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CHAPTER 5 Natural convection heat transfer Course outline 5.1 Introduction 5.2 Objectives 5.3 Free convection over flat vertical surfaces 5.3.1 Flow equations 5.3.2 Energy equations 5.4 Integral method for laminar flow 5.5 Turbulent flow 5.6 Combined free and forced convection. 5.7 Simplified correlations for air 5.8 Summary 5.1 Introduction: Welcome to chapter 6. In the previous chapters in this unit, we have considered forced convection heat transfer where the fluid motion is forced over a surface or through a tube by an external device such as a fan or pump. In this chapter we consider heat transfer by natural convection where fluid motion is due to differences in density of the fluid as a result of a heating process. Practical examples where heat transfer by natural convection occur include; cooling of transmission lines, electric transformers, rectifiers, heating rooms by radiators, heat transfer from hot pipes and ovens. We start by describing the physical mechanism of natural convection where the characteristic parameters are identified. Then we discuss free convection on a heated vertical plate by examining momentum and energy equations in laminar flow. Relations in turbulent flow are briefly highlighted and finally combined free and forced convection is briefly considered. 5.2 Objectives: The objectives of this chapter are: 1) To describe the physical mechanism of Natural convection. 2) To examine and discuss the flow equations for a vertical plate in laminar flow 3) To discuss briefly turbulent flow in free convection and the combined free and forced convection 5.3 Physical mechanism in Free convection: When a heated surface is in a fluid whose temperature is lower, the fluid adjacent to the surface gets heated. This results in thermal expansion of the fluid and reduction in density as compared to the fluid away from the surface. Because of the density differences, a buoyancy force acts on the low density fluid causing it to flow up the surface. As a result, hydrodynamic and thermal boundary layers are set up in the neighbourhood of the surface. Since the flow velocity is developed here as a result of 1 temperature differential, the two boundary layers are of the same order irrespective of the Prandtl number. The fluid motion in natural convection however is characterized by low velocities (typically less than 1 m/s) compared to forced convection. Hence, the convection heat transfer coefficients in natural (free) convection are lower. An important property in natural convection is the coefficient of thermal expansion of the fluid, . The coefficient of thermal expansion is defined by, 1 v 1 = = v T p T p (5.1) For ideal or perfect gases, R RT 1 v and = and therefore, = P T T p P The temperature being absolute v= In natural convection, Grashof number is the dimensionless grouping that plays a major role in natural convection just as Reynolds number plays in forced convection. It incorporates the coefficient of thermal expansion β, in the expression. In physical terms, the Grashof number represents the ratio of buoyancy force on the fluid to viscous force. It is expressed as; Gr = g 2 (Tw − T ) L3 (5.2) 2 L – is the characteristic length - is the fluid density - is the dynamic viscosity of the fluid. Tw – is the surface or wall temperature T - is the temperature of the surrounding fluid. For the flow of air over a flat plate, the critical Grashof number has been observed to be approximately 4 × 108. Free convection relationship is found to have the following dependence, Nu = f (Gr , Pr) In many cases, the relationship simplifies to the form (5.3) Nu = C (Gr Pr) m . (5.4) A dimensionless group that is often used is the Rayleigh number, Ra = Gr Pr (5.5) The product is also a criterion of laminar or turbulent character of the flow as determined by its values. For laminar flow 104 < GrPr < 109. For turbulent flow, GrPr > 109. 2 The values of Gr and Pr are evaluated at the mean film temperature. Tw + T (5.6) 2 When average Nusselt number and therefore average convection coefficient is known, the rate of heat transfer by natural convection from a solid surface at a uniform temperature Tw to the surrounding fluid at T∞ is expressed by the Newton’s law of cooling, Tf = Q = hAs (Tw − T ) , (5.7) where h is the average heat transfer coefficient and A s is the heat transfer surface area. 5.4 Free convection over vertical flat surface. Although we understand the mechanism of natural convection, the complexities of fluid motion make it very difficult to obtain simple analytical relations for heat transfer by solving the governing equations of motion and energy. The analytical solutions that exist are obtained for simple geometries under simplifying assumptions and lack generality and are normally for laminar flow. Turbulent flow and other geometries are too complex and empirical and relations and experimental data are used to estimate heat transfer rates. Therefore with exception of some simple cases, heat transfer relations in natural convection are based on experimental studies. In this section we briefly discuss flow over a vertical surface – a simple geometry. Consider a flat vertical surface at a temperature higher than the surrounding fluid. Because of this temperature difference, a natural convection boundary layer will be formed on the surface. The figure below shows a convection boundary layer formed on the surface. Turbulent GrPr > 109 Tw T u FB x Laminar 104 < GrPr < 109 g y Figure 5.1 Natural convection on a vertical plate. The velocity of the fluid at the surface is zero and increases to some maximum value and then decreases to zero again at the edge of the boundary. 3 The initial boundary-layer development is laminar, but some distance from the leading edge depending on the fluid properties and the temperature difference between the plate and the surrounding, turbulent eddies are formed and transition to turbulent boundary layer begins. Further up the surface the boundary layer becomes fully turbulent. 5.4.1 Flow equations: The continuity equation in the boundary layer is the same as that for forced convection. For the case of momentum equation, there is an additional body force due to gravity, g. Hence, the momentum equation becomes, u u p 2u + v = − − g + 2 y x y x u g – weight force on the element and the pressure gradient in the x- direction is hydrostatic due to the height of the plate. p = − g x u u 2u + v = g ( − ) + 2 Therefore, u (5.8) y y x Further, the density difference (-) may be expressed in terms of the volume coefficient of expansion . 1 1 − = T p T − T The momentum equation becomes, =− u u 2u + v = 2 + g (T − T ) (5.9) y y x The momentum equation is coupled to the energy equation in that the solution for velocity profile demands knowledge of the temperature distribution. u 5.4.2 Energy equation: The energy equation for natural convection is similar to that for forced convection, especially at low velocities. Hence the energy equation still remains, dT dT d 2T cp u +v =k 2 dy dy dx 1 For ideal gases, = T (5.10) 4 The viscous dissipation has been neglected in the energy equation. The coupling of the momentum and energy equations requires that both equations be solved simultaneously. Numerous solutions to laminar free convection boundary layer equations have been obtained. Free convection from an isothermal vertical surface has received special attention. In this case the boundary layer equations must be solved with boundary conditions, y = 0: u=v=0 T = Tw y→ u→0 T → T A similarity solution to the foregoing problem was obtained by Ostrach, using transformation of variables. A numerical solution gives local Nusselt number as a function of x as, 1/ 4 Gr (5.11) Nu x = x f ( Pr ) 4 The function of Prandtl number f(Pr) is determined from the results of the numerical solution, as shown in the table 5.1. Table 5.1 The function of Prandtl number Pr 0.01 0.72 1 f(Pr) 0.081 0.505 0.567 2 0.716 10 1.169 100 2.191 1000 3.966 The average coefficient for surface of length L is then, L 1 k g (Tw − T ) h = hx dx = L0 L 4 2 1/ 4 L f ( Pr ) 0 dx x1/ 4 (5.12) Integrating, 1/ 4 hL 4 GrL Nu L = = k 3 4 For x = L, NuL = f ( Pr ) (5.13) 4 Nu L 3 (5.14) The above equations are applicable whether the surface temperature is lower or higher than the surrounding fluid. Further, the results may be used with reasonable accuracy for constant heat flux as well as constant surface temperature. 5.4.3 Integral equations for laminar flow: Free convection laminar flow on a vertical flat plate has also been solved by use of integral equations. The momentum equation for flat vertical plate is derived as, 5 d 2 u dy = − + g (T − T )dy w dx 0 0 du = − + g (T − T ) dy dy y =0 0 The energy equation is, (5.15) dT d t (T − T )udy = dx 0 dy y =0 (5.16) Viscous work is neglected. For simplicity in analysis it will be assumed that the thermal and velocity layers are virtually the same ( = t). The boundary conditions are, At y = 0. u = 0, T = Tw At y = , u = 0, T = T, And at y = 0, dT du = =0 dy dy d 2u T −T = −g w 2 dy The dimensionless temperature profile is assumed to be in the form, T − T y = 1 − Tw − T 2 (5.17) The selection of the velocity profile is more difficult because an obvious characteristic velocity for non-dimensionalizing does not exist. The velocity profile is however assumed to be of the form, u y y = 1 − u* 2 (5.18) where u* is an arbitrary function with dimensions of velocity. Substitution of the temperature and velocity profiles in the integral equations, (6.15), (6.16) and with algebraic manipulation gives the boundary layer thickness as, 1/ 4 = 3.93Pr −1/ 2 ( 0.952 + Pr ) Grx−1/ 4 x The heat transfer coefficient may then be evaluated from, 6 (5.19) dT qw = −kA = hA (Tw − T ) dy y =0 h= hx 2k x or x = 2 k (5.20) Therefore the Nusselt number is given as, −1/ 4 Nu x = 0.508 Pr1/ 2 ( 0.952 + Pr ) Grx1/ 4 (5.21) The average Nusselt number for a plate of length L is then readily found as, L h= 1 4 hx dx = hL L0 3 NuL = hL −1/ 4 = 0.678GrL1/ 4 Pr1/ 2 ( 0.952 + Pr ) k NuL = 0.678Ra 1/ 4 L 0.952 1 + Pr −1/ 4 (5.22) For air, Pr = 0.7, the relations simplify to, Nu x = 0.378Grx1/ 4 and NuL = 0.504GrL1/ 4 (5.23) Properties are evaluated at mean film temperature, Eq (5.6). 5.5 Turbulence considerations: Transition in a free convection boundary layer depends on the relative magnitude of the buoyancy and viscous forces in the fluid. The transition is often correlated in terms of Rayleigh number, which is the product of Grashof and Prandtl numbers. For vertical plates, the critical Rayleigh number has been found to be, Rax ,c = Grx ,c Pr = g (Tw − T ) x 3 109 (5.24) For turbulent boundary layer flows, that is when Rayleigh number RaL 109, reliance is placed on experimental results. Some of the correlations based on experimental data include the following. Nu = 0.68 + 0.67 Ra1/L 4 1 + ( 0.492 / Pr )9 /16 -1 9 for 10 < RaL < 10 . (5.25) 4/9 7 Nu1/ 2 = 0.825 + 0.387 Ra1/L 6 1 + ( 0.492 / Pr )6 /16 (5.26) 8/ 27 For 10-1 < RaL < 1012 For the entire range. For constant wall heat flux, an integral solution obtained similar to the laminar flow is given to be, Nu x = 0.295 9 For Grx > 10 , Pr 7 /15 Gr 2 / 5 (1 + 0.492 Pr ) 2/3 2/5 (5.27) For uniform wall heat flux, the recommended correlation for vertical plates is, Nu x = 0.60 ( Grx* Pr ) 1/ 5 ( Nu x = 0.17 Grx• Pr ) 1/ 4 for 105 Grx* Pr 1011 , laminar flow. for 2 103 Grx* Pr 1016 , turbulent flow. (5.28) (5.29) Gr = Grx Nu x . * x When hx = qw g x 4 qw , then Grx* = Tw − T k 2 (5.30) Properties evaluated at the mean film temperature. Example 5.1: A vertical metallic plate exposed to the sun, receives 600 W/m2 of sun’s energy. The plate is 4m high and 2m wide. It is insulated from the back side, and is blankened on the front side so that all incident radiation is absorbed by the plate, which in turn losses heat to the surrounding air at 40°C by natural convection. Determine the equilibrium temperature of the plate. Soln. This is a problem with constant heat flux qw. To estimate film temperature and air properties, it is necessary to make an approximation of the plate temperature. First approximation: An approximation value of h for free convection in air can be taken as 10 W/m 2 K, so that we have approximately, T = q w 600 T 60 = 60 0 C. Hence, T f = T + = 40 + = 70 0 C . h 10 2 2 Properties of air at 700C, are k = 0.0295 W/mK, ν = 2.005 × 10-5 m2/s, Pr = 0.7 1 = = 2.92 10 −3 / K Tf 8 Modified Grashof number, equation 6.30, gq w x 4 9.81(2.92 10 −3 )(600)(4) 4 Grx• = = = 3.71 × 1014. Turbulent flow. 2 −5 2 k (0.0295)(2.005 10 ) ( From equation 6.29, Nu x = 0.17 3.71 1014 0.7 h= ) 1/ 4 = 682.44 k 0.0295 Nu x = 682.44 = 5.03 W/m2K. x 4 Second Approximation: q 600 119 T = w = = 119 K, T f = 40 + 100 0 C h 5.03 2 Properties of air at 1000C, k = 0.0318 W/mK ν = 2.311 ×10-5 m2/s Pr = 0.694 β = 1/373 = 2.681 × 10-3 Grx = 9.81(2.681 10 −3 )(600)(4) 4 = 2.379 1014 (0.0318)(2.311 10 −5 ) 2 0.0318 (0.17)(2.379 1014 0.694)1 / 4 = 4.84 W/m2K 4 The change is small and the value of h may be accepted. h= q w 600 = = 124 K. h 4.84 Average plate temperature, Tw = 40 + 124 = 164°C. Temperature difference, T = Tw − T = Other Relations: Vertical plate; Lc = L, Ra, 104 – 109 Ra, 10 – 10 9 13 Nu = 0.59 Ra 1L/ 4 . Nu = 0.1Ra 1L/ 3 (5.31) (5.32) Inclined plate; Lc = L, Use the vertical plate equations for the upper surface of a cold plate and the lower surface of a hot plate. Replace g by gcosθ for Ra < 109. θ is the inclined angle from the normal. Horizontal plates; A Lc = s , where surface area As and perimeter P. P Upper surface of a hot plate, Ra, 104 – 107 Nu = 0.54 Ra 1L/ 4 1/ 3 Ra, 107 – 1011 Nu = 0.15Ra L 9 (5.33) (5.34) Lower surface of a hot plate or upper surface of a cold plate; Ra, 105 – 1011 Nu = 0.27 Ra 1L/ 4 (5.35) Vertical cylinder; A vertical cylinder can be treated as a vertical plate when, D 35L . GrL1 / 4 (5.36) Horizontal cylinder; Lc = D, and Nu = 0.6 + Ra d 1012 . 2 1/ 6 0.387 Ra d . 8 / 27 0.559 9 / 16 1 + Pr Ra d 1011 , Sphere, Lc = D, and Nu = 2 + 0.589 Ra 1D/ 4 0.469 9 / 16 1 + Pr 4/9 (5.37) Pr 0.7 . (5.38) Example 5.2: A 6m long section of an 8cm diameter horizontal hot-water pipe passes through a large room whose temperature is 20°C. If the outer surface temperature of the pipe is 70°C, determine the rate of heat loss from the pipe by natural convection. Soln. Tw + T (70 + 20) = = 45 0 C . 2 2 Properties of air, k = 0.02699 W/m °C, Pr = 0.7241, ν = 1.750 × 10-5 m2/s. 1 1 = = . T 318 Lc = d = 0.08m, 1 9.81 (70 − 20) 0.08 3 0.7241 3 g (Tw − T )d Pr 318 Ra d = = 2 (1.750 10 −5 ) 2 = 1.867 × 106. Mean film temperature, T f = 10 Nusselt number from equation 5.37, Nu = 0.6 + 2 1/ 6 0.387 Ra d = 17.39 8 / 27 0.559 9 / 16 1 + Pr k 0.02699 Nu = 17.39 = 5.867 W/m 2 C . d 0.08 As = dL = 0.08 6m = 1.508 m 2 . h= Heat Transfer, Q = hAs (Tw − T ) = 5.867 1.508 (70 − 20) = 442 W. Question: If the thermal emissivity of the pipe surface is assumed ε = 1, can you show that the heat loss by radiation is 553 W. And it is more than that of convection. Activity: Consider a 0.6m × 0.6m thin square plate maintained at a temperature of 90°C, while the other side is insulated. Determine the rate of heat transfer from the plate by natural convection if the plate is a) vertical b) horizontal with hot surface facing up and c) horizontal with hot surface facing down. [115 W, 128 W, 64.2 W]. Take Note: The natural convection heat transfer is the lowest in the case of the hot surface facing down. The hot air is trapped under the plate in this case and cannot get away from the plate easily. As a result, the cooler air in the vicinity of the plate will have difficulty reaching the plate, which results in a reduced rate of heat transfer. 11 Activity: Consider a thin 16cm long and 20cm wide horizontal plate suspended in air at 20°C. The plate is equipped with electrical resistance heating elements with a rating of 20W. Now the heater is turned on and the plate temperature rises. Determine the temperature of the plate when steady operating conditions are reached. The plate has an emissivity of 0.9 and the surrounding surfaces are at 17°C. Soln: (Complete this solution) At steady operating conditions, the plate surface temperature Tw. Mean T + T film temperature, T f = w . 2 Natural convection heat transfer from upper face – Qcon1.= h1 A1 (Tw − T ) Natural convection heat transfer from lower face - Qcon2. = h2 A2 (Tw − T ) ( ) Radiation heat transfer – Qrad = AR Tw4 − Ts4 . Total heat transfer, QT = Qcon1 + Qcon2 + Qrad = 20W Take Note To obtain air properties at Tf, plate temperature Tw must be known. But this is what the question requires. We therefore take properties at an assumed temperature say at T∞ and obtain an approximation temperature of the plate Tw. We then use this to obtain improved Tf, obtain properties and get a new Tw and so on until the differences in Tw are negligible. This is iteration. 5.6 Combined free and forced convection: Buoyancy has the effect of distorting the mean velocity and temperature profiles in forced convection, which in turn affects the heat transfer coefficient. The relative importance of free and forced convection is expressed as the ratio of body to inertia forces as below. Gr 1 - Free convection predominates Re 2 Gr 1 - forced and free convection important Re 2 Gr 1 - Forced convection predominates Re 2 For vertical isothermal flat plate with both free and forced convection in the same Grx direction, neglecting free convection results in an error of about 5% or less for Re 2 given in table 5.2. 12 Table 5.2 Effect of free convection. Pr 100 10 0.24 0.13 Grx 2 Re x For flow inside horizontal tubes and for 0.72 0.08 Gr Re 0.03 – 0.003 0.056 – 0.05 1 , the recommended correlation is 4/3 Nud = 1.75 Gz + 0.012 ( GzGr1/ 3 ) 1/ 3 b w 0.14 (5.39) For laminar flow. And, D Gz = Re Pr , - D = Pipe diamter, L = pipe length L Re, Gr, and Nu are all based on D – the pipe diameter, b and w refer to bulk and wall temperatures respectively. The correlation for turbulent flow is recommended to be D Nud = 4.69 Re0.27 Pr 0.21 Gr 0.07 L 0.36 (5.40) 5.7 Simplified correlations for air: Some simplified equations for free convection from various surfaces to air at atmospheric pressure that have been found very useful are as follows. Vertical plane or cylinder to air. h = 0.95(T ) 1/ 2 1/ 4 - Laminar (5.41) - Turbulent. Horizontal cylinder to air T h = 1.24 D T h = 1.42 L (5.42) T h = 1.32 D 1/ 4 - Laminar (5.43) 1/ 3 - Turbulent. (5.44) Horizontal plate to air: Heated surface facing upward or cooled surface facing downward. T h = 1.32 L 1/ 4 - Laminar (5.45) 13 h = 1.43(T ) 1/ 2 - Turbulent (5.46) Heated facing downward or cooled surface facing upward T h= L 1/ 5 (5.47) Example 5.3: a) The suction line of a refrigeration system has refrigerant 134a flowing inside a copper tube, 1.25cm OD and 2m long. The temperature of the tube wall is 22°C. The surrounding air is at a temperature of 40°C. Calculate the heat gain of the line by natural convection. b) If air is blown across the line at a velocity of 5 m/s what will be the heat gain? Soln: a) Still air, (Natural convection) 40 − 22 = 90 C . 2 1 1 = = = 3.546 10 −3 / K 273 + 9 282 k = 0.02481 W/mK Pr = 0.714 ν = 13.39 × 10-6 m2/s. Tf = Ra = GrPr = g (T )D 3 2 Pr . (9.81)(3.546 10 −3 )(0.0125) 3 (62)(0.714) 10 9 (13.39 10 −6 ) 2 The boundary layer is therefore laminar. Applying equation 6.43, = T h = 1.32 D 1/ 4 62 = 1.32 0.0125 1/ 4 = 11.1 Q = hAT = 11.1( 0.0125 2)(62) = 54.1W Heat loss to still air, b) Blowing air (Forced convection) Reynolds number, instead of Grashof number, the criterion is, Du m (0.0125)(5) = 4668 Laminar flow. 13.39 10 −6 For flow over tubes where Re = 4000 – 40,000, Re d = = 14 Nu = 0.693 Re 0.618 Pr 1 / 3 = 0.693(4668) 0.618 (0.714)1 / 3 = 31 31 0.02481 h= = 61.5 0.0125 Heat loss to blowing air, Q = 61.5( 0.0125 2)(62) = 299W 5.8 Summary: In this chapter we have considered natural convection heat transfer where the fluid motion is caused by changes in density due to a heating process. It is the resulting buoyancy force that is responsible for fluid movement. The heat transfer coefficient (or Nusselt number) is a function of Grashof number (Gr) and the Prandtl number (Pr). The Grashof number is a dimensionless grouping that plays a major role in natural convection and incorporates the coefficient of thermal expansion – an important property in natural convection. Free convection relationship is found to have the form, Nu = C (Gr Pr) m . Due to complexity of fluid motion, it is difficult to obtain simple analytical relations. Hence most heat transfer relations are largely based on experimental studies. However, for the simple geometry such as a vertical plate and with simplifying assumptions, analytical heat transfer relations have been presented; but in laminar flow. Turbulent flow relations are based on experimental measurements. The velocities in free convection are lower when compared to forced convection. Therefore the heat transfer coefficient and thus heat transfer are lower in natural convection. Activity: 1. A vertical hot Oven door, 0.5m high is at 200C and is exposed to atmosphere at 20C. Estimate the average heat transfer coefficient at the surface of the door. 2. A 350mm long glass plate is hung vertically in the air at 24C while its temperature is maintained at 80C. Calculate the boundary layer thickness at the trailing edge of the plate. If a similar plate is placed in a wind tunnel and air is blown over it at a velocity of 5 m/s, find the boundary layer thickness at its trailing edge. Also, determine the average heat transfer coefficient for natural and forced convection for the above data. 3. Gas at 350C is conveyed through 250mm diameter pipe laid in an atmosphere of quiescent air at 20C. The convection heat transfer coefficient from a hot cylindrical surface freely exposed 1/ 4 T to still air is h = 1.55 D (i) Calculate the heat loss per m length of the bare pipe 15 (ii) Estimate the percentage reduction in heat loss if the pipe is covered with 75mm thick layer of material whose thermal conductivity is 0.072 W/m C. Any temperature drop in the metal may be neglected. 4. What heat load is generated in a restaurant by 1.0m 0.8m grill which is maintained at 134C? The room temperature is 20C. (hint: find the expression for Nusselt number for a horizontal heated plate facing up) 5. An uninsulated 10.75 in OD steam line passes through a 9ft high room in which the air is at 120F. What is the heat loss per foot of pipe if its surface is at 760F when, (a) Horizontal, (b) vertical. 6. A glass-door firescreen, used to reduce exfiltration of room air through a chimney, has a height of 0.71m and a width of 1.02m and reaches a temperature of 232C. If the room temperature is 23C, estimate the convection heat rate from the fireplace to the room. 7. A horizontal, high-pressure pipe of 0.1m outside diameter passes through a large room whose walls and air temperature are 23C. The pipe has an outside surface temperature of 165C. Estimate the heat loss from the pipe per unit length. 8. Consider a large vertical plate with uniform surface temperature of 130C suspended in quiescent air at 25C and atmospheric pressure. (a) Estimate the boundary layer thickness at a location 0.25m measured from the lower edge. (b) What is the maximum velocity in the boundary layer at this location and what position in the boundary layer does the maximum occur? (c) At what location on the plate measured from the lower edge will the boundary layer become turbulent? 9. Air flow through a long, 0.2m square air conditioning duct maintains the outer duct surface temperature at 10C. If the duct is uninsulated and exposed to air at 35C in the crawl space beneath a home, what is the heat gain to the duct per unit length of the duct? 10. A flat plate 10cm wide and 18cm high at 250C is placed vertically in still air at 1 atm and 20C. What is the total loss from the two sides of the plate? Because of changes in solar radiation, the total heat loss from the plate may reach 100 W. At this loss and assuming uniform heat flux, determine the variation of surface temperature and the average plate temperature. Natural convection over a flat plate Gr = g x 3 (Tw − T ) 2 Average Nusselt number for constant wall temperature – laminar flow 16 NuL Pr1/ 2 = GrL1/ 4 ( 2.435 + 4.884 Pr1/ 2 + 4.953Pr )1/ 4 Local Nusselt number for constant heat flux – laminar flow Nu x Pr 2 / 3 = ( Grx* ) ( 3.91 + 9.32 Pr1/ 2 + 9.95 Pr )1/ 5 Grx* = Grx Nu x 17 , where