T.C. MARMARA UNIVERSITY

ME 705 - Advanced Fluid Dynamics

Third Assignment

Mechanical Engineering Faculty524611017

Due: 30/11/2012

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Assignment - III1) Given Equation:

ρDEDt

=−∂ pui∂ x i

+∂ τ iju i

∂ x j

+ ∂∂ x j

(k ∂T∂ x j

)+giuiΦ=τ ij ( ∂ui

∂ x j)

E=e+( uiui2 )

Shear Stress at the Surface:

τ ij=μ( ∂ ui∂ x j

+∂u j

∂ x i )

e=C pT

q j=−k ( ∂T∂ x j

)

σ=σ ij , ∇ σ=∂ σ ij

∂ x j

[Ref:1]

p= ρgi

gi=0 , g j=9 . 81∴g iui=0, For steady State: Dui/Dt=0

ρDDt (e+(uiui2 ))=−ρgi

∂ ui∂ x i

+Φ−∂ q j

∂ x j

+giui

ρDeDt

+ρ DDt (u iui

2 )=−ρgi∂ ui∂ x i

+Φ−∂ q j

∂ x j

+giui

ρC pDTDt

+0=0+Φ−∂ q j

∂ x j

+0

ρC pDTDt

= ∂∂ x j

(k ∂T∂ x j

)+Φ

1

2) Assuming temperature changes only in the y-direction find and plot the temperature profile if the fluid is :

(a) Air,

(b) Water,

(c) Mercury,

(d) SAE 30 oil.

Figure 1. Upper plate is moving with a velocity of 10 m/s, while bottom plate is stationary.

Flow Assumptions: Steady-state conditions, fully developed Couette flow and incompressible fluid with constant properties.

Table 1. Properties for the fluids in the second problem. For mercury, interpolation was used at 20°C.

At 20°C, 1 atm μ, Ns/m2 k, W/mK Ref:(a) air 0.000018 0.0263 [1](b) water 0.001003 0.613 [1](c) mercury 0.001564 8.449 [2](d) SAE 30 oil 0.29 0.12 [3]

The linear velocity profile for Couette flow is τ=μ (du/dy) = μ (U/L).

(a) Air

τ air=μ( ∂ ui∂ x j

)=μUupperPlate

L=184 .6×10−710m /s

0 .2m=0. 0009 N /m2

(b) Water

τ w=μ( ∂u i

∂ x j)=μ

UupperPlate

L=1.003×10−310m /s

0 .2m=0.05015 N /m2

2

y

x

Mercury- Table A-14

(c) Mercury

τ hg=μ( ∂u i

∂ x j)=μ

UupperPlate

L=1. 5646×10−310m /s

0 . 2m=0.07823 N /m2

(d) SAE 30 oil

τ SAE 30=μ( ∂u i

∂ x j)=μ

UupperPlate

L=0 .29

10m / s0 . 2m

=14 .5N /m2

The viscous dissipation is .hence μϕ = μ(du/dy)2 = μ(U/L) 2.

(a) Air

( μϕ )air=μ ( ∂ui∂ x j

)2

=μ (UupperPlate

L )2

=184 .6×10−7 (10m /s0 . 2m )

2

=0 .045W /m3

(b) Water

( μϕ )w=μ( ∂ ui∂ x j

)2

=μ(UupperPlate

L )2

=1 .003×10−3 (10m /s0 . 2m )

2

=2.5075W /m3

(c) Mercury

( μϕ )hg=μ( ∂ui∂ x j

)2

=μ (U upperPlate

L )2

=1 .5646×10−3(10m /s0 . 2m )

2

=3.9115W /m3

(d) SAE 30 oil

( μϕ )SAE=μ ( ∂ui∂ x j

)2

=μ (U upperPlate

L )2

=0. 29(10m / s0. 2m )

2

=725W /m3

3

The location of the maximum temperature corresponds to ymax=L. Hence Tmax=To+μU2/8k and

T 0( y )=−Ay2+C3 y+C4

where:

sp stands for steel plate with a k of 1.5 W/mK,

A= μ2k fluid

(U upperPlate

L )2

dT 0

dy=−2 Ay+C3

−k fluid

dT 0

dy|y=L=

T 0 ( y )−T sp

Lsp /ksp, Lsp=0

−k fluid

Lsp

k sp(−2 AL+C3)=T 0(L)−T sp=−AL2+C3L+C4−T sp

−AL2+C3L+C4=T sp .UP

−A02+C3 0+C4=T sp .DOWN=10 °C⇒C4=10 ( for all fluids )−AL2+C3L+10=T sp .UP=20 ° C

−[μ2k fluid(U upperPlate

L )2 ](0 .2m)2+C3 (0 .2m)+10=20 °C

C3=10+A (0 . 2m )2

(0 .2m)

4

C4 is constant for all liquids, yet C3 depend on fluid properties. “A” values for each fluid type are:

At 20°C, 1 atm A(a) air 0.855513(b) water 2.045269(c) mercury 0.231486(d) SAE 30 oil 3020.833

(a) Air

C3=10+0 .856 (0 .2m)2

(0 .2m)=50 .1711⇒T 0( y )=−Ay2+C3 y+C4

T 0( y )=−0 .856 y2+50 .2 y+10

(b) Water

C3=10+2.05(0 . 2m)2

( 0. 2m)=50 . 409

T 0( y )=−2 . 05 y2+50 . 4 y+10

(c) Mercury

C3=10+0 .231(0 .2m)2

( 0. 2m)=50 .046

T 0( y )=−0 . 23 y2+50 y+10

(d) SAE 30 oil

C3=10+3020(0 .2m)2

(0 .2m)=654 .167

T 0( y )=−3020 y2+654 .2 y+10

5

Executed Temperature values depending on the layer height between planes are:

°Cy T air T water T mercury T SAE 30 oil0 10.000 10.000 10.000 10.0000.01 10.502 10.504 10.500 16.2400.02 11.003 11.007 11.001 21.8750.03 11.504 11.510 11.501 26.9060.04 12.005 12.013 12.001 31.3330.05 12.506 12.515 12.502 35.1560.06 13.007 13.017 13.002 38.3750.07 13.508 13.519 13.502 40.9900.08 14.008 14.020 14.002 43.0000.09 14.508 14.520 14.502 44.4060.1 15.009 15.020 15.002 45.2080.11 15.508 15.520 15.502 45.4060.12 16.008 16.020 16.002 45.0000.13 16.508 16.519 16.502 43.9900.14 17.007 17.017 17.002 42.3750.15 17.506 17.515 17.502 40.1560.16 18.005 18.013 18.001 37.3330.17 18.504 18.510 18.501 33.9060.18 19.003 19.007 19.001 29.8750.19 19.502 19.504 19.500 25.2400.2 20.000 20.000 20.000 20.000

0 0.05 0.1 0.15 0.20

5

10

15

20

25

f(x) = − 0.8555133079858 x² + 50.171102661597 x + 10R² = 1

T air

T airPolynomial (T air)

6

?

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.20

5

10

15

20

25

f(x) = − 2.04526916802876 x² + 50.4090538336057 x + 9.99999999999999R² = 1

T water

T waterPolynomial (T water)

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.20

5

10

15

20

25

f(x) = − 0.231486303635233 x² + 50.046297260727 x + 10R² = 1

T mercury

T mercuryPolynomial (T mercury)

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.205

101520253035404550

f(x) = − 3020.83333333333 x² + 654.166666666667 x + 9.99999999999997R² = 1

T SAE 30 oil

T SAE 30 oilPolynomial (T SAE 30 oil)

7

TEMPERATURE PROFILES

5 10 15 20 25 30 35 40 45 500.00

0.05

0.10

0.15

0.20

T air

5 10 15 20 25 30 35 40 45 500.00

0.05

0.10

0.15

0.20

T water

5 10 15 20 25 30 35 40 45 500.00

0.05

0.10

0.15

0.20

T mercury

5 10 15 20 25 30 35 40 45 500.00

0.05

0.10

0.15

0.20

T SAE 30 oil

8

?

References:

1) www.itiomar.it/pubblica/dispense/MECHANICAL%20ENGINEERING%20HANDBOOK/Ch03.pdf

2) http://highered.mcgraw-hill.com/sites/dl/free/0073398128/835451/App1.pdf

3) http://www.lytron.com/Tools-and-Technical-Reference/Thermal-Reference/Material-Properties

4) http://courses.washington.edu/overney/privateChemE340/HW340_Solutions/HW5.pdf

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