Review for Final Exam Outline Final Exam Basic Equations Heat ...

Final Review

May 16, 2006

Outline

Review for Final Exam

• • • •

Larry Caretto Mechanical Engineering 375

Basic equations, thermal resistance Heat sources Conduction, steady and unsteady Computing convection heat transfer

Heat Transfer

– Forced convection, internal and external – Natural convection

• Radiation properties • Radiative Exchange

May 16, 2007

2

Basic Equations

Final Exam • Wednesday, May 23, 3 – 5 pm • Open textbook/one-page equation sheet • Problems like homework, midterm and quiz problems • Cumulative with emphasis on second half of course • Complete basic approach to all problems rather than finishing details of algebra or arithmetic

• Fourier law for heat conduction (1D) k (T1 − T2 ) kA(T1 − T2 ) q& = or Q& = q&A = L L • Convection heat transfer Q& conv = hAs (Ts − T∞ )

• Radiation (from small object, 1, in large enclosure, 2)

(

Q& rad ,1→2 = A1ε 1σ T14 − T24

)

3

Heat Generation • Various phenomena in solids can generate heat • Define e& gen as the heat generated per unit volume per unit time

Rectangular Energy Balance

Figure 2-21 from Çengel, Heat and Mass Transfer

ρc p

Stored = energy

ρc p

5

ME 375 – Heat Transfer

∂q& y ∂q& z ∂q& ∂T =− x − − + e&gen ∂t ∂x ∂y ∂z heat inflow – heat outflow

+

heat generated

∂T ∂ ∂T ∂ ∂T ∂ ∂T = k + e&gen + k + k ∂t ∂x ∂x ∂y ∂y ∂z ∂z ∂T Uses q&ξ = −k ∂ξ Fourier Law

6

1

Final Review

May 16, 2006

Cylindrical Coordinates

Spherical Coordinates

∂T = ∂t 1 ∂ ∂T 1 ∂ ∂T kr k + r ∂r ∂r r 2 ∂φ ∂φ ∂ ∂T + k + e&gen ∂z ∂z

∂T = ∂t 1 ∂ 2 ∂T kr + ∂r r 2 ∂r 1 ∂T ∂ sin k + θ ∂θ r 2 sin θ ∂θ 1 ∂ ∂T k + e&gen 2 2 ∂φ ∂φ r sin θ ρc p

ρc p


dQ& r = q& r dA = − k

∂T rdφdz ∂r


7

8

Plot of (T - T0)/(TL - T0) for Heat Generation in a Slab

1-D, Rectangular, Heat Generation

2

1.8

T = T0 −

e&genx2 2k

+

e&genxL 2k

(T − T )x − 0 L L

⎡ e&gen 2x e&genL (T0 − TL ) ⎤ dT = −k ⎢− + − q& = −k ⎥ 2k L ⎦ dx ⎣ 2k q& =

1.6

TemperatureDifference Ratio

• Temperature profile for generation with T = T0 at x = 0 and T = TL at x = L

1.4

H=0 1.2

H = .01 H = .1 H=1

1

H=2 H=5

0.8

H = 10 0.6

2

H=

0.4

0.2

e&gen(2x − L) k (T0 − TL ) + 2 L

L e&gen

k (TL − T0 )

0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x/L

9

10

Slab With Heat Generation

Thermal Resistance

Both boundary temperatures = TB 2.4

2

H =

2.2

L e& gen

2

kTB H=0

T / TB

H = .01 1.8

H = .1 H=1

1.6

H=2 H=5

1.4

H = 10

• Conduction k A(T1 − T2 ) T −T ⇒ Q& = 1 2 Q& = L Rcond • Convection

(

Q& = hA Ts − T f

)

⇒ Q& =

Ts − T f Rconv

⇒ Rcond =

L kA

⇒ Rconv =

1 hA

• Radiation

1.2 1 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Rrad =

1

Dimensionless Distance, x/L 11


(

1

A1F12σ T13 + T23 + T22T1 + T12T2

1

)= Ah

1 rad 12

2

Final Review

May 16, 2006

Composite Cylindrical Shell

Composite Materials II


1 1 = h2 A4 h2 2πr4 L

13

1 ⎛ r2 ⎞ 1 ⎛ r3 ⎞ ln⎜ ⎟ ln⎜ ⎟ 1 1 = k1L ⎜⎝ r1 ⎟⎠ k2 L ⎜⎝ r2 ⎟⎠ h1 A1 h1 2πr1L

Fin Results ⇒ T − T∞ = (Tb − T∞ )e

− x hp kAc

for uniform cross section

Q& x =0 = Ac q& x =0 = kAc hp (Tb − T∞ )

A fin = Lc p

• Heat transfer at end (Lc = A/p) θ = T − T∞ = θb

cosh m( Lc − x ) cosh m( Lc − x) = (Tb − T∞ ) cosh mLc cosh mL

Q& x =0 = kAc hp (Tb − T∞ ) tanh mL

15

Overall Fin Effectiveness

Q& fin h(η fin A fin + Aunfin )(Tb − T∞ ) = & Qno fin hAno fin (Tb − T∞ )

Q& fin no fin

⎛ A fin Aunfin ⎞ ⎟ = ⎜η fin + ⎜ Ano fin Ano fin ⎟⎠ ⎝

Figure 3-45 from Çengel, Heat Transfer


• Compare actual heat transfer to ideal case where entire fin is at base temperature Q& fin η fin = & = Q fin,max Q& fin hA fin (Tb − T∞ )

Figure 3-39 from Çengel, Heat Transfer

16

Lumped Parameter Model

• Original area, A = (area with fins, Afin) + (area without fins, Aunfin)

ε total = & Q

14

Fin Efficiency

• Infinitely long fin

θ = θb e −mx

1 ⎛ r4 ⎞ ln⎜ ⎟ k3 L ⎜⎝ r3 ⎟⎠

17

• Assumes same temperature in solid • Use characteristic length Lc = V/A hA h b= = ρc p V ρc p Lc

(T − T∞ ) = (Ti − T∞ )e −bt

or

T = (Ti − T∞ )e −bt + T∞

• Must have Bi = hLc/k < 0.1 to use this 18

3

Final Review

May 16, 2006

Slab Center-line (x = 0) Temperature Chart

Transient 1D Convection

Figure 4-15(a) in Çengel, Heat and Mass Transfer

Figure 4-11 in Çengel, Heat and Mass Transfer

All problems have similar chart solutions 19

Θ T − T∞ = Θ 0 T0 − T∞

Chart II

Approximate Solutions

• Can find T at any x/L from this chart once T at x = 0 is found from previous chart • See basis for this chart on the next page Figure 4-15(b) in Çengel, Heat and Mass Transfer

• Plane that extends to infinity in all directions • Practical applications: large area for short times


• Valid for for τ > 0.2 • Slab • Cylinder

2 T − T∞ = A1e −λ1τ cos λ1ξ Ti − T∞ 2 ⎛ r⎞ T − T∞ Θ= = A1e −λ1τ J 0 ⎜⎜ λ1 ⎟⎟ Ti − T∞ ⎝ r0 ⎠

Θ=

• Sphere Θ =

2 ⎛ r⎞ T − T∞ r = A1e −λ1τ 0 sin ⎜⎜ λ1 ⎟⎟ Ti − T∞ λ1r ⎝ r0 ⎠

– Values of A1 and λ1 depend on Bi and are different for each geometry (as is Bi)

21

Semi-Infinite Solids


20

– Example: earth surface locally23

22

Multidimensional Solutions • Can get multidimensional solutions as product of one dimensional solutions – All one-dimensional solutions have initial temperature, Ti, with convection coefficient, h, and environmental temperature, T∞, starting at t = 0 – General rule: ΘtwoD = ΘoneΘtwo where Θone and Θtwo are solutions from charts for plane, cylinder or sphere 24

4

Final Review

May 16, 2006

Multidimensional Example

Flow Classifications

⎛ T (r , x, t ) − T∞ ⎞ ⎜⎜ ⎟⎟ ⎝ Ti − T∞ ⎠

x = a/2

finite cylinder

=

⎛ T (r , t ) − T∞ ⎞ ⎜⎜ ⎟⎟ • ⎝ Ti − T∞ ⎠ infinite cylinder

x = -a/2


⎛ T ( x, t ) − T∞ ⎞ ⎜⎜ ⎟⎟ ⎝ Ti − T∞ ⎠infinite slab

• Forced versus free • Internal (as in pipes) versus external (as around aircraft) – Entry regions in pipes vs. fully-developed

• Unsteady (changing with time) versus unsteady (not changing with time) • Laminar versus turbulent • Compressible versus incompressible • Inviscid flow regions (μ not important) • One-, two- or three-dimensional

25

Flows

26

Boundary Layer

• Laminar flow is layered, turbulent flows are not (but have some structure)

• Region near wall with sharp gradients – Thickness, δ, usually very thin compared to overall dimension in y direction

Figures 6-9 and 6-16. Çengel, Heat and Mass Transfer

27

Thermal Boundary Layer

• Nusselt number, Nu = hLc/kfluid – Different from Bi = hLc/ksolid

• Reynolds number, Re = ρVLc/μ = VLc/ν • Prandtl number Pr = μcp/k (in tables) • Grashof number, Gr = βgΔTLc3/ν2 – g = gravity, β = expansion coefficient = –(1/ρ)(∂ρ/∂T)p, and ΔT = | Twall – T∞ |

• Thermal boundary layer thickness may be less than, greater than or equal to that of the momentum boundary layer


28

Dimensionless Convection

• Thin region near solid surface in which most of temperature change occurs

Figure 6-15. Çengel, Heat and Mass Transfer


• Peclet, Pe = RePr; Rayleigh, Ra = GrPr 29

30

5

Final Review

May 16, 2006

Characteristic Length

How to Compute h

• Can use length as a subscript on dimensionless numbers to show correct length to use in a problem

• Follow this general pattern – Find equations for h for the description of the flow given • Correct flow geometry (local or average h?) • Free or forced convection

– ReD = ρVD/μ, Rex = ρVx/μ, ReL = ρVL/μ – NuD = hD/k, Nux = hx/k, NuL = hL/k – GrD = ρ2βgΔTD3/μ2, Grx = ρ2βgΔTx3/μ2, GrL = ρ2βgΔTL3/μ2

– Determine if flow is laminar or turbulent • Different flows have different measures to determine if the flow is laminar or turbulent based on the Reynolds number, Re, for forced convection and the Grashof number, Gr, for free convection

• Use not necessary if meaning is clear 31

32

How to Compute h

Property Temperature

• Continue to follow this general pattern – Select correct equation for Nu (laminar or turbulent; range of Re, Pr, Gr, etc.) – Compute appropriate temperature for finding properties – Evaluate fluid properties (μ, k, ρ, Pr) at the appropriate temperature – Compute Nusselt number from equation of the form Nu = C Rea Prb or D Rac – Compute h = k Nu / LC

• Find properties at correct temperature • Some equations specify particular temperatures to be used (e.g. μ/μw) • External flows and natural convection use film temperature (Tw + T∞)/2 • Internal flows use mean fluid temperature (Tin + Tout)/2

33

34

Key Ideas of External Flows

Flat Plate Flow Equations

• The flow is unconfined • Moving objects into still air are modeled as still objects with air flowing over them • There is an approach condition of velocity, U∞, and temperature, T∞ • Far from the body the velocity and temperature remain at U∞ and T∞ • T∞ is the (constant) fluid temperature used to compute heat transfer 35


• Laminar flow (Rex, ReL < 500,000, Pr > .6) C fx =

τ wall

ρU ∞2

Cf =

2 τwall

= 0.664 Re −x1 / 2

ρU ∞2 2

= 1.33 Re −L1/ 2

hx x = 0.332 Re1x/ 2 Pr1/ 3 k hL Nu L = = 0.664 Re1L/ 2 Pr1/ 3 k Nu x =

• Turbulent flow (5x105 < Rex, ReL < 107) C fx = Cf =

τ wall

ρU ∞2 2 τwall ρU ∞2 2

= 0.059 Re −x1 / 5

= 0.074 Re −L1 / 5

hx x = 0.0296 Re 0x.8 Pr1 / 3 k hL Nu L = = 0.037 Re 0L.8 Pr1 / 3 k

Nu x =

For turbulent Nu, .6 < Pr < 60

36

6

Final Review

May 16, 2006

Flat Plate Flow Equations II

Heat Transfer Coefficients

• Average properties for combined laminar and turbulent regions with transition at xc = 500000 ν/U∞ – Valid for 5x105 < ReL < 107 and 0.6 < Pr < 60 Cf =

τwall ρU ∞2 2

=

0.074 1742 − Re1L/ 5 Re L

Nu L =

(

)

hL = 0.037 Re 0L.8 − 871 Pr1 / 3 k


37

• Cylinder average h (RePr > 0.2; properties at (T∞ + Ts)/2

[

]

38

Tube Bank Heat Transfer

39

Table 7-2 from Çengel, Heat and Mass Transfer

Key Ideas of Internal Flows

40

Area Terms L

• The flow is confined • There is a temperature and velocity profile in the flow

• Acs is cross-sectional area for the flow

D

– Use average velocity and temperature

• Wall fluid heat exchange will change the average fluid temperature – There is no longer a constant fluid temperature like T∞ for computing heat transfer

W

L

– Acs = πD2/4 for circular pipe – Acs = WH for rectangular duct

• Aw is the wall area for heat transfer

H

41


1/ 4

⎛μ ⎞ hD Nu = = 2 + 0.4 Re1 / 2 + 0.06 Re 2 / 3 Pr 0.4 ⎜⎜ ∞ ⎟⎟ k ⎝ μs ⎠

Other Shapes and Equations

Part of Table 7-1 from Çengel, Heat and Mass Transfer

4/5

5/8 hD 0.62 Re1 / 2 Pr1 / 2 ⎡ ⎛ Re ⎞ ⎤ ⎢ ⎥ Nu = = 0.3 + + 1 ⎜ ⎟ 1/ 4 k ⎡ ⎛ 0.4 ⎞ 2 / 3 ⎤ ⎢⎣ ⎝ 282,000 ⎠ ⎥⎦ ⎢1 + ⎜ ⎟ ⎥ ⎣⎢ ⎝ Pr ⎠ ⎦⎥ • Sphere average h (3.5 ≤ Re ≤ 80,000; 0.7 ≤ Pr ≤ 380; μs at Ts; other properties at T∞)

– Aw = πDL for circular pipe – Aw = 2(W + H)L for rectangular duct


42

7

Final Review

May 16, 2006

Average Temperature Change • Let T represent the average fluid temperature (instead of Tavg, Tm or T ) • T will change from inlet to outlet of confined flow – This gives a variable driving force (Twall – Tfluid) for heat transfer – Can accommodate this by using the first & =m & cp(Tout – Tin) law of thermodynamics: Q – Two cases: fixed wall heat flux and fixed wall temperature

Fixed Wall Heat Flux • Fixed wall heat flux, q& wall, over given wall area, Aw, gives total heat input which is related to Tout – Tin by thermodynamics q& A Q& = q& wall Aw = m& c p (Tout − Tin ) ⇒ Tout = Tin + wall w m& c p • “Outlet” can be any point along flow path where area from inlet is Aw • We can compute Tw at this point as Tw = Tout + q& wall /h

43

Constant Wall Temperature (Tout − Ts ) = (Tin − Ts )e

−

Log-mean Temperature Diff • This is usually written as a set of temperature differences

hAw m& c p

& cp = NTU, the • hAw / m number of transfer units • This is general equation for computing Tout in internal flows Figure 8-14 from Çengel, Heat and Mass Transfer

LMΔT =

(Tout − Tin ) ⎛ T −T ⎞ ln⎜⎜ out s ⎟⎟ ⎝ Tin − Ts ⎠

=

(Tout − Ts ) − (Tin − Ts ) ⎛ T −T ⎞ ln⎜⎜ out s ⎟⎟ ⎝ Tin − Ts ⎠

hA (T − T ) Q& = w out in = hAw ( LMΔT ) ⎛T −T ⎞ ln⎜⎜ out s ⎟⎟ ⎝ Tin − Ts ⎠

Çengel uses ΔTlm for LMΔT

45

Developing Flows

46

Fully Developed Flow Momentum boundary layer development

Thermal boundary layer development 47


44

• Temperature profile does not change with x if flow is fully developed thermally • This means that ∂T/∂r does not change with downstream distance, x, so heat flux (and Nu) do not depend on x • Laminar entry Lh ≈ 0.05 Re D lengths • Turbulent entry lengths

Lt ≈ 0.05 Re Pr D

Lt Lh ≈ = 1.359 Re1 4 ≈ 10 D D 48

8

Final Review

May 16, 2006

Entry Region Nusselt Numbers

Internal Flow Pressure Drop • General formula: Δp = f (L/D) ρV2/2 • Friction factor, f, depends on Re = ρVD/μ and relative roughness, ε/D • For laminar flows, f = 64/Re – No dependence on relative roughness

• For turbulent flows Colebrook

⎛ε D 1 2.51 = −2.0 log10 ⎜ + ⎜ 3.7 Re f f ⎝

⎛ 6.9 ⎛ ε D ⎞1.11 ⎞ ≈ −1.8 log10 ⎜ +⎜ ⎟ ⎟ ⎜ Re ⎝ 3.7 ⎠ ⎟ f ⎝ ⎠

⎞ ⎟ ⎟ ⎠

Haaland 1

Eggs from Figure 8-9 in Çengel, Heat and Mass Transfer

49

Moody Diagram

50

Laminar Nusselt Number • Laminar flow if Re = ρVD/μ < 2,300 • Fully-developed, constant heat flux, Nu = 4.36 • Fully-developed, constant wall temperature: Nu = 3.66 • Entry region, constant wall temperature:

Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi. Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.

Nu = 3.66 +

0.065 (D L ) Re Pr

1 + 0.04[(D L ) Re Pr ]2 3

51

52

Noncircular Ducts • Define hydraulic diameter, Dh = 4A/P – A is cross-sectional area for flow – P is wetted perimeter – For a circular pipe where A = pD2/4 and P = πD, Dh = 4(πD2/4) / (πD) = D

From Çengel, Heat and Mass Transfer

• For turbulent flows use Moody diagram with D replaced by Dh in Re, f, and ε/D • For laminar flows, f = A/Re and Nu = B (all based on Dh) – A and B next slide 53


54

9

Final Review

May 16, 2006

Turbulent Flow

Free (Natural) Convection

• Smooth tubes (Gnielinski) ⎛ 0.5 ≤ Pr ≤ 2000 ⎞ ( f 8)(Re− 1000) Pr ⎜⎜ Nu = 3 6⎟ ⎟ 0.5 23 1 + 12.7( f 8) (Pr − 1) ⎝ 3 x10 < Re < 5 x10 ⎠ Petukhov :

f = [0.790 ln (Re ) − 1.64]

−2

• Flow is induced by temperature difference

3000 < Re < 5 x10

• Tubes with roughness

Free (Natural)

– Use correlations developed for this case – As approximation use Gnielinski equation with f from Moody diagram or f equation • Danger! h does not increase for f >4fsmooth

– No external source of fluid motion – Temperature differences cause density differences – Density differences induce flow

Forced 6

• “Warm air rises”

– Volume expansion coefficient: β = [–(1/ρ)(∂ρ/∂T)]

Eggs from Figure 1-33 in Çengel, Heat and Mass Transfer

• For ideal gases β = 1/T

55

56

Grashof and Rayleigh Numbers

Equations for Nu

• Dimensionless groups for free (natural) convection Ra = Gr Pr = β gΔTL3c ρ 2βgΔTL3c Gr = = βgΔTL3c ν2 μ2 να -2

• Equations have form of AGraPrb or BRac • Since Gr and Ra contain |Twall – Tfluid|, an iterative process is required if one of these temperatures is unknown • Transition from laminar to turbulent occurs at given Ra values

– g = acceleration of gravity (LT ) – β = –(1/ρ)(∂ρ/∂T) called the volume expansion coefficient (dimensions: 1/Θ)

– For vertical plate transition Ra = 109

– ΔT = |Twall – Tfluid| (dimensions: Θ) – Other terms same as previous use

• Evaluate properties at “film” (average) temperature, (Twall + Tfluid)/2 57

58

Vertical Plate Free Convection


10000

• Simplified equations on previous chart for constant wall temperature

1000

Nu = hL 100 k

⎧ ⎫ 0.387 Ra1L/ 6 ⎪ ⎪ Nu L = ⎨0.825 + ⎬ 9 / 16 8 / 27 ⎪⎩ ⎪⎭ 1 + (0.492 Pr )

109 < Ra < 1013

10

1 1.E+00

– More accurate: Churchill and Chu, any Ra

Nu = 0.10 Ra1/ 3

[

Nu = 0.59 Ra1 / 4 104 < Ra < 109 1.E+02

1.E+04

1.E+06

1.E+10

1.E+12

1.E+14

Nu L = 0.68 +


βgΔTL

3

Ra =

Any RaL

– More accurate laminar Churchill/Chu

1.E+08

Rayleigh Number Plate figure from Table 9-1 in Çengel, Heat and Mass Transfer

]

2

ν2

βgΔTL

3

Pr =

να

59

0.670 Ra1L/ 4

[1 + (0.492 Pr ) ]

9 / 16 4 / 9

0 < Ra L < 109 60

10

Final Review

May 16, 2006


Vertical Cylinder

• Constant wall heat flux

• Apply equations for vertical plate from previous charts if D/L ≥ 35/Gr1/4 • For this D/L effects of curvature are not important • Thin cylinder results of Cebeci and Minkowcyz and Sparrow available in ASME Transactions

– Use q& = hA(Tw – T∞) to compute an unknown temperature (Tw or T∞) from known wall heat flux and computed h – Tw varies along wall, but the average heat transfer uses midpoint temperature, TL/2 q& q& wall = hAwall (TL / 2 − T∞ ) ⇒ TL / 2 − T∞ = wall hAwall – Use trial and error solution with TL/2 – T∞ as ΔT in Ra used to compute h = kNu/L

Cylinder figure from Table 9-1 in Çengel, Heat and Mass Transfer

61

62

Horizontal Plate

Horizontal Plate II Cold surface

Cold surface

• Hot surface facing up or cold surface facing down • Lc = area / perimeter (As/p)

• Cold surface facing up or hot surface facing down • Lc = area / perimeter (As/p)

– For a rectangle of length, L, and width, W, Lc = (LW) / (2L + 2W) = 1 / ( 2 / W + 2 / L) – For a circle, Lc = πR2 / 2πR = R/2 = D/4 Figures from Table 9-1 in Çengel, Heat and Mass Transfer

– For a rectangle of length, L, and width, W, Lc = (LW) / (2L + 2W) = 1 / ( 2 / W + 2 / L) – For a circle, Lc = πR2 / 2πR = R/2 = D/4 Figures from Table 9-1 in Çengel, Heat and Mass Transfer

Nu = 0.54 Ra1L/c 4 10 4 < Ra < 107 Nu

= 0.15 Ra1L/c3

10 < Ra < 10 7

11

63

Sphere and Horizontal Cylinder ⎧ 0.387 Ra1D/ 6 ⎪ Nu D = ⎨0.6 + ⎪⎩ 1 + (0.559 Pr )9 / 16

[

]

⎫ ⎪ 8 / 27 ⎬ ⎪⎭

Nu D = 2 +

L

0.589 Ra1D/ 4

[1 + (0.469 Pr ) ]

Figures from Çengel, Heat and Mass Transfer


9 / 16 4 / 9

65

64

Horizontal Enclosures

2

• NuD results are average values

Nu = 0.27 Ra1L/c 4 105 < Ra < 1011

• Top side warmer: no convection • Conduction only, Nu = hL/k = 1 • Bottom warmer: convection becomes significant when RaL = (Pr)βgΔTL3/ν2 = βgΔTL3/να > 1708 Figure 9-22 in Çengel, Heat and Mass Transfer

66

11

Final Review

May 16, 2006

Horizontal Enclosures II

Vertical Enclosures Berkovsky and Polevikov, any Pr

Jakob, for 0.5 < Pr < 2

⎛ Pr ⎞ Nu L = 0.18⎜ Ra L ⎟ ⎝ 0.2 + Pr ⎠

Nu = 0.195Ra1L/ 4 10 4 < RaL < 4 x105 Nu = 0.068Ra1L/ 3

4 x105 < Ra L < 107

Globe and Dropkin for a range of liquids Nu = 0.069 Ra1L/ 3 Pr 0.074

3x105 < Ra L < 7 x109

Hollands et al. for air; also for other fluids if RaL < 105 ⎛ 1708 ⎞ ⎛ Ra ⎞ ⎟ + max⎜ 0, L − 1⎟ RaL < 108 Nu = 1 + 1.44 max⎜⎜ 0, 1 − RaL ⎟⎠ ⎝ 18 ⎠ ⎝ 67

Figure 923 in Çengel, Heat and Mass Transfer

0.29

1< H / L < 2 Ra L Pr 0.2 + Pr > 103

0.28 1/ 4 2 < H / L < 10 ⎛ Pr Ra L ⎞ ⎛ L ⎞ Nu L = 0.22⎜ ⎟ ⎜ ⎟ RaL < 1010 0 . 2 + Pr H ⎝ ⎠ ⎝ ⎠ MacGregor and Emery

⎛L⎞ Nu L = 0.42 Ra1L/ 4 Pr 0.012 ⎜ ⎟ ⎝H⎠

1 < H / L < 40 1 < Pr < 20 10 < RaL < 10 6

0.3

10 < H / L < 40

1 < Pr < 2 x10 4 10 4 < RaL < 107

Nu L = 0.46 Ra1L/ 3 9

68

Heat Exchangers • Used to transfer energy from one fluid to another • One fluid, the hot fluid, is cooled while the other, the cold fluid, is heated • May have phase change: temperature of one or both fluids is constant • Simplest is double pipe heat exchanger – Parallel flow and counter flow Figure 11-1 from Çengel, Heat and Mass Transfer 69

Compact Heat Exchangers

70

Shell-and-Tube Exchanger

• Counter flow exchanger with larger surface area; baffles promote mixing Figure 11-3 from Çengel, Heat and Mass Transfer


71


72

12

Final Review

May 16, 2006

Shell and Tube Passes

Overall U • U is overall heat transfer coefficient • Analyzed here for double-pipe heat exchanger

Tube flow has three complete changes of direction giving four tube passes Shell flow changes direction to give two shell passes 73

1 1 + Rwall + hi Ai ho Ao 1 1 1 = = = U o Ao U i Ai UA

R=

Figure 11-5(b) from Çengel, Heat and Mass Transfer

Heat Exchange Analysis • Heat transfer from hot to cold fluid

Parallel Flow • Parallel flow Q& = UAΔTlm heat exchanger

Q& = UAΔT

( (

) )

Q& = m& c c pc Tc,out − Tc,in • First law energy Q& = m& h c ph Th,in − Th,out balances • Assumes no heat loss to surroundings

ΔTlm =

ΔTlm =

– Subscripts c and h denote cold and hot fluids, respectively – Alternative analysis for phase change 75

ΔT2 = Th,out – Tc,in

Counter Flow

ΔT − ΔT1 Q& = UAΔTlm = UA 2 ⎛ ΔT ⎞ ln⎜⎜ 2 ⎟⎟ ⎝ ΔT1 ⎠

(Th,out − Tc,in ) − (Th,in − Tc,out )



⎛T −T ⎞ ln⎜ h,out c,in ⎟ ⎜T −T ⎟ ⎝ h,in c,out ⎠

(Th,out − Tc,out ) − (Th,in − Tc,in ) ⎛T −T ln⎜ h,out c,out ⎜ T −T ⎝ h,in c,in

⎞ ⎟ ⎟ ⎠


76

• With ΔTlm method we want to find U or A when all temperatures are known • If we know three temperatures, we can find the fourth by an energy balance with known mass flow rates (and cp’s)

– Difference in ΔT1 and ΔT2 definitions

ΔTlm =

ΔT2 − ΔT1 ΔT1 − ΔT2 = ⎛ ΔT ⎞ ⎛ ΔT ⎞ ln⎜⎜ 2 ⎟⎟ ln⎜⎜ 1 ⎟⎟ ⎝ ΔT1 ⎠ ⎝ ΔT2 ⎠

Heat Exchanger Problems

• Same basic equations

ΔT1 = Th,in – Tc,out

74


Q& = m& c c pc (Tc,out − Tc,in ) ) Q& = m& c (T − T h ph

77

h,in

h,out

& from two Can find Q temperatures for one stream and then find unknown temperature 78

13

Final Review

May 16, 2006

Correction Factors

Correction Factor Chart I

• Correction factor parameters, R and P – Shell and tube definitions below

Ttube,out − Ttube,in

t 2 − t1 Tshell ,in − Ttube,in T1 − t1 m& c p Tshell ,in − Ttube,in T1 − T2 = = R= Ttube,out − Ttube,in t 2 − t1 m& c p P=

=

( )tube ( )shell

P

– Correction factor charts show diagrams that illustrate the equations for P and R Figure 11-18 from Çengel, Heat and Mass Transfer

79

80

Effectiveness, ε

Effectiveness-NTU Method • Used when not all temperatures are known • Based on ratio of actual heat transfer to maximum possible heat transfer • Maximum possible temperature difference, ΔTmax is Th,in – Tc,in – Only one fluid, the one with the smaller value & cp, can have ΔTmax of m & cp)h & cp)c and Ch = (m – Define Cc = (m

ε=

Q& Q& = & ( Qmax Cmin Th.in − Tc,in )

Cmin = min (Ch , Cc )

• In effectiveness-NTU method we find ε, & & = εQ then find Q max & – Use C ΔT to find Q because C ΔT min

max

max

1

1

= C2ΔT2 or ΔT2 = C1ΔT1/C2 – If ΔT2 = ΔTmax and C1/C2 > 1, ΔT2 > ΔTmax – CminΔTmax is maximum heat transfer that can occur without impossible T < Tc,in

81

Find ε Example chart for finding effectiveness from NTU = UA/Cmin and Cmin/Cmax ratio For NTU = 1.5 and Cmin/Cmax .7 = 0.25, ε = ? Figure 11-26 from Çengel, Heat and Mass Transfer


83

82

Effectiveness Equations • Double pipe parallel flow

ε=

1 − e − NTU (1+c ) 1+ c

• Double pipe counter flow

UA Cmin

NTU =

c=

1 − e − NTU (1−c ) ε= 1 − ce − NTU (1−c )

Cmin Cmax

84

Figures from Figure 11-26 from Çengel, Heat and Mass Transfer

14

Final Review

May 16, 2006

Ultraviolet

Infrared

Black-Body Radiation

Spectral Ebλ

• Basic black body equation: Eb = σT4

• Energy (W/m2) emitted varies with wavelength and temperature • Maximum point occurs where λT = 2897.8 μm·K • T increase shifts peak shift to lower λ

– Eb is total black-body radiation energy flux W/m2 or Btu/hr·ft2 – σ is the Stefan-Boltzmann constant • σ = 5.670x10-8 W/m2·K4 • σ = 0.1714x10-8 Btu/hr·ft2·R4

– Must use absolute temperature

• Radiation flux varies with wavelength – Ebλ is flux at given wavelength, λ 85

86


Partial Black-body Power

Radiation Tables

Black body radiation between λ = 0 and λ = λ1 is Eb,0-λ1 λ 1

∫

Eb,0−λ1 = Ebλ dλ 0

Fraction of total radiation (σT4) between λ = 0 and any given λ is fλ

fλ =

λ

1 σT 4

∫ Ebλ dλ' 0

• Can show that fλ is function of λT fλ =

λ

1 σT

4

1

λ

C1

∫ Ebλ dλ = σT 4 ∫ λ5 (eC 0

2

λT

0

)

−1

dλ =

1 σ

λT

∫ (λT )5 (eC1 λT − 1) d (λT ) C

2

0

• Radiation tables give fλ versus λT – See table 12-2, page 672 in text – Extract from this table shown at right

87

88


Δλ

Emissivity • Ratio of actual emissive power to blakc body emissive power

• Radiation in finite band, Δλ f λ1 −λ2 = 1 σT 4

1 σT 4

λ2

λ2

∫ E λ dλ = λ b

1

λ

1 1 Ebλ dλ σT 4 ∫0 0 = f λ (λ2T ) − f λ (λ1T )

∫ Ebλ dλ −

f λ (0) = 0

f λ (∞ ) = 1 89

– Diffuse surface – emissivity does not depend on direction – Gray surface – emissivity does not depend on wavelength – Gray, diffuse surface – emissivity is the does not depend on direction or wavelength • Simplest surface to handle and often used in radiation calculations 90



15

Final Review

May 16, 2006

Average Emissivity • Average over all wavelengths ε=

1

σT 4

∞

∞

1

ε C1

∫ ε λ E λ dλ = σT ∫ λ (e λ λ b

4

0

5

0

C2

T

1

∞

ε λ C1

) dλ = σ ∫ (λT ) (e

−1

5

0

C 2 λT

)

−1

d (λT )

• For emissivity with constant values in a series of wavelength ranges λT λT ∞ ε 3C1d (λT ) ε 1C1d (λT ) ε 2C1d (λT ) 1 1 1 ε= ∫ + + σ 0 (λT )5 (eC λT − 1) σ λ∫T (λT )5 (eC λT − 1) σ λ∫T (λT )5 (eC λT − 1) 1

2

2

2

2

1

2

ε = ε 1 [ f λ (λ1T ) − 0] + ε 2 [ f λ (λ2T ) − f λ (λ1T )] + ε 3 [1 − f λ (λ2T )]

• Applies to other properties as well 91

92

Properties • Incoming radiation properties – Reflectivity, ρ – Absorptivity, α – Transmissivity, τ Figure 12-31 from Çengel, Heat and Mass Transfer

• Energy balance: ρ+α+τ=1

93

α Data

Kirchoff’s Law

• Solar radiation has effective source temperature of about 5800 K

• Absorptivity equals emissivity (at the same temperature) • True only for values in a given direction and wavelength • Assuming total hemispherical values of α and ε are the same simplifies radiation heat transfer calculations, but is not always a good assumption



94

95

96

16

Final Review

May 16, 2006

Effect of Temperature

View Factor, Fi→j or Fij

• Emissivity, ε, depends on surface temperature • Absorptivity, α, depends on source temperature (e.g. Tsun ≈ 5800 K) • For surfaces exposed to solar radiation

• Fi→j or Fij is the fraction of radiation, leaving surface i, that strikes surface j – AiFij = AjFji – ΣkFik = 1 (enclosure) – F1→2+3 = F12 + F13

– high α and low ε will keep surface warm – low α and high ε will keep surface cool – Does not violate Kirchoff’s law since source and surface temperatures differ

• Fk→k = 0 only if k is a flat surface • View factors from equations or charts Figure 13-1 from Çengel, Heat and Mass Transfer

97

Gray Diffuse Opaque Enclosure

All Black Surface Enclosure • Heat transfer from surface 1 reaching surface 2 is A1F12σT14 • Heat transfer from surface 2 reaching surface 1 is A2F21σT24 = A1F12σT24 • Net heat exchange between surface 1 and surface 2: A1F12σ(T14 – T24) – Negative value indicates heat into surface 1 – For multiple surfaces Q& i =

∑ Q&i→ j = ∑ Ai Fij σ(Ti4 − T j4 ) N

N

j =1

j =1

• Kirchoff’s law applies to the average: α = ε at all temperatures • For opaque surfaces τ = 0 so α + ρ = 1 • For gray, diffusive, opaque surfaces then ρ = 1 – α = 1 – ε • Define radiosity, J = εEb + ρG = emitted and reflected radiation Aε E − Ji 1 − εi Q& i = i i (Ebi − J i ) = bi where Ri = Ri Ai εi 1 − εi

99

100

Gray Diffuse Opaque II

Net Radiation Leaving Surface & = A(J – G) • Q • Can show Aε (Eb − J ) Q& = 1− ε

E − Ji Q& i = b ,i Ri Ri =

1− εi Aε i



98

101

Q& i =

N

Ji − J j

j =1

Rij

∑ Q&ij = ∑ Ai Fij (J i − J j ) = ∑ N

j =1

N

j =1

Rij =

1 Ai Fij

• Combining two equations for Q& i N J −J N J −J Ebi − J i J − Ebi i j i j = ⇒ + i =0 ∑ Rij ∑ R R Ri i ij j =1 j =1 • Solve system of N simultaneous linear equations for N values of Ji • Black or reradiating surface ( Q& i = 0) has Ji = Ebi = σTi4 102

17

Final Review

May 16, 2006

Review Circuit Analogy

Three-Surface Circuit

• Look at simple enclosure with only two surfaces • Apply circuit analogy with total resistance

• Three or more surfaces easirer by system of equations & =0 • Exception: Q 3

E − Eb 2 Eb1 − Eb 2 Q&12 = b1 = 1 − ε1 1 1− ε2 RTotal + + A1ε 1 A1 F12 A2ε 2

E − Eb 2 Eb1 − Eb 2 = Q& net ,12 = b1 RTotal R1 + Rc + R2

Rc =

1 1 1 + R12 R13 + R23

103

104

Review Three-Surface Circuit

Radiation Exchange • Two possible surface conditions: (1) & known temperature, (2) known Q i N Aε Q& i = i i Ebi − J i = Ai Fij J i − J j i = 1, K, N 1 − εi j =1 ⎛ 1− ε N ⎞ 1 − εi N i (1) ⎜1 + Fij ⎟ J i − Fij J j = Ebi = σTi4 ⎜ εi j =1, j ≠i ε i j =1, j ≠i ⎟ Solve this set ⎝ ⎠ of N N ⎛ N ⎞ Ai Fij ⎟ J i − Ai Fij J j = Q& i simultaneous (2) ⎜ ⎜ ⎟ equations for j =1, j ≠i ⎝ j =1, j ≠i ⎠

(

• If Q& 3 = 0, Q& net,1→2 can be found from circuit with two parallel resistances

) ∑

∑

E − Eb 2 Eb1 − Eb 2 Q& net ,12 = b1 = RTotal R1 + Rc + R2

Rc =

1 1 1 + R12 R13 + R23

∑

(

)

∑

∑

N values106of Ji

105

Radiation Exchange II

Numerical Heat Transfer • Finite difference expressions with truncation error • Computers give roundoff error • Convert differential equations to algebraic equations

• Once all Ji values are known we can & compute unknown values of Ti and Q i – For known Ti

(

)

(

Aε Aε Q& i = i i Ebi − J i = i i σTi4 − J i 1 − εi 1 − εi & – For known Q i

Ebi = J i +

1 − εi & Qi Ai εi

⇒ Ti =

)

– Solve system of algebraic equations to get temperatures at discrete points – Reduce step size for stability

1 1 − εi & 4 Ji + Qi Ai ε i σ

• Will not be covered on final 107


108

18