Natural Convection from Isothermal Spheres

M.M. Yovanovich

NCSPH3.MWS

Natural convection from isothermal spheres. Use the correlation equation based on sphere diameter. Consider the problem where the heat transfer rate is known along with the fluid, air, the ambient temperature , and the sphere diameter. Calculate the sphere surface temperature . The second problem finds the heat transfer rate, when the surface temperature is specified along with all other pertinent system parameters.

> restart:

Air properties correlation equations.

> k:= T->-.2571428571e-7*T^2+.9300000000e-4*T+.6685714286e-3:

> rho:= T->.8997142857e-5*T^2-.9361400000e-2*T+3.168594286:

> cp:= T->.4285714286e-6*T^2-.2260000000e-3*T+1.035857143:

> mu:=T->-.3000000000e-10*T^2+.6654000000e-7*T+.1200000000e-5:

> nu:=T->.1005714286e-9*T^2+.3444000000e-7*T-.3466857143e-5:

> alpha:=T->.1600000000e-9*T^2+.4480000000e-7*T-.5320000000e-5:

> Pr:=T-> -.1700000000e-3*T+.7601000000:

> airprops:= [k=k(T), rho=rho(T), cp=cp(T), mu=mu(T), nu=nu(T), alpha=alpha(T), Pr=Pr(T)]:

Newton's Law of Cooling.

> ncsphere:= Q-h*A*(Tw-Tinfty)=0;

> A:= Pi*Dia^2;

> h:= k(T)*NuD/Dia:

Sphere correlation equation of Churchill for laminar flow.

> NuD:= 2+0.589/(1+(0.469/Pr)^(9/16))^(4/9)*RaD^(1/4):

> Pr:= Pr(T):

> RaD:= g*beta*(Tw-Tinfty)*Dia^3/(alpha(T)*nu(T)):

Specify the system parameters.

> g:= 9.81: beta:= 1/Tinfty:

> Tinfty:= 300: T:= (Tw+Tinfty)/2: Dia:= 100/1000:

Specify the heat transfer rate and solve for the surface temperature .

> Q:= 10.5: Tw:= 'Tw':

> Tw:= evalf(fsolve(ncsphere, Tw, 0..1000), 5);

Check Rayleigh number.

> RaD;

The fluid flow is laminar.

Specify the wall temperature and solve for the heat transfer rate .

> Q:='Q': Tw:='Tw': Tw:= 385:

> Q:= evalf(fsolve(ncsphere, Q, 0..100), 5);

> RaD;

>

The fluid flow is laminar.