4.3
Heat Exchanger Duty (E-101)
Since stream (4) is vaporized so 1st we determine Bubble point then calculate energy transfer T4 to Tbubble and them add S of stream (4) T4
=
308.5 k
Tb
=
361.73 k
Tavg
=
T4 + Tb 2
Tavg
qCH3OH
=
335.115 k
=
121.7856 [40.046 – 3.8287 x 10-2 (T) +2.4529 x 10-4 (T2) -2.1679 x 10-7 (T3)]
=
121.7856 [40.046-12.83 + 27.54 – 8.1587]
=
n Cp (361.73 – 308.5)
=
5729.68 (53.23)
=
304991.06 kJ + n
=
304991.06 kJ +4682993.423
qCH3OH
=
4987984.483 kJ
qH2O
=
T = 53.23
qH2O
=
1.948 (29.526-8.8999x 10-3 T + 3.8083x 10-5T2)
=
1.948 (29.526 –2.982 + 4.2768 – 1.228) T
=
1.948 (29.59) T+ n
=
3068.24 + 80469.85
=
83538.094 kJ
qCH3OH
43