Problem 13.M

from Bird Stewart & Lightfoot, page 424

In Example 13.5-1 would the heat loss be higher or lower if the pipe-surface temperature were 200F and the air temperature were 180F?

> restart;

The fluid properties of air at 1 atm and a film temperature of 190 F (650 R).

> mu:=0.046*lbm/ft/hr; rho:=0.0723*lbm/ft^3; Cp:=0.241*Btu/lbm/R; k:=0.0152*Btu/hr/ft/R;

To is the temperature on the surface and Tinf is far from the surface. The film temperature, Tf, is the average of To and Tinf.

> To:=200*F; Tinf:=180*F; Tf:=(((To+Tinf)/2-32*F)*5/9+273.15*F)*K/F*1.8*R/K; beta:=1/Tf;

Other important information

> d:=0.5*ft; deltaT:=(To-Tinf)*R/F; g:=4.17*10^8*ft/hr^2;

Find the Prandlt number, a dimensionless quantity of importance in convective heat transfer: equation 8.3-16

> Pr:=Cp*mu/k;

Find the Grashof number, also a dimensionless quantity used with heat transfer coefficients: equation 9.9-17

> Gr:=(rho^2*beta*g*d^3*deltaT)/mu^2;

> GrPr:= Gr*Pr;

For a long horizontal cylinder in an infinite fluid, when GrPr > 10^4, equation 13.5-3 is a good approximation for the Nusselt number.

> Num:=0.518*(GrPr)^.25;

From definition of Nusselt number on page 397 of BS&L

> hm:=Num*k/d;

The rate of heat loss: equation 13.1-1

> Q:=hm*A*deltaT;

> A:=evalf(Pi*d*L);

The rate of heat loss from a unit length

> QperL:=Q/L;

So the heat loss is lower with these increased temperatures even though the difference in temperature is the same.

> difference:=21.27011205*Btu/hr/ft - 20.39962027*Btu/hr/ft;

> percent:=.87049178/21.27011205;

But the heat loss only decerases by about 0.04 %