Question 11.16: Ethylene (C2H4) gas is to be isothermally compressed from 15...

a. The change in specific enthalpy of the ethylene is given by Eq. (11.41) as

h_2 − h_1 = (h_2^*− h_1^*) -[(\frac{\overline{h}^*− \overline{h} }{T_c} )_2 – (\frac{\overline{h}^*− \overline{h} }{T_c} )_1] (\frac{T_c}{M} )

and since this is an isothermal process, T_2 = T_1 and the ideal gas portion of this equation for a constant specific heat gas is h_2^*− h_1^* = c_p(T_2 − T_1) = 0. The properties of ethylene at its critical state and its molecular mass are found in Table C.12a as T_c = 508.3  \text{R} , p_c = 742  \text{psia}, and M = 28.05  \text{lbm/lbmoles}. Then,

p_{R1} = \frac{150.}{742} = 0.202    and    T_{R1} = \frac{80.0 + 459.67}{508.3} = 1.06

p_{R2} = \frac{15.0×10^3}{742} = 0.202    and    T_{R2} =T_{R1} = 1.06

Using p_{R1} = 0.202 and T_{R1} = 1.06, Figure 11.9 gives the enthalpy correction for state 1 as

(\frac{\overline{h} ^*−\overline{h} }{T_c} )_1 = 1.50\frac{\text{kJ}}{\text{kgmole.K}} =(1.50\frac{\text{kJ}}{\text{kgmole.K}})(\frac{1 \text{Btu/(lbmole.R)}}{4.1865\text{kJ/(kgmole.K)}}) = 0.360 \frac{\text{Btu}}{\text{lbmole.R}}

and using p_{R2} = 20.2 and T_{R2} = 1.06, Figure 11.9 gives the enthalpy correction for state 2 as

(\frac{\overline{h} ^*−\overline{h} }{T_c} )_2 = 31.5\frac{\text{kJ}}{\text{kgmole.K}} =(31.5\frac{\text{kJ}}{\text{kgmole.K}})(\frac{1 \text{Btu/(lbmole.R)}}{4.1865\text{kJ/(kgmole.K)}}) = 7.52 \frac{\text{Btu}}{\text{lbmole.R}}

Then, Eq. (11.41) gives

h_2 − h_1 = (h_2^*− h_1^*) -[(\frac{\overline{h}^*− \overline{h} }{T_c} )_2 – (\frac{\overline{h}^*− \overline{h} }{T_c} )_1] (\frac{T_c}{M} )

=0− [7.52 − 0.360  \text{Btu(lbm.R)}](\frac{508.3  \text{R}}{28.05  \text{lbm/lbmole}} )

= -130.\frac{ \text{Btu}}{lbm}

Note that, since the compressibility charts cannot be read to more than two or three significant figures, our final calculation results are limited to this accuracy as well

b. The compressibility charts do not give values for the specific internal energy, so it must be calculated from the definition of enthalpy as u = h  –  p_v, or u_2  –  u_1 = h_2  –  h_1  –  (p_2 v_2  –  p_1 v_1), where v_1 = Z_1 RT_1/p_1 and v_2 = Z_2 RT_2/p_2. For p_{R1} = 0.202 and T_{R1} = 1.06, Figure 11.5 gives Z_1 = 0.940, and for p_{R2} = 20.2 and T_{R2} = T_{R1} = 1.06, Figure 11.7 gives Z_2 = 2.15. The gas constant R for ethylene can be found in Table C.13a as R = 55.1  \text{ft.lbf/(lbm.R)}. Then,

v_1 = \frac{Z_1RT_1}{p1} = \frac{0.940[55.1  \text{ft.lbf/(lbm.R)}](80.0 + 459.67  \text{R})}{(150.  \text{lbf/in}^2)(144  \text{in}^2\text{/ft}^2)} = 1.29 \frac{\text{ft}^3}{\text{lbm}}

and

v_2 = \frac{Z_2RT_2}{p2} = \frac{2.15[55.1  \text{ft.lbf/(lbm.R)}](80 + 459.67  \text{R})}{(15.0× 10^3  \text{lbf/in}^2)(144  \text{in}^2\text{/ft}^2)} = 0.030 \frac{\text{ft}^3}{\text{lbm}}

Then,

u_2 − u_1 = h_2 − h_1 −(p_2v_2 − p_1v_1)

=130. \frac{\text{Btu}}{\text{lbm}} – (15.0 × 10^3 \frac{\text{lbf}}{\text{in}^2})(144 \frac{\text{in}^2}{\text{ft}^2})(0.0300 \frac{\text{ft}^3}{\text{lbm}})(\frac{1  \text{Btu}}{778.16  \text{ft.lbf}} )

– (150.\frac{\text{lbf}}{\text{in}^2})(144 \frac{\text{in}^2}{\text{ft}^2})(1.29 \frac{\text{ft}^3}{\text{lbm}})(\frac{1  \text{Btu}}{778.16  \text{ft.lbf}} )

=−180.  \frac{\text{Btu}}{\text{lbm}}

c. Finally, from Eq. (11.42), we have the change in specific entropy as

s_2 − s_1 = [(s_2^*− s_1^*) -[(\overline{s}^*− \overline{s})_2 – (\overline{s}^*− \overline{s} )_1] (\frac{1}{M} )

where, for a constant specific heat ideal gas,

s_2^*− s_1^* =  cp  \text{ln} (T_2/T_1) − \text{R ln} (p_2/p_1)− s_1^*

and, since T_2 = T_1 here, this becomes

s_2^*− s_1^* = 0 −[\frac{55.1  \text{ft.lbf /(lbm.R)}}{778.16  \text{ft.lbf /Btu}} ]  \text{ln} (\frac{15.0 × 10^3}{150.} ) = −0.326 \frac{\text{Btu}}{\text{lbm.R}}

Using p_{R1} = 0.202 and T_{R1} = 1.06, Figure 11.11 gives the entropy correction for state 1 as

(\overline{s}^*− \overline{s})_1 = 1.50 \frac{\text{kJ}}{\text{kgmole.K}} = (1.50  \frac{\text{kJ}}{\text{kgmole.K}} )  (\frac{1 \text{Btu/(lbmole.R)}}{4.1865\text{ kJ/(kgmole.K)}})= 0.360  \frac{\text{Btu}}{\text{lbmole.R}}

and using p_{R2} = 20.2 and T_{R2} = 1.06 , Figure 11.11 gives the entropy correction for state 2 as

(\overline{s}^*− \overline{s})_2 = 22.2 \frac{\text{kJ}}{\text{kgmole.K}} = (22.2 \frac{\text{kJ}}{\text{kgmole.K}} )  (\frac{1 \text{Btu/(lbmole.R)}}{4.1865\text{ kJ/(kgmole.K)}})= 5.30  \frac{\text{Btu}}{\text{lbmole.R}}

Then, Eq. (11.42) gives

s_2 − s_1 = [(s_2^*− s_1^*) -[(\overline{s}^*− \overline{s})_2 – (\overline{s}^*− \overline{s} )_1] (\frac{1}{M} )

= − 0.326 \frac{\text{Btu}}{\text{lbm.R}} – [5.30 − 0.360  \frac{\text{Btu}}{\text{lbm.R}} ] (\frac{1}{28.05  \text{ lbm/lbmole}} )

= −0.500 \frac{\text{Btu}}{\text{lbm.R}}

The following exercises illustrate some of the elements of Example 11.16.