import math
#Operating Conditions
Tho = 60+273 ;#[K] Hot Fluid outlet Temperature
Thi = 100+273 ; #[K] Hot Fluid intlet Temperature
Tci = 30+273 ;#[K] Cold Fluid intlet Temperature
mh = .1 ;#[kg/s] Hot Fluid flow rate
mc = .2 ;#[kg/s] Cold Fluid flow rate
Do = .045 ;#[m] Outer annulus
Di = .025 ;#[m] Inner tube
#Table A.5 Engine Oil Properties T = 353 K
cph = 2131 ;#[J/kg.K] Specific Heat
kh = .138 ; #[W/m.K] Conductivity
uh = 3.25/100. ; #[N.s/m^2] Viscosity
#Table A.6 Saturated water Liquid Properties Tc = 308 K
cpc = 4178 ;#[J/kg.K] Specific Heat
kc = 0.625 ; #[W/m.K] Conductivity
uc = 725*math.pow(10,-6) ; #[N.s/m^2] Viscosity
Pr = 4.85 ;#Prandtl Number
#calculations and results
q = mh*cph*(Thi-Tho); #Heat transferred
Tco = q/(mc*cpc)+Tci;
T1 = Thi-Tco;
T2 = Tho-Tci;
Tlm = (T1-T2)/(2.30*math.log10(T1/T2)); #logarithmic mean temp. difference
#Through Tube
Ret = 4*mc/(math.pi*Di*uc);
print '%s %.2f %s' %("\n Flow through Tube has Reynolds Number as", Ret," .Thus the flow is Turbulent");
#Equation 8.60
Nut = .023*math.pow(Ret,.8)*math.pow(Pr,.4);#Nusselt number
hi = Nut*kc/Di;
#Through Shell
Reo = 4*mh*(Do-Di)/(math.pi*uh*(Do*Do-Di*Di));
print '%s %.2f %s' %("\n Flow through Tube has Reynolds Number as",Reo,". Thus the flow is Laminar");
#Table 8.2
Nuo = 5.63;
ho = Nuo*kh/(Do-Di);
U = 1./(1./hi+1./ho); #overall heat transfer coefficient
L = q/(U*math.pi*Di*Tlm); #Length
print '%s %.2f' %("\n Tube Length to achieve a desired hot fluid temperature is (m) = ",L);
#END
import math
import numpy
from numpy import linspace
import matplotlib
from matplotlib import pyplot
#Operating Conditions
Tho = 60.+273 ;#[K] Hot Fluid outlet Temperature
Thi = 100.+273 ;#[K] Hot Fluid intlet Temperature
Tci = 30.+273 ;#[K] Cold Fluid intlet Temperature
mh = .1 ;#[kg/s] Hot Fluid flow rate
mc = .2 ;#[kg/s] Cold Fluid flow rate
Do = .045 ;#[m] Outer annulus
Di = .025 ;#[m] Inner tube
#Table A.5 Engine Oil Properties T = 353 K
cph = 2131 ;#[J/kg.K] Specific Heat
kh = .138 ;#[W/m.K] Conductivity
uh = 3.25/100. ;#[N.s/m^2] Viscosity
rhoh = 852.1 ;#[kg/m^3] Density
#Table A.6 Saturated water Liquid Properties Tc = 308 K
cpc = 4178 ;#[J/kg.K] Specific Heat
kc = 0.625 ;#[W/m.K] Conductivity
uc = 725*math.pow(10,-6) ;#[N.s/m^2] Viscosity
Pr = 4.85 ;#Prandtl Number
rhoc = 994 ;#[kg/m^3] Density
#calculations
q = mh*cph*(Thi-Tho); #Heat required
Tco = q/(mc*cpc)+Tci;
T1 = Thi-Tco;
T2 = Tho-Tci;
Tlm = (T1-T2)/(2.30*math.log10(T1/T2));
N=numpy.zeros(61)
for i in range (0,60):
N[i]=i+20;
L = numpy.zeros(61)
for i in range (0,60):
a=float(N[i]);
L[i] = q/Tlm*(1./(7.54*kc/2.)+1/(7.54*kh/2.))/(a*a-a);
pyplot.plot(N,L);
pyplot.xlabel("L (m)");
pyplot.ylabel('Number of Gaps(N)')
pyplot.show()
#Close the graph to complete the execution
N2 = 60;
L = q/((N2-1)*N2*Tlm)*(1./(7.54*kc/2.)+1/(7.54*kh/2.));
a = L/N2;
Dh = 2*a ;#Hydraulic Diameter [m]
#For water filled gaps
umc = mc/(rhoc*L*L/2.);
Rec = rhoc*umc*Dh/uc;
#For oil filled gaps
umh = mh/(rhoh*L*L/2.);
Reh = rhoh*umh*Dh/uh;
print '%s %.2f %s %.2f %s' %("\n Flow of the fluids has Reynolds Number as",Reh," & ",Rec," Thus the flow is Laminar for both");
#Equations 8.19 and 8.22a
delpc = 64/Rec*rhoc/2*umc*umc/Dh*L ;#For water
delph = 64/Reh*rhoh/2*umh*umh/Dh*L ;#For oil
#For example 11.1
L1 = 65.9;
Dh1c = .025;
Dh1h = .02;
Ret = 4*mc/(math.pi*Di*uc);
f = math.pow((.790*2.30*math.log10(Ret)-1.64),-2) ;#friction factor through tube Eqn 8.21
umc1 = 4*mc/(rhoc*math.pi*Di*Di);
delpc1 = f*rhoc/2*umc1*umc1/Dh1c*L1;
Reo = 4*mh*(Do-Di)/(math.pi*uh*(Do*Do-Di*Di)); #Reynolds number
umh1 = 4*mh/(rhoh*math.pi*(Do*Do-Di*Di));
delph1 = 64/Reo*rhoh/2*umh1*umh1/Dh1h*L1;
#results
print '%s %.3f %s' %("\n Exterior Dimensions of heat Exchanger L =",L,"m");
print '%s %.3f %s' %("\n Pressure drops within the plate-type Heat exchanger with N=60 gaps\n For water = ", delpc," N/m^2")
print '%s %.3f %s' %(" For oil = ",delph," N/m^2\n ")
print '%s %.3f %s' %("Pressure drops tube Heat exchanger of example 11.1\n For water = ",delpc1 ,"N/m^2")
print '%s %.3f %s' %("\n For oil =",delph1," N/m^2");
#END
#Operating Conditions
Tho = 100+273. ;#[K] Hot Fluid outlet Temperature
Thi = 300+273. ;#[K] Hot Fluid intlet Temperature
Tci = 35+273. ;#[K] Cold Fluid intlet Temperature
Tco = 125+273. ; #[K] Cold Fluid outlet Temperature
mc = 1 ;#[kg/s] Cold Fluid flow rate
Uh = 100 ;#[W/m^2.K] Coefficient of heat transfer
#Table A.5 Water Properties T = 353 K
cph = 1000 ;#[J/kg.K] Specific Heat
#Table A.6 Saturated water Liquid Properties Tc = 308 K
cpc = 4197 ;#[J/kg.K] Specific Heat
#calculations
Cc = mc*cpc;
#Equation 11.6b and 11.7b
Ch = Cc*(Tco-Tci)/(Thi-Tho);
# Equation 11.18
qmax = Ch*(Thi-Tci); #Max. heat
#Equation 11.7b
q = mc*cpc*(Tco-Tci); #Heat available
e = q/qmax;
ratio = Ch/Cc;
#results
print '%s %.2f %s %.2f' %("\n As effectiveness is", e," with Ratio Cmin/Cmax =", ratio);
print '%s' %(", It follows from figure 11.14 that NTU = 2.1");
NTU = 2.1; #No. of transfer units
A = 2.1*Ch/Uh;
print '%s %.2f' %("\n Required gas side surface area (m^2) = ",A);
#END
#Operating Conditions
Thi = 250+273. ;#[K] Hot Fluid intlet Temperature
Tci = 35+273. ;#[K] Cold Fluid intlet Temperature
mc = 1 ;#[kg/s] Cold Fluid flow rate
mh = 1.5 ; #[kg/s] Hot Fluid flow rate
Uh = 100 ;#[W/m^2.K] Coefficient of heat transfer
Ah = 40 ; #[m^2] Area
#Table A.5 Water Properties T = 353 K
cph = 1000. ;#[J/kg.K] Specific Heat
#Table A.6 Saturated water Liquid Properties Tc = 308 K
cpc = 4197. ;#[J/kg.K] Specific Heat
#calculations
Cc = mc*cpc;
Ch = mh*cph;
Cmin = Ch;
Cmax = Cc;
NTU = Uh*Ah/Cmin; #No.of transfer units
ratio = Cmin/Cmax;
#results
print '%s %.2f' %("\n As Ratio Cmin/Cmax =", ratio)
print '%s %.2f' %("and Number of transfer units NTU =", NTU)
print '%s' %(", It follows from figure 11.14 that e = .82");
e = 0.82;
qmax = Cmin*(Thi-Tci); #Max. heat transferred
q = e*qmax; #Actual heat transferred
#Equation 11.6b
Tco = q/(mc*cpc) + Tci;
#Equation 11.7b
Tho = -q/(mh*cph) + Thi;
print '%s %.2e %s' %("\n Heat Transfer Rate =",q," W ")
print '%s %.1f %s' %("\n Fluid Outlet Temperatures Hot Fluid (Tho) =" ,Tho-273,"degC")
print '%s %.2f %s' %("Cold Fluid (Tco) =", Tco-273,"degC");
#END
import math
#Operating Conditions
q = 2*math.pow(10,9) ;#[W] Heat transfer Rate
ho = 11000. ;#[W/m^2.K] Coefficient of heat transfer for outer surface
Thi = 50+273. ;#[K] Hot Fluid Condensing Temperature
Tho = Thi ;#[K] Hot Fluid Condensing Temperature
Tci = 20+273. ;#[K] Cold Fluid intlet Temperature
mc = 3*math.pow(10,4) ;#[kg/s] Cold Fluid flow rate
m = 1 ;#[kg/s] Cold Fluid flow rate per tube
D = .025 ;#[m] diameter of tube
#Table A.6 Saturated water Liquid Properties Tf = 300 K
rho = 997 ;#[kg/m^3] Density
cp = 4179 ;#[J/kg.K] Specific Heat
k = 0.613 ;#[W/m.K] Conductivity
u = 855*math.pow(10,-6) ;#[N.s/m^2] Viscosity
Pr = 5.83 ;# Prandtl number
#calculations and results
#Equation 11.6b
Tco = q/(mc*cp) + Tci;
Re = 4*m/(math.pi*D*u);
print '%s %.2f' %("\n As the Reynolds number of tube fluid is", Re)
print '%s' %(". Hence the flow is turbulent. Hence using Diettus-Boetllor Equation 8.60");
Nu = .023*math.pow(Re,.8)*math.pow(Pr,.4);
hi = Nu*k/D; #Heat transfer coefficient
U = 1/(1/ho + 1/hi); #Overall heat transfer coefficient
N = 30000. ;#No of tubes
T1 = Thi-Tco;
T2 = Tho-Tci;
Tlm = (T1-T2)/(2.30*math.log10(T1/T2));#Logarithmic mean temp. difference
L2 = q/(U*N*2*math.pi*D*Tlm);
print '%s %.1f %s' %("\n Outlet Temperature of cooling Water = ",Tco-273," degC")
print '%s %.2f %s' %("\n Tube length per pass to achieve required heat transfer =",L2," m");
#END
import math
#Operating Conditions
hc = 1500. ;#[W/m^2.K] Coefficient of heat transfer for outer surface
hi = hc;
Th = 825. ;#[K] Hot Fluid Temperature
Tci = 290. ;#[K] Cold Fluid intlet Temperature
Tco = 370. ;#[K] Cold Fluid outlet Temperature
mc = 1 ;#[kg/s] Cold Fluid flow rate
mh = 1.25 ;#[kg/s] Hot Fluid flow rate
Ah = .20 ;#[m^2] Area of tubes
Di = .0138 ;#[m] diameter of tube
Do = .0164 ;#[m] Diameter
#Table A.6 Saturated water Liquid Properties Tf = 330 K
cpw = 4184. ;#[J/kg.K] Specific Heat
#Table A.1 Aluminium Properties T = 300 K
k = 237 ;#[W/m.K] Conductivity
#Table A.4 Air Properties Tf = 700 K
cpa = 1075 ;#[J/kg.K] Specific Heat
u = 33.88*math.pow(10,-6) ;#[N.s/m^2] Viscosity
Pr = .695 ;# Prandtl number
#calculations
#Geometric Considerations
si = .449;
Dh = 6.68*math.pow(10,-3) ;#[m] hydraulic diameter
G = mh/si/Ah;
Re = G*Dh/u; #Reynolds number
#From Figure 11.16
jh = .01;
hh = jh*G*cpa/math.pow(Pr,.66667); #Heat transfer coefficient
AR = Di*2.303*math.log10(Do/Di)/(2*k*(.143)); #Area of cross section
#Figure 11.16
AcAh = Di/Do*(1-.830);
#From figure 3.19
nf = .89;
noh = 1-(1-.89)*.83;
U = 1/(1/(hc*AcAh) + AR + 1/(noh*hh)); #Overall heat transfer coefficient
Cc = mc*cpw;
q = Cc*(Tco-Tci); #Heat released
Ch = mh*cpa;
qmax = Ch*(Th-Tci); #MAx. heat transferred
e = q/qmax;
ratio = Ch/Cc;
#results
print '%s %.2f %s %.2f' %("\n As effectiveness is",e," with Ratio Cmin/Cmax = ",ratio)
print '%s' %(", It follows from figure 11.14 that NTU = .65");
NTU = .65;
A = NTU*Ch/U; #Area of cross section
#From Fig 11.16
al = 269.; #[m^-1] gas side area per unit heat wxchanger volume
V = A/al;
#Answers may vary a bit due to rounding off errors.!
print '%s %.2f %s' %("\n Gas-side overall heat transfer coefficient.r =", U , "W/m^2.K")
print '%s %.3f %s' %(" \n Heat exchanger Volume = ",V," m^3");
#END;