Thermodynamics: Steam Required to Heat Water
Question:
Calculate the mass of steam at a pressure of 0.7 MN/m² that must be blown into 20 kg of water at 15°C to raise its temperature to 50°C:
- (a) If the steam is dry saturated.
- (b) If the steam is 0.9 dry.
Assume that all the steam condenses into the water.
Answer:
Step 1: Heat required by water (Qw)
Qw = m × c × ΔT = 20 kg × 4.18 kJ/kg·K × (50°C − 15°C) = 2926 kJ
Step 2: Steam data at 0.7 MN/m² (from steam tables)
- Latent heat of vaporization (Lv) ≈ 2065 kJ/kg
- Condensation temperature = ~164.97°C
- Cooling of condensate from 164.97°C to 50°C:
- Total heat released per 1 kg of dry saturated steam:
Q = mcΔT = 1 kg × 4.18 kJ/kg·K × (164.97 − 50) ≈ 480.5 kJ
Qsteam = Lv + mcΔT = 2065 + 480.5 = 2545.5 kJ/kg
(a) Mass of Dry Saturated Steam
m = Qw / Qsteam = 2926 / 2545.5 ≈ 1.15 kg
(b) Mass of 0.9 Dry Steam
When steam is not fully dry, only 90% contributes latent heat:
Qlat = 0.9 × 2065 = 1858.5 kJ/kg
Qcooling = 480.5 kJ/kg
Qtotal = 1858.5 + 480.5 = 2339 kJ/kg
Qcooling = 480.5 kJ/kg
Qtotal = 1858.5 + 480.5 = 2339 kJ/kg
m = Qw / Qtotal = 2926 / 2339 ≈ 1.25 kg
✅ Final Answers:
- (a) For dry saturated steam: 1.15 kg
- (b) For 0.9 dry steam: 1.25 kg