The specific enthalpy of a liquid hydrocarbon at 1atm varies linearly with tempe

Question

The specific enthalpy of a liquid hydrocarbon at 1atm varies linearly with temperature at equals 25 kJ/kg at 30C and 130 kJ/kg at 50C. Determine the equation relating specific enthalpy to temperature, and calculate the temperature reference temperature for which the given enthalpies are based. Derive an equation for U(T) at 1atm. Calculate the heat transfer rate required to cool the liquid hydrocarbon, flowing at 30 kg/min from 60C to 25C at a constant pressure of 1atm. What is the change in specific internal energy for this change?

Explanation / Answer

Now we know that heat load defined as:

Q = mCpT

Where m=mass

  Cp = specific heat

T = temperature difference

For constant pressure

Specific enthalpy H = Q =mCpT

(A)

At 300C H1 = 25 kJ/kg

At 500C H2 = 130 kJ/kg

So we have to find refrence temperature.

H1 = mCp(30 -T)=25

H2 =mCp(50 -T)=130

Now

H1/H2 = (30-T)/(50-T) = 25/130

So

30-T=9.6-0.1932T

T=18.1 0C

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