = p1v1 - p2v2 / n-1$

                                        $\alpha cut w = 0.6 * 0.00704 - 0.12 * 0.0258 / 1.239 - 1$

                                        $\alpha cut w = 4.71966 kt$

from stem table $ab$  $P2 = 0.12MPa$ in saturated state

                               $uf = 440 kt / kg$

                               $vfg = 2070 kt / kg$

                               $vf = 0.001047$

                               $vg = 1.3800$

                     $v2 = vf + x vfg = vf + x (vg - vf)$

                    $1.29 = 0.001047 + x (1.38 - 0.001047)$

                    $x = 0.935$  ( dryness fraction)

               $u2 = uf + xufg$

               $u2 = 440 + 0.935 * 2070$

              $u2 = 2374.897 kt / kg$

Change in internal energy in the process

                 $dU = m (u2-u1)$

                 $dU = 0.02(2374.897 - 2639.4)$

                 $dU = -5.29 kt$

According to the first law of thermo dynamics

                $\alpha cut Q = du + \alpha cut w$

                $\alpha cut Q = -5.29 + 4.71966$

                $\alpha cut Q = -0.57 kt$

 Negative sign indicates the heat is rejected from the system.

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