OUTCOME 2 - TUTORIAL 1
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1 Unit 4: Heat Transfer and Combustion Unit code: K/60/44 QCF level: 5 Credit value: 5 OUTCOME - TUTORIAL Heat transfer coefficients Dimensional analysis: dimensionless groups; Reynolds, Nusselt, Prandtl, Stanton, Grashof numbers Heat transfer mechanism: description of flow in tubes, ducts and across surfaces; boundary layer; laminar and turbulent; forced and natural convection; fluid properties; flow parameters; boiling and condensation Determine heat transfer coefficients: Dittus-Boelter equation for forced convection in circular ducts and tubes, for various fluids, tube dimensions and flow parameters; use of charts and data for fluid properties You should judge your progress by completing the self assessment exercises. On completion of this tutorial the student should be able to do the following. This is the third tutorial in the series on heat transfer and covers some of the advanced theory of convection. The tutorials are designed to bring the student to a level where he or she can solve problems involving practical heat exchangers. On completion of this tutorial the student should be able to do the following. Explain the use of the overall heat transfer coefficient. Explain the factors involved in heat transfer. Apply dimensional analysis to heat transfer. Solve heat transfer problems for basic shapes. D.J.Dunn
2 DERIVATION AND CALCULATION OF HEAT TRANSFER COEFFICIENTS The surface heat transfer coefficients depend on many things such as shape and orientation of the surface, the velocity of the fluid and the properties of the fluid. The fluid properties also depend on the temperature. The velocity of the fluid is a major consideration and this varies with distance from the surface in the boundary layer and distance from the leading edge. All these factors are involved when deriving formula from basic principles. Perhaps the best place to start is with boundary layers. BOUNDARY LAYER - LAMINAR AND TURBULENT When a fluid flows over a surface, the velocity grows from zero at the surface to a maximum at distance. In theory, the value of is infinity but in practice it is taen as the height needed to obtain 99% of the mainstream velocity and this is very small. The boundary layer is important for the following reason. When a fluid is convecting heat to or from a solid surface, the heat transfer close to the surface is by conduction through the boundary layer. Outside the boundary layer the temperature is the bul temperature of the fluid but inside it decreases to the surface temperature. The diagram shows how the temperature varies from the hot fluid on one side of a wall to cold fluid on the other side. There is a boundary layer on both sides. Consider the hot side only. Using the conduction law from the bul fluid : The heat transfer at θ h to the surface at θ is Φ = (/δ)a( θ h - θ ) The heat transfer using the convection law is Φ = ha( θ h - θ ) Equating to find h we have h = /δ So the surface heat transfer depends on the thicness of the boundary layer and the thermal conductivity of the fluid. The boundary layer thicness depends on several factors such as the velocity and distance from the leading edge maing the calculation of h a lot less simple than implied above. The boundary layer shape is discussed next. BOUNDARY LAYER SHAPE There are many theories and formulae that describe the shape of the boundary and this is covered in fluid mechanics. The following is a summary. When the flow is laminar the boundary layer is a curve following the precise law δ dp uo u y μ dl δ When the flow is turbulent, the shape of the boundary layers waivers and the mean shapes are usually shown. D.J.Dunn
3 Comparing a laminar and turbulent boundary layer reveals that the turbulent layer is thinner than the laminar layer. When laminar flow occurs in a round pipe the boundary layer grows in three dimensions from the wall and meets at the middle. The velocity u in a pipe of radius R at any radius r is given by Δp u R r Δp is the pressure drop. 4 μ L There are many formulae and theories for the shape of the boundary layer such as that given by Prandtl. u = u(y/) /7 This law fits the turbulent case well for Reynolds numbers below 07: FORMATION OF BOUNDARY LAYER When a fluid first encounters a solid surface, the boundary layer grows with distance until it becomes fully formed. Eventually it changes from laminar to turbulent. This also affects the heat transfer. DIMENSIONLESS GROUPS One of the best ways to analyse how all the factors come together is the use of dimensional analysis and the dimensionless groups that they form. Those required to study this topic should refer to the fluid dynamics module. A dimensionless group is a combination of variables that have no overall dimensions when grouped in a certain pattern and when evaluated yields a number with no units. D.J.Dunn
4 LIST OF VARIABLES THAT AFFECT HEAT TRANSFER Dynamic viscosity μ Kinematic viscosity ν = μ /ρ Density ρ Thermal conductivity Specific heat capacity c p Temperature θ Mean velocity u o Characteristic length l or diameter D or distance from leading edge x. Coefficient of cubical expansion β There are many dimensionless groups for heat transfer and a summary is given next. Here is a summary of the groups that occur. ρuod uod ρuox uox REYNOLDS NUMBER Red or Rex μ ν μ ν This group occurs in studies of fluid friction and is widely used in determining whether the flow is laminar or turbulent. It may be based on pipe diameter D or distance from the edge of a surface x. hx NUSSELT NUMBER Nu x This group is very important for convection. The solution of h often depends on evaluating Nux first. cpμ PRANDTL NUMBER Pr This is a group that relates some of the thermal properties. βgρ x θ GRASHOF NUMBER Grx μ This is a group is important in convection because it determines how the buoyancy of the fluid is affected by temperature. STANTON NUMBER St h ρ u c p Nu Re Pr The following demonstrates that Nu = Pr Re DIMENSIONAL ANALYSIS Tutorials on dimensional analysis may be found in the modules on fluid mechanics. The following is the application of this method to heat transfer. There are many dimensionless groups for heat transfer. A dimensionless group is a combination of variables that have no overall dimension when grouped in a certain pattern and when evaluated yields a number with no units. The surface heat transfer coefficient h is a function of the following variables. Dynamic viscosity μ, density ρ, thermal conductivity, specific heat capacity c, temperature θ, mean velocity u and Characteristic length l D.J.Dunn 4
5 The solution to this problem is much easier if we use energy E as a base unit. The dimensions used here are: Length L Mass M Time T Temperature θ Energy E First convert the units into dimensions. h Watts/m K Energy/s m K E/(T L θ) μ g/m s M/(LT) ρ g/m M/L Watts/m K Energy/s m K E/(T L θ) c J/g K E/(M θ) θ K θ u m/s L/T l m L The general relationship is This may be represented as a power series h = (μ,ρ,,c,θ,u,l) h = constant μ a ρ b c c d θ e u f l g There are 8 quantities and 5 dimensions so there will be 8 5 = dimensionless groups. The solution is easier if told in advance that we solve a, b, c,e and g in terms of d and f Put in the dimensions. Any not present are implied index 0 ET - L - θ - M 0 = constant (ML - T - ) a (ML - ) b (ET - L - θ - ) c (EM - θ - ) d θ e (LT - ) f L g Equate dimensions on left and right of equality. Energy = c + d c = d Mass 0 = a + b d Hence b = d - a or a = d - b Length - = -a - b - c + f + g substitute to get rid of b and c - = -a (d - a) ( - d) + f + g - = -a d + a + d + f + g - = a d + f + g Temperature - = -c d + e substitute for c - = -(- d) - d + e Hence e = 0 Time - = -a - c - f substitute for c and a - = -( d - b) ( - d) f hence b = f Now finish with a = d - b = d - f From - = a d + f + g substitute for unnowns - = (d - f) d + f + g - = d - f d + f + g - = - f + g g = f - Next form the groups by substituting the indices into h = constant μ a ρ b c c d θ e u f l g h = constant μ d-f ρ f -d c d θ 0 u f l f- d f μ c ρ u l h l μ c ρ u l h const const l μ μ Nu = h l/ Pr =cμ/ Re = ρ u l/μ = u l/ν The Prandtl number is a function of fluid properties only and so may be looed up in tables at the appropriate temperature. d f D.J.Dunn 5
6 SOME STANDARD CASES Studies of fluids and boundary layers have led to formulae for the surface heat transfer coefficient under specified conditions. In the following you need to be conversant with tables of fluid properties in order to loo them up. Here are some of the better nown formulae. NATURAL CONVECTION WITH A VERTICAL SURFACE Nu x Pr Pr 0.95) 4 Gr 4 This has an approximate solution of h =.4 (ΔT/D) /4 when Gr is in the range 0 4 to 0 9 and h = 0.9 (ΔT/D) /4 when Gr is in the range 0 9 to 0 FORCED CONVECTION OVER A HORIZONTAL SURFACE Nu = 0. Pr / Re / (Ta/Ts) 0.7 TURBULENT FLOW THROUGH A PIPE (Dittus-Boelter Equation) Nu = 0.0 Re 0.8 Pr 0.4 NATURAL CONVECTION FROM A HORIZONTAL CYLINDER Nu 0.57Pr This has an approximate solution of h =.(ΔT/D) /4 4 - Pr 0.95 Gr 4 x WORKED EXAMPLE No. Air at 88 K and bul velocity 6 m/s flows over a flat horizontal plate with a temperature of 8 K at all points on its surface. Given that Nu = 0. Pr / Re / (Ta/Ts) 0.7 Calculate the heat transfer rate per metre width from both sides over the first 50 mm. Ta is the bul air temperature and Ts is the surface temperature. The fluid properties should be obtained at the mean temperature. SOLUTION Nu = 0. Pr / Re / (Tw/Ts) 0.7 Mean temperature is (88 + 8)/ = K From fluids tables Pr = 0.68 υ = 4.55 x 0-5 m /s Re = v l/υ = 6 x 0.5/ 4.55 x 0-5 = 9.9 x D.J.Dunn 6 = 4.9 x 0 - W/m K 8 Nu x h l 4.9 x0 Nu h W/m K l 0.5 Heat Transfer from one side = Φ = h A (Tw-Ts) =.64 x ( x 0.5) x (8-88) = W For two sides double the answer.
7 WORKED EXAMPLE No. Calculate the heat transfer to air at 5 o C by natural convection with a vertical surface 0.6 m tall and m wide maintained at 79 o C at all points. The Nusselt Number is given by Nux Pr Pr 0.95) 4 Grx4 β = coefficient of cubical expansion given by β = /T SOLUTION β = /T = /(7 + 5) = /88 Use the fluid tables for dry air at (5 o C) 88 K and atmospheric pressure Taing atmospheric pressure as.0 bar Gas constant R = 87 J/g K ρ = p/rt =.0 x 0 5 /(87 x 88) =.6 g/m This can be also be found in the tables. μ is the dynamic viscosity μ =.788 x 0-5 g/m s Pr = 0.69 Thermal conductivity =.5 x 0 - W/ m K Taing x as the height of the wall 9.8x.6 (0.6) x 88 βgρx θ 88 9 Grashof Number Grx x 0 5 μ.788 x 0 Nux Pr Pr 0.95) 4 Grx ) x Nux h x Nux x0 h W/m K x 0.6 This the value at height 0.6 m. We need the mean over the height and this is 4/ of the value h = (4/)5.9 = W/m K Heat Transfer from one side = Φ = h A (79-5) = x (0.6 x ) x (79-5) = 8 W D.J.Dunn 7
8 WORKED EXAMPLE No. Calculate the heat transfer to air at 5 o C by natural convection with a vertical surface 0.6 m tall and m wide maintained at 79 o C at all points. The Nusselt Number is given by Nux Pr Pr 0.95) 4 Grx4 β = coefficient of cubical expansion given by β = /T SOLUTION β = /T = /(7 + 5) = /88 Use the fluid tables for dry air at (5 o C) 88 K and atmospheric pressure Taing atmospheric pressure as.0 bar Gas constant R = 87 J/g K ρ = p/rt =.0 x 0 5 /(87 x 88) =.6 g/m This can be also be found in the tables. μ is the dynamic viscosity μ =.788 x 0-5 g/m s Pr = 0.69 Thermal conductivity =.5 x 0 - W/ m K Taing x as the height of the wall 9.8x.6 (0.6) x 88 βgρx θ 88 9 Grashof Number Grx x 0 5 μ.788 x 0 Nux Pr Pr 0.95) 4 Grx ) x Nux h x Nu x0 h W/m K x 0.6 This the value at height 0.6 m. We need the mean over the height and this is 4/ of the value h = (4/)5.9 = W/m K Heat Transfer from one side = Φ = h A (79-5) = x (0.6 x ) x (79-5) = 8 W D.J.Dunn 8
9 WORKED EXAMPLE No. 4 Dry saturated steam at 77 o C flows in a pipe with a bore of 50 mm with a mean velocity of 6 m/s. The pipe has a wall 7 mm thic and is covered with a layer of insulation 50 mm thic. The surrounding atmospheric air is at 7 o C. Calculate the heat transfer rate for m length. The Nusselt number is given by Nu = 0. Re 0.8 Pr 0.4 for pipe = 50 W/m K for insulation = 0.06 W/m K The surface heat transfer coefficient for a long horizontal cylinder is h =.(ΔT/D) /4 SOLUTION From the fluids table for d.s.s. at 77 o C we find Pr =.4 and v g = 0. m /g μ = 4.88 x 0-6 g m/s = W/m K D = 0.5 m u = 6 m/s Re = u D/μ v g = 04 for pipe = 50 W/m K for insulation = 0.06 W/m K Nu = 0.0 Re 0.8 Pr 0.4 = 9.9 Nu = h D/ = h = 9.9 x /0.5 = 0.89 W/m K Heat transfer = Φ per metre length From steam to pipe Φ = h A ΔT = h πd ΔT = 0.89 π x 0.5 ΔT = ΔT Through the pipe wall Φ = πδt/ln(d /D ) = π x 50 ΔT/ln(64/50) = 5 ΔT Through the insulation Φ = πδt/ln(d /D ) = π x 0.06 ΔT/ln(64/64) = 0.79 ΔT From the outer surface h =. (ΔT/D) /4 =. (ΔT/0.64) /4 =.84(ΔT) /4 Φ = h A ΔT =.84(ΔT) /4 π x 0.64 ΔT =.58(ΔT) 5/4 Because the surface temperature is unnown we need the heat transfer between the steam and the surface and the surface and the air. Steam to Surface Φ = U (77 T s ) = U (77 T s ) U is found from. 64 U = 0.7 U U Φ = 0.78(77 T s ) Surface to Air Φ =.57(T s 7) 5/4 Equate Φ = 0.7(77 T s ) =.57(T s 7) 5/4 Possibly the best way to solve this is by plotting 0.7(77 T s ) and.57(t s 7) 5/4 against temperature to find the temperature that maes them the same. This appears to be 45 o C D.J.Dunn 9
10 Φ = 0.7(77 T s ) = 0.7(77 45) = 0.4 W Φ =.57(T s 7) 5/4 =.57(45 7) 5/4 = 0.8 W Say 0.6 W SELF ASSESSMENT EXERCISE No.. Air at 00 K and bul velocity 8 m/s flows over a flat horizontal plate with a temperature of 900 K at all points on its surface. Given that Nu = 0. Pr / Re / (Ta/Ts) 0.7 Calculate the heat transfer rate per metre of width from one side over the first 00 mm and the first 00 mm. Ta is the bul air temperature and Ts is the surface temperature. The fluid properties should be obtained at the mean temperature. (897 W and 68 W). Calculate the heat transfer to air at o C by natural convection with a vertical surface m tall and m wide maintained at 0 o C at all points. Nux Pr β = coefficient of cubical expansion given by β = /T (676 W) The Nusselt Number is given by 4 4 Pr 0.95). Dry saturated steam at 5 bar flows in a pipe with a bore of 00 mm with a mean velocity of 4 m/s. The pipe has a wall 4 mm thic and is covered with a layer of insulation 60 mm thic. The surrounding atmosphere is at 0 o C. Calculate surface temperature of the lagging and the heat transfer rate for m length. The Nusselt number is given by Nu = 0.0 Re 0.8 Pr 0.4 for pipe = 55 W/m K for insulation = 0.08W/m K The surface heat transfer coefficient for a long horizontal cylinder is h =.(ΔT/D) /4 (0 o C and 6.4 W ) Grx D.J.Dunn 0
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