Energy Balance?

[匿名]

2016-04-16 7:42 pm

The specific enthalpy of liquid n-hexane at 1 atm varies linearly with temperature and equals 25.8 kJ/kg at 30C and 129.8 kJ/kg at 50C.
Determine the equation that relates 𝑯 (kJ/kg) to T (C) and calculate the reference temperature on which the given enthalpies are based. Then derive an equation for 𝑼 (T) (kJ/kg) at 1 atm. (b) Calculate the average heat transfer rate required to cool 20.0 kg of liquid n-hexane from 80C to 20C in 5 min.

回答 (1)

az_lender

2016-04-17 12:08 am

✔ 最佳答案

H(T) = [ (129.8 - 25.8) (kJ/kg) /(20C) ] * T + Ho.
Here Ho must satisfy
25.8 kJ/kg = (104 kJ/kg)*(30/20) + Ho, so
Ho = 25.8 kJ/kg - 156 kJ/kg = -130.2 kJ/kg, and
H(T) = 5.2 kJ/(kg*degC) * T - 130.2 kJ/kg.

I don't like the question about the reference temperature; it could be many different things. But I suppose they want you to set H(T) to zero, obtaining T = 130.2C/5.2 = about 26C, use calculator.

To cool 20.0 kg by 60C in 5 minutes requires the removal of (5.2 kJ)(20.0 kg)(60C)/(kg*degC), or 62.4 MJ in 300 seconds, so the average rate of heat transfer is 208 kJ/s = 208 kW.

If the liquid is nearly incompressible, enthalpy changes and internal energy changes will be equal; that's all I can contribute on the subject of
U(T).

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