Chapter 11 : Energy Balance Thermophysics¶
Example 11.1 pageno : 405¶
In [1]:
# variables
m = 75. #kg
g = 9.81 #m**2/s
d = 10. #m
t = 2.5*60 #s
# Calculation
f = m*g
w = f * d
P = w / t
# Result
print "The work done = ",w,"Nm"
print "Power required = ",P,"W"
Example 11.2 pageno : 405¶
In [2]:
# variables
PE = 1.5*10**3 #J
m = 10. #kg
g = 9.81 #m/s**2
v = 50. #m/s
# Calculation
#PE = mgz
z = PE / (m*g)
KE = m* (v**2) / 2
# Result
print "Height of the body from the ground = %.1f"%z,"m"
print "Kinetic energy of the body = ",KE/1000,"kJ"
Example 11.3 pageno : 405¶
In [5]:
import math
# Variables
d = 100. /1000 #m
m = 50. #kg
P = 1.01325*10**5. #Pa
# Calculation
A = math.pi * (d**2)/4
Fatm = P * A
Fwt = m * g
Ftotal = Fatm + Fwt
P = Ftotal / A
# Result
print "(a)Pressure of the gas = %.4f"%(P/10**5),"bar"
z = 500/1000. #m
w = Ftotal * z
print "(b)Work done by the gas = %.2f"%w,"J"
Example 11.4 pageno : 406¶
In [6]:
# variables
Sgr = 0.879
F = 5. #m**3/h
# Calculation
D = Sgr * 1000.
m = F * D/3600. #kg/s
P = 3500. #kPa
W = P * m * 1000/ D
# Result
print "Power requirement for the pump = %d"%W,"W"
Example 11.5 pageno : 408¶
In [7]:
import math
# Variables
d = 3. #m
m = 12500. #kg
P = 7000. #kPa
# Calculation
U = 5.3*10**6 #kJ
Vtank = 4*math.pi*((d/2)**3) / 3
Vliq = Vtank / 2
H = U + P * Vliq
# Result
print "Specific enthalpy of the fluid in the tank = %.2f"%(H/m),"kJ/kg"
Example 11.6 pageno : 409¶
In [8]:
# variables
P = 101.3 #kPa
SVl = 1.04 * 10**-3 #m**3/kmol
SVg = 1.675 #m**3/kmol
Q = 1030. #kJ
# Calculation
W = P * 10**3 * (SVg - SVl)/1000
U = Q - W
H = U + P * 10**3 * (SVg - SVl)/1000
# Result
print "Change in internal energy = %.2f"%U,"kJ/kmol"
print "Change in enthalpy = ",H,"kJ/kmol"
Example 11.7 pageno : 409¶
In [9]:
# variables
#work is done on the system, hence, W is negative
W = - 2 * 745.7 #J/s
#heat is transferres to the surrounding, hence, heat transferred is negative,
Q = -3000. #kJ/h
# Calculation
U = Q*1000/3600 - W
# Result
print "Change in internal energy = %.2f"%U,"J/s"
Example 11.8 pageno : 410¶
In [1]:
# variables
#Fe(s) + 2HCl(aq) = FeCl2(aq) + H2(g)
MFe = 55.847
m = 1. #kg
# Calculation
Nfe = m * 10**3/MFe
Nh2 = Nfe #(since 1 mole of Fe produces 1 mole of H2)
T = 300. #K
R = 8.314
#the change in volume is equal to the volume occupied by hydrogen produced
PV = Nh2 * R * T
W = PV
# Result
print "Work done = %.2f"%W,"J"
Example 11.9 pageno : 412¶
In [11]:
#Cp =1.436 + 2.18*10**-3*T
from scipy.integrate import quad
# variables
m = 1000./3600 #kg/s
T1 = 380. #K
T2 = 550. #K
# Calculation
def f0(T):
return 1.436 + 2.18*10**-3*T
x = quad(f0,T1,T2)[0]
Q = m*x
# Result
print "Heat load on the heater = %.1f"%Q,"kW"
Example 11.10 page no : 413¶
In [2]:
#Cp = 26.54 + 42.454*10**-3 * T - 14.298 * 10**-6 * T**2
from scipy.integrate import quad
# variables
T1 = 300. #K
T2 = 1000. #K
m = 1. #kg
N = m/44 #kmol
# Calculation
def f3(T):
return 26.54 + 42.454*10**-3 * T - 14.298 * 10**-6 * T**2
x = quad(f3,T1,T2)[0]
Q = N*x
# Result
print "(a)Heat required = %.2f"%Q,"kJ"
#for temperature in t degree celsius
#Cp = 26.54 + 42.454*10**-3 * (t + 273.15) - 14.298 * 10**-6 * (t + 273.15)**2
#Cp = 37.068 + 34.643 * 10**-3*t - 14.298* 10**-6 * t**2 (kJ/kmolC)
#Cp = 8.854 + 8.274*10**-3*t -3.415*10**-6*t**2 ( Kcal/kmolC)
#For degree Fehreneit scale,replacet by ( t1 - 32)/18, we get
#Cp = 8.7058 + 4.6642 * 10**-3 *t1 - 1.0540 * 10**-6 * t1**2 ( Btu/lbmolF)
Example 11.11 pageno : 414¶
In [3]:
%matplotlib inline
from matplotlib.pyplot import *
# variables
T = [273, 373, 473, 573, 673, 773, 873, 973, 1073, 1173, 1273]
Cp = [33.6, 35.1, 36 ,36.6, 37, 37.3, 37.5, 37.6, 37.7, 37.8, 37.9]
# Calculation
plot(T,Cp)
suptitle("Determination of enthalpy change ")
xlabel("Temperature, K ")
ylabel("Heat capacity, kJ/kmol K")
H = 36828. #kJ/kmol
# Result
print "Enthalpy change = ",H,"kJ/kmol"
Example 11.12 page no : 415¶
In [5]:
#Cp = 26.586 + 7.582 * 10 **-3 * T - 1.12 * 10**-6 * T**2
from scipy.integrate import quad
# variables
T1 = 500. #K
T2 = 1000. #K
# Calculation
def f5(T):
return 26.586 + 7.582 * 10**-3 * T - 1.12 * 10**-6 * T**2
x = quad(f5,T1,T2)[0]
Cpm = 1 *x / ( T2 - T1 )
# Result
print "(a)Mean molal heat capacity = %.3f"%Cpm,"kJ/kmolK"
V = 500. #m**3
N = V / 22.4143
Q = N * Cpm * ( T2 - T1 )
print "(b)Heat to be supplied = %.3e"%Q,"kJ/h"
T3 = 1500. #K
Q1 = Cpm * (T3 - T1)
def f6(T):
return 26.586 + 7.582 * 10 **-3 * T - 1.12 * 10**-6 * T**2
y = quad(f6,T1,T3)[0]
Q2 = y
Perror = (Q2 - Q1) * 100 / Q2
print "(c)Percent error = %.1f"%Perror,"%"
# note : answer may vary because of rounding error.
Example 11.13 page no : 416¶
In [16]:
# variables
T1 = 1500. #K
Tr = 273. #K
T2 = 400. #K
Cpm1 = 50. #kJ/kmol
Cpm2 = 35. #kJ/mol
# Calculation
H = Cpm1 * ( T1 - Tr ) - Cpm2 * ( T2 - Tr )
# Result
print "Enthalpy change = ",H,"kJ/kmol"
Example 11.14 pageno : 417¶
In [4]:
from scipy.integrate import quad
# variables
#CO, 26.586 + 7.582*10**-3*T - 1.12*10**-6*T**2
#CO2, 26.540 + 42.454*10**-3*T - 14.298*10**-6*T**2
#O2, 25.74 + 12.987*10**-3*T - 3.864*10**-6*T**2
#N2, 27.03 + 5.815*10**-3*T - 0.289*10**-6*T**2
#Cpmix = summation ( yi*Cpi ) = summation(yi*ai + yi*bi*T + yi*ci*T**2)
xco2 = 0.09
xco = 0.02
xo2 = 0.07
xn2 = 0.82
T1 = 600. #K
T2 = 375. #K
# Calculation
sumai = xco * 26.586 +xco2 * 26.540 + xo2 * 25.74 + xn2*27.03
sumbi = xco * 7.582*10**-3 + xco2*42.454*10**-3+xo2*12.987*10**-3 + xn2*5.815*10**-3
sumci = -(xco * 1.12*10**-6 + xco2*14.298*10**-6+xo2*3.864*10**-6+xn2*0.289*10**-6)
def f8(T):
return sumai+sumbi*T+sumci*T**2
H = quad(f8,T2,T1)[0]
# Result
print "Enthalpy change = %.2f"%H,"kJ/kmol"
# note : Calculating integration using quad may vary answer. Because Python use's its own methodology to calculate quad.
Example 11.15 pageno : 420¶
In [18]:
# variables
Hna = 26.04 #J/g-atomK
Hs = 22.6 #J/g-atomK
Ho = 16.8 #J/g-atomK
Hh = 9.6 #J/g-atomK
# Calculation
Hna2so410h2o = 2*Hna + Hs + 14*Ho + 20*Hh
Hexp = 592.2 #J/molK
Deviation = (Hexp - Hna2so410h2o)*100/Hexp
# Result
print "Deviation in heat capacity = %.2f"%Deviation,"%"
Example 11.16 pageno : 421¶
In [19]:
# variables
P1 = 75. #kPa
T1 = 573. #K
Tvap = 365. #K
Tbasis = 273. #K
#Since, the boiling point of water at 75kPa is 375K, the vapour at 573K is superheated
H1 = 3075. #kJ/kg
Cliq = 4.2 #kJ/kgK
Cvap = 1.97 # kJ/kg/K
m = 1. #kg
# Calculation
#let assume converting liq. water into superheated stream occurs in 3 steps,
Hc1 = m*Cliq * ( Tvap - Tbasis )
Hc3 = m*Cvap*(T1 - Tvap)
#total enthalpy = 3075 = Hc1 + Hc2 + Hc3, therefore
Hc2 = H1 - Hc1 - Hc3
# Result
print "Heat of vapourisation = ",Hc2,"kJ/kg"
Example 11.17 pageno : 422¶
In [21]:
from scipy.integrate import quad
# variables
T1 = 250. #K
T = 273.15 #K
T2 = 400. #K
Cice = 2.037 #kJ/kgK
T3 = 373.15 #K
Cliq = 75.726 #kJ/kmolK
#Cp = 30.475 + 9.652*10**-3*T + 1.189*10**-6*T**2
Hfusion = 6012. #kJ/kmol
Hvap = 40608. #kJ/kmol
# Calculation
#1 - Heat for raising the temperature of ice, H1
H1 = Cice * (T - T1)
#2 - Latent heat of fusion of ice, Hf
Hf = Hfusion / 18.016 #kJ
#3 - Sensible heat of raising the temperature of water, H2
H2 = Cliq * ( T3 - T)/18.016
#4 - Latent heat of vaporization of water, Hv
Hv = Hvap / 18.016
#5 - Sensible heat of raising the temperature of water vapou, H3
def f4(T):
return 30.475 + 9.652*10.**-3*T + 1.189*10.**-6*T**2.
H3 = 51.243883 #quadrature(f4,T3,T2)[0]
Q = H1 + H2 + H3 + Hf + Hv
# Result
print "Heat required = %.1f"%Q,"kJ"
Example 11.18 pageno : 423¶
In [22]:
# variables
#Cp = 0.16 + 4.78 * (10**-3) * T ( organic liquid )
#Cp = 0.7935 + 1.298 * (10**-4) * T ( CCL4 )
Tb = 349.9 #K
Hv = 195. #kJ/kg
Cp = 0.4693 #kJ/kgK
# Calculation
#Let T be the final temperature
#integration(T - 650)(0.16 + 4.78 * (10**-3) * T)dt = integration(295 - T)(0.7935 + 1.298 * (10**-4) * T)dt
# the above equation yields, 2.4549*(10**-3)*T**2 + 0.9535*T - 1353.51 = 0, from this we get
T = 573.3 #K
#since this temperature is above boiling point of CCl4,
#heat balance is, integration(T - 650)(0.16 + 4.78 * (10**-3) * T)dt = integration(295 - 349.9)(0.7935 + 1.298 * (10**-4) * T)dt + Hv + integration(349.9 - T)*0.4693*dT
#solving above equation, we get,
T1 = 540.1 #K
# Result
print "equilibrium temperature of the mixture = ",T1,"K"
Example 11.19 pageno : 427¶
In [23]:
import math
# Variables
T1 = 363. #K
T2 = 373. #K
P1s = 70.11 #kPa
P2s = 101.3 #kPa
R = 8.314 #kJ/kmolK
# Calculation
# ln(P2s / P1s) = Hv / R * (1/T1 - 1/T2)
Hv = (math.log(P2s/P1s)*R)/(1/T1 - 1/T2)
Hv1 = Hv / (18)
# Result
print "Mean heat of vaporization = %.f"%Hv1,"kJ/kg"
Example 11.20 pageno : 427¶
In [24]:
# variables
T = 273.15 - 30
R = 8.314 #K
# Calculation
#lnPs = 14.2410 - 2137.72 / (T-26.72)
#dlnPs/dT = Hv / RT2
Hv = 2137.72 * R * T**2 / ( T - 26.72 )**2
# Result
print "Heat of vaporization = %.2f"%Hv,"kJ/kmol"
Example 11.21 pageno : 427¶
In [25]:
# variables
Hv1 = 2256. #kJ/kg
T1 = 373. #K
T2 = 473. #K
Tc = 647. #K
# Calculation
Tr1 = T1 / Tc
Tr2 = T2 / Tc
#Hv2 / Hv1 = ((1-Tr2)/(1-Tr1))**0.38
Hv2 = Hv1*(((1-Tr2)/(1-Tr1))**0.38)
# Result
print "Latent heat of vaporization of water at 473K = %d"%Hv2,"kJ/kg"
Example 11.22 pageno : 428¶
In [8]:
import math
from scipy.integrate import quad
# Variables
#Cp = a + b*T
T1 = 293.15 #K
Cp1 = 131.05 #J/molK
T2 = 323. #K
Cp2 = 138.04 #J/molK
# Calculation
#a + 293*b = 131.05
#a + 323*b = 138.04
b = (Cp1 - Cp2)/(T1 - T2)
a = Cp1 - b * T1
#Cp = 62.781 + 0.233*T
# Hvb / Tb = 36.63 + 8.31lnTb
Tb = 273.15 + 80.1 #K
Hvb = (36.63 + 8.31*math.log(Tb)) * Tb
m = 100. #kg
def f7(T):
return 62.781 + 0.233*T
k = quad(f7,T1,Tb)[0]
H = m*(10.**3) * ( k)/78.048 + m*(10.**3) * Hvb/78.048
# Result
print "Heat required = %.3e J" %H
Example 11.23 pageno : 430¶
In [27]:
# variables
P = 10. #kPa
T1 = 323.15 #K
T2 = 373.15 #K
T = 358.15 #K
H1 = 2592.6 #kJ/kg
H2 = 2687.5 #kJ/kg
# Calculation
#H by interpolation,
H = H1 + ((H2 - H1)/(T2 - T1))*(T - T1)
Hl = 697.061 #kJ/kg
Hg = 2762. #kJ/kg
#H = x*Hl + ( 1 - x )* Hg
x = (H - Hg)/(Hl - Hg)
Pmois = x*100
Psteam = ( 1 - x )*100
# Result
print "Percentage of moisture = %.f"%Pmois,"%"
print "Percentage of dry saturated steam = %.f"%Psteam,"%"
Example 11.24 pageno : 430¶
In [25]:
# variables
P = 3500. #kPa
T = 673.15 #K
SV = 0.08453 #m**3/kg
Vcondensed = 1/2.
m = 100. #kg
# Calculation
V = m * SV / (m/2)
#m*(Vl+Vg)*Vcondensed = m * SV
#But Vl is negligible,
Vg = m * SV / (m * Vcondensed)
#using steam table
T1 = 459.5 #K
P1 = 1158. #kPa
#internal energy of superheated steam from steam table
I = 2928.4 #kJ/kg
U1 = m * I
Ul = 790. #kJ/kg
Ug = 2585.9 #kJ/kg
U2 = m*Vcondensed*Ul + m*(1-Vcondensed)*Ug
Q = U2 - U1
# Result
print "The amount of heat removed fromt he system = ",Q,"kJ"
Example 11.25 pageno : 433¶
In [28]:
# variables
m = 1000. #kg/h ( basis mass of 10% NaOH solution )
Pfeed = 10. #%
Ppro = 50. #(Percentage NaOH in product)
# Calculation
#Taking NaOH balance,P being the weight of the product
P = Pfeed * m / Ppro
#W be the weight of water evaporized
W = m - P
#step1 - cooling 1000kg/h of 10% solution from 305K to 298K
T1 = 305. #K
T2 = 298. #K
Cliq = 3.67 #kJ/kgK
H1 = m*Cliq * (T2 - T1)
#step2 - separation into pure components
Hsolution = -42.85 #kJ/mol
H2 = -Pfeed * m *1000 *Hsolution/ (40*100)
#step3 - W kg water is converted to water vapour
Hvap = 2442.5 #kJ/kg
H3 = W * Hvap
#step4 - water vapour at 298K is heated to 373.15K
Cvap = 1.884 #kJ/kgK
T3 = 373.15 #K
H4 = W * Cvap * ( T3 - T2 )
#step5 - formation of 200kg of 50% NaOH solution at 298K
Hsolu = -25.89 #kJ/mol
H5 = Pfeed * m *1000 *Hsolu/ (40*100)
#step6 - Heating the solution from 298K to 380K
Csolu = 3.34 #kJ/kg
T4 = 380. #K
H6 = P * Csolu * (T4 - T2)
Htotal = H1 + H2 + H3 + H4 + H5 + H6
# Result
print "The enthalpy change accompanying the complete process = %d"%Htotal,"kJ"
Example 11.26 page no : 435¶
In [30]:
# variables
Nwater = 0.8 #moles
Nethanol = 0.2 #moles
T = 323. #K
Cwater = 4.18*10**3 #J/kgK
Cethanol = 2.58*10**3 #J/kgK
Hmixing1 = -758. #J/mol ( at 298K )
Hmixing2 = -415. #J/mol ( at 323K )
T1 = 298. #K
T2 = 523. #K
# Calculation
#step1 - 0.8 mol of water is cooled from 323 K to 298K
H1 = Nwater * 18 * Cwater * ( T1 - T )/ 1000
#step2 - 0.2 mol ethanol cooled from 323K to 298K
H2 = Nethanol * 46 * Cethanol * ( T1 - T )/1000
#step3 - 0.8 mol water and 0.2 mol ethanol are mixed together,
H3 = Hmixing1
#step4 solution is heated to 323K, H4 = Cpm * (T - T1)
#Hmixing2 = H1 + H2 + H3 + H4
H4 = Hmixing2 - H1 - H2 - H3
Cpm = H4 / ( T - T1 )
# Result
print "The mean heat capacity of a 20 percent solution = %.2f"%Cpm,"J/molK"
Example 11.27 page no : 437¶
In [31]:
# variables
F = 1000. #kg/h
H1 = 116.3 #kJ/kg ( enthalpy of feed solution - 10% NaOH, 305 K )
H2 = 560.57 #kJ/kg ( enthalpy of thick liquor - 50% NaOH, 380 K )
Hsteam = 2676. #kJ/kg ( 1atm , 373.15K )
#by doing material balances,
P = 200. #kg/h
mvap = 800. #kg/h
# Calculation
#Enthalpy balance gives, F*H1 + Q = mvap*Hsteam + P*H2
Q = (mvap*Hsteam + P*H2)-F*H1
# Result
print "Heat to be supplied = ",Q,"kJ/h"
Example 11.28 pageno : 439¶
In [29]:
# variables
U2 = 0.35*10**3 #kJ
U1 = 0.25*10**3 #kJ
#since the tank is rigid the volume does not change during heating, Under constant volume, the change in the internal energy is equal to the heat supplied
# Calculation
Q = U2 - U1
# Result
print "Heat transferred to the air = ",Q,"kJ"
Example 11.29 pageno : 439¶
In [32]:
# variables
W = -2.25*745.7 #W ( work done on the system and 1hp = 745.7W)
Q = -3400. #kJ/h ( Heat transferred to the surrounding )
# Calculation
U = Q*1000/3600 - W
# Result
print "Rise in the Internal energy of the system = %.3f"%U,"J/s"
Example 11.30 pageno : 440¶
In [34]:
# variables
#2Fe + 3/2O2 = Fe2O3
Hliberated = 831.08 #kJ
# Calculation
Q = -Hliberated*1000
# Result
print "Q = ",Q,"J"
#P(V) = (n)RT
#W = P(V) = (n)RT
n = -1.5
R = 8.314
T = 298. #K
W = (n) * R * T
print "W = %.1f"%W,"J"
U = Q - W
print "U = %.5e"%U,"J"
Example 11.31 pageno : 440¶
In [35]:
# variables
Vgas = 0.09 #m**3
Vliq = 0.01 #m**3
SVliq = 1.061*10**-3 #m**3/kg
SVvap = 0.8857 #m**3/kg
# Calculation
mvap = Vgas / SVvap
mliq = Vliq / SVliq
Ul = 504.5 #kJ/kg
Ug = 2529.5 #kJ/kg
U1 = Ul * mliq + Ug * mvap
SVtotal = (Vgas + Vliq)/(mvap + mliq)
#using steam table , these value of specific volume corresponds to
# pressure of 148.6bar and internal energy of 2464.6kJ/kg
U = 2464. #kJ/kg
Utotal = U * (mvap + mliq)
#Utotal - U1 = Q - W,but W = o, hence,
Q = Utotal - U1
# Result
print "Heat to be added = %.f"%Q,"kJ"
# note: answer may vary because of rounding error
Example 11.32 pageno : 443¶
In [36]:
# variables
m = 10. #kg(air)
N = m / 29 #kmol
P1 = 100. #kPa
T1 = 300. #K
R = 8.314
# Calculation
V1 = N * R * T1 / P1
V2 = V1
T2 = 600. #K
Cv = 20.785 #kJ/kmolK
Cp = 29.099 #kJ/kmolK
U = N * Cv * (T2 - T1)
Q = U
W = Q - U
H = U + N * R * ( T2 - T1 )
# Result
print "(a)Change in internal energy at constant volume = %d"%U,"kJ"
print "heat supplied at constant volume = %d"%Q,"kJ"
print "Work done at constant volume = ",W,"kJ"
print "Change in Enthalpy at constant volume = %d"%H,"kJ"
P2 = P1
H2 = N * Cp * ( T2 - T1 )
Q2 = H2
U2 = H2 - N * R * (T2 - T1)
W2 = Q2 - U2
print "(b)Change in internal energy at constant Pressure = %d"%U2,"kJ"
print "heat supplied at constant Pressure = %d"%Q2,"kJ"
print "Work done at constant Pressure = %d"%W2,"kJ"
print "Change in Enthalpy at constant Pressure = %d"%H2,"kJ"
Example 11.33 pageno : 444¶
In [11]:
# variables
Cp = 29.3 #kJ/kmol
R = 8.314
Cv = Cp - R
T1 = 300. #K
P1 = 1. #bar
P2 = 2. #bar
# Calculation
#step1 - Volume remains constant, therefore the work done is
# zero and heat supplied is Cv, Also T2/T1 = P2/P1
T2 = P2 * T1 / P1
Q1 = Cv * ( T2 - T1 )
W1 = 0
# Result
print "Work done at constant volume = ",W1,"kJ"
print "Heat supplied at constant volume = ",Q1,"kJ"
#step2 - Process is abdiabatic
Q2 = 0
r = 1.4
T3 = T2 * (( P1 / P2 )**((r - 1)/r))
W2 = Cv * ( T2 - T3 )
print "Work done in adiabatic process = %.1f"%W2,"kJ"
print "Heat supplied in adiabatic process = ",Q2,"kJ"
#step3 - process is isobaric
Q3 = Cp * (T1 - T3)
U3 = Cv * (T1 - T3)
W3 = Q3 - U3
print "Work done at constant pressure = %.2f"%W3,"kJ"
print "Heat supplied at constant pressure = %.1f"%Q3,"kJ"
# Note : Answers in book is wrong. while calculating heat supplied i.e. Cv(T2-T1), value of Cv is been taken wrongly.
Example 11.34 pageno : 445¶
In [19]:
import math
# Variables
P1 = 5. #bar
P2 = 4. #bar
T1 = 600. #K
V = 0.1 #m**3
T2 = 400. #K
T = 298. #K
Cp = 30. #J/molK
#step1 - isothermal condition
U1 = 0
H1 = 0
P = 1. #bar
R = 8.314
# Calculation
W1 = R*T1*math.log(P1/P2)
Q1 = W1
# Result
print "(a)Change in the internal energy in isothermal condition = ",U1,"kJ/kmol"
print "Change in the enthalpy energy in isothermal condition = ",H1,"kJ/kmol"
print "Work done in isothermal condition = %.2f"%W1,"kJ/kmol"
print "Heat supplied in isothermal condition = %.2f"%Q1,"kJ/kmol"
N = round(P * (1.01325 * 10**5) * V / ( R * T ),2) # answer slightly different because of rouding error.
Cv = Cp - R
U2 = Cv * (T2 - T)*N #Answer differes due to slighlty different value of N
H2 = Cp * (T2 - T)*N #Answer differes due to slighlty different value of N
W2 = 0
Q2 = U2 + W2
print
print "\n(b)Change in the internal energy at constant volume condition = %d"%U2,"J"
print "Change in the enthalpy energy at constant volume condition = %d"%H2,"kJ/kmol"
print "Work done at constant volume condition = ",W2,"kJ/kmol"
print "Heat supplied at constant volume condition = %d"%Q2,"kJ/kmol"
# note : answer varies because of rounding error.
Example 11.35 pageno : 448¶
In [39]:
# variables
m = 1. #kg
u2 = 0.5 #m/s
u1 = 60. #m/s
H = -3000. #kJ/kg
# Calculation
#KE = (u**2)/2
KE = ((u2 ** 2) - (u1**2))/2000
g = 9.81 #m/s**2
Z1 = 7.5 #m
Z2 = 2. #m
#PE = g * (Z)
PE = g * (Z2 - Z1)/1000
W = 800. #kJ/kg
Q = H + PE + KE + W
# Result
print "Heat removed from the fluid = %.2f"%Q,"kJ/kg"
Example 11.37 pageno : 449¶
In [40]:
# variables
g = 9.81 #m/s**2
z = 55.
PE = g * z
KE = 0.
T2 = 288. #K
f = 1.5*10**-2 #m**3/min
D = 1000. #kg/m**3
m = f * D
Qsupp = 500. #kJ/min
Qlost = 400. #kJ/min
# Calculation
Qnet = (Qsupp - Qlost) * D / m
W = 2*745.7 #W
Ws = -W * 0.6 / (m/60)
H = Qnet - Ws - PE - KE
Cp = 4200.
T1 = H / Cp
T = T1 + T2
# Result
print "The temperature of exit water = %.2f"%T,"K"
Example 11.38 pageno : 450¶
In [39]:
# variables
m = 1000. #kg/h (dried product)
# S be the amount of dry solid in the product stream
Pmoisture1 = 4. #%
Pmoisture2 = 0.2 #%
P = 1.
S = m *(1 - P/1000)
# Calculation
X1 = Pmoisture1/(100 - Pmoisture1)
X2 = Pmoisture2/(100 - Pmoisture2)
#let G be the weight of dry air in the air stream
Y1 = 0.01 #kg water/kg dry solid
Cp = 1.507
Cw = 4.2
T1 = 298. #K
T = 273. #K
T2 = 333. #K
Tg1 = 363. #K
Tg2 = 305. #K
Hs1 = (Cp + X1 * Cw) * (T1 - T)
Hs2 = (Cp + X2 * Cw) * (T2 - T)
#Hg = Cs(Tg - To) + Y*L
#Cs = 1.005 + 1.884*Y
L = 2502.3 #kJ/kg dry air
Hg1 = (1.005 + 1.884 * Y1)*(Tg1 - T) + Y1 * L
Q = -40000. #kJ/h
#Calculating for T2, Hg2 = 32.16 + 2562.59*Y
#change in enthalpy = Q
#H1 = S * Hs1 + G * HG1 = 37814.22 + 117.17G
#H2 = 100728.14 + G* (32.16 + 2561.59*Y)
#change in enthalpy = Q
#62913.92 + G *(-85.01 + 2561.59*Y) + 40000 = 0
#102913.92 + G *(-85.01 + 2561.59*Y) = 0 (1)
#moisture balance, S*X1 + G*Y1 = S*X2 + G*Y2
#G*(Y-0.01) = 39.62 (2)
#solving simultaneously ( 1 ) and ( 2 ),
Gdry = 3443. #kg/h
G = Gdry*(1 + Y1)
# Result
print "Air requirement = ",G,"kg/h"
Example 11.39 pageno : 452¶
In [41]:
# variables
m = 1000. #kg/h ( feed solution )
#F - mass of feed distilled, W - mass of the bottom product, D - mass of the distillate, xf, xd and xw - weight fraction of actone in feed, distillate and residue resp_
#total balance, F = D + W
#Acetone balance, F*xf = D*xd + w*xw
F = 1000
xf = 0.10
xd = 0.9
xw = 0.01
# Calculation
#substituting in above equations,
D = F * (xf - xw) / (xd - xw)
W = F - D
R = 8
L = R * D
#material balance around the condenser,G vapour reaching the condenser
G = L + D
Td = 332. #K
T2 = 300. #K
Tw = 370. #K
Tf = 340. #K
Lacetone1 = 620. #kJ/kg
Lwater1 = 2500. #kJ/kg
Ld = xd * Lacetone1 + (1 - xd) * Lwater1
Cpacetone = 2.2 #kJ/kgK
Cpwater = 4.2 #kJ/kgK
Cp = xd * Cpacetone + (1-xd)*Cpwater
H = Ld + Cp * ( Td - T2 )
Cpc = 4.2 #kJ/kg
Tc = 30. #K ( change in temperature allowable for cooling water )
m = G * H / ( Cpc * Tc )
# Result
print "(a)The circulation rate of cooling water = %2f"%m,"kg/h"
Qc = G * H
Hd = 0.
Hw = (xw * Cpacetone + (1-xw)*Cpwater)*(Tw - T2)
Hf = (xf * Cpacetone + (1-xf)*Cpwater)*(Tf - T2)
Qb = D * Hd + W * Hw + Qc - F * Hf
Hcondensation = 2730. #kJ/kg
msteam = Qb/Hcondensation
print "(b)Amount of steam supplied = %.2f"%msteam,"kg/h"