Two kg of water is contained in a piston-cylinder assembly, initially at 10 bar and 200°C. The water is slowly heated at constant pressure to a final state. If the heat transfer for the process is 1740 kJ, determine the temperature at the final state, in °C, and the work, in kJ. Kinetic and potential energy effects are negligible.

Explanation:

Given that:

mass of water = 2kg

Pressure P₁ = 10 bar

temperature T₁ = 200⁰ C

Obtaining the following value from superheated tables at P₁ = 10 bar and T₁ = 200⁰ C

v₁ = 0.2060 m³/kg

h₁ = 2827.9 kJ/kg

The heat transfer Q₁₋₂ = 1740 kJ

Using the energy balance equation:

Q₁₋₂ = m(h₂ - h₁)

1740 = 2(h₂ - 2827.9)

1740 = 2h₂ - 5655.8

1740 + 5655.8 = 2h₂

7395.8 = 2h₂

h₂ = 7395.8/2

h₂ = 3697.9 kJ/kg

Since P₁ = P₂ = 10 bar

Using the superheated table at P₂ = 10 bar and h₂ = 3697.9 kJ/kg;

v₂ = 0.4011 m³/kg

The temperature at final state = 600 °C

The work done

= (10×10² kPa)×2kg (0.4011 - 0.2060) m³/kg

= (10×10² kPa)×2kg × 0.1951 m³/kg

= 390.2 kJ