Fundamentals of heat and mass transfer f PDF

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PROBLEM 7.1 KNOWN: Temperature and velocity of fluids in parallel flow over a flat plate. FIND: (a) Velocity and thermal boundary layer thicknesses at a prescribed distance from the leading edge, and (b) For each fluid plot the boundary layer thicknesses as a function of distance. SCHEMATIC:

ASSUMPTIONS: (1) Transition Reynolds number is 5

105.

PROPERTIES: Table A.4, Air (300 K, 1 atm): = 15.89 10-6 m2/s, Pr = 0.707; Table A.6, Water (300 K): = / = 855 10-6 N s/m2/997 kg/m3 = 0.858 10-6 m2/s, Pr = 5.83; Table A.5, Engine Oil (300 K): = 550 10-6 m2/s, Pr = 6400; Table A.5, Mercury (300 K): = 0.113 10-6 m2/s, Pr = 0.0248. ANALYSIS: (a) If the flow is laminar, the following expressions may be used to compute respectively, Fluid

5x

t

1/ 2 Re x

1/ 3

Air Water Oil Mercury

Pr

where Re x

u x

1m s 0.04 m

Rex

0.04 m2 s

and

(mm)

2517 4.66 104 72.7 3.54 105

3.99 0.93 23.5 0.34

t

t,

(mm)

<

4.48 0.52 1.27 1.17

(b) Using IHT with the foregoing equations, the boundary layer thicknesses are plotted as a function of distance from the leading edge, x. 10

5

BL thickness, deltat (mm)

BL thickness, delta (mm)

8 6 4 2 0

4 3 2 1 0

0

10

20

30

Distance from leading edge, x (mm) Air Water Oil Mercury

40

0

10

20

30

40

Distance from leading edge, x (mm) Air Water Oil Mercury

COMMENTS: (1) Note that > t for water, >> t for oil, and < t for mercury. As t for air, expected, the boundary layer thicknesses increase with increasing distance from the leading edge. (2) The value of t for mercury should be viewed as a rough approximation since the expression for / was derived subject to the approximation that Pr > 0.6.

t

PROBLEM 7.2 KNOWN: Temperature and velocity of engine oil. Temperature and length of flat plate. FIND: (a) Velocity and thermal boundary layer thickness at trailing edge, (b) Heat flux and surface shear stress at trailing edge, (c) Total drag force and heat transfer per unit plate width, and (d) Plot the boundary layer thickness and local values of the shear stress, convection coefficient, and heat flux as a function of x for 0 x 1 m. SCHEMATIC:

ASSUMPTIONS: (1) Critical Reynolds number is 5

105, (2) Flow over top and bottom surfaces.

PROPERTIES: Table A.5, Engine Oil (Tf = 333 K): W/m K, Pr = 1081.

= 864 kg/m3,

10-6 m2/s, k = 0.140

= 86.1

ANALYSIS: (a) Calculate the Reynolds number to determine nature of the flow,

Re L

u L

0.1m s 1m

1161

86.1 10 6 m 2 s

Hence the flow is laminar at x = L, from Eqs. 7.19 and 7.24, and 1/ 2 1/ 2 5L ReL 5 1m 1161 0.147 m 1/ 3 1/ 3 Pr 0.147 m 1081 0.0143m t

< <

(b) The local convection coefficient, Eq. 7.23, and heat flux at x = L are

k 1/ 2 1/ 3 0.332 Re L Pr L

hL qx

h L Ts T

0.140 W m K 1/ 2 1/ 3 2 0.332 1161 1081 16.25 W m K 1m

16.25 W m

2

K 20 100 C

1300 W m

2

<

Also, the local shear stress is, from Eq. 7.20, 2 3 u 1/ 2 864 kg m 0.1m s 2 0.664 1161 1/ 2 0.664 Re s,L L

2

2

2 0.0842 N m 2 s,L 0.0842 kg m s (c) With the drag force per unit width given by D 2L s,L where the factor of 2 is included to account for both sides of the plate, it follows that

<

D

2L

u

2

1/ 2

2 1.328 Re L

1m 864 kg m

3

0.1m s

2

/ 2 1.328 1161

1/ 2

0.337 N m

For laminar flow, the average value h L over the distance 0 to L is twice the local value, hL, 2 h L 2h L 32.5 W m K The total heat transfer rate per unit width of the plate is 2 q 2Lh L Ts T 2 1m 32.5 W m K 20 100 C 5200 W m

<

< Continued...

PROBLEM 7.2 (Cont.) (c) Using IHT with the foregoing equations, the boundary layer thickness, and local values of the convection coefficient and heat flux were calculated and plotted as a function of x.

deltax*10, hx*100, -q''x

5000 4000 3000 2000 1000 0 0

0.2

0.4

0.6

0.8

1

Distance from leading edge, x (m) BL thickness, deltax * 10 (mm) Convection coefficient, hx * 100 (N/m^2) Heat flux, - q''x (W/m^2)

COMMENTS: (1) Note that since Pr >> 1, >> t. That is, for the high Prandtl liquids, the velocity boundary layer will be much thicker than the thermal boundary layer. (2) A copy of the IHT Workspace used to generate the above plot is shown below. // Boundary layer thickness, delta delta = 5 * x * Rex ^-0.5 delta_mm = delta * 1000 delta_plot = delta_mm * 10 // Scaling parameter for convenience in plotting // Convection coefficient and heat flux, q''x q''x = hx * (Ts - Tinf) Nux = 0.332 * Rex^0.5 * Pr^(1/3) Nux = hx * x / k hx_plot = 100 * hx // Scaling parameter for convenience in plotting q''x_plot = ( -1 ) * q''x // Scaling parameter for convenience in plotting // Reynolds number Rex = uinf * x / nu // Properties Tool: Engine oil // Engine Oil property functions : From Table A.5 // Units: T(K) rho = rho_T("Engine Oil",Tf) // Density, kg/m^3 cp = cp_T("Engine Oil",Tf) // Specific heat, J/kg·K nu = nu_T("Engine Oil",Tf) // Kinematic viscosity, m^2/s k = k_T("Engine Oil",Tf) // Thermal conductivity, W/m·K Pr = Pr_T("Engine Oil",Tf) // Prandtl number

// Assigned variables Tf = (Ts + Tinf) / 2 Tinf = 100 + 273 Ts = 20 + 273 uinf = 0.1 x=1

// Film temperature, K // Freestream temperature, K // Surface temperature, K // Freestream velocity, m/s // Plate length, m

PROBLEM 7.3 KNOWN: Velocity and temperature of air in parallel flow over a flat plate. FIND: (a) Velocity boundary layer thickness at selected stations. Distance at which boundary layers merge for plates separated by H = 3 mm. (b) Surface shear stress and v( ) at selected stations. SCHEMATIC:

ASSUMPTIONS: (1) Steady flow, (2) Boundary layer approximations are valid, (3) Flow is laminar. 3

PROPERTIES: Table A-4, Air (300 K, 1 atm):

-6

= 1.161 kg/m , = 15.89

10

2

m /s.

ANALYSIS: (a) For laminar flow,

5x

5

1/2 Rex

1/2

u /

x m mm

0.001 0.126

5x 1/2

x1/2

25 m/s/15,89 10 -6 m 2 / s 0.01 0.399

0.1 1.262

Boundary layer merger occurs at x = xm when

0.0015 m

1/2

xm

3.99 10

-3

m

= 1.5 mm. Hence

1/2

1/2

3.99 10 3 x1/2 .

1/2

0.376 m

<

x m 141 mm.

(b) The shear stress is s,x

0.664

u2 /2 Re1/2 x

x m

0.001

2 s,x N/m

6.07

u2 /2 u /

1 / 2 1/2 x

0.01

0.1

1.92

0.664 1.161 kg/m 3 25 m/s 25 m/s/15.89 10 -6 m 2 / s

2

1/2

/2

0.192 1/2

N/m2 .

x

x 1/2

0.61 1/2

The velocity distribution in the boundary layer is v = (1/2) ( u /x) ( df/d - f). At y = , 5.0, f 3.24, df/d 0.991. 1/2 0.5 v 15.89 10 6 m 2 / s 25 m/s 5.0 0.991 3.28 0.0167/x 1/2 m/s. 1/2 x

x m v m/s

0.001 0.528

0.01 0.167

COMMENTS: (1) v L = 0.25 m, the air flow is laminar over the entire heater. For the first strip, q1 = h1 ( L w)(Ts - T ) where h1 is obtained from

h1 h1

k 2 1/ 3 0.664 Re1/ x Pr L 0.0429 W m K 2 m s 0.01m 0.664 6 2 0.01m 43.54 10 m s

q1 53.8 W m

2

K 0.01m 0.2 m 500 25 C

q0 5 q0 4 , q 5 h 0 5 5 L w Ts T

1/ 2

0.683

1/ 3

53.8 W m 2 K

<

51.1W

For the fifth strip, q 5

q5

5h 0 5 4h 0 4

h 0 4 4 L w Ts T

L w Ts T

Hence, with x5 = 5 L = 0.05 m and x4 = 4 L = 0.04 m, it follows that h0 5 = 24.1 W/m2 K and h 0 4 = 26.9 W/m2 K and

q5

5 24.1 4 26.9 W m 2 K 0.01 0.2 m 2 500 25 K 12.2 W 2

.

<

2

Similarly, where h0 10 = 17.00 W/m K and h 0 9 = 17.92 W/m K.

q10

10h 0 10 9h 0 9

L w Ts T

q10

10 17.00 9 17.92 W m

2

K 0.01 0.2 m

2

500 25 K 8.3 W

< Continued...

PROBLEM 7.9 (Cont.) For the entire heater, h 0 25

k

1/ 2

1/ 3

0.664 Re L Pr

0.0429

L and the heat rate over all 25 strips is

0.25

1/ 2

2 0.25

0.664

43.54 10

0.683

6

2

1/ 3

10.75 W m

2

K

<

2

q 0 25 h0 25 L w Ts T 10.75 W m K 0.25 0.2 m 500 25 C 255.3 W (b,c) Using the IHT Correlations Tool, External Flow, for Laminar or Mixed Flow Conditions, and following the same method of solution as above, the heat rates for the first, fifth, tenth and all the strips were calculated for air velocities of 2, 5 and 10 m/s. To evaluate the heat rates for fully turbulent conditions, the analysis was performed setting Res,c = 1 10-6. The results are tabulated below. Flow conditions Laminar Fully turbulent

u (m/s) 2 5 10 2 5 10

q1 (W) 51.1 80.9 114 17.9 37.3 64.9

q5 (W) 12.1 19.1 27.0 10.6 22.1 38.5

q10 (W) 8.3 13.1 18.6 9.1 19.0 33.1

q0-25 (W) 256 404 572 235 490 853

COMMENTS: (1) An alternative approach to evaluating the heat loss from a single strip, for example, strip 5, would take the form q 5 h 5 L w T s T , where h 5 h x 4.5 L or h 5 h x 5 L h x 4 L 2 . (2) From the tabulated results, note that for both flow conditions, the heat rate for each strip and the entire heater, increases with increasing air velocity. For both flow conditions and for any specified velocity, the strip heat rates decrease with increasing distance from the leading edge. (3) The effect of flow conditions, laminar vs. fully turbulent flow, on strip heat rates shows some unexpected behavior. For the u = 5 m/s condition, the effect of turbulent flow is to increase the heat rates for the entire heater and the tenth and fifth strips. For the u = 10 m/s, the effect of turbulent flow is to increase the heat rates at all locations. This behavior is a consequence of low Reynolds number (Rex = 2.3 104) at x = 0.25 m with u = 10 m/s. (4) To more fully appreciate the effects due to laminar vs. turbulent flow conditions and air velocity, it is useful to examine the local coefficient as a function of distance from the leading edge. How would you use the results plotted below to explain heat rate behavior evident in the summary table above? Local coefficient, hx (W/m^2.K)

100 80 60 40 20 0 0

0.02

0.04

0.06

0.08

Distance from the leading edge, x (m) uinf uinf uinf uinf

= 2 m/s, laminar flow = 5 m/s, laminar flow = 10 m/s, laminar flow = 10 m/s, fully turbulent flow

0.1

PROBLEM 7.10 KNOWN: Speed and temperature of atmospheric air flowing over a flat plate of prescribed length and temperature. 5

FIND: Rate of heat transfer corresponding to Rex,c = 10 , 5

5

6

10 and 10 .

SCHEMATIC:

ASSUMPTIONS: (1) Flow over top and bottom surfaces. 3

PROPERTIES: Table A-4, Air (Tf = 348K, 1 atm): 0.0299 W/m K, Pr = 0.700.

= 1.00 kg/m ,

= 20.72

-6

10

2

m /s, k =

ANALYSIS: With Re L

u L

25 m/s

1m

20.72 10 -6m 2 / s

1.21 106

the flow becomes turbulent for each of the three values of Rex,c . Hence, Nu L A

0.037 Re 4/5 A Pr 1/3 L

1/2 0.037 Re4/5 x,c 0.664 Re x,c

Rex,c

10

5

5

5 10

10

6

__________________________________________________________________

A Nu L h L W/m 2 K q W/m where q

2 h LL Ts T

160

871

1671

2272

1641

931

67.9 13,580

49.1 9820

27.8 5560

is the total heat loss per unit width of plate.

COMMENTS: Note that h L decreases with increasing Rex,c , as more of the surface becomes covered with a laminar boundary layer.

PROBLEM 7.11 KNOWN: Velocity and temperature of air in parallel flow over a flat plate of 1-m length. FIND: (a) Calculate and plot the variation of the local convection coefficient, hx(x), with distance for flow conditions corresponding to transition Reynolds numbers of 5 105, 2.5 105 and 0 (fully turbulent), (b) Plot the variation of the average convection coefficient, h x x , for the three flow conditions of part (a), and (c) Determine the average convection coefficients for the entire plate, h L , for the three flow conditions of part (a). SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant surface temperature, and (3) Critical Reynolds depends upon prescribed flow conditions. PROPERTIES: Table A.4, Air (Tf = 300 K, 1 atm): 0.707.

10-6 m2/s, k = 0.0263 W/m K, Pr =

= 15.89

ANALYSIS: (a) The Reynolds number for the plate (L = 1 m) is u L 10 m s 1m 5 Re L 6.29 10 . 6 2 15.89 10 m s Hence, the boundary layer conditions are mixed with Rex,c = 5 5 5 10 x c L Re x,c Re L 1m 0.795 m 5

105,

6.29 10

Using the IHT Correlation Tool, External Flow, Local coefficients for Laminar or Turbulent Flow, hx(x) was evaluated and plotted with critical Reynolds numbers of 5 105, 2.5 105 and 0 (fully turbulent). Note the location of the laminar-turbulent transition for the first two flow conditions.

Local coefficient, hx (W/m^2.K)

100

80

60

40

20

0 0

0.2

0.4

0.6

0.8

1

Distance from the leading edge, x (m) Rexc = 5.0e5, Mixed flow Rexc = 2.5e5, Mixed flow Rexc = 0, Fully turbulent flow

Continued...

PROBLEM 7.11 (Cont.) (b) Using the IHT Correlation Tool, External Flow, Average coefficient for Laminar or Mixed Flow, h x x was evaluated and plotted for the three flow conditions. Note that the change in h x x at the

Average coefficient, hLbar (W/m^2.K)

critical length, xc, is rather gradual, compared to the abrupt change for the local coefficient, hx(x). 100 80 60 40

20 0 0

0.2

0.4

0.6

0.8

1

Distance from the leading edge, x (m) Rexc = 5.0e5, Mixed flow Rexc = 2.5e5, Mixed flow Rexc = 0, Fully turbulent flow

(c) The average convection coefficients for the plate can be determined from the above plot since h L h x L . The values for the three flow conditions are, respectively,

h L 17.4, 27.5 and 37.8 W m 2 K COMMENTS: A copy of the IHT Workspace used to generate the above plots is shown below. // Method of Solution: Use the Correlation Tools, External Flow, Flat Plate, for (i) Local, laminar or turbulent flow and (ii) Average, laminar or mixed flow, to evaluate the local and average convection coefficients as a function of position on the plate. In each of these tools, the value of the critical Reynolds number, Rexc, can be set corresponding to the special flow conditions. // Correlation Tool: External Flow, Plate Plate, Local, laminar or turbulent flow. Nux = Nux_EF_FP_LT(Rex,Rexc,Pr) // Eq 7.23,37 Nux = hx * x / k Rex = uinf * x / nu Rexc =1e-10 // Evaluate properties at the film temperature, Tf. //Tf = (Tinf + Ts) / 2 /* Correlation description: Parallel external flow (EF) over a flat plate (FP), local coefficient; laminar flow (L) for RexRexc, Eq 7.37; 0.6...