MECH 4822 (F10) Mid-Term Test Page 1 of 4

University of ManitobaDept. of Mechanical & Manufacturing Engineering

MECH 4822 Numerical Heat Transfer and Fluid Flow (F10)

Mid-Term Test 24 November 2010 Duration: 110 minutes

1. You are permitted to use the following reference material during this test:

• Incropera, F.P., and Dewitt, D.P., Bergman, T.L., and Lavine, A.S., Fundamentals of Heat

and Mass Transfer, 6th Ed. John Wiley and Sons, New York, 2007.• Ormiston, S.J.,MECH 4822 Numerical Heat Transfer and Fluid Flow Supplementary Course

Notes V2.0, Department of Mechanical & Manufacturing Engineering, University of Mani-toba, August 2010.

• Mathematical reference tables.

Extra pages, complete problem solutions, and class notes are not permitted.2. Ask for clarification if any problem statement is unclear to you.3. Clear solutions are required. Marks will not be assigned for answers that require unreasonable

effort for the instructor to decipher.4. The weight of each problem is indicated. The test will be marked out of 100. You may solve the

test problems in any order.Values

1. The governing equation for steady, one-dimensional heat conduction in the constant area, squarecross-section pin fin shown in Figure 1(a) is:50

kAc

d2T

dx2− h∞,2 P (T − T∞,2) = 0 (1)

where P is the perimeter of the fin and Ac is the cross-sectional area. The pin fin has a sidedimension, s, uniform thermal conductivity, k, and is supported by an insulating wall. The finis exposed on the right side of the wall to a fluid with an ambient temperature of T∞,2. Theconvection heat transfer coefficient between the fin surface and the fluid at T∞,2 is h∞,2. The leftend of the fin is exposed to a fluid with an ambient temperature of T∞,1. The convection heattransfer coefficient between the fin end and the fluid at T∞,1 is h∞,1.

(b) typical control volume nomenclature

fin

(a) fin geometry nomenclature

h∞,1

T∞,1

h∞,2

h∞,2

T∞,2

T∞,2

TW TP TE

L

w

s

skx x

(δx)w(δx)p

(δx)e

Ac

Ac

P

Ac = s2P = 4 s

Side ViewEnd View

insulating wall

Figure 1: Nomenclature used in Question #1

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MECH 4822 (F10) Mid-Term Test Page 2 of 4

(a) Using the governing equation (Equation (1)), the finite volume method described in thiscourse, and the nomenclature given in Figure 1(b), derive the algebraic equation for thetypical control volume centred at TP . In your derivation, state your profile assumptions.

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Perform your derivation so that you can show that the algebraic equation is:

aPTP = aWTW + aETE + bP (2)

where

aP = aE + aW + h∞,2P (δx)p aE =kAc

(δx)eaW =

kAc

(δx)wbP = h∞,2P (δx)pT∞,2

Notes: (1) Because the cross-sectional area appears in the governing equation, the integra-tion used in the finite volume method is over only the x co-ordinate direction. (2) You donot need to introduce the equations for P and Ac into the derivation; they are in Figure 1(a)for reference purposes.

(b) Using the grid and problem specifications shown in Figure 2, determine the algebraic equa-tions for nodal temperatures T1, to T4. When doing this, clearly derive and simplify theboundary condition equation for TL and substitute it into the equation for T1. Use all tem-perature values in degrees Celsius. Notes: (1) The grid spacing is uniform within the wallregion and non-uniform elsewhere in the fin. (2) The nodes are centred within each controlvolume. (3) Do not derive the algebraic equations for T5 and TR.

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(a) grid

(b) problem specifications

x

x = 0 x = w x = w + 0.2L x = w + 0.5L x = w + L

k = 150 [W/m ·K]

L = 0.25 [m]h∞,1 = 240 [W/m2·K] T∞,1 = 20 [◦C]

h∞,2 = 60 [W/m2·K] T∞,2 = 126 [◦C]s = 0.02 [m]

w = 0.05 [m]

TR

T1 T2

T3 T4 T5

TL

wall

wall

qf

Figure 2: Grid and problem specifications for Question #1

(c) Given that T5 = 122.340 [◦C], solve the equation set in part (b) for T1 to T4 using theTDMA (Tri-Diagonal Matrix Algorithm). Show your work. Keep five significant figures inyour calculations.15

(d) Using the numerical solution, calculate the heat transfer rate though the fin, qf .2

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MECH 4822 (F10) Mid-Term Test Page 3 of 4

2. Figure 3 shows a typical corner (one quarter) of a square thick-walled duct that has specified innerand outer surface temperatures. The computational grid to be used (with control volumes fornodal temperatures T1 through T7) is also shown in the figure. The material thermal conductivityis k = 10 [W/m ·K], the inner surface temperature is held at Ti = 110 [◦C], and the outer surfacetemperature is held at To = 10 [◦C]. The values of the grid spacings are shown in Figure 3.

The temperature field is governed by the equation for steady two-dimensional heat conductionwith no volumetric energy generation sources. Calculations are to be performed on a per-unit-depth basis.50

T1 T2 T3 T4

T5

T6

T7

Ti To

To

x

y

k

Ti = 110 [◦C]

To = 10 [◦C]

k = 10 [W/m · K]

∆yA

∆yA

∆yB

∆yB

∆xA∆xA

∆xB∆xB

∆xA = 0.1 [m]∆yA = 0.1 [m]

∆xB = 0.2 [m]∆yB = 0.2 [m]

symmetry

symmetry

Figure 3: Domain definition and grid system for Question #2.

(a) Use the energy balance finite difference method to obtain the appropriate algebraicequations to determine the values of all the nodal temperatures (T1 through T7) in thedomain shown in Figure 3. Show your work in the formation and in the solution of thenecessary algebraic equations.

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(b) If the value of the thermal conductivity were to change, would the values of the nodaltemperatures (T1 through T7) change? Explain why or why not.

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(c) Calculate the energy transfer rate per unit depth (in [W/m]) into the surface at Ti for thenumerical solution, Q′

num, using the nodal temperatures computed in part (a).12

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MECH 4822 (F10) Mid-Term Test Page 4 of 4

(d) Compare the value of Q′

num from part (c) with the value of total rate of energy transferper unit depth based on a conduction shape factor analysis, Q′

csf , by calculating the truerelative error (with Q′

csf (Equation (3)) as the “true” value).4

For the geometry shown in Figure 4, the total heat transfer rate per unit depth between thesurfaces at Ti and To is:

Q′

csf = S ′ k (Ti − To) (3)

where

S ′ =2 π

0.93 ln(

0.948ab

) (4)

Ti

To

a

a

b

b

Figure 4: Geometry for conduction shape factor used in Equation (3) in Question #2

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