Question

20 liters of water at 20 bar and 100 °C are in a variable volume container. 56,000 kJ of heat are added to the system at constant pressure which causes an increase in both temperature and volume.

Draw a schematic of the process

Write the energy balance for this process and clearly state any assumptions:

What is the final volume of the container?

How much work in [kJ] is completed? Pay careful attention to your negative signs!

In one sentence explain your answer from Part D in the context of the process described in the problem statement and the heat that is added to the system.

Answer 1

Schematic diagram of the process

The energy balance for constant pressure process is

mH₁ + Q =m H₂

Q = ∆H

∆H =m(H₂- H₁)

Assumption

Pressure is constant and there are no heat losses

Water conditions

P₁= 20 bar

T ₁= 100°C

V₁ = 20 L

From steam tables

At P = 20 bar , T = 100°C

H₁ = 420.59 KJ/Kg

V_L = 0.00104249 m³/Kg

Initial volume of water present = 20 L= 0.020 m³

Mass of water present =0.020/0.00104249 = 19.1848 Kg

Since process is isobaric

Q = ∆H

Q =56000 KJ

∆H = 56000 KJ

m = 19.1848 Kg

∆H = 56000/(19.1848) = 2918.977 KJ/Kg

H₁ = 420.59 KJ/Kg

H₂ = 420.59 + ∆H

H₂ = 420.59 + 2918.977

H₂ = 3339.567 KJ/Kg

Pressure is constant at 20 bar

From steam tables

At P₂= 20 bar and H₂ = 3339.567 KJ/Kg

T₂ = 441.492°C  

V_g = 0.161455 m³/Kg

Mass of water remains same

m = 19.1848 Kg

V₂ = 19.1848(0.161455) = 3.0974 m³

Final volume = 3.0974 m³

For constant pressure process

W= P∆V

P = 20 bar = 20×10⁵ Pa

W= 20×10⁵ (3.0974-0.020)

W = + 6154800 J = 6154.8 KJ

In this process water heated from 100°C and 20 bar to 441.49°C and there is phase change involved resulting in work

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