Chapter 7 Eternal Forced Convection 1 School of Mechanical Engineering
Contents Chapter 7 7-1 rag and Heat Transfer in Eternal Flow 3 page 7-2 Parallel Flow Over Flat Plates 5 page 7-3 Flow Across Cylinders and Spheres 18 page 7-4 Flow Across Tube Banks 27 page 2 School of Mechanical Engineering
7-1 rag and Heat Transfer in Eternal Flow rag The force a flowing fluid eerts on a body in the flow direction F rag coefficient: C (where A frontal area) 1 2 V A 2 rag force depends on the density of the fluid, the upstream velocity, and the size, shape, and orientation of the body. 1 Total drag coefficient: C C C,friction,pressure 2 Flat plate normal to flow direction: High pressure + + + + + + - - - - - - ow pressure C C C, C 0,pressure p,friction Wall shear 3 Flat plate parallel to flow direction : C C C, C 0,friction f,pressure rag force acting on a flat plate normal to flow depends on the pressure only and is independent of the wall shear, which acts normal to flow. 3 School of Mechanical Engineering
7-1 rag and Heat Transfer in Eternal Flow Heat transfer 1. The average Nusselt number in eperimental approach can be epressed in terms of the Reynolds and Prandtl numbers: Nu m C Re Pr n 2. The fluid properties are evaluated at the film temperature, as T f T s T 2 3. The average friction and convection coefficients for the entire surface 1 C C 0, d, and h 1 0 h d 4. The rate of heat transfer is Q has Ts T (where A surface area) s 4 School of Mechanical Engineering
7-2 Parallel Flow Over Flat Plates Parallel flow over flat pates The transition from laminar to turbulent flow is best characterized by the Reynolds number. V h C f aminar Transition Turbulent The Reynolds number at a distance from the leading edge of a flat plate. h, C f, V V Re The critical Reynolds number is V Recr 510 5 5 School of Mechanical Engineering
7-2 Parallel Flow Over Flat Plates Friction coefficient: The local friction coefficient and the boundary layer thickness at location aminar: Turbulent: C C 0.664 5 and, Re 510 f, 1/2, 1/2 Re Re 0.0592 0.382 and, 510 Re 10 5 7 f, 1/5, 1/5 Re Re The average friction coefficient over the entire plate aminar: Turbulent: 1 1.328 C C d, Re 5 10 f 0 f, 1/ 2 Re 1 0.074 C C d, 5 10 Re 10 5 7 f 0 f, 1/5 Re Combined laminar and turbulent flow ( when the length of laminar region cannot be neglected. ) 5 5 1 cr 0.074 1742 C C d C d Re f 0 f, laminar f, turbulent 1/5 cr Re 510 Re 10 cr 5 7 Recr V 6 School of Mechanical Engineering
7-2 Parallel Flow Over Flat Plates Heat transfer coefficient: The local Nusselt number at location aminar: Turbulent: The average Nusselt number over the entire plate aminar: Turbulent: h k h 0.8 1/ 3 5 7 Nu 0.0296Re Pr, 510 Re 10, 0.6 Pr 60 k 0.5 1/ 3 5 Nu 0.332Re Pr, Re 510, Pr 0.60 h k h 0.8 1/ 3 5 7 Nu 0.037 Re Pr, 510 Re 10, 0.6 Pr 60 k 0.5 1/ 3 5 Nu 0.664Re Pr, Re 510, Pr 0.60 Combined laminar and turbulent flow ( when the length of laminar region cannot be neglected. ) 1 cr h h 0, laminar h, turbulentd cr h 0.8 1/3 Nu = 0.037Re 871Pr k 510 Re 10 5 7 0.6 Pr 60 7 School of Mechanical Engineering
7-2 Parallel Flow Over Flat Plates Flat Plate with Unheated Starting ength The local Nusselt number at location T T for aminar: Turbulent: Nu Nu The average convection coefficients for ξ= 0, aminar: Turbulent: h aminar: Turbulent: Nu 0.332 Re Pr 1 / 1 / 0.5 1/ 3 for =0 3/ 4 1/ 3 3/ 4 1/ 3 Nu 0.0296 Re Pr 1 / 1 / 3/ 4 21 / 1 / h h s 0.8 1/ 3 for =0 9 /10 1/ 9 9 /10 1/ 9 51 / 4 1 / ave h ave 2 h 5 4 h 9 /10 h h T V Velocity boundary layer Thermal boundary layer T s 8 School of Mechanical Engineering
7-2 Parallel Flow Over Flat Plates Uniform Heat Flu aminar: Turbulent: Nu 0.453Re Pr 0.5 1/ 3 Nu 0.0308Re Pr 0.8 1/ 3 When heat flu is prescribed, the rate of heat transfer to or from the plate and the surface temperature at a distance are determined from Q q A s s and q h s qs h T s T Ts T where As is the heat transfer surface area. 9 School of Mechanical Engineering
7-2 Parallel Flow Over Flat Plates E 7.1 Engine oil at 60 flows over a 5-m-long flat plate whose temperature is 20 with a velocity of 2 m/s. etermine the total drag force and the rate of heat transfer per unit width of the entire plate. u T Oil 2 m/s o 60 C A Q o Ts 20 C 5 m Solution 1. Given Engine oil flow over an isothermal flat plate. = 5 m, T = 60, T s = 20, u = 2 m/s 2. Find The total drag force The rate of heat transfer per unit width of the entire plate. Continue 10 School of Mechanical Engineering
7-2 Parallel Flow Over Flat Plates 3. Schematic u 2 m/s T o 60 C Q Oil A Ts o 20 C 5 m 4. Assumption 1. Steady state condition 2. Re cr =5 10 5 3. Negligible radiation heat transfer 5. Properties o - The film temperature of the engine oil: Tf ( Ts T ) / 2 40 C 3 - The density of the oil: 876kg/m - The Prandtl number: Pr 2870 o - The thermal conductivity of the oil: k 0.144 W/m C 6 2 - The kinematic viscosity: 24210 m /s Continue 11 School of Mechanical Engineering
7-2 Parallel Flow Over Flat Plates 6. Solve (a) The total drag force To determine the friction coefficient, the Reynolds number must first be determined. The Reynolds number at the end of the plate is u v (2 m/s)(5 m) 24210 m /s 4 5 Re 5 m 4.1310 < Re ( 5 10 ) 6 2 cr Thus we have laminar flow over the entire plate, and the average friction coefficient is determined from 0.5 4 0.5 C f 5 m 1.328Re 1.328 (4.1310 ) 0.00653 Then the drag force acting on the plate becomes F 2 3 2 u 2 (876 kg/m )(2 m/s) C f A 0.00653 (51 m ) 57.2 N 2 2 Continue 12 School of Mechanical Engineering
7-2 Parallel Flow Over Flat Plates (b) The rate of heat transfer per unit width of the entire plate From Newton s law of cooling that the rate of convection heat transfer to the plate is Q ha T T conv To determine the average Nusselt number by using the laminar flow relations for a flat plate Then, k h 5 m o 0.144 W/m C 2 o Nu= (1,918) 55.2 W/m C and the rate of heat transfer per unit width of the entire plate is Q ha T T conv h k 2 o 2 o ( s ) (55.2 W/m C)(5 1 m )(60 20) C =11,040 W=11.04 kw s 1/3 0.5 1/3 4 0.5 Nu 0.664 Re Pr 0.664 (4.1310 ) 2,870 1,918 13 School of Mechanical Engineering
7-2 Parallel Flow Over Flat Plates E 6.2 The local atmospheric pressure in enver, Colorado (elevation 1610 m), is 83.4 kpa. Air at this pressure and 20 flows with a velocity of 8 m/s over a 1.5 m 6 m flat plate whose temperature is 134. etermine the rate of heat transfer from the plate if the air flows parallel to the 6m long side. Solution 1. Given Airflow over an isothermal flat plate = 6 m, T = 20, T s = 134, u = 8 m/s, P = 83.4 kpa 2. Find The rate of heat transfer from the plate. Continue 14 School of Mechanical Engineering
7-2 Parallel Flow Over Flat Plates 3. Schematic T o 20 C, V 8 m/s Q Patm 83.4 kpa air T s o 134 C 1.5 m 6 m 4. Assumption 1. Steady state condition 2. Re cr =5 10 5 3. Radiation effects are negligible. 4. Air is an ideal gas. Continue 15 School of Mechanical Engineering
7-2 Parallel Flow Over Flat Plates 5. Properties The properties of air at the film temperature of T f = (T s +T )/2 = 77 The Prandtl number: Pr = 0.706 The thermal conductivity of the air: k = 0.0297 W/m K The kinematic viscosity at atmospheric pressure: ν = 2.06 10-5 m 2 /s 2 p1 p1 From the ideal gas law, 6. Solve When air flow is parallel to the long side, we have c = m, and the Reynolds number at the end of the plate becomes V v (8 m/s)(6 m) 2.5010 m /s p p 2 1 1 2 2 The kinematic viscosity in enver: ν = 2.50 10-5 m 2 /s c 6 5 Re 1.9210 Re (5 10 ) 5 2 cr Thus, we have combined laminar and turbulent flow, and the average Nusselt number for the entire plate is determined from h k 1/3 0.8 1/3 6 0.8 Nu Pr (0.037 Re 871) 0.706 [0.037(1.9210 ) 871] 2, 727 Continue 16 School of Mechanical Engineering
7-2 Parallel Flow Over Flat Plates Convection heat transfer coefficient Convection heat transfer area k h 6 m o 0.0297 W/m C 2 o Nu (2, 727) 13.5 W/m C A w 2 (1.5 m)(6 m) = 9 m Heat transfer rate 7. Comment Q ha T T 2 o 2 o ( s ) (13.5 W/m C)(9 m )(134 20) C 13,850 W If we disregarded the laminar region and assumed turbulent flow over entire plate, we would get Nu = 3,520 from h k 4 5 1 3 5 7 Nu 0.037 Re Pr (0.6 Pr 60, 510 Re 10 ) which is 28% higher than value we calculated above. Note that, in this eample, the length of laminar region is cr = Re cr ν/v = 1.563 m, which is not negligible. 17 School of Mechanical Engineering
7-3 Flow Across Cylinders and Spheres rag Friction drag : This is due to the boundary layer surface shear stress. Form (or pressure) drag : This is due to a pressure gradient in the flow direction resulting from formation of the wake. F 2 u CA 2 where A = frontal area Cylinder: A = Sphere: A = π 2 Favorable pressure gradient Adverse pressure gradient u ( ) p 0 u p 0 midplane wake Separation point Wake Stagnation point Separation point Flow reversal Note : With increasing Reynolds Number, the effect of separation, and therefore form drag, becomes more important. u 5 The critical Reynolds number : Re, Recr 210 v Vortices 18 School of Mechanical Engineering
7-3 Flow Across Cylinders and Spheres Average drag coefficient for cross flow over a smooth circular cylinder and sphere Re 1 : There is no flow separation in this regime. Re 10 : Separation starts occurring on the rear of the body. 3 10 Re 10 : The region of separation increases with increasing Re. 3 5 10 Re 10 : The drag co efficient remains relatively constant, 5 6 10 Re 10 : There is a sudden drop in the drag coefficient due to the transition to turbulent regime. 19 School of Mechanical Engineering
7-3 Flow Across Cylinders and Spheres Heat transfer coefficient at cylinder & sphere Cylinder Churchill & Bernstein 0.5 1/3 5/8 0.62Re Pr d Red Nud 0.3 1 2/3 1/4 0.4 282,000 1 Pr Where all properties are evaluated at the film temperature. 4/5 100 Re 10 d Red Pr 0.2 7 Sphere Whitaker h Nu 2 0.4Re 0.06Re Pr 1/2 2/3 0.4 spd k s 1/4 3.5 Re 80,000 d 0.7 Pr 380 Where all properties are evaluated at the free-stream temperature T ecept for μ s 20 School of Mechanical Engineering
7-3 Flow Across Cylinders and Spheres Compact form of the average Nusselt number for flow across cylinders Cross section Re f C n Circle 0.4~4 0.989 0.33 4~40 0.911 0.385 40~4,000 0.683 0.466 Nu f n C Re Pr f 1/3 4,000~40,000 0.193 0.618 40,000~400,000 0.0266 0.805 (where Re V / ) f Square 5000~100,000 0.102 0.675 Square (titled 45º) 5000~100,000 0.246 0.588 Table 7-1 Constants for C and n 21 School of Mechanical Engineering
7-3 Flow Across Cylinders and Spheres Cross section Re f C n Heagon 5000~100,000 0.153 0.683 Nu f n C Re Pr f 1/3 Heagon (titled 45º) 5000~19,500 19,500~100,000 0.160 0.0385 0.638 0.782 (where Re V / ) f Vertical plate 4000~15,000 0.228 0.731 Ellipse 2500~15,000 0.248 0.612 22 School of Mechanical Engineering
7-3 Flow Across Cylinders and Spheres E 7.3 Assume that a person can be approimated as a cylinder of 0.3-m diameter and 1.8-m height with a surface temperature of 24. Calculate the body heat loss while this person is subjected to a 15 m/s wind whose temperature is -5. Solution 1. Given Person can be approimated as a cylinder = 0.3 m, l = 1.8 m, T = -5, T s = 24, u = 15 m/s 2. Find The body heat loss Continue 23 School of Mechanical Engineering
7-3 Flow Across Cylinders and Spheres 3. Schematic Air V T 15 m/s o 5 C 0.3 m 1.8 m 4. Assumption Ts 1. Steady state condition 2. Re cr =5 10 5 3. Negligible radiation heat transfer o 24 C 5. Properties The properties of air at the film temperature of T f = (T s +T )/2 = 9.5 The Prandtl number: Pr = 0.7337 The thermal conductivity of the air: k = 0.024352 W/m K The kinematic viscosity at atmospheric pressure: ν = 1.4216 10-5 m 2 /s Continue 24 School of Mechanical Engineering
7-3 Flow Across Cylinders and Spheres 6. Solve The heat transfer rate from the cylinder, approimating the person, is given as q has( Ts T ) where A s = πl Reynolds number V (15 m/s)(0.3 m) Red 313,548 6 2 v 14.21610 m /s h must be estimated from a correlation appropriate. Churchill and Bernstein: this relation appropriates under the below condition 1/2 1/3 5/8 0.62 Re Pr d Red Nu d 0.3 1 2/3 1/4 1 (0.4 / Pr) 282,000 4/5 100 Re 10 d Red Pr 0.2 7 Continue 25 School of Mechanical Engineering
7-3 Flow Across Cylinders and Spheres hence, 1/2 1/3 5/8 0.62 313,548 0.7337 313,548 Nu d 0.3 1 493.176 2/3 1/4 1 0.4 / 0.7337 282,000 Convection heat transfer coefficient is 1/4 h k 0.3 m 3 24.35210 W/m K 2 Nu d 493.176 40.03 W/m K The heat transfer rate is 7. Comment 2 Q 40.03 W/m K 0.3 m1.8 m 24 5 K 1969.4 W Properties are obtained from Table A-15 by using the interpolation method. The wind-chill factor will make you feel colder. In an engineering sense, the increased Reynolds number results in the enhanced Nu number & h. 26 School of Mechanical Engineering
7-4 Flow Across Tube Banks Flow across Tube Banks VT, 1 S T A 1 S u ma efficiency < noise < VT, A 1 T 1 S T A A T S A A S In-line tube Staggered tube The Reynolds number Vma Vma Re The maimum velocity for in-line arrangement. VA V A or VS V S 1 ma T T ma T V ma ST S T V where S S S A 2 T 1 T S T T A S A S /2 2 27 School of Mechanical Engineering
7-4 Flow Across Tube Banks The maimum velocity for staggered arrangement. If 2A > A T : If 2A < A T : V ma ST S Zukauskas correlations (Table 7-2) : T V ST Vma V S ST / 2 2 S N 16 and 0.7 Pr 500 row 0.25 m n Nu C Re Pr Pr/ Pr s Arrangement Range of Re C m n 0~100 0.9 0.4 0.36 In-line 100~1000 0.52 0.5 0.36 1000~2 10 5 0.27 0.63 0.36 2 10 5 ~2 10 6 0.033 0.8 0.4 0~500 1.04 0.4 0.36 Staggered 500~1000 0.71 0.5 0.36 1000~2 10 5 0.35(S T /S ) 0.2 0.6 0.36 2 10 5 ~2 10 6 0.031(S T /S ) 0.2 0.8 0.36 All properties ecept Pr are evaluated at T T T / 2. Pr is evaluated at T. s m inlet outlet s s 28 School of Mechanical Engineering