Chapter 20: Additoinal Applications

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
print"\t unsteady state \t"
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
print"\t steady state \t"
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 	
	 unsteady state 	
	 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
	 steady state 	
	 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
print"\t unsteady state \t"
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 	
	 unsteady state 	
	 weight of plate :  lb 	81.25
	 heat :  Btu 	2429.375
	 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 	
	 radiation loss :  kw 	0.7722
	 total requirement :  kw 	1.48420908558
	 staedy state 	
	 heat :  Btu 	3542.0
	 total requirement :  kw 	1.83810082063
	 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 	
	 delt is :  F 	264.0
	 The sheath temperature is then 300 + 264 = 564DegF, which is satisfactory for steel sheathed elements with a 750F maximum.