ME 353 Heat Transfer 1

M.M. Yovanovich

FCINT2.MWS

Internal forced convection in a square duct.

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Laminar forced convection in a square tube.
The wall condition is isoflux.

The flow is fully-developed hydraulically and thermally

developing.

The fluid is engine oil at temperature T = 340 K.

The side dimension of the square tube is 15 mm.

The mean velocity is 0.5 m/s and the distance from

the tube inlet is 4 m.

Determine the average Nusselt number and the average

heat transfer coefficient.

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> restart:

> oilprops:=
(rho = 859.9, cp = 2.076, mu = 5.31e-2, nu = 61.7e-6,
kf = 0.139, alpha = 0.779e-7, Pr = 793);

System parameters.

> syspar:= (s = 15/1000, Um = 0.5, x = 4, Tw = 380, Tinlet = 320);

Shah (1975) correlation equations for laminar internal flow.

Isothermal wall.

> h:= kf*Nu_Dh/Dh;

> Nu_m_UWF1:= 1.953/xstar^(1/3); #For xstar < 0.03

> Nu_m_UWF2:= 4.354 + 0.0722/xstar; #For xstar => 0.03

> Dh:= 4*A/P; #Hydraulic diameter

> A:= s^2; P:= 4*s;

Dimensionless axial distance from tube entrance.

> xstar:= x/(Dh*ReDh*Pr);

> ReDh:= Um*Dh/nu;

Compute the hydraulic diameter, the mass flow rate, the Reynolds number,

the dimensionless distance, the Nusselt number and the heat transfer coefficient.

> Dh1:= evalf(subs(syspar, Dh), 4);

> mdot1:= evalf(subs(oilprops, syspar, rho*Um*s^2), 5);

> ReDh1:=
evalf(subs(syspar, oilprops, ReDh), 6); #Laminar flow

> xstar1:= evalf(subs(syspar, oilprops, xstar), 6);

> NuDh1:= evalf(subs(xstar = xstar1, Nu_m_UWF1), 6);

> h1:= evalf(subs(oilprops, Nu_Dh = NuDh1, syspar, h), 5);

>

The mean Nusselt number is much larger than the fully-developed value

which is equal to 3.61 for an isoflux square tube (see Table 8.1).

The thermal boundary layer is developing; therefore it is still quite small

relative to the hydraulic diameter.