# Chapter 17: HEAT TRANSFER BY CONVECTION

## Example 17.01, page: 408

In :
from __future__ import division
import math
from scipy import integrate

# Initialization  of  Variable

#calculations:
x2 = lambda x: x**-0.1
ratio = integrate.quad(x2,0,1)

#Results
print  "The ratio of the average heat transfer coefficient is", round(ratio,2)

The ratio of the average heat transfer coefficient is 1.11


## Example 17.02, page: 415

In :
from __future__ import division
import math

# Initialization  of  Variable
Tinf = 300 #degC
uinf = 10 #m/s
L = 0.5 #m
Ts = 27 #degC

#calculations:
#from Table HT-3
Tf = 437 #K
P = 1 #atm
v = 30.84E-6 #m2/s
k = 36.4E-3 #W/m.K
Pr = 0.687
#Reynolds number
ReL = uinf*L/v
#Correlation
NuL = 0.664*ReL**0.5*Pr**(1/3)
#average convection coefficient
hbar = NuL*k/L
#cooling rate per unit width of plate
q1 = hbar*L*(Tinf - Ts)

#Result
print "cooling rate per unit width of plate is", round(q1,0),"W/m"
#answer wrong in book

cooling rate per unit width of plate is 2344.0 W/m


## Example 17.03, page: 416

In :
from __future__ import division
import math

# Initialization  of  Variable
Tinf = 25 #degC
uinf = 60 #m/s
Ts = 230 #degC
d = 50 #mm
L4 = 200 #mm
L5 = 250 #mm

#calculations:
#from Table HT-3
Tf = 400 #K
P = 1 #atm
v = 26.41E-6 #m2/s
k = 0.0338 #W/m
Pr = 0.690
#reynolds num
Re4 = uinf*L4*1E-3/v
Re5 = uinf*L5*1E-3/v
#
h14bar = 0.664*Re4**0.5*Pr**(1/3)*k/(L4*1E-3)
h15bar = k*(0.037*Re5**0.8 - 871)*Pr**(1/3)/(L5*1E-3)
#the electrical power required
qconv5 = (h15bar*L5*1E-3 - h14bar*L4*1E-3)*1*(Ts - Tinf)

#Results
print "Electrical power required for the fifth heater is", round(qconv5,0),"W"
#answer wrong in book

Electrical power required for the fifth heater is 1018.0 W


## Example 17.04, page: 419

In :
from __future__ import division
import math

#Initialization  of  Variable
Tinf = 26.2 #degC
uinf = 10 #m/s
Ts = 128.4 #degC
Pe = 46.0 #W
qconv = 0.85*Pe
D = 12.7 #mm
L = 94 #mm

#calculations:
#From Table HT-3
Tf = 350 #K
v = 20.92E-6 #m2/s
k = 30E-3 #W/m.K
Pr = 0.7
#Area
A = math.pi*D*L*1E-6
#convection heat transfer coefficient
hbar1 = qconv/(A*(Ts - Tinf))
#reynolds num
ReD = uinf*D*1E-3/v
#Churchill-Bernstein correlation
NuDbar = 0.3 + 0.62*ReD**0.5*Pr**(1/3)/(1 + (0.4/Pr)**(2/3))**0.25*(1 + (ReD/282000)**(5/8))**0.8
#the convection coefficient
hbar2 = NuDbar*k/(D*1E-3)

#Results
print  "Convection coefficient associated with the operating condition is",round(hbar1,0),"W/m2.K"
print  "Convection coefficient from an appropriate correlation is", round(hbar2,0),"W/m2.K"

Convection coefficient associated with the operating condition is 102.0 W/m2.K
Convection coefficient from an appropriate correlation is 96.0 W/m2.K


## Example 17.05, page: 421

In :
from __future__ import division
import math

# Initialization  of  Variable
Pinf = 1 #atm
uinf = 10 #m/s
Tinf = 23 #degC
D = 10 #mm
Ti = 75 #degC
Tt = 35 #degC

#calculations:
#from Table HT-1
pcu = 8933 #kg/m3
ccu = 387 #J/kg.K
kcu = 399 #W/m.K
#from Table HT-3
muTinf = 181.6E-7 #Ns/m2
v = 15.36E-6 #m2/s
k = 0.0258 #W/m.K
Pr = 0.709
muTs = 197.8E-7 #N.s/m2
#reynolds Num
ReD = uinf*D*1E-3/v
#the average convection coefficient
NuDbar = 2 + (0.4*ReD**0.5 + 0.06*ReD**(2/3))*Pr**0.4*(muTinf/muTs)**0.25
hbar = NuDbar*k/D/1E-3
#time
t = pcu*ccu*D*1E-3/6/hbar*math.log((Ti - Tinf)/(Tt - Tinf))

#Results
print  "cooling  time  required is", round(t,1),"s"

cooling  time  required is 69.1 s


## Example 17.06, page: 429

In :
from __future__ import division
import math

# Initialization  of  Variable
mdot = 0.1 #kg/s
D0 = 0.040 #m
Di = 0.020 #m
Tmi = 20 #degC
qdot = 1E6 #W/m3
Ts0 = 70 #degC
Tm0 = 60 #degC

#calculations:
#from Table HT-5
Cp = 4179 #J/kg.K
mu = 6.57E-4 #N.s/m2
#required tube length
L = 4*mdot*Cp*(Tm0 - Tmi)/(math.pi*(D0**2 - Di**2)*qdot)
#Reynolds num
ReD = 4*mdot/math.pi/Di/mu
#local coefficient at the outlet
qs2 = qdot*(D0**2 - Di**2)/4/Di
h0 = qs2/(Ts0 - Tm0)

#Results
print  "tube  length is", round(L,1),"m"
print  "reynolds  number is", round(ReD,0),",the flow is fully turbulant"
print  "local  convection  coefficient at outlet is", round(h0,0),"W/m2.K"

tube  length is 17.7 m
reynolds  number is 9690.0 ,the flow is fully turbulant
local  convection  coefficient at outlet is 1500.0 W/m2.K


## Example 17.07, page: 432

In :
from __future__ import division
import math

# Initialization  of  Variable
mdot = 0.25 #kg/s
Tmi = 15 #degC
D = 0.05 #m
Ts = 100 #degC
Tm0 = 57 #degC
L = 6 #m

#calculations:
#from Table HT-5
Cp = 4178 #J/kg.K
#log mean temp diff
dTlm = ((Ts - Tm0) - (Ts - Tmi))/math.log((Ts-Tm0)/(Ts-Tmi))
#avg Convection Coeff
hbar = mdot*Cp*(Tm0 - Tmi)/math.pi/D/L/dTlm

#Results
print  "Average convection heat transfer coefficient is",round(hbar,0),"W/m2.K"

Average convection heat transfer coefficient is 755.0 W/m2.K


## Example 17.08, page: 433

In :
from __future__ import division
import math

# Initialization  of  Variable
Tinf = 37 #degC
h0bar = 2 #W/m2.K
Tmi = 7 #degC
D = 0.3 #m
L = 15 #m
mdot = 0.05 #kg/s
hibar = 3 #W/m2.K

#calculations:
Cp = 1007 #J/kg.K
#overall heat transfer
Ubar = (1/hibar + 1/h0bar)**-1
#Area
As = math.pi*D*L
#Temp
Tm0 = Tinf - (Tinf - Tmi)*math.e**(-1*Ubar*As/(mdot*Cp))
#heat transfer rate
q = mdot*Cp*(Tm0 - Tmi)

#Results
print  "Chilled air exit temperature and the heat transfer rate are", round(Tm0,1),"degC and",round(q,0),"W respectively"

Chilled air exit temperature and the heat transfer rate are 15.6 degC and 432.0 W respectively


## Example 17.09, page: 434

In :
from __future__ import division
import math

# Initialization  of  Variable
D = 0.06 #m
mdot = 0.01 #kg/s
Tmi = 20 #degC
Tmo = 80 #degC
qs2 = 2000 #W/m2

#calculations:
Cp = 4181 #J/kg.K
k = 0.670 #W/m.K
mu = 352E-6 #Ns/m2
Pr = 2.2
#Length
L = mdot*Cp*(Tmo - Tmi)/qs2/math.pi/D
#Reynolds num
ReD = 4*mdot/math.pi/D/mu
#local coeff
h = 4.36*k/D
#surface temp at outlet
Tso = qs2/h + Tmo

#Results
print  "a)tube  length", round(L,2),"m"
print  "b)surface  temperature at outlet section is", round(Tso,0),"degC"

a)tube  length 6.65 m
b)surface  temperature at outlet section is 121.0 degC


## Example 17.10, page: 436

In :
from __future__ import division
import math

# Initialization  of  Variable
mdot = 2 #kg/s
Tmi = 25 #degC
Ts = 95 #degC
L = 4 #m
D = 0.04 #m

#calculations:
Cp = 4178 #J/kg.K
mu = 695E-6 #Ns/m2
k = 0.628 #W/m.K
Pr = 4.62
mus = 296E-6 #Ns/m2
#Reynolds NUm
ReD = 4*mdot/math.pi/D/mu
#avg coeff
hbar = (0.023*ReD**0.8*Pr*0.4)*k/D
#Temp
Tmo = Ts - (Ts - Tmi)*math.e**(-1*math.pi*D*L*hbar/mdot/Cp)
#heat rate
q = mdot*Cp*(Tmo - Tmi)

#Results
print  "outlet water temperature is", round(Tmo,1),"degC"
print  "rate  of  heat  transfer to the water is", round(q/1000,0),"kW"

outlet water temperature is 46.9 degC
rate  of  heat  transfer to the water is 183.0 kW


## Example 17.11, page: 441

In :
from __future__ import division
import math

# Initialization  of  Variable
L = 0.71 #m
w = 1.02 #m
Ts = 232 #degC
Tinf = 23 #degC
g = 9.8 #m/s2

#calculations:
k = 33.8E-3 #W/m.K
v = 26.4E-6 #m2/s
a = 38.3E-6 #m2/s
Pr = 0.69
b = 0.0025 #1/K
#Rayleigh num
RaL = g*b*(Ts - Tinf)*L**3/a/v
#Churchill-Chu correlation
NuLbar = (0.825 + 0.387*RaL**(1/6)/(1 + (0.492/Pr)**(9/16))**(8/27))**2
#average convection coefficient
hbar = NuLbar*k/L
#the heat transfer by free convection between the firescreen and room air is
q = hbar*L*w*(Ts - Tinf)

#Results
print  "the heat transfer by free convection between the firescreen and room air is", round(q,0),"W"

the heat transfer by free convection between the firescreen and room air is 1060.0 W


## Example 17.12, page: 443

In :
from __future__ import division
import math

# Initialization  of  Variable
Tinf = 300 #K
Tsur = 300 #K
Ts = 350 #K
e = 0.25
s1 = 1.2 #m
s2 = 1.2 #m
Tf = 325 #K
g = 9.8 #m/s2
sigma = 5.67E-8 #W/m2.K4

#calculations:
v = 18.4E-6 #m2/s
k = 0.028 #W/m.K
a = 26.2E-6 #m2/s
# Area
As = s1*s2
#perimeter
P = 4*s1
#characteristic length
L = As/P
#Rayleigh num
b = 1/Tf
RaL = g*b*(Ts - Tinf)*L**3/v/a
#avg conv coeff
hbar = (0.15*RaL**(1/3))*k/L
#allowable electrical power
Pe = hbar*As*(Ts - Tinf) + e*As*sigma*(Ts**4 - Tsur**4)

#Results
print  "Maximum allowable electrical power dissipation is", round(Pe,0),"W"
#answer wrong in book

Maximum allowable electrical power dissipation is 583.0 W


## Example 17.13, page: 445

In :
from __future__ import division
import math

#  Initialization  of  Variable
D = 0.1 #m
Tinf = 23 #degC
Tsur = 23 #degC
Ts = 165 #degC
e = 0.85
g = 9.8 #m/s2
sigma = 5.67E-8 #W/m2.K4

#calculations:
k = 0.0313 #W/m.K
v = 22.8E-6 #m2/s
a = 32.8E-6 #m2/s
Pr = 0.697
b = 2.725E-3 #1/K
#Rayleigh number
RaD = g*b*(Ts - Tinf)*D**3/v/a
#average convection coefficient
hbar = (0.6 + 0.387*RaD**(1/6)/(1 + (0.559/Pr)**(9/16))**(8/27))**2*k/D
#total heat transfer rate from the pipe is
q1 = hbar*math.pi*D*(Ts - Tinf) + e*math.pi*D*sigma*((Ts + 273)**4 - (Tsur+273)**4)

#Results
print  "Heat transfer from the pipe per unit length is", round(q1,0),"W/m"
#anwer wrong in book

Heat transfer from the pipe per unit length is 763.0 W/m


## Example 17.14, page: 452

In :
from __future__ import division
import math

# Initialization  of  Variable
ho = 40 #W/m2.K
Do = 0.045 #m
Di = 0.025 #m
Thi = 100 #degC
mhdot = 0.1 #kg/s
Tho = 60 #degC
mcdot = 0.2 #kg/s
Tci = 30 #degC

#calculations:
Cpc = 4178 #J/kg.K
mu = 725E-6 #Ns/m2
k = 0.625 #W/m.K
Pr = 4.85
Cph = 2131 #J/kg.K
#heat transfer rate
q = mhdot*Cph*(Thi - Tho)
#water outlet temperature
Tco = q/mcdot/Cpc + Tci
#log mean temperature difference
dTlm = ((Thi - Tco) - (Tho - Tci))/math.log((Thi-Tco)/(Tho-Tci))
#reynolds num
ReD = 4*mcdot/math.pi/Di/mu
#conv coeff
hi = (0.023*ReD**0.8*Pr**0.4)*k/Di
#overall heat transfer coefficien
U = 1/(1/hi + 1/ho)
#the required length of the exchanger
L = q/U/math.pi/Di/dTlm

#Results
print  "Tube length to achieve a desired hot fluid outlet temperature is", round(L,1),"m"

Tube length to achieve a desired hot fluid outlet temperature is 63.9 m


## Example 17.15, page: 454

In :
from __future__ import division
import math

# Initialization  of  Variable
ho = 400 #W/m2.K
n = 10
D = 0.025 #m
Thi = 160 #degC
Tho = 100 #degC
Tco = 85 #degC
Tci = 15 #degC
mcdot = 2.5 #kg/s

#calculations:
Cph = 2350 #J/kg.K
Cpc = 4181 #J/kg.K
mu = 548E-6 #Ns/m2
k = 0.643 #W/m.K
Pr = 3.56
#heat transfer rate
q = mcdot*Cpc*(Tco - Tci)
#required flow rate
mhdot = q/Cph/(Thi - Tho)
#m1dot
m1dot = mcdot/n
#reynolds num
ReD = 4*m1dot/math.pi/D/mu
#conv coeff
hi = (0.023*ReD**0.8*Pr**0.4)*k/D
#overall Coeff
U = 1/(1/ho + 1/hi)
#
R = (Thi - Tho)/(Tco - Tci)
P = (Tco - Tci)/(Thi - Tci)
F = 0.87
#log mean temperature difference
dTlm = ((Thi - Tco) - (Tho - Tci))/math.log((Thi-Tco)/(Tho-Tci))
#the required tube length
L = q/U/n/math.pi/D/F/dTlm

#Results
print  "Oil flow rate required is", round(mhdot,2),"kg/s"
print  "Tube length required is", round(L,1),"m"

Oil flow rate required is 5.19 kg/s
Tube length required is 37.9 m