Question 3.13: In the high-pressure process for the production of polyethyl...

As the calculations will be repetitive, use a spreadsheet.
Data        T_{in} 288 K P_{in} 1 bar P_{out} 250 ba R = 8.1345 J/mol K
T_{c} 282.4 K P_{c} 50.4 bar M 28.05 C_{p} data from Appendix D

First stage
Intermediate pressure P_{2} = 15.811388 bar eqn 3.39
Compression ratio = P_{2}/P_{1} = 15.814
C_{p} for ethylene T_{in} = 288 use eqn 3.11a

sum,C_{p} = 41.535011 kJ/kmol K
gamma = C_{p}/(C_{p} – R) = 1.2502821

From Figure 3.7, extrapolated, Ep = 0.86.

m = 0.232768 eqn 3.36a
T_{2} = 547.52197 eqn 3.35

Mean temp = (T_{1} + T_{2})/2 = 417.76099

C_{p} at mean temp of 419.6 K

sum, C_{p} = 54.851135 kJ/kmol K

new gamma = 1.1786657
revised m = 0.1811049
revised T_{2} = 474.76117
revised mean temp = 381.38058
little change so leave T_{mean} at 419.6 K
T_{r} = T_{mean}/T_{c} = 1.4858357 (1.5)

P_{mean} = (P_{1} * P_{2})/2 = 0.5
P_{t} = P_{mean}/P_{c} = 0.0099206 (0.17)
From Figure 3.2 correction to C_{p} for pressure is negligible.
From Figures 3.8, 3.9, 3.10
Z = 1.0 X = 0 Y = 0
Essentially ideal at this pressure

m = 0.1762593 eqn 3.36a
n = 1.2139743 eqn 3.38a
-W = 303.47285 kJ/kmol eqn 3.31

Actual work required = polytropic work/efficiency = 352.87541
Say \underline{\underline{353} } kJ/kmol

Second stage work
As the intermediate pressure was selected to give equal work in each stage the second stage work could be taken as equal to the first stage work. This will be checked.
T_{in} = 298 K

compression ratio = P_{3}/P_{2} = 15.822785, i.e. same as first stage So, take gamma and efficiency as for first stage
m = 0.1811049
T_{3} = 491.24593 K
T_{mean} = 394.62296 K
C_{p} at mean temp

sum, C_{p} = 52.494599 kJ/kmol K

Little change from first stage, so use same gamma and T_{mean}

T_{r} = 1.5
P_{mean} = 20.4 bar
P_{r} = 0.4047619 (0.4)
From Figure 3.2 correction to C_{p} for pressure is approximately 2.5 J/mol.
This is less than 5 per cent, so neglect.

From Figures 3.8, 3.9, 3.10

Z = 1.0 X = 0.1 approx. Y = 0

So, gas can be taken as ideal
-W = 314.01026 slightly higher as T_{in} is higher

Actual work = 365.12821 \underline{\underline{365 kJ/kmol} }
Total work required first step = \underline{\underline{718} } kJ/kmol
The spreadsheet used for this example was Microsoft Works. A copy of the solution using Microsoft Excel can be found on the Butterworth-Heinemann web site: bh.com/companions/ 0750641428.