Asked by jordan
How much heat must be supplied to 200 kg of water at 20°C to make steam at 850 kPa which is 87% dry?
All Answers 1
Answered by
GPT-5 mini
AI
Steps:
1. Heat the water from 20°C to the saturation temperature at 850 kPa.
2. Vaporize enough to reach quality x = 0.87.
From steam tables: Tsat(850 kPa) ≈ 170.4°C.
Take cp (liquid) ≈ 4.18 kJ/kg·K.
Take hfg at 850 kPa ≈ 2030 kJ/kg (table value; using 2030 kJ/kg gives a good estimate).
Calculate:
- Sensible heat to reach saturation:
Q1 = m cp (Tsat − T1) = 200 kg · 4.18 kJ/kg·K · (170.4 − 20) K
≈ 200 · 4.18 · 150.4 ≈ 125.7 × 10^3 kJ ≈ 125.7 MJ.
- Latent heat to produce 87% quality:
Q2 = m x hfg = 200 kg · 0.87 · 2030 kJ/kg ≈ 353.2 × 10^3 kJ ≈ 353.2 MJ.
Total heat required:
Q = Q1 + Q2 ≈ 125.7 MJ + 353.2 MJ ≈ 478.9 MJ ≈ 4.79 × 10^5 kJ.
Answer: Approximately 4.79 × 10^5 kJ (≈ 479 MJ).
(If you want greater accuracy, use the exact hfg and hf values from your steam tables for p = 850 kPa.)
1. Heat the water from 20°C to the saturation temperature at 850 kPa.
2. Vaporize enough to reach quality x = 0.87.
From steam tables: Tsat(850 kPa) ≈ 170.4°C.
Take cp (liquid) ≈ 4.18 kJ/kg·K.
Take hfg at 850 kPa ≈ 2030 kJ/kg (table value; using 2030 kJ/kg gives a good estimate).
Calculate:
- Sensible heat to reach saturation:
Q1 = m cp (Tsat − T1) = 200 kg · 4.18 kJ/kg·K · (170.4 − 20) K
≈ 200 · 4.18 · 150.4 ≈ 125.7 × 10^3 kJ ≈ 125.7 MJ.
- Latent heat to produce 87% quality:
Q2 = m x hfg = 200 kg · 0.87 · 2030 kJ/kg ≈ 353.2 × 10^3 kJ ≈ 353.2 MJ.
Total heat required:
Q = Q1 + Q2 ≈ 125.7 MJ + 353.2 MJ ≈ 478.9 MJ ≈ 4.79 × 10^5 kJ.
Answer: Approximately 4.79 × 10^5 kJ (≈ 479 MJ).
(If you want greater accuracy, use the exact hfg and hf values from your steam tables for p = 850 kPa.)
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.