In [1]:

```
print"\t example 19.1 \t"
# For orientation purposes, one can make an estimate of the number of tubes required in the radiant section by assuming avg flux is 12000 Btu/(hr)*(ft**2)
# from Fig.19.14 it can be seen that with a tube temperature of 800DegF, an exit-gas temperature of l730DegF will be required to effect such a flux.
print"\t approxiate values are mentioned in the book \t"
Q=50000000; # Btu/hr
QF=(Q/0.75); # efficiency of tank is 75%
print"\t heat liberated by the fuel : Btu/hr \t",QF
w1=(QF/17130); # heating value of fuel is 17130Btu/lb
print"\t fuel quantity : lb/hr \t",w1
w2=(w1*17.44); # lb of fuel fired with 17.44lb of air
print"\t air required : lb/hr \t",w2
w3=(w1*0.3); # 0.3 lb of air is used for atomizing lb of fuel
print"\t steam for atomizing : lb/hr \t",w3
QA=(w2*82); # heating value at 400F is 82Btu/lb
print"\t QA is : Btu/hr \t",QA
print"\t QS is negligible \t"
QW=(0.02*QF);
print"\t QW is : Btu/hr \t",QW
Qnet=(QF+QA-QW);
print"\t Qnet is : Btu/hr \t",Qnet
#Heat out m gases at 1730DegF, 25 per cent excess air, 476 Btu/lb of flue gas
QG=(476*(w1+w2+w3));
print"\t QG is : Btu/hr \t",QG
Q1=(Qnet-QG);
print"\t Q1 is : Btu/hr \t",Q1 # calculation mistake in book
A=(3.14*38.5*(5./12.)); # area of tube
print"\t area of tube is : ft**2 \t",A
Nt=(Q1/(12000*A)); # 12000 is avg flux
print"\t estimated number of tubes : \t",Nt
# The layout of the cross section of the furnace may be as shown m Fig. 19.16.
# center to center distance is 8(1/2)in
Acp=(8.5*38.5/12);
print"\t cold plane surface per tube : ft**2 \t",Acp # calculation mistake in book
a=0.937; # a=alpha, from fig 19.11 as Ratio of center-to-center/OD is 1.7
Acp1=(Acp*a);
print"\t Acp1 is : ft**2 \t",Acp1
Acpt=(Acp1*Nt);
print"\t total cold plane surface is : ft**2 \t",Acpt
A1=(2*20.46*14.92); # from fig 19.16
print"\t surface of end walls : ft**2 \t",A1
A2=(38.5*14.92); # from fig 19.16
print"\t surface of side wall : ft**2 \t",A2
A3=(38.5*9.79); # from fig 19.16
print"\t surface of bridge walls : ft**2 \t",A3
A4=(2*20.46*38.5); # from fig 19.16
print"\t surface of floor and arch : ft**2 \t",A4
AT=(A1+A2+A3+A4);
print"\t AT is : ft**2 \t",AT
AR=(AT-Acpt);
print"\t AR is : ft**2 \t",AR
Ar=(AR/Acpt);
print"\t ratio of areas is : \t",Ar
print"\t dimension ratio is 3:2:1 \t"
L=((2/3)*(38.5*20.46*14.92)**(1/3));
print"\t length is : ft \t",L
print"\t gas emissivity \t"
# From the analysis of the fuel, the steam quantity, and the assumption that the humidity of the air is 50 per cent of saturation at 60F, the partial pressures of CO2 and H2O in the combustion gases with 25 per cent excess air are
pCO2=0.1084;
pH2O=0.1248
pCO2L=1.63; # pCO2L=(pCO2*L)
pH2OL=1.87;
P=((pCO2)/(pCO2+pH2O));
print"\t percentage correction at P : \t",P
Pt=pCO2L+pH2OL;
print"\t Pt is : \t",Pt
# %correction estimated to be 8%
eG=(((6500+14500)-(650+1950))/(39000-4400))*((100-8)/100); # values from fig 19.12 and 19.13, eq 19.5
print"\t eG is : \t",eG
f=0.635; # from fig 19.15 as (AR/Acpt)=1.09 and eG=0.496
print"\t overall exchange factor : \t",f
Z=(Q1/(Acpt*f));
print"\t Z is : \t",round(Z)
print"\t TG required (at Ts = 800F) = 1670F compared with 1730DegF assumed in heat balance) \t"
# end
```

In [4]:

```
print"\t example 19.2 \n"
QF=50000000.;
G=22.36;
Acpt=1500.;
print"\t approxiate values are mentioned in the book "
Q=(QF/(1+(G/4200)*(QF/Acpt)**(1/2)))/2; # eq 19.15
print"\t Q is : Btu/hr ",Q
print"\t The radiant-section average rate will be 8350 Btu/(hr) (ft2), and the exit-flue-gas temperature 1540DegF by heat balance. \n"
# end
```

In [5]:

```
print"\t example 19.3 \n"
Qr=1.5; # Qr=(QF2/QF1)
Cr=1.5; # Cr=(CR2/CR1)
Gr=140/125; # Gr=(G2/G1)
Qr1=0.38; # Qr1=(Q1/QF1)
print"\t approxiate values are mentioned in the book \n"
a1=1.63; # a1=(G1*(CR1/27)^(1/2)), from eq 19.17
print"\t a1 is : ",a1
a2=1.37*(a1); # a2=(G2*(CR2/27)^(1/2))
print"\t a2 is : ",a2
Qr2=(1/(1+a2)); # Qr2=(Q2/QF2),from eq 19.15
print"\t Qr2 is : ",Qr2
Q21=(Qr2/Qr1)*(Qr); # Q21=(Q2/Q1)
print"\t ratio of heats is : ",round(Q21,2)
print"\t Hence the radiant absorption will be increased only 22 per cent for an increase of 50 per cent in the heat liberated. \n"
# end
```

In [6]:

```
print"\t example 19.4 \t"
eS=0.9; # assumed
TG=1500;
TS=650;
pCO2=0.1084;
pH2O=0.1248;
print"\t approxiate values are mentioned in the book \t"
L=(0.4*8.5)-(0.567*5); # table 19.1
print"\t L is : ft \t",L
pH2OL=0.1248*L;
pCO2L=0.1084*L;
print"\t pH2OL is : atm-ft \t",pH2OL
print"\t pCO2L is : atm-ft \t",pCO2L
qH2O=1050; # at TG, from fig 19.12 ana 19.13
qCO2=1700; # at TG, from fig 19.12 ana 19.13
qTG=(qH2O+qCO2);
print"\t qTG is : \t",qTG
qsH2O=165; # at TS, from fig 19.12 ana 19.13
qsCO2=160; # at TS, from fig 19.12 ana 19.13
qTS=(qsH2O+qsCO2);
print"\t qTG is : \t",qTS
q=(0.9*(qTG-qTS)); # q=(QRC/A)
print"\t q is : \t",q
P=((pCO2)/(pCO2+pH2O));
print"\t percentage correction at P :\t",P
Pt=pCO2L+pH2OL;
print"\t Pt is : \t",Pt
# %correction estimated to be 2%
q1=(q*0.98); # # q1=(QRC/A)
print"\t q1 is : \t",q1
hr=(q1/(TG-TS));
print"\t radiation coefficient is : Btu/(hr)*(ft^2)*(F) \t",round(hr,2)
#end
```

In [10]:

```
print"\t example 19.5 \t"
Q=500000;
print"\t approxiate values are mentioned in the book \t"
a=(3.5+(3.14*4*(120/360)))/(1); # a=(alpha*Acp) from fig 19.17
AR=(3+3.6+3);
print"\t a is : ft**2/ft \t",a
print"\t AR is : ft**2/ft \t",AR
# Arbitrarily neglecting end wa.lls and also .the side wall refractory over 3'0" above the floor
R=(AR/a);
print"\t ratio of two areas is : \t",R
eG=0.265;
TG=1174; # F
TS=500; # F
f=0.56; # from fig 19.15 as (AR/Acpt)=2.49 and eG=0.265
q=15300; # at TG and TS,q=(Q/(a*f))
# However, the convection coefficient is small, 1.0 +or- Btu/(hr)(ft2)("F), and AR/a is not 2.0 as in the assumptions for the Lobo and Evans equation.
q1=(q)-(7*(TG-TS)); # q1=(Q/(a*f))
print"\t q1 is : Btu/(hr)*(ft**2) \t",q1
q2=(q1*f); # q2=(Q/(a))
print"\t q2 is : Btu/(hr)*(ft**2) \t",q2
print"\t convection rate basis \t"
q3=(1*(TG-TS)*(4.2/a)); # q2=(Q/(a))
print"\t q3 is : Btu/(hr)*(ft**2) \t",q3 # calculation mistake in book
qt=(q2+q3); # qt=(Q/(a))
print"\t qt is : Btu/(hr)*(ft**2) \t",qt
ar=(Q/qt);
print"\t required a is : ft**2 \t",ar
L=(ar/a)-2;
print"\t length required is : ft \t",round(L,1)
# end
```