# Heat Exchangers¶

## Example 11.1 Page 680¶

In [1]:
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

 Flow through Tube has Reynolds Number as 14049.54  .Thus the flow is Turbulent

Flow through Tube has Reynolds Number as 55.97 . Thus the flow is Laminar

Tube Length to achieve a desired hot fluid temperature is (m) =  65.71


## Example 11.2 Page 683¶

In [2]:
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

 Flow of the fluids has Reynolds Number as 1.57  &  140.77  Thus the flow is Laminar for both

Exterior Dimensions of heat Exchanger L = 0.131 m

Pressure drops within the plate-type Heat exchanger with N=60 gaps
For water =  3.768  N/m^2
For oil =  98.523  N/m^2

Pressure drops tube Heat exchanger of example 11.1
For water =  6331.255 N/m^2

For oil = 18287.329  N/m^2


## Example 11.3 Page 692¶

In [3]:
#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

 As effectiveness is 0.75  with Ratio Cmin/Cmax = 0.45
, It follows from figure 11.14 that NTU = 2.1

Required gas side surface area  (m^2) =  39.66


## Example 11.4 Page 695¶

In [4]:
#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

 As Ratio Cmin/Cmax = 0.36
and Number of transfer units NTU = 2.67
, It follows from figure 11.14 that e = .82

Heat Transfer Rate  = 2.64e+05  W

Fluid Outlet Temperatures Hot Fluid (Tho) = 73.7 degC
Cold Fluid (Tco) = 98.01 degC


## Example 11.5 Page 696¶

In [5]:
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

 As the Reynolds number of tube fluid is 59566.76
. Hence the flow is turbulent. Hence using Diettus-Boetllor Equation 8.60

Outlet Temperature of cooling Water =  36.0  degC

Tube length per pass to achieve required heat transfer = 4.51  m


## Example 11.6 Page 702¶

In [6]:
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;

 As effectiveness is 0.47  with Ratio Cmin/Cmax =  0.32
, It follows from figure 11.14 that NTU = .65

Gas-side overall heat transfer coefficient.r = 95.55 W/m^2.K

Heat exchanger Volume =  0.034  m^3