## Example 20.1 pgno:719¶

In [19]:
print"\t example 20.1 \t"
print"\t approximate values are mentioned in the book \t"
T=150; # F
L=0.6; # ft
N=7500; # rev/hr
row=62.5; # lb/ft**3
mu=1.06; # at 150 F and from fig 14, lb/ft*hr
k=0.38; # Btu/(hr)*(ft**2)*(F/ft), from table 4
c=1; # Btu/(lb)*(F)
Rej=(L**2)*(N)*(row)/(mu);
print"\t Rej is :  \t",Rej
Z=1; # Z=(mu/muw)**(0.14), regarded as 1 for water
Dj=1.01; # ft, from table 11
j=1100; # fig 20.2
hi=((j)*(k/Dj)*((c*mu/k)**(1/3))*(Z)**(0.14));
print"\t hi is :  Btu/(hr)*(ft**2)*(F) \t",hi
hoi=1500; # Btu/(hr)*(ft**2)*(F)
Uc=((hi*hoi)/(hi+hoi)); # from eq 6.38
print"\t Uc is :  Btu/(hr)*(ft**2)*(F) \t",Uc
Rd=0.005;
hd=(1/Rd);
print"\t hd is :  \t",hd
UD=((Uc*hd)/(Uc+hd));
print"\t UD is :  Btu/(hr)*(ft**2)*(F) \t",UD
A=3.43; # ft**2
Q=32600;
delt=(Q/(UD*A));
print"\t temperature difference is : F \t",delt
Ts=(T+delt)-6;
print"\t temperature of the steam :  F \t",round(Ts)
# end

	 example 20.1
approximate values are mentioned in the book
Rej is :  	159198.113208
hi is :  Btu/(hr)*(ft**2)*(F) 	413.861386139
Uc is :  Btu/(hr)*(ft**2)*(F) 	324.366270047
hd is :  	200.0
UD is :  Btu/(hr)*(ft**2)*(F) 	123.717442778
temperature difference is : F 	76.8232268984
temperature of the steam :  F 	221.0


## Example 20.2 pgno:723¶

In [20]:
print"\t example 20.2 \t"
print"\t approximate values are mentioned in the book \t"
T1=150; # F
T2=220; # F
L=0.6; # ft
N=7500; # rev/hr
row=62.5; # lb/ft**3
mu=1.06; # at 150 F and from fig 14, lb/ft*hr
k=0.38; # Btu/(hr)*(ft**2)*(F/ft), from table 4
c=1; # Btu/(lb)*(F)
Rej=(L**2)*(N)*(row)/(mu);
print"\t Rej is :  \t",Rej
Z=1; # Z=(mu/muw)**(0.14), regarded as 1 for water
Dj=1.01; # ft, from table 11
j=1700; # fig 20.2
hi=((j)*(k/Dj)*((c*mu/k)**(1/3))*(Z)**(0.14));
print"\t hi is :  Btu/(hr)*(ft**2)*(F) \t",hi
hoi=1500; # Btu/(hr)*(ft**2)*(F)
Uc=((hi*hoi)/(hi+hoi)); # from eq 6.38
print"\t Uc is :  Btu/(hr)*(ft**2)*(F) \t",Uc
Rd=0.005;
hd=(1/Rd);
print"\t hd is :  \t",hd
UD=((Uc*hd)/(Uc+hd));
print"\t UD is :  Btu/(hr)*(ft**2)*(F) \t",UD
Q=32600;
A=(Q/(UD*(T2-T1)));
print"\t Area is :  ft**2 \t",A
a=0.1309; # ft**2/ft
a1=(3.14*0.8*a);
print"\t area per turn is :  ft**2 \t",a1
n=(A/a1);
print"\t number of turns :  \t",round(n)
# end

	 example 20.2
approximate values are mentioned in the book
Rej is :  	159198.113208
hi is :  Btu/(hr)*(ft**2)*(F) 	639.603960396
Uc is :  Btu/(hr)*(ft**2)*(F) 	448.40351689
hd is :  	200.0
UD is :  Btu/(hr)*(ft**2)*(F) 	138.310019983
Area is :  ft**2 	3.36717676544
area per turn is :  ft**2 	0.3288208
number of turns :  	10.0


## Example 20.3 pgno:725¶

In [21]:
print"\t example 20.3 \t"
print"\t approximate values are mentioned in the book \t"
T1=675; # inlet hot fluid,F
T2=200; # outlet hot fluid,F
t1=120; # inlet cold fluid,F
t2=140; # outlet cold fluid,F
W=33100; # lb/hr
w=510000; # lb/hr
print"\t 1.for heat balance \t"
print"\t for oil \t"
c=0.64; # Btu/(lb)*(F)
Q=((W)*(c)*(T1-T2)); # Btu/hr
print"\t total heat required for oil is : Btu/hr \t",Q
print"\t for water \t"
c=1; # Btu/(lb)*(F)
Q=((w)*(c)*(t2-t1)); # Btu/hr
print"\t total heat required for water is : Btu/hr \t",Q
delt1=T2-t1; #F
delt2=T1-t2; # F
print"\t delt1 is : F \t",delt1
print"\t delt2 is : F \t",delt2
LMTD=230;
print"\t LMTD is  F \t",LMTD
Tc=((T2)+(T1))/(2); # caloric temperature of hot fluid,F
print"\t caloric temperature of hot fluid is : F \t",Tc
tc=((t1)+(t2))/(2); # caloric temperature of cold fluid,F
print"\t caloric temperature of cold fluid is : \t",tc
print"\t hot fluid:inner tube side, oil \t"
at=0.0458; # flow area, ft**2, table 11
print"\t flow area is : ft**2 \t",at
Gt=(W/(at)); # mass velocity,lb/(hr)*(ft**2)
print"\t mass velocity is : lb/(hr)*(ft**2) \t",Gt
mu2=5.56; # at 400F,lb/(ft)*(hr)
D=0.242; # ft, table 11
Ret=((D)*(Gt)/mu2); # reynolds number
print"\t reynolds number is : \t",Ret
jH=100; # from fig.24
Z=0.245; # Z=(k(c*mu/k)**(1/3)), Btu/(hr)*(ft)*(F/ft), fig 16
hi=((jH)*(Z/D)); #Hi=(hi/phyp),using eq.6.15,Btu/(hr)*(ft**2)*(F)
print"\t hi is : Btu/(hr)*(ft**2)*(F) \t",hi
ID=2.9; # ft
OD=3.5; # ft
hio=((hi)*(ID/OD)); # using eq.6.5
print"\t Correct hio to the surface at the OD is : Btu/(hr)*(ft**2)*(F) \t",hio
ho=150; # Btu/(hr)*(ft**2)
tw=(tc)+(((hio)/(hio+ho))*(Tc-tc)); # from eq.5.31
print"\t tw is : F \t",tw
tf=(tw+tc)/2;
print"\t tf is : F \t",tf
delt=110; # F
d0=3.5; # in, fig 10.4
Uc=((ho*hio)/(ho+hio)); # from eq 6.38
print"\t Uc is : Btu/(hr)*(ft**2)*(F) \t",Uc
Rd=0.01;
hd=(1/Rd);
print"\t hd is : \t",hd
UD=((Uc*hd)/(Uc+hd));
print"\t UD is : Btu/(hr)*(ft**2)*(F) \t",UD
A=(Q/(UD*(LMTD)));
print"\t Area is : ft**2 \t",A
a=0.917; # ft**2/ft, table 11
L=(A/(a*24));
print"\t pipe length : \t",round(L)
# end

	 example 20.3
approximate values are mentioned in the book
1.for heat balance
for oil
total heat required for oil is : Btu/hr 	10062400.0
for water
total heat required for water is : Btu/hr 	10200000
delt1 is : F 	80
delt2 is : F 	535
LMTD is  F 	230
caloric temperature of hot fluid is : F 	437
caloric temperature of cold fluid is : 	130
hot fluid:inner tube side, oil
flow area is : ft**2 	0.0458
mass velocity is : lb/(hr)*(ft**2) 	722707.423581
reynolds number is : 	31455.9705947
hi is : Btu/(hr)*(ft**2)*(F) 	101.239669421
Correct hio to the surface at the OD is : Btu/(hr)*(ft**2)*(F) 	83.8842975207
tw is : F 	240.107773852
tf is : F 	185.053886926
Uc is : Btu/(hr)*(ft**2)*(F) 	53.7985865724
hd is : 	100.0
UD is : Btu/(hr)*(ft**2)*(F) 	34.9798966111
Area is : ft**2 	1267.80895267
pipe length : 	58.0


## Example 20.4 pgno:729¶

In [22]:
print"\t example 20.4 \t"
print"\t approximate values are mentioned in the book \t"
T1=450.; # inlet hot fluid,F
T2=150.; # outlet hot fluid,F
t1=85.; # inlet cold fluid,F
t2=100.; # outlet cold fluid,F
W=3360; # lb/hr
w=11100; # lb/hr
from math import log10
print"\t 1.for heat balance \t"
print"\t for SO2 \t"
c=0.165; # Btu/(lb)*(F)
Q=((W)*(c)*(T1-T2)); # Btu/hr
print"\t total heat required for SO2 is :  Btu/hr \t",Q
print"\t for water \t"
c=1; # Btu/(lb)*(F)
Q=((w)*(c)*(t2-t1)); # Btu/hr
print"\t total heat required for water is :  Btu/hr \t",Q
delt1=T2-t1; #F
delt2=T1-t2; # F
print"\t delt1 is :  F \t",delt1
print"\t delt2 is :  F \t",delt2
LMTD=((delt2-delt1)/((2.3)*(log10(delt2/delt1))));
print"\t LMTD is : F \t",LMTD
R=20;
S=0.0412;
FT=0.98; # fig 18
delt=(FT*LMTD);
print"\t delt is :  F \t",delt
Tc=((T2)+(T1))/(2); # caloric temperature of hot fluid,F
print"\t caloric temperature of hot fluid is :  F \t",Tc
tc=((t1)+(t2))/(2); # caloric temperature of cold fluid,F
print"\t caloric temperature of cold fluid is :  \t",tc
print"\t hot fluid:inner tube side, SO2 \t"
at=0.0512; # flow area, ft**2, table 11
print"\t flow area is :  ft**2 \t",at
Gt=(W/(at)); # mass velocity,lb/(hr)*(ft**2)
print"\t mass velocity is :  lb/(hr)*(ft**2) \t",Gt
mu2=0.041; # at 300F,lb/(ft)*(hr), fig 15
D=0.256; # ft, table 11
Ret=((D)*(Gt)/mu2); # reynolds number
print"\t reynolds number is :  \t",Ret
jH=790; # from fig.24
Z=0.006831; # Z=(k(c*mu/k)**(1/3)), Btu/(hr)*(ft)*(F/ft)
hi=((jH)*(Z/D)); #Hi=(hi/phyp),using eq.6.15,Btu/(hr)*(ft**2)*(F)
print"\t hi is :  Btu/(hr)*(ft**2)*(F) \t",hi
ID=3.068; # ft
OD=3.5; # ft
hio=((hi)*(ID/OD)); # using eq.6.5
print"\t Correct hio to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) \t",hio
print"\t cold fluid water \t"
L=8; # ft
G=(w/(2*L));
print"\t G :  lb/(hr)*(ft) \t",G
mu1=1.94; # at 92.5F, lb/(ft)*(hr)
Re=(4*G/mu1);
print"\t Re is : \t",Re
Do=0.292; # ft
ho=(65*(G/Do)**(1/3));
print"\t ho is :  Btu/(hr)*(ft**2)*(F) \t",ho
Uc=((ho*hio)/(ho+hio)); # from eq 6.38
print"\t Uc is :  Btu/(hr)*(ft**2)*(F) \t",Uc
Rd=0.01;
hd=(1/Rd);
print"\t hd is :  \t",hd
UD=((Uc*hd)/(Uc+hd));
print"\t UD is :  Btu/(hr)*(ft**2)*(F) \t",UD
A=(Q/(UD*(LMTD)));
print"\t Area is :  ft**2 \t",A # calculation mistake in book
a=0.917; # ft**2/ft, table 11
l=(A/(a*8))-1.9;
print"\t pipe length :  \t",round(l,2)
# end

	 example 20.4
approximate values are mentioned in the book
1.for heat balance
for SO2
total heat required for SO2 is :  Btu/hr 	166320.0
for water
total heat required for water is :  Btu/hr 	166500.0
delt1 is :  F 	65.0
delt2 is :  F 	350.0
LMTD is : F 	169.475824421
delt is :  F 	166.086307932
caloric temperature of hot fluid is :  F 	300.0
caloric temperature of cold fluid is :  	92.5
hot fluid:inner tube side, SO2
flow area is :  ft**2 	0.0512
mass velocity is :  lb/(hr)*(ft**2) 	65625.0
reynolds number is :  	409756.097561
hi is :  Btu/(hr)*(ft**2)*(F) 	21.0800390625
Correct hio to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	18.4781599554
cold fluid water
G :  lb/(hr)*(ft) 	693
Re is : 	1428.86597938
ho is :  Btu/(hr)*(ft**2)*(F) 	65.0
Uc is :  Btu/(hr)*(ft**2)*(F) 	14.3879596501
hd is :  	100.0
UD is :  Btu/(hr)*(ft**2)*(F) 	12.5782116353
Area is :  ft**2 	78.10657332
pipe length :  	8.75


# Example 20.5 pgno:736¶

In [23]:
print"\t example 20.5 \t"
print"\t approximate values are mentioned in the book \t"
Nt=25.; # number of tubes
A=50.; # total projected area
Tav=100.; # F
s=28.; # assumption spray, lb/(min)*(ft**2)
Do=0.0833; # ft
PH=0.1562;
Y=0.874;
Z=0.466;
E=(0.171*(Do*Y*Z)**0.1); # (E/(Do*Y*Z)**0.1)=0.171, from fig 20.10
from math import log10
print"\t evaporation percentage is :  \t",E
Q=(295*500*(143-130));
print"\t heat load is :  Btu/hr \t",Q
Q1=(Q*(1-0.12));
print"\t sensible heat is :  Btu/hr \t",Q1
t2=(90)+(Q1/(28*60*50));
print"\t final spray temperature is :  F \t",t2
w=(s*60*50);
print"\t total spray :  lb/hr \t",w
m=(w/(2*4*12));
print"\t m is :  lb/(hr)*(ft**2) \t",m
mu=1.84; # lb/(ft)*(hr)
Z=((m**0.3)*Do*Y*Z/(mu*0.125));
print"\t Z is :  \t",Z
N=3; # assume 3 horizontal rows
ho=300*(N**0.05); # (ho/(N**0.05))=300, from fig 20.11
print"\t ho is :  Btu/(hr)*(ft**2)*(F) \t",ho
print"\t tube side coefficient \t"
print"\t assuming even number of passes and tube side velocity about 8fps \t"
at=0.0775; # ft**2
Gt=(295*500/(at)); # mass velocity,lb/(hr)*(ft**2)
print"\t mass velocity is :  lb/(hr)*(ft**2) \t",Gt
V=(Gt/(3600*62.5));
print"\t velocity is :  fps \t",V
hi=2140; # Btu/(hr)*(ft**2)*(F), fig 25
ID=0.87; # ft
OD=1; # ft
hio=((hi)*(ID/OD)); # using eq.6.5
print"\t Correct hio to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) \t",hio
Uc=((ho*hio)/(ho+hio)); # from eq 6.38
print"\t Uc is :  Btu/(hr)*(ft**2)*(F) \t",Uc
a=0.2618; # ft**2, table 11
A1=(2*3*25*12*a);
print"\t total surface is :  ft**2 \t",A1
T1=143; # inlet hot fluid,F
T2=130; # outlet hot fluid,F
t1=90; # inlet cold fluid,F
t2=110; # outlet cold fluid,F
delt1=T2-t1; #F
delt2=T1-t2; # F
delt1=40.;
delt2=33.;
print"\t delt1 is :  F \t",delt1
print"\t delt2 is :  F \t",delt2
LMTD=((delt2-delt1)/((2.3)*(log10(delt2/delt1))));
print"\t LMTD is : F \t",LMTD # calculation mistake in book
R=0.65;
S=0.377;
FT=0.97; # fig 18
delt=(FT*LMTD);
print"\t delt is :  F \t",delt
UD=(Q/(A1*(delt)));
print"\t UD is :  Btu/(hr)*(ft**2)*(F) \t",UD
Rd=((Uc-UD)/((UD)*(Uc))); # (hr)*(ft**2)*(F)/Btu
print"\t actual Rd is :  (hr)*(ft**2)*(F)/Btu \t",round(Rd,4)
print"\t The assumption of three horizontal rows is satisfactory, since a dirt factor of 0.004 was required \t"
# end

	 example 20.5
approximate values are mentioned in the book
evaporation percentage is :  	0.121913091814
heat load is :  Btu/hr 	1917500
sensible heat is :  Btu/hr 	1687400.0
final spray temperature is :  F 	110.088095238
total spray :  lb/hr 	84000.0
m is :  lb/(hr)*(ft**2) 	875.0
Z is :  	1.12568513365
ho is :  Btu/(hr)*(ft**2)*(F) 	316.940192565
tube side coefficient
assuming even number of passes and tube side velocity about 8fps
mass velocity is :  lb/(hr)*(ft**2) 	1903225.80645
velocity is :  fps 	8.45878136201
Correct hio to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	1861.8
Uc is :  Btu/(hr)*(ft**2)*(F) 	270.83506906
total surface is :  ft**2 	471.24
delt1 is :  F 	40.0
delt2 is :  F 	33.0
LMTD is : F 	36.4287504681
delt is :  F 	35.3358879541
UD is :  Btu/(hr)*(ft**2)*(F) 	115.153519517
actual Rd is :  (hr)*(ft**2)*(F)/Btu 	0.005
The assumption of three horizontal rows is satisfactory, since a dirt factor of 0.004 was required


## Example 20.6 pgno:745¶

In [24]:
print"\t example 20.6 \t"
print"\t approximate values are mentioned in the book \t"
T1=200.; # inlet hot fluid,F
T2=100.; # outlet hot fluid,F
t1=50.; # inlet cold fluid,F
t2=100.; # outlet cold fluid,F
from math import log10
R=((T1-T2)/(t2-t1));
print"\t R is :  \t",R
V=((T1+T2-t1-t2)/(t2-t1))/(2);
print"\t V is :  \t",V
u=120;
U=60;
F=((u*1)/(U*2));
print"\t F is :  \t",F
E=1.1; # In Fig.20.18b for R = 2.0and F = l.O,the abscissa and ordinate intersect at E =1.10.
Z=(E/V);
print"\t Z is :  \t",Z
deltD=0.783*V; # deltD/V=0.783, from fig 20.17
print"\t deltD is :  \t",deltD
delt=(deltD*(t2-t1));
print"\t delt is :  \t",delt
delt1=T2-t1; #F
delt2=T1-t2; # F
print"\t delt1 is :  F \t",delt1
print"\t delt2 is :  F \t",delt2
LMTD=((delt2-delt1)/((2.3)*(log10(delt2/delt1))));
print"\t LMTD is : F \t",round(LMTD)
# end

	 example 20.6
approximate values are mentioned in the book
R is :  	2.0
V is :  	1.5
F is :  	1
Z is :  	0.733333333333
deltD is :  	1.1745
delt is :  	58.725
delt1 is :  F 	50.0
delt2 is :  F 	100.0
LMTD is : F 	72.0


## Example 20.7 pgno:752¶

In [25]:
print"\t example 20.7 \t"
print"\t approximate values are mentioned in the book \t"
T1=284.; # inlet hot fluid,F
T2=104.; # outlet hot fluid,F
t1=86.; # inlet cold fluid,F
t2=104.; # outlet cold fluid,F
W=1000; # lb/hr
k=0.15; # thermal conductivity
L=10;
Beta=((2*k)/(500*(2./12.))); # hoi=500Btu/(hr)*(ft^2)*(F) for water
print"\t beta is : \t",Beta
print"\t for sand \t"
C=0.2; # Btu/(lb)*(F)
Q=((W)*(C)*(T1-T2)); # Btu/hr
print"\t total heat required for sand is : Btu/hr \t",Q
c=1;
w=(Q/(t2-t1));
print"\t w is : lb/hr \t",w
R=((W*C)/(w*c));
print"\t R is : \t",R
S=((T2-T1)/(t1-T1));
print"\t S is : \t",S
W1=(8.33*(k*L)/C); # ((W1*C)/(k*L))=8.33 from fig 20.20b for Beta=0
print"\t rate per tube is : lb/hr \t",W1
N1=(W/W1);
print"\t number of tubes : \t",N1
print"\t for air assume hoi=9 and Beta=0.2 \t"
c1=0.25;
w1=(Q/(c1*(t2-t1)));
print"\t w1 is : lb/hr \t",w1
W2=(5.23*(k*L)/C); # ((W1*C)/(k*L))=5.23 from fig 20.20b for Beta=0.2
print"\t rate per tube is : lb/hr \t",round(W2)
N2=(W/W2);
print"\t number of tubes : \t",round(N2)
# end

	 example 20.7
approximate values are mentioned in the book
beta is : 	0.0036
for sand
total heat required for sand is : Btu/hr 	36000.0
w is : lb/hr 	2000.0
R is : 	0.1
S is : 	0.909090909091
rate per tube is : lb/hr 	62.475
number of tubes : 	16.006402561
for air assume hoi=9 and Beta=0.2
w1 is : lb/hr 	8000.0
rate per tube is : lb/hr 	39.0
number of tubes : 	25.0


## Example 20.8a pgno758¶

In [26]:
print"\t example 20.8a \t"
print"\t approximate values are mentioned in the book \t"
L=3; # ft
B=2; # ft
h=18/12; # ft , height of water present in tank
m=(L*B*h*62.5);
print"\t Lb of water is :  lb \t",m
t1=50;
t2=150;
c=1;
Q=(m*c*(t2-t1))/(2*3412); # kwhr
print"\t heat to be supplied :  kwhr \t",Q
print"\t losses \t"
Q1=(L*B*260)/(1000); # from fig 20.25c
print"\t from surface of water :  kwhr \t",Q1
Q2=(5.5*((2*B*2)+(2*L*B))/(1000)); # from fig 20.25c
print"\t from sides of vessel :  kwhr \t",Q2
print"\t losses from bottom are negigible \t"
Qt=(Q+Q1+Q2);
print"\t total requirement :  kwhr \t",Qt
m1=8; # gal/hr
Qs=(m1*8.33*c*(t2-t1))/(3412); # kwhr
print"\t heat to be supplied :  kwhr \t",Qs
Qts=(Qs+Q1+Q2);
print"\t total requirement :  kwhr \t",round(Qts,2)
# end

	 example 20.8a
approximate values are mentioned in the book
Lb of water is :  lb 	375.0
heat to be supplied :  kwhr 	5.49531066823
losses
from surface of water :  kwhr 	1
from sides of vessel :  kwhr 	0.11
losses from bottom are negigible
total requirement :  kwhr 	6.60531066823
heat to be supplied :  kwhr 	1.9531066823
total requirement :  kwhr 	3.06


## Example 20.8b pgno:760¶

In [27]:
print"\t example 20.8b \t"
print"\t approximate values are mentioned in the book \t"
m=100; # lb
t1=70;
t2=370;
L=4;
B=3;
n=4; # number of air changers
c1=0.12
Q1=(m*c1*(t2-t1));
print"\t heat to steel charge :  Btu \t",Q1
c2=0.25
Q2=(n*L*B*2*0.075*c2*(t2-t1));
print"\t heat to air :  Btu \t",Q2
print"\t From Fig. 20.25a for 52ft^2 of oven outside surface and a temperature rise of 300F the loss is 5kw for 1 in.thick insulations.For 2 in.thick insulation the loss is 2.5kw \t"
Qt=((Q1+Q2)/(3412))+(2.5)
print"\t total requirement :  kw \t",round(Qt,2)
# end

	 example 20.8b
approximate values are mentioned in the book
heat to steel charge :  Btu 	3600.0
heat to air :  Btu 	540.0
From Fig. 20.25a for 52ft^2 of oven outside surface and a temperature rise of 300F the loss is 5kw for 1 in.thick insulations.For 2 in.thick insulation the loss is 2.5kw
total requirement :  kw 	3.71


## Example 20.8c pgno:762¶

In [28]:
print"\t example 20.8c \t"
print"\t approximate values are mentioned in the book \t"
m=270; # cfm
t1=70;
t2=120;
L=1.5; # ft
B=1.5; # ft
c=0.25
row=0.075; # lb/ft^3
Q=(m*row*60*c*(t2-t1));
print"\t heat :  Btu \t",Q
V=(m/(L*B*60)); # fps
print"\t velocity is :  fps \t",V
print"\t Refer to Fig.20.22a.The air is capable of removing 33watts/in which is the maximum dissipation which may be expected. Any group of heaters providing 5 kw which do not require a dissipation of more than 33 w/in. and which will fit into the duct will be satisfactory \t"
print"\t Thus in Table 20.3 elements of 350 watts with a total length each of 18 in. each are satisfactory \t"
# end

	 example 20.8c
approximate values are mentioned in the book
heat :  Btu 	15187.5
velocity is :  fps 	2.0
Refer to Fig.20.22a.The air is capable of removing 33watts/in which is the maximum dissipation which may be expected. Any group of heaters providing 5 kw which do not require a dissipation of more than 33 w/in. and which will fit into the duct will be satisfactory
Thus in Table 20.3 elements of 350 watts with a total length each of 18 in. each are satisfactory


## Example 20.8d pgno:762¶

In [29]:
print"\t example 20.8d \t"
print"\t approximate values are mentioned in the book \t"
t1=70;
t2=300;
L=26; # in
B=12; # in
H=1; # in
c1=0.13
# specific gravity of cast iron is 7.2
m=(L*B*H*62.5*7.2/1728); # lb
print"\t weight of plate :  lb \t",m
Q1=(m*c1*(t2-t1));
print"\t heat :  Btu \t",Q1
print"\t From Figure 20.25b for a black body the radiation is 1.5w/in^2.The radiation from the top is actually 110 per cent of this value, and from the bottom of the plate it is 55 per cent for an average of 82.5 per cent is taken \t"
Q2=(2*26*12*1.5*0.825/1000); # ke
print"\t radiation loss :  kw \t",Q2
Qt=((Q1)/(3412))+(Q2);
print"\t total requirement :  kw \t",Qt
print"\t staedy state \t"
m2=70;
c2=0.22;
Qs=(m2*c2*(t2-t1));
print"\t heat :  Btu \t",Qs
Ql=0.8; # kw
Qts=((Qs)/(3412))+(Ql);
print"\t total requirement :  kw \t",Qts
print"\t The steady state is controlling.The requirements are satisfied, by four 24-in. strip heaters, but the sheath temperature must now be checked. Since the temperature drop per unit flux density is 14 to 19F, assume an average of 16.5DegF. For clamp-on strips 24 in. long the watts per square inch deliverable are 16 \t"
delt=(16*16.5);
print"\t delt is :  F \t",delt
print"\t The sheath temperature is then 300 + 264 = 564DegF, which is satisfactory for steel sheathed elements with a 750F maximum. \t"
# end

	 example 20.8d
approximate values are mentioned in the book