# Chapter 9: Heat Exchanger¶

### Example 9.1 Page No.458¶

In [2]:
T1=100.0
T2=75.0
t1=5.0
t2=50.0

import math
LMTD_counter=((T1-t2)-(T2-t1))/(math.log((T1-t2)/(T2-t1)))
LMTD_parallel=((T1-t1)-(T2-t2))/(math.log((T1-t1)/(T2-t2)))

print"The LMTD for counter flow configuration is  ",round(LMTD_counter,1),"C"
print"The LMTD for parallel flow configuration is ",round(LMTD_parallel,2),"C"

The LMTD for counter flow configuration is   59.4 C
The LMTD for parallel flow configuration is  52.43 C


### Example 9.2 Page No. 459¶

In [8]:
T1=250
T2=150
t1=100
t2=150

import math
LMTD_counter=((T1-t2)-(T2-t1))/(math.log((T1-t2)/(T2-t1)));
LMTD_parrelel=0

print"The LMTD for counter flow configuration is",round(LMTD_counter,1),"C"
print"if parallel flow is to give  equal  outlet  temperatures,then the area  needed  must be infinite which is not feasible economically."
print"The LMTD for parrelel flow configuration is",LMTD_parrelel,"C"

The LMTD for counter flow configuration is 72.1 C
if parallel flow is to give  equal  outlet  temperatures,then the area  needed  must be infinite which is not feasible economically.
The LMTD for parrelel flow configuration is 0 C


### Example 9.3 Page No.463¶

In [14]:
rou_1= 0.985*62.4        # density in lbm/ft**3
cp_1=0.9994              # specific heat BTU/(lbm-degree Rankine)
v_1= 0.514e-5            # viscosity in ft**2/s
kf_1 = 0.376             # thermal conductivity in BTU/(hr.ft.degree Rankine)
a_1 = 6.02e-3            # diffusivity in ft**2/hr
Pr_1 = 3.02              #  Prandtl Number
m_1=5000                 # mass flow rate in lbm/hr
T_1=195                  # temperature in degree F
rou_2= 1.087*62.4        # density in lbm/ft**3
cp_2=0.612               # specific heat BTU/(lbm-degree Rankine)
v_2= 5.11e-5             # viscosity in ft**2/s
kf_2 = 0.150             # thermal conductivity in BTU/(hr.ft.degree Rankine)
a_2 = 3.61e-3            # diffusivity in ft**2/hr
Pr_2 = 51                # Prandtl Number
m_2=12000                # mass flow rate in lbm/hr
T_2=85                   # temperature in degree F
ID_a=0.1674
ID_p=0.1076
OD_p=1.375/12
A_p=math.pi*ID_p**2/4
A_a=math.pi*((ID_a)**2-(OD_p)**2)/4

D_h=ID_a-OD_p
D_e=(ID_a**2-OD_p**2)/(OD_p)

Re_1=(m_1/3600.0)*(ID_p)/(v_1*rou_1*A_p)
Re_2=(m_2/3600.0)*(D_e)/(v_2*rou_2*A_a)

Nu_1=0.023*(Re_1)**(4/5.0)*(Pr_1)**0.3
Nu_2=0.023*(Re_2)**(4/5.0)*(Pr_2)**0.4

h_1i=Nu_1*kf_1/ID_p
h_1o=h_1i*ID_p/OD_p
h_2=Nu_2*kf_2/D_e

Uo=1/((1/h_1o)+(1/h_2))
R=(m_2*cp_2)/(m_1*cp_1)
L=20
A=math.pi*OD_p*L
T1=195
t1=85
T2=((T1*(R-1))-(R*t1*(1-exp((Uo*A*(R-1))/(m_2*cp_2)))))/(R*exp(Uo*A*(R-1)/(m_2*cp_2))-1)
t2=t1+(T1-T2)/R
print"The outlet temperature of Ethylene glycol is %.1f degree F",round(t2,1),"F"

The outlet temperature of Ethylene glycol is %.1f degree F 99.4 F


### Example 9.4 Page No. 467¶

In [24]:
rou_1= 1.088 		# density in kg/m**3
cp_1= 1007		 # specific heat in J/(kg*K)
v_1= 18.2e-6		 # viscosity in m**2/s
Pr_1 =0.703 		# Prandtl Number
kf_1= 0.02814 		# thermal conductivity in W/(m.K)
a_1 = 0.26e-4 		# diffusivity in m**2/s
m_1=100 		                # mass flow rate in kg/hr
t1_air=20+273
t2_air=80+273
rou_2= 1.0732		 # density in kg/m**3
cp_2= 1013 		# specific heat in J/(kg*K)
v_2= 21.67e-6 	               #viscosity in m**2/s
Pr_2 =0.702 		# Prandtl Number
kf_2= 0.03352 		# thermal conductivity in W/(m.K)
a_2 = 0.3084e-4 		# diffusivity in m**2/s
m_2=90			 # mass flow rate in kg/hr
T1_CO2=600
ID_a=.098
ID_p=.07384
OD_p=.07938

import math
A_p=math.pi*ID_p**(2)/4.0
A_a=math.pi*((ID_a)**2-(OD_p)**2)/4.0

q_air=(m_1/3600.0)*(cp_1)*(t2_air-t1_air)
T2_CO2=T1_CO2-(q_air/(m_2*cp_2/3600.0))

LMTD_counter=((T1_CO2-t2_air)-(T2_CO2-t1_air))/(log((T1_CO2-t2_air)/(T2_CO2-t1_air)))
D_h=ID_a-OD_p
D_e=(ID_a**2-OD_p**2)/(OD_p)

Re_1=(m_1/3600.0)*(ID_p)/(v_1*rou_1*A_p)
Re_2=(m_2/3600.0)*(D_e)/(v_2*rou_2*A_a)

Nu_1=0.023*(Re_1)**(0.8)*(Pr_1)**0.3
Nu_2=0.023*(Re_2)**(0.8)*(Pr_2)**0.4

h_1i=Nu_1*kf_1/ID_p
h_1o=h_1i*ID_p/OD_p
h_2=Nu_2*kf_2/D_e

Rd_air=0.0004
Rd_CO2=0.002

Uo=1/((1/h_1o)+(1/h_2))
Uo=1/((1/Uo)+Rd_air+Rd_CO2)

A=q_air/(Uo*LMTD_counter)

L=(A/(math.pi*OD_p))   # length of each exchanger
L_available=2          # available exchanger length
N=L_available/L        # no. of exchangers

fp=0.0245              #friction factor for air fom figure 6.14 corresponding to Re
fa=0.033               #friction factor for cCO2fom figure 6.14 corresponding to Re
V_air=(m_1/3600.0)/(rou_1*A_p)
V_CO2=(m_2/3600.0)/(rou_2*A_a)

dP_p=(fp*L_available*rou_1*V_air**2)/(ID_p*2)
dP_a=((rou_2*V_CO2**2)/2.0)*((fa*L_available/D_h)+1)

print"(a)The number of exchangers is ",round(N,0)
print"(b)The overall exchanger coefficient is ",round(Uo,1)," W/(sq.m.K)"
print"(c)The pressure drop for tube side is  ",round(dP_p,2),"Pa"
print"The pressure drop for shell side is ",round(dP_a,1),"Pa"

(a)The number of exchangers is  1.0
(b)The overall exchanger coefficient is  14.2  W/(sq.m.K)
(c)The pressure drop for tube side is   12.83 Pa
The pressure drop for shell side is  196.7 Pa


### Example 9.5 Page No. 484¶

In [ ]:
rou_1= 0.994*62.4         # density in lbm/ft**3
cp_1=0.998             # specific heat BTU/(lbm-degree Rankine)
v_1= 0.708e-5          # viscosity in ft**2/s
kf_1 = 0.363          # thermal conductivity in BTU/(hr.ft.degree Rankine)
a_1 = 5.86e-3          # diffusivity in ft**2/hr
Pr_1 = 4.34          # Prandtl Number
m_1=170000           # mass flow rate in lbm/hr
T1=110.0               # temperature in degree F
rou_2= 62.4           # density in lbm/ft**3
cp_2=0.9988           # specific heat BTU/(lbm-degree Rankine)
v_2= 1.083e-5         # viscosity in ft**2/s
kf_2 = 0.345          # thermal conductivity in BTU/(hr.ft.degree Rankine)
a_2 = 5.54e-3         # diffusivity in ft**2/hr
Pr_2 = 7.02           # Prandtl Number
m_2=150000            # mass flow rate in lbm/hr
t1=65                 # temperature in degree F
OD=3/(4*12.0)
ID=0.652/12.0
OD_p=1.375/12.0
Nt=224.0                # from table 9.3
Np=2                  # no. of tube passes
Ds=17.25/12.0
Nb=15.0                # no. of baffles
B=1
sT=15/(16*12.0)
C=sT-OD

import math
At=(Nt*math.pi*ID**2)/(4*Np)
As=(Ds*C*B)/sT

De=4*((sT/2.0)*(0.86*sT)-(math.pi*OD**2/8.0))/(math.pi*OD/2.0)

Re_s=(m_1/3600.0)*(De)/(v_1*rou_1*As)
Re_t=(m_2/3600.0)*(ID)/(v_2*rou_2*At)

Nu_t=0.023*(Re_t)**(0.8)*(Pr_2)**0.4
Nu_s=0.36*(Re_s)**(0.55)*(Pr_1)**(1/3.0)
h_ti=Nu_t*kf_2/ID
h_to=h_ti*ID/OD
h_s=Nu_s*kf_1/De

Uo=1/((1/h_to)+(1/h_s))
R=(m_2*cp_2)/(m_1*cp_1)
L=16
Ao=Nt*math.pi*OD*L
UoAo_mccp=(Uo*Ao)/(m_2*cp_2)
S=0.58              #value of S from fig. 9.13 Ten Broeck graph corresponding to the value of (UoAo)/(McCpc)
t2=S*(T1-t1)+t1
T2=T1-R*(t2-t1)
ft=0.029             #friction factor for raw water fom figure 6.14 corresponding to Reynolds Number calculated above
fs=0.281             #friction factor for distilled water fom figure 6.14 corresponding to Reynolds Number calculated above

V_t=(m_2/3600.0)/(rou_2*At)
V_s=(m_1/3600.0)/(rou_1*As)

gc=32.2
dP_t=(rou_2*V_t**2)*((ft*L*Np/ID)+4*Np)/(2*gc)
dP_s=((rou_1*V_s**2)*(fs*Ds*(Nb+1)))/(2*gc*De)

print"Outlet Temperatures of raw water is ",round(t2,1),"F"
print"Outlet Temperatures of distilled water is ",round(T2,1),"F"
print"\nThe pressure drop for tube side is",round(dP_t/147,1),"psi"
print"The pressure drop for shell side is",round(dP_s/147,1),"psi"


### Example 9.6 Page No.492¶

In [39]:
m_1=170000 		# mass flow rate in lbm/hr
T1=110 			# temperature in degree F
cp_1=0.998		 # specific heat BTU/(lbm-degree Rankine)
m_2=150000 		# mass flow rate in lbm/hr
t1=65 			# temperature in degree F
cp_2=0.9988	                # specific heat BTU/(lbm-degree Rankine)
Uo=350 			# exchanger coefficient
Ao=703.7
mcp_raw=m_2*cp_2
mcp_distilled=m_1*cp_1

mcp_min_max=mcp_raw/mcp_distilled
UA_mcpmin=(Uo*Ao)/(mcp_raw)
effectiveness=0.58 		#value of effectiveness from figure 9.15 corresponding to the above calculated values of capacitance ratio and (UoAo/mcp_min)
qmax=mcp_raw*(T1-t1)
q=effectiveness*qmax 	# actual heat transfer
t2=(q/mcp_raw)+t1
T2=T1-(q/mcp_distilled)

print"The Outlet temperature is Raw Water is",round(t2,1),"F"
print"The Outlet temperature is disilled Water is",round(T2,1),"F"

The Outlet temperature is Raw Water is 91.1 F
The Outlet temperature is disilled Water is 87.0 F


### Example 9.7 Page No. 499¶

In [21]:
rou_1= 0.852*62.4		 # density in lbm/ft**3
cp_1=0.509		 # specific heat BTU/(lbm-degree Rankine)
v_1=0.404e-3 		# viscosity in ft**2/s
kf_1=0.08 		# thermal conductivity in BTU/(hr.ft.degree Rankine)
a_1=2.98e-3   	# diffusivity in ft**2/hr
Pr_1=490.0 		# Prandtl Number
m_1=39.8 		  # mass flow rate in lbm/min
T1=190.0
T2=158.0
rou_2= 0.0653		 # density in lbm/ft**3
cp_2=0.241		 # specific heat BTU/(lbm-degree Rankine)
v_2= 20.98e-5 		# viscosity in ft**2/s
kf_2 = 0.01677 		 # thermal conductivity in BTU/(hr.ft.degree Rankine)
a_2 = 1.066 		# diffusivity in ft**2/hr
Pr_2 = 0.706 		# Prandtl Number
m_2=67.0 			# mass flow rate in lbm/min
t1=126.0
t2=166.0
q_air=m_2*cp_2*60*(t2-t1)
q_oil=m_1*cp_1*60*(T1-T2)

import math
LMTD=((T1-t2)-(T2-t1))/(math.log((T1-t2)/(T2-t1)))
Area_air=(9.82*8)/144.0
Area_oil=(3.25*9.82)/144.0

S=(t2-t1)/(T1-t1)
R=(T1-T2)/(t2-t1)
F=0.87              #value of correction factor from figure 9.21a corresponding to above calculated values of S and R
UA=q_air/(F*LMTD)
mcp_air=m_2*cp_2*60
mcp_oil=m_1*cp_1*60

mcp_min_max=mcp_air/mcp_oil
NTU=(UA/mcp_air)
effectiveness=0.62 		#effectiveness from fig 9.21b corresponding to the values of capacitance ratio
t2_c=(T1-t1)*effectiveness+t1
T2_c=T1-(mcp_air)*(t2_c-t1)/(mcp_oil)

print"The Overall Coefficient is ",round(UA,0)," BTU/(hr. degree R)"
print"Calculated outlet temprature are:"
print"Outlet temprature for air",round(t2_c,1),"F"
print"Outlet temprature for Engine Oil",round(T2_c,0),"F"

The Overall Coefficient is  1602.0  BTU/(hr. degree R)
Calculated outlet temprature are:
Outlet temprature for air 165.7 F
Outlet temprature for Engine Oil 158.0 F

In [ ]: