Question

A piston-cylinder assembly contains 5kg of water that undergoes a series of processes to form a thermodynamic cycle. Process 1à2: Constant pressure cooling from p1=20bar and T1=360°C to saturated vapor Process 2à3: Constant volume cooling to p3=5 bar Process 3à4: Constant pressure heating Process 4à1: Polytropic process following Pv =constant back to the initial state Kinetic and potential energy effects are negligible. Calculate the net work for the cycle in kJ.

Answer 1

Answer:

The net work done is 272.38 kJ

Explanation:

The parameters given are;

Mass of water = 5 kg

p₁ = 20 bar

T₁ = 360°C

v₁ = 0.141147 m³/kg  

Process 1 to 2 = Constant pressure process

p₂ = 20 bar

Process 2 to 3 = Constant volume process

p₃ = 5 bar

Process 3 to 4 = Constant pressure process

Process 4 to 1 = Polytropic process pv = Constant

For Stage 1 to 2, we have;

p₂ = 20 bar

From the steam tables for superheated steam, we have;

T₂ = 212.385°C

v₂ = 0.0995805 m³/kg

Work done = p₂×(v₂ - v₁) = 2×10⁶ × (0.0995805 - 0.141147 ) = -83133 J/kg

For the 5 kg, we have;

$W_{1-2}$ = -83133 J/kg × 5 = -415,665 J

Stage 2 to 3: Constant volume cooling

v₂ = v₃ = 0.0995805 m³/kg

p₃ = 5 bar

T₃ = 151.836°C

(0.0995805 - 0.00109256)/(0.374804 - 0.00109256) = 0.2635 liquid vapor mixture

Work done, $W_{2-3}$ = 0

Stage 3 to 4: Constant pressure heating

p₃ = p₄ = 5 bar

v₄/T₄ = v₃/T₃

v₄ =  0.374804 m³/kg

T₄ = v₄×T₃/v₃ = 0.374804*(273.15 + 151.836)/0.0995805 = 1599.6 K = 1326.4 °C

Work done = p₄×(v₄ - v₃) = 5×10⁵ × (0.374804  - 0.0995805 ) = 137611.75 J/kg

For the 5 kg, we have;

$W_{3-4}$ = 137,611.75  J/kg × 5 = 688,058.75 J

Stage 4 to 1: Polytropic process    

$\dfrac{p_{4}}{p_{1}} = \left (\dfrac{V_{1}}{V_{4}} \right )^{n} = \left (\dfrac{T_{4}}{T_{1}} \right )^{\dfrac{n}{n-1}}$

Which gives;

$\dfrac{5}{20} = \left (\dfrac{0.141147 }{0.374804} \right )^{n}$

n = log(5/20) ÷log(0.141147/0.374804) = 1.42

Work done, $W_{pdv}$, is given as follows;

$W_{pdv} = \dfrac{p_4 \times v_4 -p_4 \times v_4 }{n-1}$

Which gives;

$W_{pdv} = \dfrac{5\times 0.374804 -20\times 0.141147 }{1.42-1} = -2.259 \, J$

For the 5 kg, we have;

$W_{4-1}$ = -2.259 J/kg × 5 = -11.2967 J

The net work done, $W_{Net}$, is therefore;

$W_{Net}$ = $W_{1-2}$  + $W_{3-4}$ + $W_{4-1}$

-415,665  + 688,058.75 -11.2967 = 272,382.45 J = 272.38 kJ.