Fundamentals of heat and mass transfer 7th edition incropera solutions manual

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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): ν = 15.89 × 10-6 m2/s, Pr = 0.707; Table A.6, Water (300 K): ν = μ/ρ = 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): ν = 550 × 10-6 m2/s, Pr = 6400; Table A.5, Mercury (300 K): ν = 0.113 × 10-6 m2/s, Pr = 0.0248.

ANALYSIS: (a) If the flow is laminar, the following expressions may be used to compute δ and δt, respectively,

δ=

5x

δt =

2 Re1/ x

δ

Fluid

Rex

δ (mm)

δt (mm)

Pr1/ 3

Air Water Oil Mercury

2517 4.66 × 104 72.7 3.54 × 105

3.99 0.93 23.5 0.34

4.48 0.52 1.27 1.17

where u x 1m s ( 0.04 m ) 0.04 m 2 s Re x = ∞ = =

ν

ν

ν

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

5

8

BL thickness, deltat (mm)

BL thickness, delta (mm)

10

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 δ ≈ δt for air, δ > δt for water, δ >> δt for oil, and δ < δt for mercury. As expected, the boundary layer thicknesses increase with increasing distance from the leading edge.

(2) The value of δt for mercury should be viewed as a rough approximation since the expression for δ/δt was derived subject to the approximation that Pr > 0.6.

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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): ρ = 864 kg/m3, ν = 86.1 × 10-6 m2/s, k = 0.140 W/m⋅K, Pr = 1081.

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

u L ReL = ∞ =

ν

0.1m s ×1m 86.1× 10−6 m 2 s

= 1161

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

< <

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

hL =

k 2 Pr1/ 3 = 0.140 W m ⋅ K 0.332 1161 1/ 2 1081 1/ 3 = 16.25 W m 2 ⋅ K 0.332 Re1/ ( ) ( ) L L 1m o

q′′x = h L ( Ts − T∞ ) = 16.25 W m 2 ⋅ K ( 20 − 100 ) C = −1300 W m 2

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Also, the local shear stress is, from Eq. 7.20, 3 ρu2 −1/ 2 = 864 kg m 0.1m s 2 0.664 1161 −1/ 2 τ s,L = ∞ 0.664 Re L ( ) ( )

2

2

<

τ s,L = 0.0842 kg m ⋅ s 2 = 0.0842 N m 2

(c) With the drag force per unit width given by D′ = 2Lτ s,L where the factor of 2 is included to account

for both sides of the plate, it follows from Eq. 7.29 that

(

)

2 D′ = 2L ρ u ∞ 2 1.328 Re −L1/ 2 = (1m ) 864 kg m3 ( 0.1m s ) 1.328 (1161) 2

−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, h L = 2h L = 32.5 W m 2 ⋅ K The total heat transfer rate per unit width of the plate is o q′ = 2Lh L ( Ts − T∞ ) = 2 (1m ) 32.5 W m 2 ⋅ 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, δ >> δt. 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(δ) at selected stations.

SCHEMATIC:

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

3 -6 2 PROPERTIES: Table A-4, Air (300 K, 1 atm): ρ = 1.161 kg/m , ν = 15.89 × 10 m /s.

ANALYSIS: (a) For laminar flow,

δ=

5x Re1/2 x

=

x (m) δ ( mm )

5

( u ∞ /ν )

1/ 2

0.001 0.126

x1/2 = 0.01 0.399

5x1/2

( 25 m/s/15.89 ×10-6 m2 / s )

1/ 2

= 3.99 × 10−3 x1/2 .

0.1 1.262

Boundary layer merger occurs at x = xm when δ = 1.5 mm. Hence 0.0015 m x1/2 = 0.376 m1/2 x m = 141 mm. m = -3 1/2 3.99 × 10 m

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(b) The shear stress is 2 2 /2 0.664 × 1.161 kg/m3 ( 25 m/s ) / 2 ρu2 / 2 ρ u∞ 0.192 = = N/m 2 . τ s,x = 0.664 ∞ = 0.664 1/2 1/ 2 1/ 2 1/2 1/2 Re x x ( u ∞ /ν ) x 25 m/s/15.89 × 10-6 m 2 / s x1/2

(

x (m)

0.001

(

τ s,x N/m 2

) 6.07

0.01

0.1

1.92

0.61

)

(

1/2 The velocity distribution in the boundary layer is v = (1/2) (νu∞/x) (ηdf/dη - f). At y = δ, η ≈ 5.0, f ≈ 3.24, df/dη ≈ 0.991.

1/ 2 0.5 1/2 −6 2

v=

x1/2 x (m) v ( m/s )

(15.89 ×10 m / s × 25 m/s ) (5.0 × 0.991 − 3.28) = ( 0.0167 / x ) m/s.

0.001 0.528

0.01 0.167

0.1 0.053

COMMENTS: (1) v << u∞ and δ << x are consistent with BL approximations. Note, v → ∞ as x → 0 and approximations breakdown very close to the leading edge. (2) Since Re x = 2.22 × 105 , m laminar BL model is valid. (3) Above expressions are approximations for flow between parallel

plates, since du∞/dx > 0 and dp/dx < 0.

)

PROBLEM 7.4 KNOWN: Liquid metal in parallel flow over a flat plate.

FIND: An expression for the local Nusselt number.

SCHEMATIC:

ASSUMPTIONS: (1) Steady, incompressible flow, (2) δ << δt, hence u(y) ≈ u∞, (3) Boundary layer approximations are valid, (4) Constant properties.

ANALYSIS: The boundary layer energy equation is ∂ T ∂ T ∂ 2T

u

∂ x

+ v

∂ y

=α

∂ y2

.

Assuming u(y) = u∞, it follows that v = 0 and the energy equation becomes ∂ T ∂ 2T ∂ T α ∂ 2T u∞ or . =α = ∂ x ∂ x u ∞ ∂ y2 ∂ y2 T(x,0) = Ts, T(x,∞) = T∞.

Boundary Conditions:

Initial Condition: T(0,y) = T∞.

The differential equation is analogous to that for transient one-dimensional conduction in a plane wall, and the conditions are analogous to those of Fig. 5.7, Case (1). Hence the solution is given by Eqs. 5.60 and 5.61. Substituting y for x, x for t, T∞ for Ti, and α/u∞ for α, the boundary layer temperature and the surface heat flux become

T ( x,y ) − Ts T∞ − Ts

q′′s = Hence, with

find

⎡

⎤ ⎥ ⎢⎣ 2 (α x/u ∞ )1/ 2 ⎥⎦ y

= erf ⎢

k ( Ts − T∞ )

(π α x/u ∞ )1/ 2

.

Nu x ≡

hx k

=

q′′s x

( Ts − T∞ ) k

( xu ∞ )1/ 2

1/ 2

⎡ ρ u ∞ x cp μ ⎤ = = ⋅ Nu x = ⎥ 1/ 2 1/ 2 1/ 2 ⎢ μ 1/ 2 k ⎦ π ⎣ π α x/u ( π ∞) ( k/ρ cp ) x

Nu x = 0.564 ( Re x Pr )

1/ 2

= 0.564 Pe1/2

1

<

where Pe = Re ⋅ Pr is the Peclet number.

COMMENTS: Because k is very large, axial conduction effects may not be negligible. That is, the α 2 2 ∂ T/∂x term of the energy equation may be important.

PROBLEM 7.5 KNOWN: Form of velocity profile for flow over a flat plate.

FIND: (a) Expression for profile in terms of u∞ and δ, (b) Expression for δ(x), (c) Expression for Cf,x.

SCHEMATIC:

ASSUMPTIONS: (1) Steady state conditions, (2) Constant properties, (3) Incompressible flow, (4) Boundary layer approximations are valid.

ANALYSIS: (a) From the boundary conditions

u ( x,0 ) = 0 → C1 = 0

u ( x,δ ) = u ∞ → C2 = u ∞ / δ .

and

u = u ∞ ( y/δ ) .

Hence,

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(b)From the momentum integral equation for a flat plate d δ ∫ ( u ∞ − u ) u dy = τ s / ρ dx 0 ν u∞ 2 d ∫ δ ⎛1 − u ⎞ u dy = μ ∂ u ⎞ = u∞ ⎜ ⎟ ⎟ ρ ∂ y ⎠ y =0 δ dx 0 ⎝ u ∞ ⎠ u ∞ 2 d ∫ δ ⎛1 − y ⎞ y dy = ν u ∞ u∞ ⎜ ⎟

dx

2 u∞

0 ⎝

δ⎠ δ

δ ⎤

δ

d ⎡⎢⎛ y2 y3 ⎞ ⎥ = μ u∞ ⎜ ⎟ − δ dx ⎢⎜⎝ 2δ 3δ 2 ⎟⎠ 0 ⎥ ⎣ ⎦

or

ν u ∞ dδ = . δ 6 dx

Separating and integrating, find δ

6ν x ∫ δ dδ = ∫ dx 0 u∞ 0

1/ 2

⎛ 12 ν x ⎞ δ = ⎜ ⎟ ⎝ u∞ ⎠

1/ 2

⎛ ν ⎞ = 3.46 x ⎜ ⎟ ⎝ u∞ x ⎠

= 3.46 x Re-1/2 x .

<

(c) The shear stress at the wall is

τs = μ

∂ u⎞ u μ u ∞ 1/2 Re x =μ ∞ = ⎟ ∂ y ⎠ y =0 δ 3.46 x

and the friction coefficient is

Cf,x =

τs

2 /2 ρ u∞

=

μ

2 −1/2 Re1/2 x = 0.578 Re x . ρ u ∞ x 3.46

COMMENTS: The foregoing results underpredict those associated with the exact solution -1/2 and the cubic profile δ = 4.64 x Re−1/2 , δ = 4.96 x Re-1/2 x , Cf,x = 0.664 Re x x

(

)

Cf,x = 0.646 Re−x1/2 .

)

(

<

PROBLEM 7.6 KNOWN: Velocity and temperature profiles and shear stress-boundary layer thickness relation for turbulent flow over a flat plate.

FIND: (a) Expressions for hydrodynamic boundary layer thickness and average friction coefficient, (b) Expressions for local and average Nusselt numbers.

SCHEMATIC:

ASSUMPTIONS: (1) Steady flow, (2) Constant properties, (3) Fully turbulent boundary layer, (4) Incompressible flow, (5) Isothermal plate, (6) Negligible viscous dissipation, (7) δ ≈ δt.

ANALYSIS: (a) The momentum integral equation is 2 d ∫ δ ⎛ 1 − u ⎞ u dy = τ . ρ u∞ ⎜ ⎟ s dx 0 ⎝ u ∞ ⎠ u ∞ Substituting the expression for the wall shear stress 1/ 7 1/ 7 −1/ 4 ⎛y⎞ ⎤ ⎛y⎞ ⎛ u ∞δ ⎞ 2 ⎢ ⎥ ∫ 1− ⎜ ⎟ ⎜ ⎟ dy = 0.0228 ρ u ∞ ⎜ dx 0 ⎢ ⎝ δ ⎠ ⎥ ⎝ δ ⎠ ν ⎟⎠ ⎝ ⎣ ⎦ 1/ 7 2/7 ⎤ d δ ⎡⎛ y ⎞ d ⎛ 7 y8/7 7 y9/7 ⎞ δ ⎛y⎞ ⎥ dy = ⎜ ⎟ ∫ ⎢⎜ ⎟ −⎜ ⎟ − dx 0 ⎢⎝ δ ⎠ dx ⎜⎝ 8 δ 1/7 9 δ 2/7 ⎟⎠ 0 δ⎠ ⎥ ⎝ ⎣ ⎦

2 d ρ u∞

δ ⎡

d ⎛7 7 ⎞ ⎛ u ∞δ ⎞ ⎜ δ − δ ⎟ = 0.0228 ⎜ ⎟ dx ⎝ 8 9 ⎠ ⎝ ν ⎠ 1/ 4

⎛ ν ⎞ 7 dδ = 0.0228 ⎜ ⎟ 72 dx ⎝ u∞ ⎠

δ −1/ 4

1/ 4

⎛ ν ⎞ 4 × δ 5 / 4 = 0.0228 ⎜ ⎟ 72 5 ⎝ u∞ ⎠ 7

−1/ 4 1/ 4 x

⎛ ν ⎞ 7 δ 1/ 4 dδ = 0.0228 ⎜ ∫ δ ⎟ 72 0 ⎝ u∞ ⎠ 1/ 5

x,

⎛ ν ⎞ δ = 0.376 ⎜ ⎟ ⎝ u∞ ⎠

x 4/5 ,

δ x

∫

dx

0

= 0.376Re −x1/5 .

<

Knowing δ, it follows 2 ⎛ u∞ ⎞ τ s = 0.0228 ρ u ∞ ⎜ ⎟ ⎝ ν ⎠

−1/ 4

⎡0.376 x Re−1/5 ⎤ x ⎥⎦ ⎢⎣

−1/ 4

−1/ 5 ⎡ ⎤ u ⎛u ⎞ = 0.0456 ⎢0.376 ∞ ⎜ ∞ ⎟ Cf,x = x x −1/5 ⎥ 2 /2 ν ⎝ ν ⎠ ⎢⎣ ⎥⎦ ρ u∞

τs

−1/ 4

= 0.0582 Re-1/5 x .

Continued …

PROBLEM 7.6 (Cont.)

The average friction coefficient is then 1 x 1 ⎛u ⎞ ∫ Cf,x dx = 0.0582 ⎜ ∞ ⎟ Cf,x = x 0 x ⎝ ν ⎠ 1 ⎛u ⎞ Cf,x = 0.0582 ⎜ ∞ ⎟ x ⎝ ν ⎠

−1/ 5

−1/ 5 x

∫

0

x −1/5 dx

⎛5⎞ x 4/5 ⎜ ⎟ = 0.073 Re−x 1/5 . ⎝4⎠

<

(b) The energy integral equation for turbulent flow is q′′s d δt h u ( T∞ − T ) dy = ∫ =− ( Ts − T∞ ) . dx 0 ρ cp ρ cp Hence, u∞

d δ t u T − T∞ d δt h 1/ 7 ⎡ 1/ 7 ∫ dy = u ∞ ∫ ( y/δ ) 1 − ( y/δ t ) ⎤ dy = ⎢⎣ ⎥⎦ dx 0 u ∞ Ts − T∞ dx 0 ρ cp

d ⎡ 7 δ t8/7 7 δ t8/7 ⎤ h − ⎢ ⎥ = ρ cp dx ⎢ 8 δ 1/7 9 δ 1/7 ⎥ ⎣ ⎦ or, with ξ ≡ δ t / δ , u∞

u∞

d ⎡7 7 δξ 8/7 − δξ8/7 ⎢ dx ⎣ 8 9

h ⎤ ⎥= ρ c ⎦ p

u∞

d ⎡ 7 8/7 ⎤ h . δξ = ⎢ ⎥ dx ⎣ 72 ρ cp ⎦

Hence, with ξ ≈ 1 and δ /x = 0.376 Re−x 1/5 , 7 ⎛u ⎞ u ∞ ( 0.376 ) ⎜ ∞ ⎟ 72 ⎝ ν ⎠

( )= h

−1/ 5 d x 4/5

ρ cp

dx

k ν u∞ x Re−x 1/5 x α ν

h = 0.0292 ρ cp u ∞ Re−x1/5 = 0.0292 Nu x =

hx = 0.0292 Re4/5 x Pr. k

<

Hence, 1 x 0.0292 Pr ⎛ u ∞ ⎞ ∫ h dx = hx = k⎜ ⎟ x 0 x ⎝ ν ⎠

4/5

∫

x 0

k ⎛ u∞ x ⎞ x −1/5 dx = 0.0292 Pr

h x Nu x = x = 0.037 Re 4/5 x Pr. k

x

⎜ ⎟ ⎝ ν ⎠

4/5

5 4

<

COMMENTS: (1) The foregoing results are in excellent agreement with empirical 1/3 correlations, except that use of Pr instead of Pr, would be more appropriate. This result arose because of the assumption δ ≈ δ t , which is only valid for Pr ≈ 1.

(2) Note that the 1/7 profile breaks down at the surface. For example, ∂ ( u/u ∞ ) ⎞ 1 = δ −1/7 y−6/7 = ∞ ⎟ 7 ∂ y ⎠ y =0

or τs = ∞. Despite this unrealistic characteristic of the profile, its use with integral methods provides excellent results.

PROBLEM 7.7 KNOWN: Parallel flow over a flat plate and two locations representing a short span x1 to x2 where (x2 - x1) << L. FIND: Three different expressions for the average heat transfer coefficient over the short span x1 to x2, h1− 2 .

SCHEMATIC:

ASSUMPTIONS: (1) Parallel flow over a flat plate.

ANALYSIS: The heat rate per unit width for the span can be written as q1′ − 2 = h1− 2 ( x 2 − x1 )( Ts − T∞ )

(1)

where h1− 2 is the average heat transfer coefficient over the span and can be evaluated in either of the following three ways: (a) Local coefficient at x = ( x1 + x 2 ) / 2. If the span is very short, it is reasonable to assume that h1− 2 ≈ h x where h x is the local convection coefficient at the mid-point of the span.

<

(2)

(b) Local coefficients at x1 and x2. If the span is very short it is reasonable to assume that h1− 2 is the average of the local values at the ends of the span, h1− 2 ≈ [ h x1 + h x2 ] / 2.

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(c) Average coefficients for x1 and x2. The heat rate for the span can also be written as q1′ − 2 = q′0 − 2 − q′0 −1

(3) (4)

where the rate q0-x denotes the heat rate for the plate over the distance from 0 to x. In terms of heat transfer coefficients, find h1− 2 ⋅ ( x 2 − x1 ) = h 2 ⋅ x 2 − h1 ⋅ x1 h1− 2 = h 2

x2 x1 − h1 x 2 − x1 x 2 − x1

<

(5)

where h1 and h 2 are the average coefficients from 0 to x1 and x2, respectively.

COMMENTS: Eqs. (2) and (3) are approximate and work better when the span is small and -0.2 -0.5 the flow is turbulent rather than laminar (hx ~ x vs hx ~ x ). Of course, we require that

xc < x1, x2 or xc > x1, x2; that is, the approximations are inappropriate around the transition region. Eq. (5) is an exact relationship, which applies under any conditions.

PROBLEM 7.8 KNOWN: Flat plate comprised of rectangular modules of surface temperature Ts, thickness a and length b cooled by air at 25°C and a velocity of 30 m/s. Prescribed thermophysical properties of the module material.

FIND: (a) Required power generation for the module positioned 700 mm from the leading edge of the plate and (b) Maximum temperature in this module.

SCHEMATIC:

ASSUMPTIONS: (1) Laminar flow at leading edge of plate, (2) Transition Reynolds number of 5 × 5 10 , (3) Heat transfer is one-dimensional in y-direction within each module, (4) q& is uniform within module, (5) Negligible radiation heat transfer.

3 PROPERTIES: Module material (given): k = 5.2 W/m⋅K, cp = 320 J/kg⋅K, ρ = 2300 kg/m ; Table -6 2 A-4, Air ( Tf = ( Ts + T∞ ) / 2 = 360 K, 1 atm ) : k = 0.0308 W/m⋅K, ν = 22.02 × 10 m /s, Pr = 0.698.

ANALYSIS: (a) The module power generation follows from an energy balance on the module surface,

q′′conv = q′′gen

h ( Ts − T∞ ) = q& ⋅ a

q& =

or

h ( Ts − T∞ ) a

.

To select a convection correlation for estimating h, first find the Reynolds numbers at x = L.

u L 30 m/s × 0.70 m ReL = ∞ = = 9.537 ×105.

-6 2 ν 22.02 × 10 m / s Since the flow is turbulent over the module, the approximation h ≈ h x ( L + b/2 ) is appropriate, with

ReL + b/2 =

30 m/s × ( 0.700 + 0.050/2 ) m 22.02 × 10-6 m 2 / s

= 9.877 ×105.

Using the turbulent flow correlation with x = L + b/2 = 0.725 m,

h x 1/3 Nu x = x = 0.0296Re4/5 x Pr k

(

)

4/5

Nu x = 0.0296 9.877 × 105 ( 0.698) = 1640 Nu x k 1640 × 0.0308 W/m ⋅ K h ≈ hx = = = 69.7 W/m 2 ⋅ K. x 0.725 1/ 3

Continued …

PROBLEM 7.8 (Cont.)

Hence,

q& =

69.7 W/m 2 ⋅ K (150 − 25 ) K 0.010 m

= 8.713 × 105 W/m3.

<

(b) The maximum temperature within the module occurs at the surface next to the insulation (y = 0).

For one-dimensional conduction with thermal energy generation, use Eq. 3.42 to obtain

8.713 × 105 W/m3 × ( 0.010 m ) & 2 qa T (0) = + Ts = + 150o C = 158.4oC. 2k 2 × 5.2 W/m ⋅ K 2

<

COMMENTS: An alternative approach for estimating the average heat transfer coefficient for the module follows from the relation

q module = q 0→ L + b − q 0→ L h ⋅ b = hL+b ⋅ ( L + b ) − hL ⋅ L

or

h = hL+ b

L+b L − hL . b b

Recognizing that laminar and turbulent flow conditions exist, the appropriate correlation is

(

)

1/3 Nu x = 0.037Re 4/5 x − 871 Pr With x = L + b and x = L, find

h L + b = 54.79 W/m 2 ⋅ K

and

h L = 53.73 W/m2 ⋅ K.

Hence,

0.750 0.700 ⎤ ⎡ h = ⎢54.79 W/m 2 ⋅ K = 69.7 W/m 2 ⋅ K. − 53.73 ⎥ 0.050 0.05 ⎦ ⎣ which is in excellent agreement with the approximate result employed in part (a).

PROBLEM 7.9 KNOWN: Dimensions and surface temperature of electrically heated strips. Temperature and velocity of air in parallel flow.

FIND: (a) Rate of convection heat transfer from first, fifth and tenth strips as well as from all the strips, (b) For air velocities of 2, 5 and 10 m/s, determine the convection heat rates for all the locations of part (a), and (c) Repeat the calculations of part (b), but under conditions for which the flow is fully turbulent over the entire array of strips.

SCHEMATIC:

ASSUMPTIONS: (1) Top surface is smooth, (2) Bottom surface is adiabatic, (3) Critical Reynolds number is 5 × 105, (4) Negligible radiation.

PROPERTIES: Table A.4, Air (Tf = 535 K, 1 atm): ν = 43.54 × 10-6 m2/s, k = 0.0429 W/m⋅K, Pr = 0.683.

ANALYSIS: (a) The location of transition is determined from ν 43.54 × 10−6 m 2 s x c = 5 × 105 = 5 ×105 = 10.9 m u∞ 2m s Since xc >> L = 0.25 m, the air flow is laminar over the entire heater. For the first strip, q1 = h1 (ΔL × w)(Ts - T∞ ) where h1 is obtained from

h1 =

k 2 1/ 3 0.664 Re1/ x Pr ΔL 1/ 2

⎛ 2 m s × 0.01m ⎞ 0.0429 W m ⋅ K × 0.664 ⎜ ⎟ h1 = ⎜ 43.54 ×10−6 m 2 s ⎟ 0.01m ⎝ ⎠

( 0.683)1/ 3 = 53.8 W m2 ⋅ K

o

q1 = 53.8 W m 2 ⋅ K ( 0.01m × 0.2 m )( 500 − 25 ) C = 51.1W

<

For the fifth strip, q5 = q 0 −5 − q 0 − 4 ,

q5 = h 0 −5 ( 5ΔL × w )( Ts − T∞ ) − h 0 − 4 ( 4ΔL × w )( Ts − T∞ )

q5 = ( 5h 0 −5 − 4h 0 − 4 ) ( ΔL × w )( Ts − T∞ )

Hence, with x5 = 5ΔL = 0.05 m and x4 = 4ΔL = 0.04 m, it follows that h 0 −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 .

<

Similarly, where h 0 −10 = 17.00 W/m2⋅K and h 0 −9 = 17.92 W/m2⋅K.

q10 = (10h 0 −10 − 9h 0 −9 ) ( ΔL × 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,

h0 − 25 =

k

⎛ 2 1/ 3 0.0429 0.664 Re1/ = × 0.664 ⎜ L Pr

L and the heat rate over all 25 strips is

1/ 2

⎞ ⎟ ⎝ 43.54 × 10−6 ⎠

0.25

2 × 0.25

( 0.683)1/ 3 = 10.75 W m 2 ⋅ K

q 0 − 25 = h0 − 25 ( L × w )( Ts − T∞ ) = 10.75 W m 2 ⋅ K ( 0.25 × 0.2 ) m 2 ( 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 Rex,c = 1 × 10-6. The results are tabulated below. o

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 = h5 ( ΔL × w )( Ts − T∞ ) , where h 5 ≈ h x = 4.5ΔL or h5 ≈ ( h x = 5ΔL + h x = 4 ΔL ) 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) 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 = 2 m/s, laminar flow uinf = 5 m/s, laminar flow uinf = 10 m/s, laminar flow uinf = 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

5

6

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

SCHEMATIC:

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

3

-6

2

PROPERTIES: Table A-4, Air (Tf = 348K, 1 atm): ρ = 1.00 kg/m , ν = 20.72 × 10 m /s, k = 0.0299 W/m⋅K, Pr = 0.700.

ANALYSIS: With

u L ReL = ∞ =

ν

25 m/s × 1m 20.72 × 10-6 m 2 / s

= 1.21×106

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

)

(

1/3 Nu L = 0.037 Re 4/5 L − A Pr 1/2 A = 0.037 Re4/5 x,c − 0.664 Re x,c

Rex,c A Nu L

5

6

5×10

10

160 2267

871 1635

1671 926

48.9

27.7

9780

5530

h L W/m 2 ⋅ K

) 67. 8

q′ ( W/m )

13,560

(

5

10

where q′ = 2 h L L ( Ts − T∞ ) 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: Length of isothermal flat plate in parallel flow, L. FIND: Expression for the average heat transfer coefficients for N plates each of length LN = L/N to the average coefficient for the single plate.

SCHEMATIC:

T∞, u∞

T∞, u∞ L LN = L/N (N = 3 is shown)

ASSUMPTIONS: (1) Laminar flow, (2) Constant properties.

ANALYSIS: For the single plate, Equation 7.30 applies Nu L =

h L,1L 2 1/ 3 = 0.664 Re1/ or L Pr k

2 1/ 3 h L,1 = ( k / L ) 0.664 Re1/ L Pr

(1)

For the multiple plates, Nu L, N =

h L, N LN 2 1/ 3 = 0.664 Re1/ where LN = L/N and ReLN = ReL /N LN Pr k

(2a, b, c)

Combining Equations 2a, 2b and 2c yields h L, N =

kN 1/2 0.664 ( ReL /N ) Pr1/3 L

(3)

Dividing Equation 3 by Equation 1 yields

h L, N / h L,1 = N 1/ 2

<

COMMENTS: (1) By breaking the single plate into shorter segments, the average boundary layer thickness is reduced, resulting in an increase of the average heat transfer coefficient. This is an effective strategy for heat transfer enhancement. (2) If the boundary layer over the single plate is not completely laminar, breaking it into shorter segments may or may not result in an increase in the average heat transfer coefficient since the turbulent section of the boundary layer over the single plate may be eliminated. (3) The relationship for completely turbulent flow is h LN / h L = N 1/5 , revealing less sensitivity to the plate length than for laminar conditions.

PROBLEM 7.12

KNOWN: Length of isothermal flat plate in parallel flow, L. FIND: Expression for the average heat transfer coefficients for N plates each of length LN = L/N to the average heat transfer coefficient for the single plate.

SCHEMATIC:

T∞, u∞

T∞, u∞ L LN = L/N (N = 3 is shown)

ASSUMPTIONS: (1) Turbulent flow, (2) Constant properties.

ANALYSIS: Since Rex,c = 0, Equation 7.39 yields A = 0. Therefore, for the single plate, Nu L =

h L,1L = 0.036 ReL4 / 5 Pr1/ 3 or k

h L,1 = ( k / L ) 0.036 ReL4 /5 Pr1/ 3

(1)

For the multiple plates,

Nu L, N =

h L, N LN = 0.037 ReL4N/ 5 Pr1/ 3 where LN = L/N and ReLN = ReL /N k

(2a, b, c)

Combining Equations 2a, 2b and 2c yields h L, N =

kN 4/5 0.037 ( ReL /N ) Pr1/3 L

(3)

Dividing Equation 3 by Equation 1 yields

h L, N / h L,1 = N 1/ 5

<

COMMENTS: (1) By breaking the single plate into shorter segments, the average boundary layer thickness is reduced, resulting in a slight increase of the average heat transfer coefficient. Hence, breaking the plate into shorter lengths results in modest heat transfer enhancement. (2) The relationship for laminar flow is h L, N / h L = N 1/ 2 , revealing more sensitivity to the plate length for laminar conditions.

PROBLEM 7.13 KNOWN: Dimensions and surface temperatures of a flat plate. Velocity and temperature of air and water flow parallel to the plate.

FIND: (a) Average convective heat transfer coefficient, convective heat transfer rate, and drag force when L = 2 m, w = 2 m. (b) Average convective heat transfer coefficient, convective heat transfer rate, and drag force when L = 0.1 m, w = 0.1 m.

Ts = 50°C or 80°C

SCHEMATIC:

u∞ = 5 m/s

u∞ = 5 m/s T∞ = 20°C

w=2m

Air, 1 atm

T∞ = 20°C

w = 0.1 m

Water

L=2m

L = 0.1 m

(a) Air

(b) Water

ASSUMPTIONS: (1) Steady-state conditions, (2) Boundary layer assumptions are valid, (3) Constant properties, (4) Transition Reynolds number is 5 × 105.

PROPERTIES: Using IHT, Air (p = 1 atm, Tf = 35°C = 308 K): Pr = 0.706, k = 26.9 × 10-3 W/m⋅K, ν = 1.669 × 10-5 m2/s, ρ = 1.135 kg/m3. Air (p = 1 atm, Tf = 50°C = 323 K): Pr = 0.704, k = 28.0 × 10-3 W/m⋅K, ν = 1.82 × 10-5 m2/s, ρ = 1.085 kg/m3. Water (Tf = 308 K): Pr = 4.85, k = 0.625 W/m⋅K, ν = 7.291 × 10-7 m2/s, ρ = 994 kg/m3. Water (Tf = 323 K): Pr = 3.56, k = 0.643 W/m⋅K, ν = 5.543 × 10-7 m2/s, ρ = 988 kg/m3.

ANALYSIS: (a) We begin by calculating the Reynolds numbers for the two different surface temperatures: u L 5 m/s × 2 m Re L1 = ∞ = = 5.99 × 105 -5 2 ν1 1.669 × 10 m /s

u∞L 5 m/s × 2 m = = 5.49 × 105 -5 2 ν2 1.82 × 10 m /s Therefore, in both cases the flow is turbulent at the end of the plate and the conditions in the boundary layer are “mixed.”

The average drag coefficient can be calculated from Equation 7.40. For the first case, -1 Cf,L1 = 0.074 Re-1/5 L1 - 1742 Re L1 Re L2 =

= 0.074(5.99 × 105 )-1/5 - 1742(5.99 × 105 )-1 = 2.27 × 10-3

Then FD1 = Cf,L1

1 2 ρu ∞ As 2

= 2.27 × 10-3 ×

= 0.257 N

1 × 1.135 kg/m3 × (5 m/s)2 × 8 m 2 = 0.257 N 2

< Continued…

PROBLEM 7.13 (Cont.)

The average Nusselt number is calculated from Equation 7.38, with A = 871 for a transition Reynolds number of 5 × 105.

1/3 Nu L1 = (0.037 Re4/5 L - 871) Pr

= ⎡0.037(5.99 × 105 ) 4/5 - 871⎤ (0.706)1/3 = 604. ⎣ ⎦ Then h L1 = Nu L1k/L = 604 × 26.9 × 10-3 W/m ⋅ K/2 m = 8.13 W/m 2 ⋅ K

<

q1 = h L1As (Ts - T∞ ) = 8.13 W/m 2 ⋅ K × 8 m 2 × (50°C - 20°C) = 1950 W

<

and

Similarly for Ts = 80°C we find FD2 = 0.227 N, h L2 = 7.16 W/m2·K, q2 = 3440 W

<

(b) Repeating the calculations for water u L 5 m/s × 0.1 m Re L1 = ∞ = = 6.86 × 105 ν 7.291 × 10-7 m 2 /s Re L2 = 9.02 × 105 The flow is turbulent at the end of the plate in both cases.

Cf,L1 = 0.074(6.86 × 105 )-1/5 - 1742(6.86 × 105 )-1 = 2.49 × 10-3 FD1 = 2.49 × 10-3 × 1/2 × 994 kg/m3 × (5 m/s) 2 × 0.02 m 2 = 0.620 N

<

Nu L = ⎡ 0.037(6.86 × 105 ) 4/5 - 871⎤ (4.85)1/3 = 1450 ⎣ ⎦ h L1 = 1450 × 0.625 W/m ⋅ K/0.1 m = 9050 W/m 2 ⋅ K

q1 = 9050 W/m 2 ⋅ K × 0.02 m 2 × (50°C - 20°C) = 5430 W

< <

For the higher surface temperature, FD2 = 0.700 N, h L2 = 12,600 W/m2·K, q2 = 15,100 W

<

COMMENTS: (1) For air, kinematic viscosity increases with increasing temperature. This decreases the Reynolds number which causes the transition to turbulence to move downstream, thereby decreasing the drag force and average heat transfer coefficient. The heat transfer rate increases for the higher surface temperature, however, because of the greater temperature difference between the surface and air. (2) For water, kinematic viscosity decreases with increasing temperature, causing the opposite trends as for air. The heat transfer rate increases dramatically for the higher surface temperature because of the increases in both the heat transfer coefficient and temperature difference. (3) Even though the water flows over a plate that is 400 times smaller, the drag force and heat transfer rate are larger than for air because of the smaller viscosity and greater density, thermal conductivity, and Prandtl number. The discrepancy is particularly great for the hear transfer rate. (4) The problem highlights the importance of carefully accounting for the temperature dependence of thermal properties.

PROBLEM 7.14 KNOWN: Velocity and temperature of water 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, 3 × 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.6, Water (300 K): ρ = 997 kg/m3, μ = 855 × 10-6 N⋅s/m2, ν = μ/ρ = 0.858 × 10-6 m2/s, k = 0.613 W/m⋅K, Pr = 583.

ANALYSIS: (a) The Reynolds number for the plate (L = 1 m) is

u L ReL = ∞ =

ν

2 m s × 1m 0.858 ×10−6 m 2 s

= 2.33 ×106 .

and the boundary layer is mixed with Rex,c = 5 × 105,

(

)

x c = L Re x,c ReL = 1m

5 × 105 2.33 × 106

= 0.215 m

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, 3.0 × 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)

8000

6000

4000

2000

0 0

0.2

0.4

0.6

0.8

1

Distance from the leading edge, x (m) Rexc = 5.0e5, Mixed flow Rexc = 3.0e5, Mixed flow Rexc = 0, Fully turbulent

Continued...

PROBLEM 7.14 (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 critical length, xc, is rather gradual, compared to the abrupt change for the local coefficient, hx(x).

Average cofficient, hxbar (W/m^2.K)

10000

8000

6000

4000

2000

0 0

0.2

0.4

0.6

0.8

1

Distance from the leading edge, x (m) Rexc = 5e5, Mixed flow Rexc = 3e5, 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

h L = 4110, 4490 and 5072 W m 2 ⋅ K COMMENTS: A copy of the IHT Workspace used to generate the above plot 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 7.23,36 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 Rex<Rexc, Eq 7.23; turbulent flow (T) for Rex>Rexc, Eq 7.36; 0.6<=Pr<=60. See Table 7.9. */ // Correlation Tool: External Flow, Plate Plate, Average, laminar or mixed flow.

NuLbar = NuL_bar_EF_FP_LM(Rex,Rexc,Pr) // Eq 7.30, 7.38, 7.39 NuLbar = hLbar * x / k // Changed variable from L to x //ReL = uinf * x / nu //Rexc = 5.0E5 /* Correlation description: Parallel external flow (EF) over a flat plate (FP), average coefficient; laminar (L) if ReL<Rexc, Eq 7.30; mixed (M) if ReL>Rexc, Eq 7.38 and 7.39; 0.6<=Pr<=60. See Table 7.9. */ // Properties Tool - Water: // Water property functions :T dependence, From Table A.6 // Units: T(K), p(bars); xf = 0 // Quality (0=sat liquid or 1=sat vapor); "x" is used as spatial coordinate p = psat_T("Water", Tf) // Saturation pressure, bar nu = nu_Tx("Water",Tf,x) // Kinematic viscosity, m^2/s k = k_Tx("Water",Tf,x) // Thermal conductivity, W/m·K Pr = Pr_Tx("Water",Tf,x) // Prandtl number // Assigned Variables: x=1 uinf = 2 Tf = 300

// Distance from leading edge; 0 <= x <= 1 m // Freestream velocity, m/s // Film temperature, K

<

PROBLEM 7.15 KNOWN: Two plates of length L and 2L experience parallel flow with a critical Reynolds 5 number of 5 × 10 .

FIND: Reynolds numbers for which the total heat transfer rate is independent of orientation.

SCHEMATIC:

ASSUMPTIONS: (1) Plate temperatures and flow conditions are equivalent.

ANALYSIS: The total heat transfer rate would be the same (qL = q2L), if the convection coefficients were equal, h L = h 2L . Conditions for which such an equality is possible may be inferred from a sketch of h L versus ReL.

(

)

For laminar flow ReL < Re x,c , h L α L−1/2 , and for mixed laminar and turbulent flow

( ReL >Rex,c ) , h L = C1L−1/5 − C2L−1. Hence h L varies with ReL as shown, and two possibilities are suggested.

Case (a): Laminar flow exists on the shorter plate, while mixed flow conditions exist on the longer plate.

Case (b): Mixed boundary layer conditions exist on both plates.

In both cases, it is required that h L = h 2L and Re2L = 2 ReL .

Continued …

PROBLEM 7.15 (Cont.)

Case (a): From expressions for h L in laminar and mixed flow

)

(

k 1/2 1/3 k 1/3 Re Pr = 0.037 Re 4/5 2L − 871 Pr L L 2L 4/5 0.664 Re1/2 L = 0.032 Re L − 435.

0.664

5

5

Since ReL < 5 × 10 and Re2L = 2 ReL > 5 × 10 , the required value of ReL may be narrowed to the range 5

2.5 × 10

<

ReL

<

5

5 × 10 .

From a trial-and-error solution, it follows that ReL ≈ 3.2 ×105.

<

Case (b): For mixed flow on both plates

)

(

(

)

k k 4/5 − 871 Pr1/3 0.037 Re4/5 − 871 Pr1/3 = 0.037 Re 2L L L 2L or 4/5 0.037 Re4/5 L − 871 = 0.032 ReL − 435

0.005 Re4/5 L = 436 ReL ≈ 1.50 ×106.

COMMENTS: (1) Note that it is impossible to satisfy the requirement that h L = h 2L if 5

ReL < 0.25 × 10 (laminar flow for both plates).

(2) The results are independent of the nature of the fluid.

<

PROBLEM 7.16 KNOWN: Dimensions of and heat generation rate in thin membrane. Velocity and temperature of air flow parallel to membrane. Air streams above and below membrane are in same, opposite, or orthogonal directions.

FIND: (a) Minimum and maximum local membrane temperatures. Flow configuration that minimizes the membrane temperature. (b) Plot the surface temperature distribution for flow in the same and opposite directions. Find configuration that minimizes spatial temperature gradients. q& ′′s = 100 W/m2

SCHEMATIC:

L

Air u∞ = 2 m/s T∞ = 25°C

L = 150 mm y x

(i) same

(ii) opposite

(iii) orthogonal

ASSUMPTIONS: (1) Steady-state conditions, (2) Boundary layer assumptions hold, (3) Constant properties, (4) Solutions are bounded by constant surface temperature and constant heat flux cases for the opposite and orthogonal flow configurations.

PROPERTIES: Table A-4, Air (Tf ≈ 323 K): ν = 18.20 × 10-6 m2/s, k = 0.0280 W/m⋅K, Pr = 0.704.

ANALYSIS: (a) We begin by calculating the Reynolds number u L 2 m/s × 0.15 m Re L = ∞ = = 1.65 × 104 -6 2 ν 18.20 × 10 m /s Therefore, the flow is laminar.

(a) Top and bottom flows in same direction.

By symmetry, the heat flux from the membrane to the air is 50 W/m2 everywhere for the top and bottom air flows. Since the heat flux is uniform, the local Nusselt number is given by Equation 7.45,

Thus where

1/3 Nu x = 0.453 Re1/2 x Pr hx = cqx-1/2

cq = 0.453(u ∞ /ν)1/2 Pr1/3 k

Then

= 0.453(2 m/s/18.2 × 10-6 m 2 /s)1/2 ×(0.704)1/3 × 0.028 W/m ⋅ K = 3.74 W/m3/2 ⋅ K q′′ q′′ (1) q′′s = h x (Ts - T∞ ) and Ts - T∞ = s = s x1/2 hx cq Continued…

PROBLEM 7.16 (Cont.)

Clearly the minimum temperature occurs at x = 0 and is

<

Tmin = T∞ = 25°C The maximum temperature occurs at x = L and is

<

Tmax = 25°C + 50 W/m2 × (0.15 m)1/2 /3.74 W/m3/2 ⋅ K = 30.2°C

(ii) Top and bottom flows in opposite direction.

The heat flux entering each of the top and bottom flows will no longer be uniform. Near x = 0, where the top flow first encounters the plate, the heat transfer coefficient on the top surface is theoretically infinite, and all the generated heat will enter the top flow. The opposite situation will occur at x = L. We bound the solution by considering Nusselt number correlations for uniform surface temperature and uniform surface heat flux, Equations 7.23 and 7.45. In both cases, the heat transfer coefficient varies as x-1/2, where x is the distance from the leading edge, thus for the top and bottom, hx,t = cx-1/2, hx,b = c(L - x)-1/2 And all of the generated heat is removed by the top and bottom flows: q& ′′ = (h x,t + h x,b )(Ts - T∞ ) q& ′′ q& ′′ = h x,t +h x,b c ⎡ x -1/2 + (L - x)-1/2 ⎤ ⎣ ⎦ The minimum temperature occurs at x = 0 or x = L, and is

Thus

Ts - T∞ =

(2)

Tmin = T∞ = 25°C

<

The maximum temperature occurs where the denominator is minimum: d ⎡ -1/2 x + (L - x)-1/2 ⎤ = 0 ⎦ dx ⎣ 1 1 - x -3/2 + (L - x)-3/2 = 0 2 2 x=L-x x = L/2 At that location Tmax = T∞ +

q& ′′ c2(L/2)-1/2

For uniform surface temperature, cT = 0.332(u ∞ /ν)1/2 Pr1/3 k

= 0.332(2 m/s/18.2 × 10-6 m 2 /s)1/2 × (0.704)1/3× 0.0280 W/m ⋅ K = 2.74 W/m3/2 ⋅ K And

Tmax = 25°C +

100 W/m 2 2.74 W/m3/2 ⋅ K × 2 × (0.15 m/2)-1/2

= 30.0°C

Continued…

PROBLEM 7.16 (Cont.)

For uniform surface heat flux, we previously found cq = 3.74 W/m3/2·K, thus, Tmax = 28.7°C.

<

Therefore, for the opposite flow case, 28.7°C ≤ Tmax ≤ 30.0°C (iii) Top and bottom flows in orthogonal directions.

Here the heat transfer coefficients are given by hx,t = cx-1/2, hy,b = cy-1/2 And So

q& ′′ = c(x -1/2 + y-1/2 )(Ts - T∞ ) q& ′′ Ts - T∞ = -1/2 c x + y-1/2

(

)

The temperature will be minimum along x = 0 or y = 0, where

<

Tmin = T∞ = 25°C The temperature will be maximum along x = y = L, where q& ′′ Tmax = T∞ + c 2 L-1/2

The values of c are the same as previously, therefore we find 30.2°C ≤ Tmax ≤ 32.1°C

<

The surface temperature is minimized when the air streams are in opposite directions, because a small heat transfer coefficient on the top is paired with a large heat transfer coefficient on the

<

bottom, and vice versa. (b) IHT was used to plot Equation (1) and (2) for c = cT or cq. The result is shown below.

31 30

Same direction

Opposite direction, uniform T

T (deg. C)

29 28

Opposite direction, uniform q

27 26 25

0

0.03

0.06

0.09

0.12

0.15

x (m)

The spatial temperature gradients are somewhat less for the opposite flow case.

COMMENTS: To correctly treat the convective heat transfer would require a coupled numerical solution of the thermal energy equation for both boundary layers simultaneously.

PROBLEM 7.17 KNOWN: Temperature, pressure and Reynolds number for air flow over a flat plate of uniform surface temperature.

FIND: (a) Rate of heat transfer from the plate, (b) Rate of heat transfer if air velocity is doubled and pressure is increased to 10 atm.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform surface temperature, (3) Negligible radiation, (4) Re x = 5 × 105. c PROPERTIES: Table A-4, Air (Tf = 348K, 1 atm): k = 0.0299 W/m⋅K, Pr = 0.70.

ANALYSIS: (a) The heat rate is

q = h L ( w × L ) ( Ts − T∞ ) .

4 Since the flow is laminar over the entire plate for ReL = 4 × 10 , it follows that

h L 1/ 2 1/ 3 Nu L = L = 0.664 Re1/2 Pr1/3 = 0.664 ( 40, 000 ) ( 0.70 ) = 117.9.

L k k 0.0299 W / m ⋅ K Hence h L = 117.9 = 117.9 = 17.6 W / m2 ⋅ K L 0.2m and

q = 17.6

W m2 ⋅ K

( 0.1m × 0.2m ) (100 − 50 )o C = 17.6 W.

<

(b) With p2 = 10 p1, it follows that ρ2 = 10 ρ1 and ν2 = ν1/10. Hence

⎛u L⎞ ⎛u L⎞ ReL,2 = ⎜ ∞ ⎟ = 2 ×10 ⎜ ∞ ⎟ = 20 ReL,1 = 8 × 105 ⎝ ν ⎠2 ⎝ ν ⎠1

and mixed boundary layer conditions exist on the plate. Hence

)

(

(

)

h L 1/3 = ⎡0.037 × 8 × 105 4 / 5 − 871⎤ 0.70 1/ 3 Nu L = L = 0.037 Re4/5 − 871 Pr ) ⎢ ⎥ ( L k ⎣ ⎦ Nu L = 961.

Hence,

h L = 961 q = 143.6

W m2 ⋅ K

0.0299 W / m ⋅ K = 143.6 W / m2 ⋅ K 0.2m

( 0.1m × 0.2m ) (100 − 50 )o C = 143.6 W.

<

COMMENTS: Note that, in calculating ReL,2, ideal gas behavior has been assumed. It has also been assumed that k, μ and Pr are independent of pressure over the range considered.

PROBLEM 7.18 KNOWN: Solar cell material dimensions and properties, solar-to-electrical conversion efficiency dependence on silicon temperature, solar irradiation and location where the irradiation is absorbed, air velocity and temperature.

FIND: (a) Electrical power produced and silicon temperature for a L = 1 m long, w = 0.1 m wide solar cell with G = 700 W/m2 with tripped boundary layer, (b) Same as Part (a) but with L = 0.1 m, w = 1 m, (c) Plot of the electrical power produced and the silicon temperature for air velocities in the range 0 ≤ um ≤ 10 m/s for the L = 0.1 m configuration.

SCHEMATIC: Tsur = 25°C 20°C

G = 700 W/m2

0 ≤ um ≤ 10 m/s η = 0.28 – 0.001°C-1 (Tsi)

20°C T∞ = 25°C um, T∞

0.1 G εg = 0.9 T1

Lg = 3 mm

Glass, kg = 1.4 W/m2•K

T2

La = 0.1 mm

T3 = Tsi

Adhesive, ka = 145 W/m2•K Silicon, Tsi

Ls = 0.1 mm

Solder, ks = 50 W/m2•K

Lan = 2 mm

Aluminum nitride, kan = 120 W/m2•K 0.83 G

L = 1 m, w = 0.1 m or L = 0.1 m, w = 1 m

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) One-dimensional heat transfer, (4) Tripped and turbulent boundary layer, (5) Large surroundings, (6) Negligible contact resistances.

PROPERTIES: Table A.4, air (assume Tf = 308 K, p = 1 atm): k = 0.0269 W/m⋅K, ν = 1.669 × 10-5 m2/s, Pr = 0.706.

ANALYSIS: (a) We begin by drawing the thermal circuit for the problem, recognizing that there is no heat transfer downward from the thin silicon layer.

Continued…

PROBLEM 7.18 (Cont.)

Tsur

T∞ Rt,conv = 1/ h A

Rt,rad = 1/hradA

T1 0.1 GA Rt,g = Lg/kgA

T2 Rt,a = La/kaA

0.83 GA ( 1 – η)

T3 = Tsi

The thermal resistances are

R t,g = Lg /k g A = 3 × 10-3 m (1.4 W/m ⋅ K × 1 m × 0.1 m) = 21.43 × 10-3 K/W R t,a = La /k a A = 0.1 × 10-3 m (145 W/m ⋅ K × 1 m × 0.1 m) = 6.897 × 10-6 K/W 2 h rad = ε g σ (T1 + Tsur )(T12 + Tsur )

R t,rad =

1 0.9 × 5.67 × 10 W/m ⋅ K × (T1 + 295 K) × (T12 + (295K) 2 ) × 1 m × 0.1 m -8

2

4

(1)

For the tripped boundary layer, u L 4m/s × 1m Re L = m = = 239.7×103 ν 1.669 × 10-5 m 2 /s From Equation 7.38 1/3 Nu L = 0.037Re 4/5 = 0.037 × ⎡ 239.7 × 103 ⎤ L Pr ⎣ ⎦

0.8

× 0.7061/3 = 662.8

h = Nu L k/L = 662.8 ×0.0269 W/m ⋅ K/1m = 17.82 W/m 2 ⋅ K R t,conv = 1/hA =

1 2

17.82 W/m ⋅ K × 1 m × 0.1 m

= 561.2 × 10−3 K / W

From the thermal circuit, 0.83GA (1 - η) = (T3 - T1 )/(R t,g + R t,a ) or T3 - T1 = (R t,g + R t,a ) 0.83GA (1 - η) T3 – T1 = (21.43 × 10-3 K/W + 6.897 × 10-6 K/W)× 0.83 × 700 W/m2 × 1m × 0.1m × (1-η)

T3 - T1 = 1.245(1 - η)

(2)

We also note from the thermal circuit, Continued…

PROBLEM 7.18 (Cont.)

0.83GA (1 - η) + 0.1G = (T1 - Tsur )/R t,rad + (T1 - T∞ )/R t,conv Since T∞ = Tsur ⎡ 1 1 ⎤ 0.83GA (1 - η) + 0.1G = (T1 - Tsur ) ⎢ + ⎥ R t,conv ⎥⎦ ⎢⎣ R t,rad 0.83GA (1 - η) + 0.1GA T1 - Tsur = ⎡ 1 ⎤ 1 ⎢ R t,rad + R t,conv ⎥ ⎣ ⎦ 0.83 × 700 W/m 2 × 1 m × 0.1 m × (1 - η) + 0.1 × 700 W/m 2 × 1 m × 0.1 m ⎡ 1 ⎤ ⎢ R t,rad + 1.7819 W/K ⎥ ⎣ ⎦ 58.1 W (1 - η) + 7 W T1 - Tsur = (3) ⎡ 1 ⎤ ⎢ R t,rad + 1.7819 W/K ⎥ ⎣ ⎦ T1 - Tsur =

where η = 0.28 – 0.001°C-1 × (T3 – 273)°C

(4)

Equations (1) – (4) may be solved simultaneously to yield η = 0.2353, T3 = Tsi = 44.7°C, T1 = 43.7°C, R t,rad = 1.71 K/W The electric power is P = 0.83GAη = 0.83 × 700 W/m2 × 1 m × 0.1 m × 0.2353 = 13.67 W

< <

(b) For the tripped boundary layer u L 4/ms × 0.1m Re L = m = = 239.7×102 -5 2 ν 1.669 × 10 m /s From Equation 7.38 Nu L = 0.037Re 4/5 Pr1/3 = 0.037 × ⎡ 239.7 × 102 ⎤ ⎣ ⎦

h = Nu L k/L = 105 × 0.0269 W/m ⋅ K R t,conv = 1/hA =

1m

× 0.7061/3 = 105

= 28.25 W/m 2 ⋅ K

1 2

0.8

28.25 W/m ⋅ K × 1 m × 0.1 m

= 354.0 × 10−3 K / W

Proceeding as in Part (a) we find η = 0.242, T3 = Tsi = 38.0°C, T1 = 37.1°C, R t,rad = 1.768 K/W, P = 14.06 W

<

(c) Solving Equations 1 through 4 over the velocity range 0 ≤ um ≤ 10 m/s yields the following behavior Continued…

PROBLEM 7.18 (Cont.)

Silicon Temperature vs. Air Velocity

Silicon Temperature (C)

100

80

60

40

20 0

2

4

6

8

10

8

10

Air Velocity (m/s)

Electric Power vs Air Velocity 15

Power (W)

14

13

12

11

10 0

2

4

6

Air Velocity (m/s)

COMMENTS: (1) Changing the orientation of the solar panel to the L = 0.1 m configuration reduces the temperature of the silicon semiconductor significantly. The influence of the orientation on the electric power would be more pronounced for warmer air temperatures. (2) Decreasing the air velocity results in significantly diminished power output. At very low cross flow velocities, natural convection would become significant and would lead to slightly improved power output relative to that predicted here. (3) Film temperatures for Parts (a) and (b) are 31.9 °C and 28.6 °C, respectively. The assumed value of the film temperature is good.

PROBLEM 7.19 KNOWN: Dimensions of a photovoltaic cell, cooling air velocity and temperature, size of concentrating lens, photovoltaic construction and properties.

FIND: (a) The electric power output and silicon temperature of a system consisting of a 400 mm × 400 mm concentrating lens and a 100 mm × 100 mm photovoltaic cell, (b) Variation of the electric power output and silicon temperature for 100 mm ≤ Llens ≤ 600 mm.

SCHEMATIC: Solar irradiation, G

Llens Tsur = 25°C

Focusing lens

Concentrated irradiation, Gc

L = 100 mm Air um = 5 m/s T∞ = 25°C

Photovoltaic cell

See Problem 7.18 Tsur = 25°C

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) One-dimensional heat transfer, (4) Tripped and turbulent boundary layer, (5) Large surroundings, (6) Negligible contact resistances.

PROPERTIES: Given, Glass: kg = 1.4 W/m·K, εg = 0.90, Adhesive: ka = 145 W/m·K, Solder, ks = 50 W/m·K, Aluminum nitride: kan = 120 W/m·K, Table A.4, air (assume Tf = 70°C): k = 0.02948 W/m⋅K, ν = 2.022 × 10-5 m2/s, Pr = 0.701.

ANALYSIS: (a) We begin by drawing the thermal circuit,

Continued…

PROBLEM 7.19 (Cont.)

Tsur

T∞

Rt,rad,tp = 1/(hrad,tpA)

Rt,conv = 1/(hconvA)

T1

Rt,g = Lg/(kgA)

T2 Rt,a = La/(kaA) qt 0.83 GcA ( 1 – η)

T3 qb Rt,s = Ls/(ksA) T4

Rt,an = Lan/(kanA)

T5 Rt,rad,b = 1/(hrad,bA)

Rt,conv = 1/(hconvA) Tsur

T∞

The thermal resistances are R t,g = Lg /(k g A) = 3 × 10-3 m (1.4 W/m ⋅ K × 0.1 m × 0.1 m ) = 0.2143 K/W

R t,a = La /(k a A) = 0.1 × 10-3 m (145 W/m ⋅ K × 0.1 m × 0.1 m ) = 6.897 × 10-5 K/W R t,s = Ls /(k s A) = 0.1 × 10-3 m ( 50 W/m ⋅ K × 0.1 m × 0.1 m ) = 200 × 10-6 K/W R t,an = Lan /(k a n A) = 2 × 10-3 m (120 W/m ⋅ K × 0.1 m × 0.1 m ) = 1.67 × 10-3 K/W 2 h rad,tp = ε g σ (T1 + Tsur )(T12 + Tsur )

R t,rad,tp =

1 -8

2

4

0.9 × 5.67 × 10 W/m ⋅ K × (T1 + 298 K) × (T12 + (298 K)2 )

(1)

2 h rad,b = ε b σ (T5 + Tsur )(T52 + Tsur )

R t,rad,b =

1 -8

2

4

0.95 × 5.67 × 10 W/m ⋅ K × (T5 + 298 K) × (T52 + (298 K) 2 )

(2)

For the top and bottom tripped boundary layers, u∞L 5 m/s × 0.1 m = = 24.73×103 -5 2 ν 2.022 × 10 m /s From Equation 7.38 Re L =

Continued…

PROBLEM 7.19 (Cont.)

1/3 Nu L = 0.037Re 4/5 = 0.037 × ⎡ 24.73 × 103 ⎤ L Pr ⎣ ⎦

h = Nu L k/L =

0.8

× 0.7011/3 = 107.5

107.5 × 0.02948 W/m ⋅ K = 31.69 W/m 2 ⋅ K 0.1 m

R t,conv = 1/hA = 1

(31.69 W/m 2 ⋅ K × 0.1 m × 0.1 m)

= 3.155K / W

From the thermal circuit, 0.83G c A (1 - η) = 0.83G (Llens /L) 2 A(1 - η) = q t + q b 0.83 × 700 W/m 2 × 42 × 0.1 m × 0.1 m × (1 - η) = q t + q b

9296 W = (q t + q b )/(1 - η)

(3)

qt = (T3 – T1) /(Rt,a + Rt,g) = (T3 – T1) /0.2144 K/W

(4)

where and qb = =

(T1 - Tsur )

R t,rad,tp

(T1 - 298 K)

+

R t,rad,tp

(T1 - T∞ )

+

R t,conv

(T1 - 298 K)

3.155 K/W

Likewise qb = (T3 – T5) /(Rt,s + Rt,an) = (T3 – T5) /1.87 × 10-3 K/W

(5) (6)

and qb =

(T5 - Tsur )

R t,rad,b

+

(T5 - T∞ )

R t,conv

(T - 298 K) + 5 R t,rad,b 3.155 K/W The solar-to-electrical conversion efficiency is =

(T5 - 298 K)

η = 0.28 - 0.001°C-1 × (T3 - 273)°C

(7)

(8)

Equations (1) through (8) may be solved simultaneously to yield Tsi = T3 - 273 = 125.9°C, η = 0.1541, T1 = 390.9 K, T5 = 398.8 K, R t,rad,tp = 11.78 K/W, R t,rad,b = 10.75 K/W, q t = 37.32 W, q b = 41.32 W.

The electric power produced is P = 0.83G (Llens /L) 2 Aη = 0.83 × 700 W/m 2 × 42 × 0.1 m × 0.1 m × 0.1541 = 14.33 W

< Continued…

PROBLEM 7.19 (Cont.)

(b) The IHT Software may be used to investigate the sensitivity of the silicon temperature and the electric power produced in response to the concentrating lense size. Results are shown below.

<

Electric Power vs Concentrator Size 16

Power (W)

12

8

4

0 100

200

300

400

500

600

Concentrator Dimension (mm)

Silicon Temperature vs Concentrator Size 250

Silicon Temperature (C)

200

150

100

50

0 100

200

300

400

500

600

Concentrator Size (mm)

COMMENTS: (1) The electric power output is highly sensitive to the size of the concentrating lens The concentrated irradiation continually increases as the concentrator is made larger until, eventually, the silicon temperature becomes very high and the solar-to-electrical conversion efficiency becomes small. (2) The electric power could be increased if heat sinks and/or liquid cooling could be applied to the solar cell, keeping the silicon temperature low and the conversion efficiency relatively high. (3) The assumed film temperature is a good estimate since Tf,tp = 71.4 °C and Tf,bot = 75.4 °C.

PROBLEM 7.20 KNOWN: Material properties, inner surface temperature and dimensions of roof of refrigerated truck compartment. Truck speed and ambient temperature. Solar irradiation.

FIND: (a) Outer surface temperature of roof and rate of heat transfer to compartment, (b) Effect of changing radiative properties of outer surface, (c) Effect of eliminating insulation.

SCHEMATIC:

ASSUMPTIONS: (1) Negligible irradiation from the sky, (2) Turbulent flow over entire outer surface, (3) Average convection coefficient may be used to estimate average surface temperature, (4) Constant properties.

-6

2

PROPERTIES: Table A-4, air (p = 1 atm, Tf ≈ 300K): ν = 15.89 × 10 m /s, k = 0.0263 W/m⋅K, Pr = 0.707.

ANALYSIS: (a) From an energy balance for the outer surface,

αSGS + q′′conv − E = q′′cond =

Ts,o − Ts,i

R ′′tot 4 = Ts,o − Ts,i αSGS + h T∞ − Ts,o − εσ Ts,o 2R ′′p + R ′′i

(

)

where R ′′p = ( t1 / k p ) = 2.78 × 10 −5 m 2 ⋅ K / W, R i′′ = ( t 2 / k i ) = 1.923 m 2 ⋅ K / W, and with Re L = u ∞ L / ν = 29.2 m / s × 10m / 15.89 × 10 h=

k L

4/5

−6

2

1/ 3

0.037 Re L

Pr

)

2

7

m / s = 1.84 × 10 , =

0.0263 W / m ⋅ K

(

0.037 1.84 × 10

10m

)

7 4/5

( 0.707 )1/ 3 = 56.2 W / m 2 ⋅ K

Hence,

(

2

(

)

0.5 750 W / m ⋅ K + 56.2 W / m ⋅ K 305 − Ts,o − 0.5 × 5.67 × 10

−8

2

4

4

W / m ⋅ K Ts,o =

(

Ts,o − 263K 5.56 × 10

−5

)

2

+ 1.923 m ⋅ K / W

Solving, we obtain

<

Ts,o = 306.8K = 33.8°C Hence, the heat load is

q = ( W ⋅ L ) q′′cnd = ( 3.5m × 10m )

( 33.8 + 10 ) °C = 797 W

<

1.923m 2 ⋅ K / W

(b) With the special surface finish (αS = 0.15, ε = 0.8 ) , Continued …

PROBLEM 7.20 (Cont.)

Ts,o = 300.1K = 27.1°C

<

q = 675.3W

<

(c) Without the insulation (t2 = 0) and with αS = ε = 0.5, Ts,o = 263.1K = −9.9°C

<

q = 90, 630W

<

COMMENTS: (1) Use of the special surface finish reduces the solar input, while increasing radiation emission from the surface. The cumulative effect is to reduce the heat load by 15%. (2) The thermal resistance of the aluminum panels is negligible, and without the insulation, the heat load is enormous.

PROBLEM 7.21 KNOWN: Surface characteristics of a flat plate in an air stream.

FIND: Orientation which minimizes convection heat transfer.

SCHEMATIC:

ASSUMPTIONS: (1) Surface B is sufficiently rough to trip the boundary layer when in the upstream position (Configuration 2).

-6

2

-3

PROPERTIES: Table A-4, Air (Tf = 333K, 1 atm): ν = 19.2 × 10 m /s, k = 28.7 × 10 W/m⋅K, Pr = 0.7.

ANALYSIS: Since Configuration (2) results in a turbulent boundary layer over the entire surface, the lowest heat transfer is associated with Configuration (1). Find

u L ReL = ∞ =

ν

20 m/s ×1m 19.2 × 10-6 m 2 / s

= 1.04 × 106.

Hence in Configuration (1), transition will occur just before the rough surface (xc = 0.48m).

Note that

)

(

4/5 ⎡ ⎤ Nu L,1 = ⎢ 0.037 1.04 ×106 − 871⎥ 0.71/3 = 1366 ⎣ ⎦

(

Nu L,2 = 0.037 1.04 × 106 For Configuration (1): Hence

(

)

4/5

h L,1L k

( 0.7 )1/ 3 = 2139 > Nu L,1 .

= Nu L,1 = 1366.

)

h L,1 = 1366 28.7 ×10−3 W/m ⋅ K /1m = 39.2 W/m 2 ⋅ K and q1 = h L,1A ( Ts − T∞ ) = 39.2 W/m 2 ⋅ K ( 0.5m × 1m )(100 − 20 ) K q1 = 1568 W.

<

PROBLEM 7.22 KNOWN: Dimensions and orientation of a flat plate placed in atmospheric airstream, airstream velocity and temperature, plate temperature.

FIND: The average convection coefficient for the entire surface.

SCHEMATIC: A (smooth)

u∞ = 20 m/s T∞ = 20°C

Ts = 100°C

B (rough) L = 0.5m

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Boundary layer on rough plate is tripped at leading edge, (4) Transition occurs at Rex,c = 500,000.

PROPERTIES: Table A.4, air (Tf = 60°C = 333K, p = 1 atm): ν = 19.24×10-6 m2/s, Pr = 0.702, k = 0.0288 W/m⋅K.

ANALYSIS: For both the smooth and rough sections of the plate, ReL =

u∞ L

ν

=

20m/s × 0.5m = 520,000 19.24 × 10−6 m 2 /s

Therefore, flow over portion A of the plate is nearly all laminar except near the trailing edge.

For portion A, Eq. 7.38 yields hA =

k⎡ 0.0288W/m ⋅ K ⎡ 0.037 ReL4 /5 − 871⎤ Pr1/ 3 = 0.037 × 520,0004 / 5 − 871⎤ 0.7021/ 3 = 26.2 W/m 2 ⋅ K ⎣ ⎦ ⎣ ⎦ L 0.5m

For portion B, Eq. 7.38 yields hB =

k⎡ 0.0288W/m ⋅ K ⎡ 0.037 ReL4 / 5 ⎤ Pr1/ 3 = 0.037 × 520,0004 / 5 ⎤ 0.7021/ 3 = 70.8 W/m2 ⋅ K ⎣ ⎦ ⎣ ⎦ L 0.5m

Therefore, for the entire plate consisting of half portion A and half portion B, h=

hA + hB = 48.5 W/m 2 ⋅ K 2

<

COMMENTS: (1) The value of the average coefficient for orientations 1 and 2 of Problem 7.21 are h1 = 39.2 W/m 2 ⋅ K and h2 = 61.4 W/m 2 ⋅ K , respectively. The average heat transfer coefficient found here is approximately midway between the values associated with the other two orientations. This is coincidental, since the nature of the flow is very different for the three different orientations. (2) In reality, boundary layer development cannot proceed on the smooth and rough parts of the plate independent of the other part of the plate, as implied in the solution above. Rather, there will be an intermediate region that will be characterized by complex flow and heat transfer phenomena. The solution should be viewed as a first approximation of actual heat transfer conditions.

PROBLEM 7.23 KNOWN: Design of an anemometer comprised of a thin metallic strip supported by stiff rods serving as electrodes for passage of heating current. Fine-wire thermocouple on trailing edge of strip.

FIND: (a) Relationship between electrical power dissipation per unit width of the strip in the transverse direction, P' (mW/mm), and airstream velocity u ∞ when maintained at constant strip temperature, Ts; show the relationship graphically; (b) The uncertainty in the airstream velocity if the accuracy with which the strip temperature can be measured and maintained constant is ±0.2°C; (c) Relationship between strip temperature and airstream velocity u ∞ when the strip is provided with a constant power, P' = 30 mW/mm; show the relationship graphically. Also, find the uncertainty in the airstream velocity if the accuracy with which the strip temperature can be measured is ±0.2°C; (d) Compare features associated with each of the operating nodes.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Strip has uniform temperature in the midspan region of the strip, (4) Negligible conduction in the transverse direction in the midspan region, and (5) Airstream over strip approximates parallel flow over two sides of a smooth flat plate.

ANALYSIS: (a) In the midspan region of uniform temperature Ts with no conduction in the transverse direction, all the dissipated electrical power is transferred by convection to the airstream,

P′ = 2h L L ( Ts − T∞ )

(1)

where P′ is the power per unit width (transverse direction). Using the IHT Correlation Tool for External Flow-Flat Plate the power as a function of airstream velocity was determined and is plotted below. The IHT tool uses the flat plate correlation, Eq. 7.30 since the flow is laminar over this velocity range.

Power dissipation, P' (mW/mm)

80

60

40

20

0 0

10

20

30

40

50

Airstream velocity, uinf (m/s)

Continued...

PROBLEM 7.23 (Cont.) (b) By differentiation of Eq. (1), the relative uncertainties of the convection coefficient and strip temperature are, assuming the power remains constant,

ΔTs Δh L =− h Ts − T∞

(2)

Since the flow was laminar for the range of airstream velocities, Eq. 7.30, 2 h L ~ u1/ ∞

Δh L Δu = 0.5 ∞ hL u∞

or

(3)

Hence, the relative uncertainty in the air velocity due to uncertainty in Ts, ΔTs = ±0.2 o C ΔTs Δu ∞ ±0.2o C =2 =2 = ±4% o u∞ Ts − T∞

<

(4)

( 35 − 25) C

(c) Using the IHT workspace setting P ′ = 30 mW/mm, the strip temperature Ts as a function of the airstream velocity was determined and plotted. Note that the slope of the Ts vs. u ∞ curve is steep for low velocities and relatively flat for high velocities. That is, the technique is more sensitive at lower velocities. Using Eq. (4), but with Ts dependent upon u ∞ , the relative uncertainty in u ∞ can be determined.

6

Rel uncertainty in unif (%)

Strip temperature, Ts (C)

55

45

35

4

2

0

25 0

10

20

30

Airstream velocity, uinf (m/s)

40

50

0

10

20

30

40

50

Airstream velocity, uinf (m/s)

(d) For the constant power mode of operation, part (a), the uncertainty in u ∞ due to uncertainty in temperature measurement was found as 4%, independent of the magnitude u ∞ . For the constanttemperature mode of operation, the uncertainty in u ∞ is less than 4% for velocities less than 30 m/s, with a value of 1% around 2 m/s. However, in the upper velocity range, the error increases to 5%.

PROBLEM 7.24 KNOWN: Plate dimensions and initial temperature. Velocity and temperature of air in parallel flow over plates.

FIND: Initial rate of heat transfer from plate. Rate of change of plate temperature.

SCHEMATIC:

ASSUMPTIONS: (1) Negligible radiation, (2) Negligible effect of conveyor velocity on boundary layer development, (3) Plates are isothermal, (4) Negligible heat transfer from sides of plate, (5) 5

Re x,c = 5 × 10 , (6) Constant properties.

PROPERTIES: Table A-1, AISI 1010 steel (573K): kp = 49.2 W/m⋅K, c = 549 J/kg⋅K, ρ = 7832 3 -6 2 kg/m . Table A-4, Air (p = 1 atm, Tf = 433K): ν = 30.4 × 10 m /s, k = 0.0361 W/m⋅K, Pr = 0.688.

ANALYSIS: The initial rate of heat transfer from a plate is q = 2 h As ( Ti − T∞ ) = 2 h L2 ( Ti − T∞ )

With Re L = u ∞ L / ν = 10 m / s × 1m / 30.4 × 10 −6 m 2 / s = 3.29 × 105 , flow is laminar over the entire surface and

(

2 1/ 3 = 0.664 3.29 ×105 Nu L = 0.664 Re1/ L Pr

) (0.688)1/ 3 = 336 1/ 2

h = ( k / L ) Nu L = ( 0.0361W / m ⋅ K /1m ) 336 = 12.1W / m 2 ⋅ K Hence,

q = 2 × 12.1W / m 2 ⋅ K (1m ) ( 300 − 20 ) °C = 6780 W 2

<

Performing an energy balance at an instant of time for a control surface about the plate, − E& out = E& st , we obtain,

ρ δ L2c

dT = − h 2L2 ( Ti − T∞ ) dt i

(

)

2 12.1W / m 2 ⋅ K ( 300 − 20 ) °C dT =− = −0.26°C / s dt i 7832 kg / m3 × 0.006m × 549 J / kg ⋅ K

<

COMMENTS: (1) With Bi = h ( δ / 2 ) / k p = 7.4 × 10 −4 , use of the lumped capacitance method is

appropriate. (2) Despite the large plate temperature and the small convection coefficient, if adjoining plates are in close proximity, radiation exchange with the surroundings will be small and the assumption of negligible radiation is justifiable.

PROBLEM 7.25 KNOWN: Prevailing wind with prescribed speed blows past ten window panels, each of 1-m length, on a penthouse tower.

FIND: (a) Average convection coefficient for the first, third and tenth window panels when the wind speed is 5 m/s; evaluate thermophysical properties at 300 K, but determine suitability when ambient air temperature is in the range -15 ≤ T∞ ≤ 38°C; (b) Compute and plot the average coefficients for the same panels with wind speeds for the range 5 ≤ u ∞ ≤ 100 km/h; explain features and relative magnitudes.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Wind over panels approximates parallel flow over a smooth flat plate, and (4) Transition Reynolds number is Res,c = 5 × 105.

PROPERTIES: Table A.4, Air (Tf = 300 K, 1 atm): ν = 15.89 × 10-5 m2/s, k = 26.3 × 10-3 W/m⋅K, Pr = 0.707.

ANALYSIS: (a) The average convection coefficients for the first, third and tenth panels are

h x − h2x2 h 2 −3 = 3 3 x3 − x 2

h1

h x − h9 x9 h9 −10 = 10 10 x10 − x 9

(1,2,3)

where h 2 = h 2 ( x 2 ) , etc. If Rex,c = 5 × 105, with properties evaluated at Tf = 300 K, transition occurs at

xc =

ν u∞

Re x,c =

15.89 × 10−6 m 2 s × 5 ×105 = 1.59 m 5m s

The flow over the first panel is laminar, and h1 can be estimated using Eq. (7.30).

hx 2 1/ 3 Nu x1 = 1 1 = 0.664 Re1/ x Pr k

(

h1 = ( 0.0263 W m ⋅ K × 0.664 lm ) 5 m s × lm 15.89 × 10−6 m 2 s

) ( 0.707 )1/ 3 = 8.73 W m2 ⋅ K < 1/ 2

The flow over the third and tenth panels is mixed, and h 2 , h 3 , h 9 and h10 can be estimated using Eq. (7.41). For the third panel with x3 = 3 m and x2 = 2 m,

(

)

h x Nu x3 = 3 3 = 0.037 Re4x / 5 − 871 Pr1/ 3 k h3 = ( 0.0263 W m ⋅ K 3m ) 4/5 ⎡ ⎤ 1/ 3 × ⎢ 0.037 5 m s × 3m 15.89 ×10−6 m 2 s − 871⎥ ( 0.707 ) = 10.6 W m 2 ⋅ K ⎣ ⎦

(

)

Continued...

PROBLEM 7.25 (Cont.)

h 2 = ( 0.0263 W m ⋅ K 2m ) 4/5 ⎡ ⎤ 1/ 3 × ⎢ 0.037 5 m s × 2m 15.89 × 10−6 m 2 s − 871⎥ ( 0.707 ) = 8.68 W m 2 ⋅ K ⎣ ⎦

)

(

From Eq. (2),

h 2 −3 =

10.61W m 2 ⋅ K × 3m − 8.68 W m 2 ⋅ K × 2m = 14.5 W m 2 ⋅ K (3 − 2) m

<

Following the same procedure for the tenth panel, find h10 = 11.64 W/m2⋅K and h 9 = 11.71 W/m2⋅K, and h9 −10 = 11.1W m 2 ⋅ K Assuming that the window panel temperature will always be close to room temperature, Ts = 23°C = 296 K. If T∞ ranges from -15 to 38°C, the film temperature, Tf = (Ts + T∞ )/2, will vary from 275 to 310 K. We’ll explore the effect of Tf subsequently.

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(b) Using the IHT Tool, Correlations, External Flow, Flat Plate, results were obtained for the average coefficients h . Using Eqs. (2) and (3), average coefficients for the panels as a function of wind speed were computed and plotted.

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

60

40

20

0 0

20

40

60

80

100

Wind speed, uinf (km/h) First panel Third panel Tenth panel

COMMENTS: (1) The behavior of the panel average coefficients as a function of wind speed can be explained from the behavior of the local coefficient as a function of distance for difference velocities as plotted below.

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

60

40

20

0 0

2

4

6

8

10

Distance from leading edge, x (m) uinf = 5 km/h uinf = 15 km/h uinf = 25 km/h uinf = 50 km/h

Continued...