ME 353 Heat Transfer 1

M.M. Yovanovich

FEF97P3.MWS

Solution of Problem 3 of Final Examination, Fall 1997.

Assumptions:

1. Forced convection, fully-developed internal flow.

2. Isothermal wall.

3. Constant properties.

> restart:

System parameters

> sys:= (D = 25/1000, U = 1, Tb = 30, Ts = 50,
mu = 0.798e-3, mu_s = 0.547e-3, nu = 0.8012e-6,
k = 0.615, Pr = 5.42);

(a) Compute Reynolds nunber

> ReD:= U*D/nu;
ReDval:= evalf(subs(sys, ReD), 4);

The flow is turbulent because ReD > 2300.

(b) Compute Nusselt number and heat transfer coefficient using Petukov relations

> unassign('f');
h:= NuD*k/D;
NuD_Petukov:= (f/8)*ReD*Pr/(1.07 + 12.7*sqrt(f/8)*(Pr^(2/3) - 1))*(mu/mu_s)^n;
n:= 0.11;

> f:= (1.82*log10(ReD) - 1.64)^(-2);

> NuD1:=
evalf(subs(ReD = 31200, NuD_Petukov), 5);

> NuD1val:= evalf(subs(sys, ReD = 31200, NuD1), 5);

> h1:= evalf(subs(sys, NuD = NuD1val, h), 5);

(c) Compute Nusselt number and heat transfer coefficient using Dittus-Boelter relation

> unassign('n');
NuD_DittusBoelter:= 0.023*ReD^(0.8)*Pr^n;
n:= 0.4;

> NuD2val:=
evalf(subs(ReD = 31200, sys, NuD_DittusBoelter), 5);

> h2:= evalf(subs(sys, NuD = NuD2val, h), 5);

>

Comparison of the two values of the heat transfer coefficient.

> percent_diff:= evalf((NuD2val - NuD1val)/NuD1val*100, 4);

>

The values of the heat transfer coefficient differ by approximately - 13.5%. The Dittus-Boelter value is smaller than the Petukov value.

(d) Compute the heat transfer rate to the water

> unassign('A, P, h, rho, mu, cp, alpha, mdot'):
Q:= mdot*cp*(Ts - Tbi)*(1 - exp(-h*P*L/(mdot*cp)));

> mdot:= rho*U*A; A:= Pi/4*D^2; P:= Pi*D;

> Qval:=
evalf(subs(sys, h = h1, L = 3, Tbi = 20, Q), 6);

Require the properties rho and cp which are not given. Use other properties

such as nu = mu/rho, alpha = k/(rho cp), Pr = nu/alpha to get rho cp product.

> rho:= mu/nu; alpha:= nu/Pr; cp:= k/(rho*alpha);

> Qval:= evalf(subs(sys, Qval), 6);

>