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? in kj
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Answered by
GPT-5 mini
AI
Procedure:
- Heat required Q = m [hf(p) + x·hfg(p) − h_initial].
- m = 200 kg, x = 0.87. Take h_initial (water at 20°C) ≈ cp·T = 4.186·20 = 83.72 kJ/kg.
- From steam tables at p = 850 kPa (saturated): hf ≈ 721.1 kJ/kg, hfg ≈ 2032.0 kJ/kg.
Compute:
- h_final = hf + x·hfg = 721.1 + 0.87·2032.0 = 2488.9 kJ/kg.
- Δh = h_final − h_initial = 2488.9 − 83.72 = 2405.2 kJ/kg.
- Q = m·Δh = 200·2405.2 = 481,040 kJ ≈ 4.81×10^5 kJ (≈ 481 MJ).
Answer: approximately 4.81×10^5 kJ. (Slightly different steam-table values will change the result a little; tell me if you want it recalculated with a specific table.)
- Heat required Q = m [hf(p) + x·hfg(p) − h_initial].
- m = 200 kg, x = 0.87. Take h_initial (water at 20°C) ≈ cp·T = 4.186·20 = 83.72 kJ/kg.
- From steam tables at p = 850 kPa (saturated): hf ≈ 721.1 kJ/kg, hfg ≈ 2032.0 kJ/kg.
Compute:
- h_final = hf + x·hfg = 721.1 + 0.87·2032.0 = 2488.9 kJ/kg.
- Δh = h_final − h_initial = 2488.9 − 83.72 = 2405.2 kJ/kg.
- Q = m·Δh = 200·2405.2 = 481,040 kJ ≈ 4.81×10^5 kJ (≈ 481 MJ).
Answer: approximately 4.81×10^5 kJ. (Slightly different steam-table values will change the result a little; tell me if you want it recalculated with a specific table.)
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