Question

A piston-cylinder assembly contains ammonia, initially at a temperature of-20°C and a quality of 70%. The ammonia is slowly heated to a final state where the pressure is 6 bar and the temperature is 180°C. While the ammonia is heated, its pressure varies linearly with specific volume. For the ammonia, determine the work and heat transfer, each in kJ/kg.

Answer 1

Answer:

w =  -28.8 kJ/kg

q = 723.13 kJ/kg

Explanation:

Given :

Initial properties of piston  cylinder assemblies

Temperature, $T_{1}$ = -20°C

Quality, x = 70%

           = 0.7

Final properties of piston  cylinder assemblies

Temperature, $T_{2}$ = 180°C

Pressure, $P_{2}$ = 6 bar

From saturated ammonia tables at $T_{1}$ = -20°C  we get

$P_{1}$ = $P_{sat}$ = 1.9019 bar

$v_{f}$ = 0.001504 $m^{3}$ / kg

$v_{g}$ = 0.62334 $m^{3}$ / kg

$u_{f}$ = 88.76 kJ/kg

$u_{g}$ = 1299.5 kJ/kg

Therefore, for initial state 1 we can find

$v_{1}$ = $v_{f}$+x ($v_{g}$-$v_{f}$

                       = 0.001504+0.7(0.62334-0.001504)

                       = 0.43678 $m^{3}$ / kg

$u_{1}$ = $u_{f}$+x ($u_{g}$-$u_{f}$

                       = 88.76+0.7(1299.5-88.76)

                       =936.27 kJ/kg

Now, from super heated ammonia at 180°C, we get,

$v_{2}$ = 0.3639 $m^{3}$ / kg

$u_{2}$ = 1688.22 kJ/kg

Therefore, work done, W = area under the curve

           $w = \left (\frac{P_{1}+P_{2}}{2}  \right )\left ( v_{2}-v_{1} \right )$

           $w = \left (\frac{1.9019+6\times 10^{5}}{2} \right )\left ( 0.3639-0.43678\right )$

           $w = -28794.52$ J/kg

                       = -28.8 kJ/kg

Now for heat transfer

$q = (u_{2}-u_{1})+w$

$q = (1688.2-936.27)-28.8$

          = 723.13 kJ/kg