Question

1.5 kg of water is in saturated liquid state at 2 bar (v_f =  0.001061 m³/kg, u_f = 504.0 kJ/kg. h_f = 505 kJ/kg). Heat is added in a constant pressure process till the temperature of water Reaches 400°C (v = 1.5493 m³/kg, u = 2968.0 kJ/kg, h = 3277.0 kJ/kg). The heat added (in kJ) in the process is _______

Answer 1

Concept:

From the first law of thermodynamics,

δQ = dU + W

δq = du + p(dV)

δq = (u₂ - u₁) + p(v₂ - v₁)

Given:

m = 1.5 kg, p = 2 bar = 200 kPa, v1 = 0.001061 m3/kg,

v2 = 1.5493 m3/kg, u1 = 504 kJ/kg, u2 = 2968 kJ/kg

h2 = 3277 kJ/kg, h1 = 505 kJ/kg

Calculation: 

δq = (2968 - 504) + 200 (1.5493 - 0.001061) = 2777.64 kJ/kg

For 1.5 kg of water,

Q = 2777.64 × 1.5 = 4160.47 kJ

Alternate Way: Q = m (h₂ – h₁) = 1.5 (3277 - 505) = 4158 kJ