# Chapter 5 : Energy Balances¶

### Example 5.1 Page 186¶

In :
# solution

# Variables
# basis pumping of 1 l/s of water
Hadd = 52.                      # kW
Hlost = 21.                     # kW

# Calculation
fi = Hadd - Hlost               # kW
p1 = 101325.                    # Pa
p2 = p1
Z1 = -50                        # m
Z2 = 10                         # m
g = 9.80665                     # m/s sq
gc = 1                          # kg.m/(N.s sq)
row = 1                         #   kg/l
W = 1.5*.55                     # kW
# energy balance b/w A and B
# dE = E2-E1 = W + Q + (Z1-Z2)*(g/gc)*qm
dE = 31.237                     # kW

# Result
print "Increase in internal energy between the storage tank and the bottom of the well = ",dE," kW."

Increase in internal energy between the storage tank and the bottom of the well =  31.237  kW.


### Example 5.2 Page 197¶

In :
# solution

# Variables
# using table 5.1
# basis 1 kmol of methane
T1 = 303.15             # K
T2 = 523.15             # K

# Calculation
# using eq 5.17
H = 19.2494*(T2-T1) + 52.1135*10**-3*(T2**2-T1**2)/2 + \
11.973*10**-6*(T2**3-T1**3)/3 - 11.3173*(T2**4-T1**4)*10**-9/4      # kJ

# Result
print " Heat added = ",H," kJ/kmol methane."

 Heat added =  9243.82741713  kJ/kmol methane.


### Example 5.3 Page 198¶

In :
# solution

# Variables
# basis 1 kmol methane at 25 bar
Pc = 46.04              # bar
Tc = 190.5              # K
Pr = 25/Pc

# Calculation
# H-Ho = intgr(from303.15 to 523.15){CmpR dT}
# solving it by simpson's rule
HE = 255.2              # kJ/kmol
H = 9175.1 + HE

# Result
print " Heat added = ",H," kJ/kmol of methane."

 Heat added =  9430.3  kJ/kmol of methane.


### Example 5.4 Page 206¶

In :
# solution

# Variables
# using table 5.3
# .25 kg/s toulene heated from 290.15K to 350.15K
qm = .25/92             # kmol/s

# Calculation
# reference 7
fi = 2.717*10**-3*(1.8083*(350.15-290.15) + 812.223*10**-3*(350.15**2-290.15**2)/2 \
- 1512.67*10**-6*(350.15**3-290.15**3)/3 + 1630.01*10**-9*(350.15**4-290.15**4)/4)

# Result
print " Heat required to be added to toulene = ",fi," kW."

 Heat required to be added to toulene =  26.1324337975  kW.


### Example 5.5 Page 206¶

In :
# solution

# Variables
# basis 1kg of 20% NaOH sol
# referring to fig 5.4
C11 = 3.56                      # kJ/kg.K   at 280.15K
C12 = 3.71                      # kJ/kg.K   at 360.15K

# Calculation
C1m = (C11+C12)/2
H = 1*C1m*(360.15-280.15)       # kJ

# Result
print " Heat required to be added = ",H,"kJ."

 Heat required to be added =  290.8 kJ.


### Example 5.6 Page 207¶

In :
# solution

# Variables
# basis 1kg Diphyl A-30
Q = .7511*(553.15-313.15) + 1.465*10**-3*(553.15**2-313.15**2)/2    # kJ/kg
fi = Q*4000                                 # kJ/h    for mass flowrate 4000 kg/h

# Calculation
Clm = (1.1807+1.5198)/2
fi1 = Clm*(553.15-313.15)*4000/3600.        # kJ/h
err = (fi1-Q)*100/Q

# Result
print " Heat to be supplied = ",fi1," kW  Percent error = ",err,"."

 Heat to be supplied =  360.066666667  kW  Percent error =  8.27133892074 .


### Example 5.7 Page 208¶

In :
# solution

# Variables
T1 = 298.15                 # K
T2 = 775.15                 #K

# Calculation
# using eq 5.17
Q = 28.839*(T2-T1)+2.0395*10**-3*(T2**2-T1**2)/2 + \
6.9907*10**-6*(T2**3-T1**3)/3 - 3.2304*10**-9*(T2**4-T1**4)/4       # kJ/kmol

# Result
print " Heat content of 1 kmol of gas mixture at 298K = ",Q," kJ/kmol."

 Heat content of 1 kmol of gas mixture at 298K =  15016.6500785  kJ/kmol.


### Example 5.8 Page 210¶

In :
# solution

# Variables
# basis 8000 kg/h mixture is to be cooled
qn1m = .118*8000                    # kg/h
qn1 = qn1m/93.1242                  # kmol/h
qn2m = 8000-qn1m                    # kg/h
qn2 = qn2m/18                       # kmol/h
T1 = 373.15                         #K
T2 = 313.15                         #K

# Calculation
fi = qn1*(206.27*(T1-T2)-211.5065*10**-3*(T1**2-T2**2)/2+564.2902*10**-6*(T1**3-T2**3)/3) \
+ qn2*(50.845*(T1-T2)+213.08*10**-3*(T1**2-T2**2)/2-631.398*10**-6*(T1**3-T2**3)/3 \
+648.746*10**-9*(T1**4-T2**4)/4)   # kJ/h

# Result
print " Heat removal rate of subcooling zone of the condenser = ",fi," kJ/h."

 Heat removal rate of subcooling zone of the condenser =  1905578.44431  kJ/h.


### Example 5.9 Page 220¶

In :
# solution

# Variables
# (a)
T = 305.15                                      #K

# Calculation
Pv1 = 10**(4.0026-(1171.530/(305.15-48.784)))   # bar
# (b)
T = 395.15
Pv2 = 10**(3.559-(643.748/(395.15-198.043)))     # bar

# Result
print " a   V.P. of n-hexane at 305.15K = ",Pv1,\
" bar.    \nb   V.P. of water at 395.15K = ",Pv2," bar."

 a   V.P. of n-hexane at 305.15K =  0.270921991993  bar.
b   V.P. of water at 395.15K =  1.96343968214  bar.


### Example 5.10 Page 225¶

In :
import math
# solution

# Variables
# (a)
Pc = 3732.              # kPa
Tc = 630.3              # K
Tb = 417.6              # K
TBr = Tb/Tc

# Calculation
lambdav = 8.314472*417.6*(1.092*(math.log(3732)-5.6182)/(.930-.6625))
# (b)
T1 = 298.15  #K
lambdav1 = 36240*((630.3-298.15)/(630.3-417.6))**.38

# Result
print " a   Latent heat of vaporization at Tb using Riedel eq is ",lambdav,\
" kJ/kmol.    \nb   Latent heat of vaporizaation at 298.15 K using Watson eq is ",\
lambdav1," kJ/kmol."

 a   Latent heat of vaporization at Tb using Riedel eq is  36944.6599382  kJ/kmol.
b   Latent heat of vaporizaation at 298.15 K using Watson eq is  42928.2871556  kJ/kmol.


### Example 5.11 Page 225¶

In :
import math
# solution

# Variables
# (a)
Pc = 61.37                  # bar
Tc = 514.                   #K
Tb = 351.4
P = 1                       # atm
TBr = Tb/Tc

# Calculation
# Riedel eq
lambdav1 = 8.314472*Tb*1.092*(math.log(6137)-5.6182)/(.930-TBr)

# NIST eq
lambdav2 = 50430*math.exp(-(-.4475*TBr))*(1-TBr)**.4989

# (b)
T1 = 298.15
TBr1 = T1/Tc

# Watson eq
lambdav21 = 38563*((514-298.15)/(514-351.4))**.38

# NIST eq
lambdav22 = 50430*math.exp(-(-.4475*TBr1))*(1-TBr1)**.4969

# Result
print " a   \nLatent heat of vaporization at Tb using  Riedel eq is ",lambdav1,\
" kJ/kmol  \nNIST eq is ",lambdav2,\
" kJ/kmol    \nb   Latent heat of vaporization at 298.15 K using  Watson eq is ",lambdav21,\
" kJ/kmol  \nNIST eq is ",lambdav22," kJ/kmol"

 a
Latent heat of vaporization at Tb using  Riedel eq is  40200.030668  kJ/kmol
NIST eq is  38564.1849773  kJ/kmol
b   Latent heat of vaporization at 298.15 K using  Watson eq is  42946.0066334  kJ/kmol
NIST eq is  42479.9527916  kJ/kmol


### Example 5.12 Page 227¶

In :
import math

# solution
# Variables
# using Appendix IV.2
Ps1 = 75.
Ps2 = 80.
T1 = 563.65
T2 = 568.12
T = 565.15

# Calculation
Ps = 75*math.exp((T2*(T-T1)*math.log(80./75)/(T*(T2-T1))))

# Result
print " Saturation Pressure of steam at 565.15K is ",Ps," bar."

 Saturation Pressure of steam at 565.15K is  76.6507315525  bar.


### Example 5.13 Page 236¶

In :
# solution

# Variables
# basis  1 kmol equimolar mix
npent = .5                  # kmol
nhex = .5                   # kmol
P = 101.325                 # kPa
x1 = .5
x2 = x1
Ts1 = 309.2                 # K
Ts2 = 341.9                 # K

# Calculation
T1 = (Ts1+Ts2)/2.

# using these data, we get table 5.10 and 5.11
Tbb = 321.6                 # K
Tdp = 329.9                 # K

# Result
print " Bubble point = ",Tbb,\
" K and  Dew point = ",Tdp," K."

 Bubble point =  321.6  K and  Dew point =  329.9  K.


### Example 5.14 Page 237¶

In :
from numpy import poly1d,roots
# solution

# Variables
# basis 1000 kg/h of condensate at the saturation temperature corresponding to 8 bar a
# using Appendix IV.2
H = 720.94                  # kJ/kg
Hm = 419.06                 # kJ/kg

# Calculation
x = poly1d(0,'x')
condensate = 1000-x
Hcondensate1 = 1000*H
Hcondensate2 = condensate*419.06
Ht = x*2676.
p = Hcondensate2+Ht-Hcondensate1

# Result
print " The quantity of flash steam produced = ",roots(p)," kg/h."

 The quantity of flash steam produced =  133.756324936  kg/h.


### Example 5.15 Page 238¶

In :
# solution
from numpy import poly1d, roots

# Variables
qv1 = 50.                           # l/s
qm = qv1*1.08                       # kg/s
fi = qm*3.08*(263.15-258.15)        # kW
lv = 384.19-168.7                   # kJ/kg

# Calculation
qm2 = fi/lv
H = 256.35                          # kJ/kg
x = poly1d(0,'x')
p = H*(qm2+x) - 168.7*qm2-x*384.19
a = qm2+roots(p)

# Result
print " Flow of vapor from he chiller = ",a," kg/s."

 Flow of vapor from he chiller =  6.50500625782  kg/s.


### Example 5.16 Page 238¶

In :
# solution

# Variables
# basis liquifaction capacity = 0.116 kg/s
p1 = 101.                   # kPa
Ts1 = 239.15
lv1 = 288.13                # kJ/kg
p2 = 530.                   # kPa
Ts2 = 290.75                # K
lv2 = 252.93                # kJ/kg

# Calculation
# referring to table 5.3 and using eq 5.21
H1 = -39.246*(Ts2-Ts1)+1401.223*10**-3*(Ts2**2-Ts1**2)/2-6047.226* \
10**-6*(Ts2**3-Ts1**3)/3+8591.4*10**-9*(Ts2**4-Ts1**4)/4        # kJ/kmol
T3 = 313.15
H2 = (28.5463*(T3-Ts1)+23.8795*10**-3*(T3**2-Ts1**2)/2-21.3631* \
10**-6*(T3**3-Ts1**3)/3+6.4726*10**-9*(T3**4-Ts1**4)/4)/70.903  # kJ/kg
fi2 = .116*H2
Cl2evp = fi2/lv1            # kg/s
Cl2recy = Cl2evp/(1-.185)
R = Cl2recy/.116            # kg/kg fresh feed

# T4/T1 = (p2/p1)**[(gamma-1)/gamma]
gm = 1.355
p22 = 326.3
p21 = 101.
T4 = Ts1*(p2/p1)**((gm-1)/gm)
T5 = 313.15
fi3 = 1.88*10**-3*(343.1+91.6-26.2+2.5)         # kW
Fwater1 = fi3/(8*4.1868)                        # kg/s

# similarly
T6 = 379.9
fi4 = 1.88*10**-3*(28.5463*(T6-T5)+23.8795*10**-3*(T6**2-T5**2)/2-21.3631 \
*10**-6*(T6**3-T5**3)/3+6.4726*10**-9*(T6**4-T5**4)/4)  # kW
Fwater2 = fi4/(8*4.1868)                        # kg/s
Wreq = Fwater1+Fwater2
fi5 = 1.88*10**-3*(28.5463*(T5-Ts2)+23.8795*10**-3* \
(T5**2-Ts2**2)/2-21.3631*10**-6*(T5**3-Ts2**3)/3+6.4726 \
*10**-9*(T5**4-Ts2**4)/4) +.1333*252.93         # kW

# Result
print " a   Recycle ratio = ",R," kg Cl2/kg fresh feed    \nb   Cooling water required at  interface = "\
,Fwater1," kg/s  after cooler = ",Wreq," kg/s    \nc   Refrigiration load of chiller = ",fi5," kW."

 a   Recycle ratio =  0.149514564975  kg Cl2/kg fresh feed
b   Cooling water required at  interface =  0.0230689309258  kg/s  after cooler =  0.152400566068  kg/s
c   Refrigiration load of chiller =  35.1468085758  kW.


### Example 5.17 Page 242¶

In :
# solution

# Variables
# basis 100 kg of tin
T1 = 303.15
T2 = 505.15
n = 100/118.7                   # kmol

# Calculation
# Q1 = n*[intgr from T1 to T2 (Cms dT)]
Q1 = 4973.3                     # kJ
lf = 7201.
Q2 = n*lf                       # kJ
Q = Q1+Q2
lv = 278.                       # kJ/kg
vp = Q/lv                       # kg

# Result
print " Quantity of eutectic mixture condensed = ",vp,\
" kg per 100 kg of tin melted at its melting point."

 Quantity of eutectic mixture condensed =  39.7117062542  kg per 100 kg of tin melted at its melting point.


### Example 5.18 Page 243¶

In :
# solution

# Variables
Ts1 = (438.2+436)/2.
Ta = 300.

# Calculation
fi1 = .045*(Ts1-Ta)*3600.
theta1 = 307293/fi1                 #h
Ts2 = (436+434)/2.
fi2 = .045*(Ts2-Ta)*3600.
theta2 = 302415./fi2
Ts3 = (434+432.1)/2
fi3 = .045*(Ts3-Ta)*3600.
theta3 = 313859/fi3
theta = theta1+theta2+theta3

# Result
print " total time required = ",theta," hrs."

 total time required =  42.2249688079  hrs.


### Example 5.19 Page 245¶

In :
from numpy import poly1d, roots

# solution

# Variables
H1 = 482.9                      # kJ/kg
H2 = 273.4
fi1 = 100*(H1-H2)               # kJ/h
T1 = 313.15
T2 = 403.15

# Calculation
fi11 = 21.3655*(T2-T1)+64.2841*10**-3*(T2**2-T1**2)/2-41.0506*10**-6 \
*(T2**3-T1**3)/3+9.7999*10**-9*(T2**4-T1**4)/4          # kJ/h

# at 20 MPa
h1 = 211.1
Ts = 277.6
H11 = 427.8
x = poly1d(0,'x')
p = x*h1+(100-x)*H11-100*H2
a = roots(p)
fi2 = (100-a)*(H11-h1)          # kJ/h
h2 = -148.39
H3 = 422.61
y = poly1d(0,'y')
p1 = 100*176.18-(100-y)*H3+h2*y
b = roots(p1)
fi3 = 100*(h1-176.8)
H = fi3+24021
H4 = H/(100-43.16)
# from ref 23
T = 262.15

# Result
print "a   Yield of dry ice = ",b,\
" kg.    \nb   Percent liquifaction = ",a,\
".    \nc   Temp of vented gas = ",T," K."

a   Yield of dry ice =  89.8658011815  kg.
b   Percent liquifaction =  71.2505768343 .
c   Temp of vented gas =  262.15  K.


### Example 5.20 Page 247¶

In :
# solution

# Variables
# basis 200 kg/h of Sulphur firing
F = 200./32                     # kmol/h
O2req = 6.25*1.1
airin = O2req/.21

# Calculation
N2in = airin-O2req
T1 = 1144.15
T2 = 463.15
fi = 788852.2                   # kJ/h
H = 15*4.1868+1945.2
qm = fi*.9/2008                 # kg/h

# Result
print " Amount of steam produced = ",qm," kg/h."

 Amount of steam produced =  353.569213147  kg/h.


### Example 5.21 Page 248¶

In :
# solution

# Variables
# enthalpy at Tbb
Tbb = 321.6
T1 = 298.15

# Calculation
H1 = 65.4961*(Tbb-T1)+628.628*10**-3*(Tbb**2-T1**2)/2-1898.8*10**-6*(Tbb**3-T1**3)/3 \
+3186.51*10**-9*(Tbb**4-T1**4)/4                                # kJ/kmol
H2 = 31.421*(Tbb-T1)+976.058*10**-3*(Tbb**2-T1**2)/2-2353.68* \
10**-6*(Tbb**3-T1**3)/3+3092.73*10**-9*(Tbb**4-T1**4)/4         # kJ/kmol
Hsol = (H1+H2)/2                                                # kJ/kmol

# enthalpy at Tdp
lv1 = 25790*((469.7-329.9)/(469.7-309.2))**.38
lv2 = 28850*((507.6-329.9)/(507.6-341.9))**.38
Tdp = 329.9
H21ig = 65.4961*(Tdp-T1)+628.628*10**-3*(Tdp**2-T1**2)/2-1898.8*10**-6* \
(Tdp**3-T1**3)/3+3186.51*10**-9*(Tdp**4-T1**4)/4 + lv1          # kJ/kmol
H22ig = 31.421*(Tdp-T1)+976.058*10**-3*(Tdp**2-T1**2)/2-2353.68* \
10**-6*(Tdp**3-T1**3)/3+3092.73*10**-9*(Tdp**4-T1**4)/4 +lv2    # kJ/kmol
Hmixig = (H21ig+H22ig)/2

# Result
print " a   H = ",Hsol," kJ/kmol    b   H = ",Hmixig," kJ/kmol"

 a   H =  4370.454066  kJ/kmol    b   H =  33019.341801  kJ/kmol


### Example 5.22 Page 252¶

In :
# solution

# Variables
H1 = 23549.                 #kJ/kmol
H2 = 16325.
H3 = 28332.

# Calculation
H4 = .4*H2+.6*H3

# Result
print "Enthalpy of vapor-liquid mixture after flashing = ",H4," kJ/mol."

Enthalpy of vapor-liquid mixture after flashing =  23529.2  kJ/mol.


### Example 5.23 Page 253¶

In :
# solution

# Variables
# basis feed gas = 12000 Nm**3 = 535.4 kmol/h
T1 = 147.65                         # K
n1 = 535.4*.3156                    # kmol/h  HP tail gas stream
T = 118.5                           # K

# Calculation
n2 = (535.4-n1)*.0602               # kmol/h    LP tail stream
n3 = 535.4-n2-n1                    # kmol/h  product H2 stream
p = 315.35*100/n3

# Result
print " Purity of product H2 stream = ",p," percent."

 Purity of product H2 stream =  91.5733341399  percent.


### Example 5.24 Page 256¶

In :
# solution

# Variables
# fi1 = integr (from 304.15 to 313.15) {11831.6+24997.4*10T**-3-5979.8*10**-6T**2-31.7*10**-9T3}dt
fi1 = 170787.7 # kJ/h

# Calculation
fi2 = 535.4*12086 - (344.36*8743.2+168.97*18036+22.07*15892)          # kJ/h

# Result
print " a   Refrigiration requirement = ",fi1,\
" kJ/h    b   Refrigiration requirement based on real enthalpies = ",fi2," kJ/h."

 a   Refrigiration requirement =  170787.7  kJ/h    b   Refrigiration requirement based on real enthalpies =  61756.688  kJ/h.


### Example 5.25 Page 257¶

In :
# solution

# Variables
# basis 100 kmol/h of benzene feed rate
Cl2 = .4*100
HClp = 40.
Benzenecon = 37.
MCBp = 100*.37*.9189

# Calculation
DCBp = Benzenecon-MCBp
unreactBenzene = 100-Benzenecon
Nt = HClp + MCBp + DCBp + unreactBenzene
# using eq      xi = Ni/(L(1-K1)+NtKi)  and sigma xi = 1
L = 89.669                  # kmol/h
V = Nt - L

# Result
print " Liquid product stream = ",L," kmol/h  Vapor product stream = ",V," kmol/h"

 Liquid product stream =  89.669  kmol/h  Vapor product stream =  50.331  kmol/h


### Example 5.26 Page 260¶

In :
# solution

# Variables
# 2C + 2O2 = 2CO2                  A
# 2H2 + O2 = 2H2O                  B
# C2H4 + 3O2 = 2CO2 + 2H2O         C
# A+B-C  gives
# 2C(g) + 2H2 = C2H4(g)            D

# Calculation
H = -2*393.51-2*241.82+1323.1           # kJ/mol

# Result
print " Heat of formation of Ethylene is ",H," kJ/mol."

 Heat of formation of Ethylene is  52.44  kJ/mol.


### Example 5.27 Page 260¶

In :
# solution

# Calculation

Hc = 2*(-393.51)-887.811+2*(-285.83)-(-73.6+0)              #kJ/mol

# Result
print " Heat of combustion of ethyl mercaptan = ",Hc," kJ/mol."

 Heat of combustion of ethyl mercaptan =  -2172.891  kJ/mol.


### Example 5.28 Page 261¶

In :
# solution
# Variables
lv1 = 26694.                # kj/kmol
Tc = 466.74

# Calculation
lv2 = lv1*((Tc-298.15)/(Tc-307.7))**.38/1000                # kJ/mol
Hf = -252.                  # kJ/mol
Hf1 = Hf-lv2                # kJ/kmol

# Result
print "Heat of formation of liquid di ethyl ether = ",Hf1," kJ/mol."

Heat of formation of liquid di ethyl ether =  -279.292123302  kJ/mol.


### Example 5.29 Page 261¶

In :
# solution

# Variables
# basis 1 kg motor spirit
G = 141.5/(131.5+64)

# Calculation
# r = C/H
r = (74.+15*G)/(26-15*G)
C = r/6.605                     # C content of motor spirit
H2 = 1-C
O2req = C+H2
Hf = 44050-27829-18306.         # kJ/kg

# Result
print " Heat of formation of motor spirit = ",Hf," kJ/kg."

 Heat of formation of motor spirit =  -2085.0  kJ/kg.


### Example 5.30 Page 267¶

In :
# solution
# Variables
# basis 1 kmol of styrene
dH = 241749-189398.                     # kJ/mol

# Calculation
Cmpn = dH/(600-298.15)                  # kJ/kmol K

# Result
print " Mean heat capacity between 600K and 298.15 K is ",Cmpn," kJ/kmol K."

 Mean heat capacity between 600K and 298.15 K is  173.433824747  kJ/kmol K.


### Example 5.31 Page 269¶

In :
# solution

# Calculation
# basis 1 mol of SiO2 reacted
Hf = (-2879+3*(-296.81)+3*0/2)-(3*(-1432.7)+1*(-903.5))  # kJ/mol SiO2

# Result
print " Heat of reaction = ",Hf," kJ/mol SiO2."

 Heat of reaction =  1432.17  kJ/mol SiO2.


### Example 5.32 Page 269¶

In :
# solution

# Variables
# basis 100 kg of 2% ammonia solution
NH3 = 2.                # kg
H2O = 98.               # kg

# Calculation
Hr = -361.2-(-45.94-285.83)             # kJ/mol NH3 dissolved
Hd = -(Hr*2*1000/17.0305)               # kJ/100 kg sol.

# Result
print " heat of reaction = ",Hd," kJ/100 kg solution."

 heat of reaction =  3456.15219753  kJ/100 kg solution.


### Example 5.33 Page 272¶

In :
# solution

# Variables
# basis 1 kmol of SO2 reacted
a = 22.036-24.771-.5*(26.026)
b = (121.624-62.948-.5*11.755)
c = (-91.876+44.258-.5*(-2.343))
d = (24.369-11.122-.5*(-.562))
Hr = -395720+296810.                # kJ/kmol

# Calculation
Hro = Hr-a*298.15-b*10**-3*298.15**2/2-c*10**-6*298.15**3/3-d*10**-9*298.15**4/4.
T = 778.15
Hrt = -Hro-15.748*T+26.4*10**-3*T**2-15.48*10**-6*T**3+3.382*10**-9*T**4

# Result
print " Heat of reaction at 775K is ",Hrt," kJ/kmol."

 Heat of reaction at 775K is  93855.2952943  kJ/kmol.


### Example 5.34 Page 272¶

In :
# solution

# Calculation
Hr = -480-285.83+277.2+484.2                            # kJ/mol
Hrt1 = Hr*1000 + (146.89+75.76-119.55-129.70)*75        # kJ/kmol
a = 4.2905+50.845-100.92-155.48
b = 934.378+213.08+111.8386+326.5951
c = -2640-631.398-498.54-744.199
d = 3342.58+648.746
Hro = Hr*1000+a*(-298.15)+b*10**-3*(-298.15**2)/2+ \
c*10**-6*(-298.15**3)/3+d*10**-9*(-298.15**4)/4
T = 373.15
Hrt = Hro+a*T+792.949*10**-3*T**2-1504.712*10**-6*T**3+997.832*10**-9*T**4

# Result
print " Heat of reaction at 373 K is ",Hrt," kJ/kmol reactant."

 Heat of reaction at 373 K is  -6441.65717017  kJ/kmol reactant.


### Example 5.35 Page 273¶

In :
# solution

# Variables
T2 = 800.
T1 = 298.15

# Calculation
fi1 = 3614.577*(T2-T1)+305.561*10**-3*(T2**2-T2**2)/2+836.881*10**-6*(T2**3-T1**3)/3-393.707*10**-9*(T2**4-T1**4)/4  # kW
T3 = 875.
fi2 = 3480.737*(T3-T1)+754.347*10**-3*(T3**2-T2**2)/2+442.159*10**-6*(T3**3-T1**3)/3-278.735*10**-9*(T3**4-T1**4)/4  # kW
Hr = -98910.                            # kJ/kmol SO2 reacted     by eg 5.33
fi3 = (8.8511-.351)*Hr/3600.            # kW
dH = fi2/3600+fi3-fi1/3600.

# Result
print " Net enthalpy change = ",dH," kW."

 Net enthalpy change =  -178.010235047  kW.


### Example 5.36 Page 275¶

In :
# solution

# Variables
# basis 100 kmol outgoing gas mixture from scrubber
moistin = 3127.7*.015/18                        # kmol
waterin = 40.2+moistin                          # kmol

# Calculation
# using tables 5.29 and 5.30
Hr = -27002658-(-26853359.)
Hr1 = Hr/246.4493                               # kJ/kmol total reactants

# Result
print " Heat of reaction = ",Hr1," kJ/kmol total reactants."

 Heat of reaction =  -605.80005705  kJ/kmol total reactants.


### Example 5.37 Page 276¶

In :
# solution

# Variables
fi3 = 15505407.                     # kJ/h
lv = 296.2                          # from table 5.6
Ht = 17131551.                      # kJ/h

# Calculation
r = Ht/lv                           # kg/h

# Result
print " Downtherm circulation rate = ",r," kg/h."

 Downtherm circulation rate =  57837.7819041  kg/h.


### Example 5.38 Page 279¶

In :
import math
# solution

# Variables
F = 100.                        # kmol/h    feed rate of ethylene
Econ = .99*F

# Calculation and Result
Econ1 = Econ*.998
Econ2 = Econ-Econ1
Cl2con = Econ1+2*Econ2
Cl2in = F*1.1
Cl2s3 = Cl2in-Cl2con
HCls3 = Econ2
TCEp = Econ2
EDCp = Econ1
nC2H4 = 1
T = 328.15
pv1 = math.exp(4.58518-1521.789/(T-24.67))               # bar
pv2 = math.exp(4.06974-1310.297/(T-64.41))               # bar
xEDC = Econ1/(Econ1+Econ2)
xTEC = 1-xEDC
pEDC = 37.2*xEDC
pTEC = 12.64*xTEC
pCl2HClC2H4 = 1.6*100-pEDC-pTEC
yEDC = pEDC/160
yTEC = pTEC/160
nt = (Cl2s3+Econ2+1)*160/pCl2HClC2H4
nEDC = yEDC*nt
nTEC = yTEC*nt
print " Compositions of gas streams :   "
print "Component        Stream 3    Stream 5    Stream 4      Stream 6 "
print " Cl2           ",Cl2s3,"     ",Cl2s3
print "  HCl          ",HCls3,"     ",HCls3
print " C2H4          ",nC2H4,"     ",nC2H4
print " EDC           ",nEDC,"       0.2355      3.3947         98.5665"
print " TEC           ",nTEC,"        Nil       ",nTEC,"      ",TCEp
fi1 = (10.802*33.9+.198*29.1+1*43.6+3.6302*17.4+.0025*85.3)*(328.15-273.15)
fi2 = 35.053*1000*3.3947+39.58*1000*.0025
fi3 = (3.3947*129.4+.0025*144.4)*55/2
fi = fi1+ fi2+ fi3                                      # kJ/h
print " Heavy duty of Overhead condenser = ",fi," kJ/h.   "
fi5 = (100*43.6+110*33.9)*(328.15-273.15)
fi6 = 3.6302*1000*33.6+.0025*1000*38.166
fi7 = (98.5665*129.4+.1988*144.4)*(328.15-273.15)
fi8 = 216845.5*98.802+392394.5*.198
ficol = fi5+fi8-fi1-fi6-fi7
print " Heavy duty of external cooler = ",ficol," kJ/h."

 Compositions of gas streams :
Component        Stream 3    Stream 5    Stream 4      Stream 6
Cl2            10.802       10.802
HCl           0.198       0.198
C2H4           1       1
EDC            3.6264582115        0.2355      3.3947         98.5665
TEC            0.00246937055797         Nil        0.00246937055797        0.198
Heavy duty of Overhead condenser =  157524.3947  kJ/h.
Heavy duty of external cooler =  21095870.3588  kJ/h.


### Example 5.39 Page 284¶

In :
# solution

# Variables
To = 298.15
T1 = 483.15

# Calculation
# fi1 = intgr(from To to T1){12199.5+2241.4*10**-3*T+1557.7*10**-6*T**2-671.3*10**-9*T**3}dT
fi1 = 2455874.6                 # kJ/h
dHr = 2*(-45.94)                # kJ/mol N2 reacted
fi2 = 91.88*1000*23.168
fi3 = fi1+fi2
# fi3 = intgr(from To to T2){10713.9+3841*10**-3*T+1278.8*10**-6*T**2-752.6*10**-9*T**3}dT
# solving it
T2 = 657.41                     # K

# Result
print "Temperature of the gas mixture leaving the reactor = ",T2," K."

Temperature of the gas mixture leaving the reactor =  657.41  K.


### Example 5.40 Page 292¶

In :
# solution

# Variables
# basis 4 kmol of HCl gas
O2req = 1.                  # kmol
O2spply = 1.35*1

# Calculation
N2 = 1.35*79/21.
air = O2spply+N2
HClbrnt = .8*4
HCl = 4-HClbrnt
O2 = O2spply-.8
Cl2 = .8*2
H2O = .8*2

# Result
print " a)   Composition of dry product gas stream : "
print " Component       Dry product gas stream,kmol"
print " HCl                 ",HCl
print "  O2                 ",O2
print " Cl2                 ",Cl2
print " H2O                 ",H2O
print "  N2                 ",N2
print "b)   "
H2 = 114.4*1000*.8
# H2 = intgr(from 298.15 to T){286.554+12.596*10**-3*T+63.246*10**-6*T**2-25.933*10**-9*T**3}dT
# solving it
T = 599.5                   # K
print " Adiabatic reaction temperature of product gas stream = ",T," K."

 a)   Composition of dry product gas stream :
Component       Dry product gas stream,kmol
HCl                  0.8
O2                  0.55
Cl2                  1.6
H2O                  1.6
N2                  5.07857142857
b)
Adiabatic reaction temperature of product gas stream =  599.5  K.


### Example 5.41 Page 294¶

In :
# solution

# Variables
# 1 kmol of EB vapors entering the reactor at 811.15 K
# (from 811.15 to T1)intgr{-36.72+671.12*10**-3*T-422.02*10**-6*T**2+101.15*10**-9*T**3}dT = (from T1 to 978.15)intgr{487.38+1.19*10**-3*T+198.16*10**-6*T**2-68.21*10**-9*T**3}dT
# we get
T1 = 929.72                     # K
To = 298.15
H1 = 493405.                    # kJ
EBr = .35

# Calculation
Styrenep = EBr*.9
Benzeneb = EBr*.03
Ethyleneb = Benzeneb
Cb = EBr*.01
Toulened = EBr*.06
Hr1 = 147.36-29.92              # kJ/mol EB
Hr2 = 82.93+52.5-29.92
Hr3 = -29.92
Hr4 = 50.17-74.52-147.36        # kJ/mol styrene
dHr = 1000*(Hr1*(Styrenep+Toulened)+Hr2*Benzeneb+Hr3*Cb+Hr4*Toulened)
H2 = H1-dHr

# H2 = (from To t0 T2)intgr{Comp2dT
# we get

T2 = 798.79                     # K

# Result
print " Adiabatic reaction T at the outlet of the reactor is ",T2," K."

 Adiabatic reaction T at the outlet of the reactor is  798.79  K.


### Example 5.42 Page 297¶

In :
# solution

# Variables
Hsol = 62.86                    # kJ/mol solute
Mcrystal = 286.1414

# Calculation
Hcry = Hsol*1000/Mcrystal       # kJ/kg solute

# Result
print " Heat of crystallization of 1 kg crystal is ",Hcry," kJ."

 Heat of crystallization of 1 kg crystal is  219.681597979  kJ.


### Example 5.43 Page 297¶

In :
# solution

# Variables
Hf = -285.82                        # kJ/mol    of H2O

# Calculation
Hcryst = -4327.26-(-1387.08+10*Hf)

# Result
print " Heat of crystallization = ",Hcryst," kJ/mol."

 Heat of crystallization =  -81.98  kJ/mol.


### Example 5.44 Page 297¶

In :
# solution

# Variables
Hfs = -1094.33
Hfao = -1072.32

# Calculation
Hsol = Hfao-Hfs

# Result
print " Heat of solution of Boric acid = ",Hsol," kJ/mol."

 Heat of solution of Boric acid =  22.01  kJ/mol.


### Example 5.45 Page 297¶

In :
# solution

# Variables
# (a)
Hf = -982.8
Hfcryst = -1053.904

# Calculation
Hdis = Hfcryst-Hf
# (b)
Hfcr = -3077.75
Hsol = Hfcryst+7*(-285.83)-(-3077.75)

# Result
print " a   Hdissolulition = ",Hdis," kJ/mol ZnSO4.    b   Hsolution = ",Hsol," kJ/kmol."

 a   Hdissolulition =  -71.104  kJ/mol ZnSO4.    b   Hsolution =  23.036  kJ/kmol.


### Example 5.46 Page 300¶

In :
# solution

# Variables
# using chart 5.16 we get
T = 329.5                       # K

# Result
print " T = ",T," K."

 T =  329.5  K.


### Example 5.47 Page 300¶

In :
# solution

# Variables
# basis 100(m1) kg 46% sol
NaOH = 46.                  # kg
H2O = 54.                   # kg

# Calculation
m2 = NaOH/.25
NaOHo = 25.                 # kg
H2Oo = 75.                  # kg
Hf1 = -453.138              # kJ/mol
Hf2 = -467.678              # kJ/mol
Hs = Hf2-Hf1
Hg = -Hs*1000*1.501

# using Appendix IV.1
Hw1 = 146.65
Hw2 = 104.9
Hadd = 84*(Hw1-Hw2)
H = Hg+Hadd
C1 = 3.55
T2 = 298.15+H/(184*C1)      # K

# Result
print " Final sol T = ",T2," K."

 Final sol T =  336.930679731  K.


### Example 5.48 Page 301¶

In :
# solution

# Variables
# basis 100 kg of sol with 32% N
MNH4NO3 = 80.0434
MNH2CONO2 = 60.0553
MN2 = 28.0134

# Calculation
na = 32/(60.9516)
Ureadis = 1.1758*na*MNH2CONO2                   # kg
water = 100-(na*MNH4NO3+Ureadis)
ndis = 525.
m = ndis/water
HE1 = 40.3044-2.5962*m+.1582*m**2-3.4782*10**-3*m**3
HE = HE1*ndis

# Result
print "Heat effect of the sol = ",HE," kJ."

Heat effect of the sol =  10388.7187316  kJ.


### Example 5.49 Page 302¶

In :
# solution

# Variables
Hmix = 896.
M1 = 88.                          # molar mass of n-amyl alcohol
M2 = 78.                          # molar mass of benzene

# Calculation
B = .473*M2
A = .527*M1
Ha = Hmix/A
Hb = Hmix/B

# Result
print " Integral heat of sol of n-amyl alcohol = ",Ha,\
" kJ/kg n-amyl alcohol and of benzene = ",Hb," kJ/kg benzene."

 Integral heat of sol of n-amyl alcohol =  19.3203381059  kJ/kg n-amyl alcohol and of benzene =  24.2857917277  kJ/kg benzene.


### Example 5.50 Page 302¶

In :
from numpy import poly1d, roots
# solution

# Variables
# from fig 5.18
Ta = 379.5                              # K
dH = -274-(-106.5)                      # kJ/kg sol
Cm = 2.05                               # kJ/kg K
dHc = Cm*(Ta-298.15)

# Calculation
# basis 100 kg of 93 % acid
# acid balance
x = poly1d(0,'x')
p = .93*100+x*.15-(100+x)*.77
y = roots(p)

#from fig
y1 = 25.3

# Result
print "a   Resultant T of 77 percent sol = ",Ta, \
" K.    \nb   Heat to be removed to cool it to 298.15 K = ",dH,\
" kJ/kg sol    \nc   By mean heat capacity method : ",dHc,\
" kJ/kg sol    \nd   Quantity of 15 percent acid to be mixed = ",y,\
" kg.    \ne   from fig : ",y1," kg."

a   Resultant T of 77 percent sol =  379.5  K.
b   Heat to be removed to cool it to 298.15 K =  -167.5  kJ/kg sol
c   By mean heat capacity method :  166.7675  kJ/kg sol
d   Quantity of 15 percent acid to be mixed =  25.8064516129  kg.
e   from fig :  25.3  kg.


### Example 5.51 Page 304¶

In :
# solution

# Variables
# basis 100 kg of 93% acid and 25.8 kg of 15% acid
Hfp = -814.
Hf1 = -830.
HE1 = Hf1-Hfp
Hf2 = -886.2
HE2 = Hf2-Hfp
Hf3 = -851.
HE3 = Hf3-Hfp

# Calculation
Hsol = .9876*1000*(-37)-(.9482*1000*(-16)+.0394*1000*(-72.2))
Hev = 100*(30-25)*1.6
Hcon = 25.8*25*3.7
netHev = -Hsol-Hcon+Hev
T = 298.15+netHev/(125.8*2.05)

# Result
print " Temp of sol = ",T," K."

 Temp of sol =  363.832345186  K.


### Example 5.52 Page 306¶

In :
# solution

# Variables
# basis 1000 kg of mixed acid
C11 = 2.45

# Calculation
H1 = -296.7+C11*(308.15-273.15)
C12 = 2.2
H2 = -87.8+C12*(308.15-273.15)
C13 = 1.45
H3 = -35.5+C13*(308.15-273.15)
C14 = 1.8
H4 = -148.9+C14*(308.15-273.15)
Hmix = 1000*H4-(76.3*H1+345.9*H2+577.7*H3)

# Result
print " Heat of mixing = ",Hmix," kJ."

 Heat of mixing =  -74878.72  kJ.


### Example 5.53 Page 308¶

In :
# solution

# Variables
F = 1135.
Benzenef = 400*.993

# Calculation
HNO3con = Benzenef*63/78.
H1 = -186.5
C11 = 1.88
H11 = H1+C11*(298.15-273.15)
H2 = -288.9
C12 = 1.96
H22 = H2+C12*(298.15-273.15)
H3 = 0
C13 = 1.98
H33 = C13*(298.15-273.15)
Hr = -285.83+12.5-(-174.1+49.08)
Benzener = Benzenef/78.1118
fi = 903.84*H22+HNO3con*H33-F*H11+Benzener*Hr*1000              # kJ/h

# Result
print " Total heat exchanged = ",fi," kJ/h."

 Total heat exchanged =  -796777.546799  kJ/h.


### Example 5.54 Page 311¶

In :
# solution

# Variables
# from ref 24
H = 1600.83
To = 273.15
h = 200.
Hf1 = -79.3                 # table 5.59
Hf2 = -46.11
Hsol = Hf1-Hf2

# Calculation
Hg = Hsol*1000*140./17.0305
Raq = 140/.15               # kg/h
dT = Hg/(4.145*Raq)
T = -dT+303.

# Result
print " Temp of resultant sol = ",T," K."

 Temp of resultant sol =  373.525565624  K.


### Example 5.55 Page 311¶

In :
# solution

# Variables
Hf1 = -80.14
Hf2 = -46.11

# Calculation
Hsol = Hf1-Hf2
Hg = Hsol*1000.*2/17.0305

# Result
print " Heat generated for making 2 percent solution = ",Hg," kJ/100 kg sol."

 Heat generated for making 2 percent solution =  -3996.35947271  kJ/100 kg sol.


### Example 5.56 Page 312¶

In :
# solution

# Variables
fi3 = 15505407.
fi4 = 11395056.
fi5 = fi3-fi4  # kJ/h

# Calculation
fi6 = 111.375*62.75*1000
fi7 = 1063379.
fi8 = 5532.15*4.1868*(303.15-298.15)
fi9 = 9030.4*3.45*(323.15-298.15)
fi = fi5+fi6+fi8-fi7-fi9

# Result
print " Heat removal in the cooler = ",fi," kJ/h."

 Heat removal in the cooler =  9372691.2781  kJ/h.


### Example 5.57 Page 314¶

In :
%matplotlib inline
# solution
from numpy import linspace
from matplotlib.pyplot import plot, show

# Variables
To = 273.15
T1 = 308.15
H1 = 124.8*(T1-To)                  # kJ/kmol
H2 = 134.9*(T1-To)                  # kJ/kmol

# Calculation
HE1 = .1*.9*(542.4+55.4*(.9-.1)-132.8*(.9-.1)**2-168.9*(.9-.1)**3)  # kJ/kmol of mix
Ha = HE1+H1*.1+H2*.9
HE2 = .2*.8*(542.4+55.4*(.8-.2)-132.8*(.8-.2)**2-168.9*(.8-.2)**3)  # kJ/kmol of mix
Hb = HE2+H1*.2+H2*.8
HE3 = .3*.7*(542.4+55.4*(.7-.3)-132.8*(.7-.3)**2-168.9*(.7-.3)**3)  # kJ/kmol of mix
Hc = HE3+H1*.3+H2*.7
HE4 = .4*.6*(542.4+55.4*(.6-.4)-132.8*(.6-.4)**2-168.9*(.6-.4)**3)  # kJ/kmol of mix
Hd = HE4+H1*.4+H2*.6
HE5 = .5*.5*(542.4+55.4*(.5-.5)-132.8*(.5-.5)**2-168.9*(.5-.5)**3)  # kJ/kmol of mix
He = HE5+H1*.5+H2*.5
HE6 = .6*.4*(542.4+55.4*(.4-.6)-132.8*(.4-.6)**2-168.9*(.4-.6)**3)  # kJ/kmol of mix
Hf = HE6+H1*.6+H2*.4
HE7 = .7*.3*(542.4+55.4*(.3-.7)-132.8*(.3-.7)**2-168.9*(.3-.7)**3)  # kJ/kmol of mix
Hg = HE7+H1*.7+H2*.3
HE8 = .8*.2*(542.4+55.4*(.2-.8)-132.8*(.2-.8)**2-168.9*(.2-.8)**3)  # kJ/kmol of mix
Hh = HE8+H1*.8+H2*.2
HE9 = .9*.1*(542.4+55.4*(.1-.9)-132.8*(.1-.9)**2-168.9*(.1-.9)**3)  # kJ/kmol of mix
Hi = HE9+H1*.9+H2*.1
HE10 = .0*1.*(542.4+55.4*(.0-1.)-132.8*(.0-1.)**2-168.9*(.0-1.)**3)  # kJ/kmol of mix
Hj = HE10+H1+H2*0
x = linspace(0,1,100)
y = linspace(4300,5000,100)
y = 4721.5-57.4*x+1137.7*x**2-3993.6*x**3+3909.2*x**4-1351.2*x**5

# Result
plot(x,y)
show()
#title("H vs x1")
#xlabel("x1")
#ylabel("H (kJ/kg sol.)")
print "                     Enthalpy, kJ/kmol mix "
print "  x1         HE             H"
print "  0          0            ",H2
print "  0.1       ",HE1,"       ",Ha
print "  0.2       ",HE2,"       ",Hb
print "  0.3       ",HE3,"       ",Hc
print "  0.4       ",HE4,"       ",Hd
print "  0.5       ",HE5,"       ",He
print "  0.6       ",HE6,"       ",Hf
print "  0.7       ",HE7,"       ",Hg
print "  0.8       ",HE8,"       ",Hh
print "  0.9       ",HE9,"       ",Hi
print "  1.0       ",HE10,"      ",Hj Enthalpy, kJ/kmol mix
x1         HE             H
0          0             4721.5
0.1        37.372608         4723.522608
0.2        78.615936         4729.415936
0.3        111.825504         4727.275504
0.4        131.236032         4711.336032
0.5        135.6         4680.35
0.6        126.566208         4635.966208
0.7        107.058336         4581.108336
0.8        79.653504         4518.353504
0.9        44.960832         4448.310832
1.0        0.0        4368.0


### Example 5.59 Page 318¶

In :
# solution
# Variables

# from graph drawn in 5.57 we can see
H1E1 = 300.
H1E2 = 63.
H2E1 = 30.
H2E2 = 214.

# Result
print " H1 at x1=0.3 is ",H1E1," kJ/kg sol  \nH2 at x1=0.3 is ",H2E1,\
" kJ/kg sol  \nH1 at x1=0.6 is ",H1E2," kJ/kg sol  \nH2 at x1=0.6 is ",H2E2," kJ/kg sol."

 H1 at x1=0.3 is  300.0  kJ/kg sol
H2 at x1=0.3 is  30.0  kJ/kg sol
H1 at x1=0.6 is  63.0  kJ/kg sol
H2 at x1=0.6 is  214.0  kJ/kg sol.


### Example 5.60 Page 320¶

In :
# solution
# Variables
# basis 100 kg 96.1% H2SO4
# from table 5.64
m1SO3 = 78.4                        # kg
m1H2O = 21.6
n1SO3 = m1SO3/80.063
n1H2O = m1H2O/18.015

# Calculation
# resultant sol has 23.2% H2SO4
m2SO3 = 19.
m2H2O = 81.
Mrsol = m1SO3*100./m2SO3
Mw = Mrsol-100.
w = Mrsol-m1SO3/18.015              # kmol
HEosol = n1SO3*(-56940.)+n1H2O*(-32657.)           # kJ
HErsol = n1SO3*(-156168.)+w-(-335.)
HE = HErsol-HEosol                  # kJ/kg original acid
C = 3.43                            # kJ/kg K
dT = -HE/(Mrsol*C)
T = 291.15+dT                       # K

# Result
print " Heat of dilution = ",HE," kJ/kg original solution   \nFinal T of resultant solution = ",T," K."

 Heat of dilution =  -57267.8712082  kJ/kg original solution
Final T of resultant solution =  331.612662617  K.


### Example 5.61 Page 321¶

In :
# solution

# Variables
# basis 100 kg of original acid
lv = 333.7                      # kJ/kg

# Calculation
H = -lv-18*4.1868
HE = (-64277-H*312.63)/100      # kJ/kg

# Result
print " Heat of dilution = ",HE," kJ/kg."

 Heat of dilution =  636.08178112  kJ/kg.