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- 2069 Poush · 7 marks
Rigid container heated from 1 bar to 1.5 bar
Short answer
At 1 bar, v₁ = 0.8475 m³/kg, T₁ = 99.6 °C, mass = 0.5 / 0.8475 = 0.590 kg, vapour mass = 0.295 kg. At 1.5 bar, constant v, T₂ = 111.4 °C, quality x₂ = 0.731, vapour mass = 0.431 kg.
A closed, rigid container of volume 0.5 m³ is placed on a hot plate. Initially, the container holds two phase mixture of saturated liquid water and saturated water vapor at P1 = 1 bar with a quality of 0.5. After heating, the pressure in the container is P2 = 1.5 bar. Indicate the initial and final states on a T-v diagram, and determine: a) The temperature, in °C, at each state b) The mass of the vapor present at each state, in kg. [Refer the attached table for properties of steam]
Given: V = 0.5 m³ (rigid, constant v); state 1: P₁ = 1 bar = 100 kPa, x₁ = 0.5; state 2: P₂ = 1.5 bar = 150 kPa.
Steam table values (pressure table):
| P | T_sat (°C) | v_f | v_g (m³/kg) |
|---|---|---|---|
| 100 kPa | 99.6 | 0.001043 | 1.6940 |
| 150 kPa | 111.35 | 0.001053 | 1.1594 |
State 1
v₁ = v_f + x(v_g − v_f) = 0.001043 + 0.5 × 1.6930 = 0.8475 m³/kg
T₁ = T_sat(1 bar) = 99.6 °C
m = V/v₁ = 0.5/0.8475 = 0.590 kg
m_g1 = x₁ m = 0.5 × 0.590 = 0.295 kg
State 2
v₂ = v₁ = 0.8475. At 150 kPa, v_f < 0.8475 < v_g (1.1594), so it is still wet:
T₂ = T_sat(1.5 bar) = 111.35 °C
x₂ = (0.8475 − 0.001053)/(1.1594 − 0.001053) = 0.731
m_g2 = 0.731 × 0.590 = 0.431 kg
T| .--*--. CP
| / \
| / 2 \ 2: 111.35 C
| / | \
| / 1 \ 1: 99.6 C
| / | \
| / v = 0.8475 \
+---------------------------- v
On the T-v diagram, 1 and 2 lie on a vertical line (v = 0.8475) inside the dome; 2 is above 1 since the temperature has increased.
Answer: T₁ = 99.6 °C, T₂ = 111.4 °C; vapour mass 0.295 kg (state 1) and 0.431 kg (state 2).
✓ Answer checked by Claude Opus 5.5 (high effort)
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