Problem 13.G: Free Convection Heat Transfer from an Isolated Sphere

from Bird Stewart & Lightfoot, page 422

A solid sphere 1 in. in diameter is suspended in still air at 1 atm pressure and 100F ambient air temperature and the sphere surface is maintained at 200F, what rate of electric heating is required. Neglect radiation.

> restart;

The fluid properties of air at 1 atm and a film temperature of 150 F

> mu:=.04894*lbm/ft/hr; rho:=0.0723*lbm/ft^3; Cp:=0.241*Btu/lbm/R; k:=0.0169*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:=100*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:=1/12*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;

> GrPr14:= Gr^.25*Pr^.25;

For a sphere in a large body of fluid, equation 13.5-2 agrees well with data for Gr^1/4 Pr^1/4 less than 200.

> Num:=2+0.59*GrPr14;

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(4*Pi*(d/2)^2);

> heat:=Q *hr/(3600*sec) *251.9608*cal/Btu;

This is close to BS&L's answer of 0.332 cal/sec, but not exactly the same. Perhaps some of the fluid properties used were not exactly the same.