Sm8-008.pdf

PROBLEM 8.8 KNOWN: Velocity and temperature profiles for laminar flow in a parallel plate channel. FIND: Mean velocity, um, and mean (or bulk) temperature, Tm, at this axial position. Plot the velocity and temperature distributions. Comment on whether values of um and Tm appear reasonable. SCHEMATIC:

ASSUMPTIONS: (1) Laminar incompressible flow, (2) Constant properties. ANALYSIS: The prescribed velocity and temperature profiles (m/s and °C, respectively) are

2 u ( y ) = 0.75 ⎡1 − ( y yo ) ⎤ ⎢⎣ ⎥⎦

T ( y ) = 5.0 + 95.66 ( y yo ) − 47.83 ( y yo ) (1,2) 2

4

The mean velocity, um, follows from its definition, Eq. 8.7,

& = ρ Ac u m = ρ ∫ m

Ac

u ( y ) ⋅ dAc

where the flow cross-sectional area is dAc = 1⋅dy, and Ac = 2yo, 1 1 +y um = u ( y ) ⋅ dy = u ( y )dy Ac Ac 2yo − yo

∫

um =

∫

(3)

+1 1 2 ⋅ yo ∫ 0.75 ⎡1 − ( y yo ) ⎤ d ( y yo ) −1 ⎣⎢ ⎦⎥ 2yo

{

}

+1

3 u m = 1 2 0.75 ⎡( y yo ) − 1 3 ( y yo ) ⎤ ⎣⎢ ⎦⎥ −1

u m = 1 2 × 0.75{[1 − 1 3] − [ −1 + 1 3]} = 1 2 × 0.75 × 4 3 = 2 3 × 0.75 = 0.50 m s

<

The mean temperature, Tm, follows from its definition, Eq. 8.25, & v Tm & = ρ Ac u m E& t = mc where m

u ( y ) ⋅ T ( y ) dAc Ac Hence, substituting velocity and temperature profiles, + yo 1 Tm = u ( y ) ⋅ T ( y ) dy u m A c − yo

ρ A c u m cp Tm = ρ cp ∫

∫

Tm =

Tm =

1

( 0.5 m s ) 2yo

yo

∫−1 {0.75 ⎡⎣1 − ( y +1

{⎡5 ( y y ) + 31.89 ( y y ) 0.5 × 2 ⎣ 0.75

o

Tm =

o

3

yo )

− 9.57 ( y y o )

(4)

}{5.0 + 95.66 ( y y )

2⎤

⎦

5⎤

o

− 47.83 ( y y o )

4

}d (y y

− ⎡1.67 ( y y o ) + 19.13 ( y y o ) − 6.83 ( y y o ) ⎦ ⎣ 3

[ 27.32 − 13.97 ] − [ −27.32 − ( −13.97 )]} = 20.0o C { 0.5 × 2 0.75

2

5

⎦}

7⎤

o)

+1 −1

< Continued...

PROBLEM 8.8 (Cont.)

Temperature, T(y) (C)

Velocity, u(y) (m/s)

The velocity and temperature profiles along with the um and Tm values are plotted below. 1 0.8 0.6 0.4 0.2 0 -1

-0.6

-0.2

0.2

0.6

1

Dimensionless coordinate, x/xo Velocity profile, u(y) Mean velocity, um = 0.5 m/s

60 50 40 30 20 10 0 -1

-0.6

-0.2

0.2

0.6

1

Dimensionless coordinate, y/yo Temperature profile, T(y) Mean temperature, Tm = 20 C

For the velocity profile, the mean velocity is 2/3 that of the centerline velocity, um = 2u(0)/3. Note that the areas above and below the um line appear to be equal. Considering the temperature profile, we’d expect the mean temperature to be closer to the centerline temperature since the velocity profile weights the integral toward the centerline. COMMENTS: The integrations required to obtain um and Tm, Eqs. (3) and (4), could also be performed using the intrinsic function INTEGRAL (y,x) in the IHT Workspace.