Question 5 of 7: Heat Duty for Real-Gas Ethylene Heating

Question 5: Heat Duty for Real-Gas Ethylene Heating (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Ethylene, state 1: $T_1=100\,°\text{C}=373.15$ K, $P_1=30$ bar → adiabatic (isenthalpic) valve → heat exchanger (negligible $\Delta P$) → state 2: $T_2=150\,°\text{C}=423.15$ K, $P_2=20$ bar. Ideal-gas heat capacity (A, B terms only) $C_p^{ig}/R = A+BT$ with $A=1.424$, $B=14.394\times10^{-3}\ \text{K}^{-1}$. Critical constants $T_c=282.3$ K, $P_c=50.4$ bar, $\omega=0.087$.

Find. The heat added per mole of ethylene in the exchanger, $Q$ (no flow rate is given, so on a molar basis).

Approach. The insulated valve is isenthalpic, so the exchanger duty equals $H_2-H_1$ between the given end states; evaluate that with an ideal-gas $C_p$ term plus generalized second-virial residual enthalpies at each state.

1. Key simplification — the valve is isenthalpic. An adiabatic throttle does no work and adds no heat, so $H_{1'}=H_1$. The exchanger energy balance (no shaft work) is $Q=H_2-H_{1'}=H_2-H_1$: the intermediate valve-outlet temperature cancels and we only need the two given end states.
2. Ideal-gas enthalpy change. Using $C_p^{ig}=R(A+BT)$ from $T_1$ to $T_2$: $$\Delta H^{ig}=R\!\left[A(T_2-T_1)+\tfrac{B}{2}(T_2^2-T_1^2)\right]=8.314\big[1.424(50)+7.197\times10^{-3}(423.15^2-373.15^2)\big]=2974\ \tfrac{\text{J}}{\text{mol}}.$$
3. Residual enthalpies (generalized second virial). With $H^R/(RT_c)=P_r\big[B^0-T_r\tfrac{dB^0}{dT_r}+\omega(B^1-T_r\tfrac{dB^1}{dT_r})\big]$, $B^0=0.083-0.422/T_r^{1.6}$, $B^1=0.139-0.172/T_r^{4.2}$: $$\text{State 1 }(T_r=1.322,P_r=0.595):\ H_1^R=-882\ \tfrac{\text{J}}{\text{mol}};\qquad \text{State 2 }(T_r=1.499,P_r=0.397):\ H_2^R=-459\ \tfrac{\text{J}}{\text{mol}}.$$ Both are negative (attractive forces lower the real-gas enthalpy), and the higher-temperature, lower-pressure state 2 is closer to ideal.
4. Assemble the duty. Since $H=H^{ig}+H^R$, $$Q=H_2-H_1=\Delta H^{ig}+\big(H_2^R-H_1^R\big)=2974+\big(-459-(-882)\big)=2974+423.$$ $$\boxed{Q \approx +3.40\ \text{kJ per mol of ethylene (heat added)}.}$$ The positive sign confirms heat is added; the residual correction adds ~14% to the ideal-gas estimate.