CHAPTER 4 BOUNDARY LAYER FLOW APPLICATION TO EXTERNAL FLOW 4.1 Introduction Boundary layer concept (Prandtl 1904): Eliminate selected terms in the governing equations Two key questions (1) What are the conditions under which terms in the governingequations can be dropped? () What terms can be dropped? 1
Answer: By two approaches Intuitive arguments Scale analysis 4. The Boundary Layer Concept: Simplification of Governing Equations 4..1 Qualitative Description
Under certain conditions the action of viscosity is confined to a thin region near the surface called the viscous or velocity boundary layer Under certain conditions thermal interaction between moving fluid and a surface is confined to a thin region near the surface called the thermal or temperature boundary layer Conditions for viscous boundary layer: (1) Slender body without flow separation () High Reynolds number ( Re > 100) Conditions for thermal boundary layer: 3
(1) Slender body without flow separation () High product of Reynolds and Prandtl ( RePr > 100) numbers ρv L c pµ ρc pv L Peclet Number Pe RePr (4.1) µ k k (1) Fluid velocity at surface vanishes () Rapid changes across BL to V (3) Rapid changes temperature across BL from T s to T () Boundary layers are thin: For air at 10 m/s parallel to 1.0 m long plate, δ 6 mm at end (3) Viscosity plays negligible role outside the viscous BL (4) Boundary layers eist in both forced and free convection flows 4
4.. The Governing Equations Simplified case: Assumptions: (1) steady state () two-dimensional (3) laminar (4) constant properties (5) no dissipation (6) no gravity Continuity: u + v y 0 (.) 5
-direction: y-direction: + + + w w w w u w ρ v + + + + + + z u y u u p g z u w y u u u t u µ ρ ρ v (.10) + + + + + + z w y w w z p g z w y u t z µ ρ ρ v (.10y) Energy: + + Ρ y T T k y T T u ρc v (.19) 6
4..3 Mathematical Simplification 4..4 Simplification of the Momentum Equations (i) Intuitive Arguments Two viscous terms in (.10): is one smaller than the other? u u + yy 7
Insect dilemma: Too windy at position 0, where to go? Move to position 4! Conclusion: Changes in u with respect to y are more pronounced than changes with respect to u Neglect in (.10) u u << y Pressure terms in (.10) and (.10y): Slender body Streamlines are nearly parallel Small vertical velocity (4.) 8
p y 0 p depends on only, i.e. p p() p dp d (4.) and (4.4) into (.10) gives: dp d Boundary layer -momentum equation Continuity equation (.) and the -momentum boundary layer equation (4.5) contain three unknowns: u, v, and p (4.3) (4.4) u 1 dp u v + (4.5) u u + y ρ ν y 9
p is pressure at edge of BL (y δ), obtained from solution of inviscid flow outside BL (ii) Scale Analysis Use scaling to arrive at BL approimations. Assign a scale to each term in an equation Slender body Free stream velocity Length L BL thickness Postulate: δ V δ << 1 L (4.6) 10
If (4.6) is valid, we pose three questions: (1) What terms in the governing equations can be dropped? () Is normal pressure gradient negligible compared to aial pressure gradient? (3) Under what conditions is (4.6) valid? Assign scales: u V V y δ Apply (4.7) to continuity (.) L (4.7a) (4.7b) (4.7c) v y u 11
Using (4.7) Solving for v v << V v δ V L Conclusion: Order of magnitude of inertia and viscous terms -momentum equation (.10) First inertia term: u u Second inertial term: v V δ L V V L (4.7d) (a) u V v v y δ 1
Use (4.7d) u y V V L Conclusion: inertia terms are of the same order Eamine viscous terms in (.10) First viscous term: Second viscous term: Conclusion: u / v y u u u << Neglect in (.10) V L V δ y u (b) (c) (d) (4.) 13
Eamine viscous terms in (.10y) v y v << (4.8) Simplify (.10) and (.10y) Using (4.) and (4.8) 1 y u p y u u u + + ν ρ v (4.9) 1 y y p y u + + v v v v ν ρ (4.9y) This answers first question Second question: pressure gradient p y p Scale and 14
Balance aial pressure with inertia in (4.9) Scale using (4.7) p u ρ u p V ρ L Balance pressure with inertial in (4.9y) p y V ρ L Compare (e) and (f) using (4.6) δ L (e) (f) p y << p (4.10) 15
Since or Scale dy d (e)-(g) into (4.11) Invoke(4.6) dp d dp dp d p p p p p(, y) d + p dy y ( p / y) + ( p / ) dy d 1 (4.11) dy d δ L [ 1 + ( δ / L) ] (g) (h) dp d p (i) 16
Conclusion Boundary layer pressure depends on Variation with y is negligible only. p () Pressure p() inside BL pressure at edge (4.1) into (4.9) p(, y) p ( ) p p dp d (4.13) is -momentum eq. for BL flow. Result is based on key assumption that / L << 1. δ << (j) (4.1) u 1 dp u v + (4.13) u u + y ρ d ν y 17
Third question: condition for validity of (4.6) δ << 1 L Balance inertia with viscous force in (4.13) Inertia: Viscous: Equate Rearrange u u V V L u V ν ν y δ V L δ L V ν δ ν V L (4.6) (a) (b) (4.14a) 18
or where δ 1 L Re L V L ν δ << 1 when ReL >> L 1 (4.14b) Re L (4.15) Generalized (4.14) δ 1 Re (4.16) 19
4..5 Simplification of the Energy Equation Simplify (.19) T ρc u + v (i) Intuitive Arguments T y Two conduction terms in (.19): T k Ρ + y T (.19) T + y T is one smaller than the other? 0
Insect dilemma: Too hot at position 0, where to go? Move to position! Conclusion: Changes in T with respect to y are more pronounced than with respect to T << y T (4.17) 1
T Neglect in (.19): T T u + α y y (4.18) is the boundary layer energy equation. (ii) Scale Analysis Use scaling to arrive at BL approimations T v (4.18) Assign a scale to each term in an equation Slender body Free stream velocity V Free stream temperature T Length L BL thickness δ t
Postulate: If (4.19) is valid, we pose two questions: (1) What terms in (.19) can be dropped? () Under what conditions is (4.19) valid? Answer first question Assign scales: δ. δ t << 1 (4.19) L y δ t (4.0) T T L s (4.1) Scales for u and v depend on whether is larger or smaller than T δ t (4.7b) 3
Two cases, Fig. 4.4: Case (1): δ t > δ Scaling of continuity: δ v V t L Scales for convection terms in (.19): u V (4.) (4.3) 4
Use (4.7b) and (4.0-4.3) and T u T v y V V T L T L Conclusion: the two terms are of the same order Scale for conduction terms: and y T T T L T δ t (a) (b) (c) (d) Compare (c) with (d), use : δ t L << 1 5
Energy equation simplifies to T u + Second question: Under what conditions is (4.19) valid? T << y T (e) T T v α (4.18) δ t L y y Balance between convection and conduction: Scaling V T u T L << 1 T α y T α δ t (4.19) 6
or or or Conclusion: δ t L <<1 δt L δt L Define Peclet number Pe δ L α V L k ρ c V p t 1 L PrRe L PrRe L >>1 (4.4) when (4.5) Pe PrRe L (4.6) Eample: For Pe 100, δ t L 0.1 7
When is δ t > δ? Take ratio of (4.4) to (4.14b) δ δ Criterion for : δ t > δ δ > δ t 1 Pr t Pr << 1 (4.7) when (4.8) Case (): δ < δ t Fig. 4.4 u V δ t δ 8
u within the thermal boundary layer is smaller than free stream velocity Similarity of triangles Scaling of continuity v u δ V t (4.9) δ V (4.30) δ t Lδδ Use (4.9), (4.30) and follow procedure of case (1): conclusion: (1) The two terms are of the same order () Aial conduction is negligible compared to normal conduction Second question: Under what conditions is (4.19) valid? 9
Balance between convection and conduction: y T T u α Use (4.9) for u, scale each term t t T L T V δ α δ δ t or However Substitute into (f) ( ) L L V c k L p t δ ρ δ 3 / (f) Re L L 1 δ (4.14b) 30
Conclusion: δ t 1/3 L Pr 1 Re L (4.31) δ t <<1 L δ t < δ When is? Take ratio of (4.31) to (4.14b) when 1/3 (4.3) Pr Re L >> 1 Criterion for : δ t < δ δ < δ δ t δ 1 1/3 Pr (4.33) t when Pr 1/3 >> 1 (4.34) 31
4.3 Summary of Boundary Layer Equations for Steady Laminar Flow Assumptions: (1) Newtonian fluid () two-dimensional (3) negligible changes in kinetic and potential energy (4) constant properties Assumptions leading to boundary layer model (5) slender surface (6) high Reynolds number (Re > 100) (7) high Peclet number (Pe > 100) 3
Introduce additional simplifications: (8) steady state (9) laminar flow (10) no dissipation (Φ 0) (11) no gravity and (1) no energy generation ( q 0 ) Governing boundary layer equations: Continuity: -Momentum: u u v + 0 y u 1 dp u v + (4.13) u + y ρ d ν y (.) 33
Energy: T u + T y T α y v (4.18) Note the following: (1) Continuity is not simplified for boundary layer flow () Pressure in (4.13) is obtained from inviscid solution outside BL. Thus (.) and (4.13) have two unknowns: u and v ρβ g( T T ) (3) To include buoyancy, add to right (4.13) (4) Recall all assumptions leading the 3 equations 34
4.4 Solutions: Eternal Flow Streamlined body in an infinite flow Eamine thermal interaction Need temperature distribution T Temperature depends on velocity distribution For constant properties, velocity distribution is independent of temperature 4.4.1 Laminar Boundary Layer Flow over Semi-infinite Flat Plate: Uniform Surface Temperature 35
Plate is at temperature T s T Upstream temperature is Upstream velocity uniform and parallel For assumptions listed in Section 4.3 the continuity, momentum and energy are given in (.), (4.13) and (4.18) Transition from laminar to turbulent at: Re V / ν 500,000 t t 36
(i) Velocity Distribution Find: Velocity distribution Boundary layer thickness δ ( ) τ o () Wall shearing stress (a) Governing equations and boundary conditions: Continuity and -momentum: u u v + 0 y u u 1 dp u u + + y ρ d ν y The velocity boundary conditions are: (.) v (4.13) u (,0) v (,0) 0 0 (4.35a) (4.35b) 37
u ) δ () τ o () (b) Scale analysis: Find and Result of Section 4..4: τ o Wall stress : τ y τ V (, (4.35c) u 0, y) V ( (4.35d) δ 1 Re y y 0 (,0) 0 At wall, v Scales for u and y : o v µ + u(,0) µ y u y (4.16) (.7a) τ (4.36) 38
(4.36) is scaled using (4.7) Use (4.16) for δ τ Friction coefficient C f : Use (b) for τ o C τ u V L o µ V δ µ V o Re f C f τ o ( 1/ ) ρv 1 Re (4.7a) (4.7c) (a) (b) (4.37a) (4.37b) 39
(c) Blasius solution: similarity method Solve (.) and (4.13) for the u and v Equations contain 3 unknowns: u, v, and Pressure is obtained from the inviscid solution outside BL Inviscid solution: Uniform inviscid flow over slightly curved edge BL Neglect thickness δ Model: uniform flow over a flat plate of zero thickness Solution: V Thus the pressure gradient is p p u, v 0, p constant (4.38) dp d 0 (4.39) 40
(4.39) into (4.13) (4.40) is nonlinear u u + u u v ν (4.40) y y Must be solved simultaneously with continuity (.) Solution was obtained by Blasius in 1908 using similarity transformation: Combine and y ito a single variable η(, y) Postulate that (, u V y) y V ν η (4.41) depends onη(, y) only 41
f f (η) to be determined NOTE: (1) Including V /ν in definition of η, is for convenience only η (, y ) u V df (1) in (4.41) is arrived at by formal procedure Continuity (.) gives v: v y Multiplying by dy, integrate v u u dy (4.4) (a) 4
dy u / Use (4.41) and (4.4) to epress and in terms of the variable η Chain rule: dy u Use (4.41) and (4.4) into above (b) and (c) into (a) v V u 1 V (b) ν ν V du d V d d f η d η f η (c) 43
Integration by parts gives v V f (η) 1 ν df η V Need function, use momentum equation First determine u / y and u / y f (4.43) u u du d η V y dy d f V ν (d) y u V d 3 f 3 V ν (e) (4.4), (4.43) and (c)-(e) into (4.40) 44
3 d f d f + f ( η) 3 0 (4.44) Partial differential equations are transformed into an ordinary differential equation NOTE: and y are eliminated in (4.44) Transformation of boundary conditions df ( ) f ( 0) 1 0 (4.45a) (4.45b) df (0) 0 (4.45c) 45
df ( ) 1 (4.45d) Equations (4.44) is third order. How many boundary conditions? Difficulty: (4.44) is nonlinear Solution by power series (Blasius) Result: Table 4.1 46
η y Table 4.1 Blasius solution [1] V v f df u V 3 0.0 0.0 0.0 0.3306 0.4 0.0656 0.1377 0.33147 0.8 0.10611 0.6471 0.3739.4 0.930 o.7899 0.809.8 1.3099 0.8115 0.18401 3. 1.56911 0.87609 0.13913 3.6 1.9954 0.9333 0.09809 4.0.30576 0.9555 0.0644 4.4.6938 0.97587 0.03897 4.8 3.08534 0.98779 0.0187 5.0 3.839 0.99155 0.01591 5. 3.48189 0.9945 0.01134 5.4 3.68094 0.99616 0.00793 5.6 3.88031 0.99748 0.00543 47 d f 3
Find δ () wall stress τ o () Define δ as the distance y from the plate where u/v 0.994, Table 4.1 gives or Scaling result: Wall stress τ o : use δ 5. ν V δ 5. Re (4.46) δ 1 Re o u(,0) µ y (4.16) τ (4.36) 48
(d) into (4.36), use Table 4.1 τ o V d f µ V (0) 0. 3306 µ V ν V ν (4.47) Friction coefficient C f : (4.47) into (4.37a) Scaling result: 0. 664 C f Re C (4.48) f 1 Re (4.37b) 49
(ii) Temperature Distribution Isothermal semi-infinite plate Determine: δ t, h() and Nu Need temperature distribution (a) Governing equation and boundary conditions Assumption: Listed in Section 4.3 50
Energy equation T u + The boundary condition are: δ T y T α y v (4.18) T (,0) T s (4.49a) T ) (, T (4.49b) T 0, y) ( (4.49c) T Nu (b) Scale analysis: t, h() and From Section 4..5: Set L (4.4) and (4.31) Case (1): δ t > δ ( Pr <<1) δ t 1 PrRe (4.50) 51
Case (): δ < δ t (Pr >>1) δ t 1 Pr 1/3 Heat transfer coefficient h() R e (4.51) h k T(,0) y T T Use scales of (4.0) and (4.1) into above s (1.10) Where δ t is given by (4.50) and (4.51). h k (4.5) δ t 5
Case (1): δ t > δ ( Pr <<1), (4.50) into (4.5) h PrRe, for Pr <<1 k Nu (4.53) h Nu k (4.54) Local Nusselt number (4.53) into (4.54) Case (): δ << δ Nusselt number: Nu PrRe, for Pr <<1 (4.55) e t ( Pr >>1). Substituting (4.51) into (4.5) k 1/3 h Pr Re, for Pr >>1 (4.56) Nu 1/3 Pr Re, for Pr >>1 (4.57) 53
(c) Pohlhausen s solution: δ T(,y),, h(), Nu Energy equation (4.18) is solved analytically Solution by Pohlhausen (191) using similarity transformation Defined θ (4.58) into (4.18) B.C. θ u + T T t s θ (4.58) T Ts θ θ v α (4.59) y θ (,0) θ (,0) θ (,0) 0 1 1 y (4.60a) (4.60b) (4.60c) 54
Solve (4.59) and (4.60) using similarity Introduce transformation variable η Assume V η (, y) y (4.41) ν θ (, y) θ ( η) Blasius solution gives u and v v V u V df ν df η V (4.41)-(4.43) into (4.59) and noting that 1 f (4.4) (4.43) 55
η θ η η η θ θ d d d d η θ ν η η θ θ d d V y d d y η θ ν θ d d V y ν y (4.59) becomes Result: 0 ) ( + η θ η η θ d d f Pr d d (4.61) transformed equation is differential Partial equation into an ordinary differential 56
NOTE: (1) One parameter: Prandtl number Pr () (4.61) is linear, nd order ordinary D.E. f (η) (3) in (4.61) represents the effect motion Transformation of B.C.: θ ( ) 1 θ ( 0) 0 θ ( ) 1 Solution: Separate variables, integrate twice, use B.C. (4.6) (Details in Appendi) θ ( η) 1 η 0 d f ) d f Pr Pr (4.6a) (4.6b) (4.6c) (4.63) 57
Surface temperature gradient: dθ (0) η [ 0.33] d f Pr Pr (4.64) Integrals are evaluated numerically d f is obtained from Blasius solution Results are presented graphically in Fig. 4.6 58
1.0 100 10 1 0.7( air) 0.8 T T Ts T s 0.6 0.4 0.1 0. Pr 0.01 Determine: δ 0 Fig. 4.6 t δ 4 6 8 10 1 14, h() and t η V y ν Pohlhausen's solution Nu T δ t T Fig. 4.6 gives. At y,, or 59
θ T T δ t Ts T () 1 Fig. 4.6 shows that depends on Pr Local heat transfer coefficient h(): use (1.10) where s T(,0) y h k T s T T(,0) y y δ, at (4.65) dt dθ (0) dθ η y t (1.10) Use (4.41) and (4.58) into above T,0) y V ν ( ( T Ts ) dθ (0) 60
Substitute into (1.10) V h( ) k ν Average heat transfer coefficient: h Use (4.66) and integrate h dθ (0) (4.66) 1 L k L 0 L h( ) d Re dθ (0) (.50) L (4.67) Local Nusselt number: (4.66) into (4.54) Nu dθ (0) Re (4.68) 61
Average Nusselt number: Nu dθ (0) L Re L (4.69) Total heat transfer rate q T : Plate length L and width W. Apply Newton s law or L qt h( )( Ts T ) W d 0 L ( Ts T ) W h( ) d ( Ts T ) WLh q 0 T ( s ) T T Ah (4.70) s Heat transfer coefficient and Nusselt number dθ (0) depend on surface temperature gradient 6
d θ (0) depends on Pr It is determined from (4.64) Values in Table 4. Approimate values of d θ (0) Table 4. d θ (0) Pr 0.001 0.0173 0.01 0.0516 0.1 0.140 0.5 0.59 are given by: 0.7 0.9 1.0 0.33 7.0 0.645 10.0 0.730 15.0 0.835 50 1.47 100 1.57 1000 3.387 63
dθ (0) dθ (0) dθ (0) 0.33 Pr Compare with scaling: 0.564 Pr 1/ 3 0.339 Pr 1/, 0.6 < Pr < 10 (4.71b) 1/ 3 Two cases: Pr << 1 and Pr >> 1 Combine (4.71a) and (4.71c) with (4.68) Nu 1/ 0.564 Pr Re, for Pr, Pr < 0.05 (4.71a), Pr >10 (4.71c) < 0.05 (4.7a) Nu 1/ 3 0.339 Pr Re, for Pr > 10 (4.7c) 64
Scaling results: Nu PrRe, for Pr <<1 (4.55) Nu Fluid properties: 1/3 Pr Re, for Pr >>1 (4.57) Evaluated at the film temperature T f T f T s + T (4.73) 65
4.4. Applications: Blasius Solution, Pohlhausen s Solutions and Scaling Three eamples Eample 4.1: Insect in Search of Advice Air at 30 o C, V 4 m/s Insect at 0 Determine velocity u at locations 0, 1,, 3, 4. Is insect inside BL? (1) Observations. Eternal forced convection boundary layer problem 66
Changes in velocity between 1 and 3 should be small compared to those between and 4 Location 4 should have the lowest velocity If the flow is laminar Blasius applies The flow is laminar if Reynolds number is less than 500,000 () Problem Definition. Determine u at the five locations (3) Solution Plan. Check the Reynolds number for BL approimations and if the flow is laminar If laminar, use Blasius solution, Table 4.1, to determine u and δ (4) Plan Eecution 67
(i) Assumptions. All assumptions leading to Blasius solution: These are: Newtonian fluid steady state constant properties two-dimensional laminar flow (Re < 5 10 5 ) viscous boundary layer flow (Re > 100) (7) uniform upstream velocity flat plate negligible changes in kinetic and potential energy no buoyancy (β 0 or g 0) 68
(ii) Analysis Re V ν (a) V upstream velocity 4 m/s 6 ν kinematic viscosity 16.01 10 m /s Transition Reynolds number: Laminar flow if Re < Re t 5 10 5 (b) Re t Viscous BL approimations are valid for At 151 mm: Re 4( m/s)0.151(m) 6 16.01 10 > 100 Re 37,76 (m /s) (c) 69
BL flow is laminar. Use Blasius solution Determine V η y δ ν 5. Re (iii) Computations. Calculate η at each location, use Table 4.1 to find u/v. Results: location (m) y (m) η u/v u(m/s) 0 0.150 0.00.581 0.766 3.064 1 0.151 0.00.573 0.765 3.06 0.150 0.003 3.87 0.945 3.78 3 0.149 0.00.59 0.768 3.07 4 0.150 0.001 1.91 0.4 1.688 (d) (4.46) 70
δ 0. 151 Re 37, 76 Use (4.46) to determine at m and 5. δ Re 5. 37,76 0.151(m) Thus the insect is within the boundary layer (iv) Checking. Dimensional check: 0.004m Equations (a) and (d) are dimensionally correct 4 mm Qualitative check: u at the five locations follow epected behavior (5) Comments. The insect should move to location 4 Changes in u with respect to are minor Changes in u with respect to y are significant 71
What is important for the insect is the magnitude of the velocity vector V (u + v ) 1/ and not u. However, since v << u in boundary layer flow, using u as a measure of total velocity is reasonable Eample 7.: Laminar Convection over a Flat Plate Water 0.5 m/s V T 35 C T s 85 C L 75 cm 7
() [a] Find equation for δ t [b] Determine h at 7.5 cm and 75 cm [c] Determine q T for a plate 50 cm wide [d] Can Pohlhausen's solution be used to at the trailing end of the plate if its length is doubled? q (1) Observations Eternal forced convection over a flat plate δ () t increases with Newton s law of cooling gives q and h() decreases with qt Pohlhausen's solution is applies laminar flow and all other assumptions made Doubling the length doubles the Reynolds number 73
() Problem Definition. Determine temperature distribution (3) Solution Plan Compute the Reynolds and Peclet numbers to establish if this is a laminar boundary layer problem Use Pohlhausen's solution to determine, h(), and q (4) Plan Eecution δ q (i) Assumptions. All assumptions leading to Blasius solution: These are: Newtonian fluid two-dimensional negligible changes in kinetic and potential energy constant properties boundary layer flow t T 74
steady state laminar flow no dissipation no gravity no energy generation flat plate negligible plate thickness uniform upstream velocity V uniform upstream temperature T uniform surface temperature Ts no radiation 75
(ii) Analysis and Computations Are BL approimations valid? Calculate the Reynolds and Peclet. Condition: Re > 100 and Pe Pr > 100 (a) Transition Reynolds number: Properties at T s 85 o C T 35 o C T f T T (85+ 35)( o C)/ 60 o C f f Re V Re ν Re t 5 Re 5 10 ( T + T )/ s (b) (c) 76
k 0.6507 W/m- o C Pr 3.0 ν 0.4748 10 6 m /s. at 7.5 cm Re and Pe are Re V ν 0.5(m/s)0.075(m) 4 3.949 10 6 0.4748 10 (m / s) Pe Re Pr 4 3.949 10 3 11.85 10 4 BL approimations are valid, flow is laminar Pohlhausen's solution is applicable. [a] Determine δ t : y δ t T T At, ( T ) Ts ( ) θ ( ηt ) T T s 1 77
ηt θ ( η t ) 1 From Fig. 4.6: Value of at and Pr 3 at is approimately.9 or η t ηt.9 δ V / ν.9 V ν [b] Heat transfer coefficient: d θ (0) : t V h( ) k ν dθ (0) 0.33 Pr.9 Re dθ (0) (d) (4.66) 1/ 3, 0.6 < Pr < 10 (4.71b) 78
Pr 3 dθ ( 0) 1/ 3 0.33(3) 0.4788 Substituting into (4.66) for 0.075 m At 0.75 m [c] Heat transfer rate: W h 85.5 o m C q h W 61 m o C T ( s ) L length of plate 75 cm 0.75 m T T Ah (4.70) s W width of plate 50 cm 0.5 m 79
Re L 3.949 10 5 h Substitute into (4.70) k L Re dθ (0) L (4.67). Substitute into the above h W 5.1 m o C q T 9789W [d] Doubling the length of plate: Re L > Re t Re L (3.949 10 5 ) 7.898 10 5 Flow is turbulent, Pohlhausen's solution is not applicable 80
(iii) Checking. Dimensional check: Reynolds number is dimensionless and that units of h and are h correct Qualitative check: As is increased h decreases Quantitative check: Computed values of h are within the range of Table 1.1 (5) Comments Check Reynolds number before applying Pohlhausen's solution Velocity boundary layer thickness is given by δ 5. Re Compare (d) with equation (4.46): δ (4.46) δ < δ t 81
Eample 7.3: Scaling Estimate of Heat Transfer Rate Use scaling to determine the total heat transfer rate for conditions described in Eample 7. (1) Observation Newton s law gives heat transfer rate The heat transfer coefficient can be estimated using scaling () Problem Definition. Determine the heat transfer coefficient h (3) Solution Plan. Apply Newton s law of cooling and use scaling to determine h (4) Plan Eecution 8
(i) Assumptions Newtonian fluid two-dimensional negligible changes in kinetic and potential energy constant properties boundary layer flow steady state no dissipation no gravity no energy generation no radiation (ii) Analysis. Application of Newton s law of cooling gives q T ( s ) T T Ah (4.70) s 83
A surface area LW, m h average heat transfer coefficient, W/m - o C L length of plate 75 cm 0.75 m q T total heat transfer rate from plate, W T s surface temperature 85 o C T free stream temperature 35 o C W width of plate 50 cm 0.5 m h by (1.10) h k T(,0) y T T s (1.10) k thermal conductivity 0.6507 W/m- o C 84
Follow analysis of Section 4.41, scale of h for Pr >>1 h h, k 1/3 h Pr Re, for Pr >>1 (4.56) V Re and Pr 3 ν Set L, A WL and substitute (4.56) into (4.70) q ( T T ) W k T ( s Pr 1/3 Re L (a) (iii) Computations Substitute into (a) q T (85 35)( o Re L 5 3.949 10 C)0.5(m)0.6507(W/m qt 14740 W o C)3 1/3 394900 85
q T 9789 Using Pohlhausen s solution gives W (iv) Checking. Dimensional Check:Solution (a) is dimensionally correct (5) Comments. Scaling gives an order of magnitude estimate of the heat transfer coefficient. In this eample the error using scaling rate is 50% 86
4.4.3 Laminar Boundary Layer Flow over Semi-infinite Flat Plate: Variable Surface Temperature Consider uniform flow over plate Surface temperature varies with as: n s( ) T C (4.7) T 87
C and n, constants T is free stream temperature Determine,, and T (, y) h() Nu qt Assumptions: summarized in Section 4.3 (i) Velocity Distribution For constant properties velocity is independent of the temperature distribution Blasius solution is applicable: u V df (4.4) v V 1 ν V df η f (4.43) 88
(, y) y V ν η (4.41) (ii) Governing Equations for Temperature Distribution Based on assumptions OF Section 4.3: T u + Boundary condition T y T α y v (4.18) n (,0) Ts T C (4.73a) T + T ) (, T (4.73b) T 0, y) ( (4.73c) T 89
(iii) Solution Solution to (4.18) is by similarity transformation Define :θ Assume T T s θ (4.58) T Ts θ (, y) θ ( η) (4.75) Use (4.41)-(4.43), (4.58), (4.7), (4.75), energy (4.18) transforms to (Appendi C) B.C. (4.73): d θ df + npr (1 θ ) + θ ( ) θ ( 0) θ ( ) Pr 1 0 1 dθ f ( η) 0 (4.76) (4.76a) (4.76b) (4.76c) 90
Note: Two B.C. coalesce into one Heat transfer coefficient and Nusselt number: Use (1.10) where h k T (,0) y T(,0) y T T s dt d θ (0) η dθ y Use (4.41),(4.58) and (4.7) into the above Substitute into (1.10) T(,0) n V dθ (0) C y ν (1.10) 91
V h( ) k ν dθ (0) (4.78) Average heat transfer coefficient: Use (.50) h 1 L 0 L h( ) d Substitute (4.78) into (.50) and integrate (.50) k dθ (0) h Re (4.79) L L Local Nusselt number: (4.78) into (4.54) Nu Average Nusselt number: dθ (0) Re (4.80) 9
(ii) Results: Nu dθ (0) L Re (4.81) L Heat transfer coefficient and Nusselt number dθ (0) depend on surface temperature gradient Equation (4.76) subject to boundary conditions (4.77) is solved numerically Solution depends on two parameters: the Prandtl number Pr and the eponent n in (4.7) d θ (0)/ is presented in Fig. 4.8 for three Prandtl numbers. 93
.0 30 d θ (0) 1.0 10 Pr 0.7 Fig. 4.8 n 0 0.5 1.0 1.5 d θ ( 0) for plate with varying surface temperature n T ( ) T C s 4.4.3 Laminar Boundary Layer Flow over a Wedge: Uniform Surface Temperature 94
Symmetrical flow over a wedge of angle Uniform surface temperature β π Uniform upstream velocity, pressure and temperature Both pressure and velocity outside the viscous BL vary with distance along wedge For assumptions of Section 4.3, the -momentum eq. is u 1 dp u v + (4.13) u u + y ρ d ν y m C is a constant and describes wedge angle: 95
β β dp Apply (4.13) at edge of BL to determine : d Flow is inviscid v ν 0 u V () Substitute into (4.13) u u + m (4.83) 1 dp ρ d V V u V u v V + ν (4.84) y y 96
The B.C. are (i) Velocity Solution: m (, ) V ( ) C (4.84a) u u (,0) v (,0) By similarity transformation (follow Blasius approach) η Define a similarity variable : 0 0 (4.84b) (4.84c) η(, y) y V ( ) ν Assume u(, y) to depend on η : V u ( ) C y ν df ( m 1) / (4.86) (4.87) 97
Continuity (.), (4.86) and (4.87) give v v V ( ) ν V ( ) m + 1 F Substitute (4.8) and (4.86)-(4.88) into (4.84) d 3 F 3 1 1 + m m m + 1 d F df + F m + m df η This is the transformed momentum equation B. C. (4.85) transform to df(0) F ( 0) df( ) 0 0 1 0 (4.88) (4.89) (4.89a) (4.89b) (4.89c) 98
Note the following regarding (4.89) and (4.90): and y do not appear Momentum eq. (4.89) is 3rd order non-linear m β 0 m 0 Special case: represents a flat plate Setting in (4.89) and (4.90) reduces to Blasius problem (4.44) & (4.45), (4.89) is integrated numerically F(η) F ( η) f ( η) df / u v Solution gives and. These give and (ii) Temperature Solution: Energy equation: θ u + θ θ v α (4.59) y y 99
Boundary conditions: where θ (,0) θ (,0) θ (,0) T 0 1 1 T (4.60a) (4.60b) (4.60c) s θ (4.58) T T s Same energy equation and B.C. as the flat plate. Is temperature distribution the same? Equation (4.59) is solved by similarity transformation. Assume: where θ (, y) θ ( η) (4.75) 100
η(, y) y V ( ) ν y C ν ( m 1) / Substitute (4.86)-(4.88) and (4.75) into (4.59) and (4.60) d θ + Pr ( m dθ + 1) F( η) 0 θ ( 0) θ ( ) θ ( ) Partial differential equations is transformed into ordinary equation Two governing parameters: Prandtl number Pr and the wedge size m (491) a linear second order equation requiring two B.C. 0 1 1 (4.86) (4.91) (4.9a) (4.9b) (4.9c) 101
F (η) in (4.91) represents effect of fluid motion B.C. (4.60b) and (4.60c) coalesce into a single condition m β 0 m 0 Special case: represents flat plate. Set in (4.91) reduces to Pohlhausen s problem (4.61) Solution: (Details in Appendi B) Separate variables in (4.91) Integrate twice Applying B.C. (4.9), gives θ ( η) 1 η (m + 1) Pr ep (m + 1) Pr ep 0 0 η 0 η F( η) F( η) (4.93) 10
d θ (0) Temperature gradient at surface : Differentiate (4.93), evaluate at η 0 dθ (0) (m + 1) Pr ep 0 0 F( η) η η d 1 (4.94) F (η) is given in the velocity solution Evaluate integrals in (4.93)&(4.94) numerically d θ (0) F (0) Results for and are in Table 4.3 103
d θ (0) Table 4.3 Surface temperature gradient and velocity gradient F (0) for flow over an isothermal wedge wedge angle. m πβ F (0) d θ (0)/ at five values of Pr 0.7 0.8 1.0 5.0 10.0 0 0 0.306 0.9 0.307 0.33 0.585 0.730 π / 5 0.111 (36 o ) 0.510 0.331 0.348 0.378 0.669 0.851 0.333 π / (90 o ) 0.7575 0.384 0.403 0.440 0.79 1.013 1.0 π (180 o ) 1.36 0.496 0.53 0.570 1.043 1.344 h() Use Table 4.3 to determine and Nu 104
where h k T(,0) y T(,0) y T T s dt dθ (0) dθ Use (4.58),(4.75) and (4.86) into above T,0) y Substitute into (1.10) ( ( T Ts ) V ( ) h( ) k ν η y V ( ) ν dθ (0) dθ (0) (1.10) (4.95) Local Nusselt number: substitute (4.95) into (4.54) 105
where Nu dθ (0) Re (4.96) Re V () ν (4.97) h() Nu d θ (0) Key factor in determining and : Surface temperature gradient is, listed in Table 4.3. 106