Conduction:FOURIER LAW OF HEAT CONDUCTION Q = = -- K A i) area, A, normal to the direction of heat flow in m2. ii) the temperature gradient at the section, , in 0K/mt ie: the rate of change of temperature with reference to the distance in the direction of heat flow.Also temperature gradient is expressed as negative since heat flows in decrease of temperature iii)Thermal conductivity , K, which is the property of the material in KW/m 0K. or Q = -- Temperature distribution in slab = for conditions when one of its faces away from origin. ie At x = x1 ; t = t1 and x = x2 ; t = t2 ( If the entering face is coinciding with Y-axis, ie origin, then x1 = 0 and x2 = x1)Slope of temperature profile. Slope of temp. distribution ie: is large for insulators. Slope of temperature distribution ie is small for the good conductors.Temperature distribution in Solids is shown belowT Direction of heat flowTDirection of heat flowin direction increasing xin direction decreasing x

Tx is + ve is ve. -dT T1 + dT q TT+dx+dxT2

L x increasing xx Fig-(a) Fig-(b) Fig-(c) Fig(a)-Temperature distribution for Fig-(b), (c)- Sign convection for conduction flow. steady state condition through plane wall. = - ; analyzing this equation, for the same value of heat , q , if K is low (ie for insulators), is large (slope of temperature distribution is large) and there will be a large temperature difference across the wall. If K is high (ie for conductors), is small (slope of temperature distribution is small ) and there will be a small temperature difference across the wall.

t1 t1Less slope

More slopet3

t2

L L

a) Insulator, b) Conductor Slope of temp. distribution is large. Slope of temperature distribution is small Fig-: Temperature distribution in insulator and conductorVariation of thermal conductivityThe variation of thermal conductivity is linear, for most cases, and varies with the temperature as per relation, K =K0 (I t) . Its temperature profile is shown in figure below. Where K0 = Thermal conductivity at Zero temperature. and = Coefficient of thermal conductivity. It is negative for metallic conductors and positive for insulating material.

0 ie is positive,(for Insulators)t1 = 0

t2

0, ie is negative,(for Conductors)

Temperature profile in conduction with variable conductivity Thermal resistance and Thermal conductance in conduction Heat transfer and Electrical Analogy with Equivalent electrical circuit. Q = ( where T1 > T2) = This can be written as: Q = = KConduction (T1 T2) ---------------------------1.11.1Hence, we can take the terms of above equation as :- Thermal Resistance of the slab = Rth = () ----------------------------1.11.2 Its unit is 0K / wattThermal conductance is reciprocal of Resistance Thermal Conductance in conduction= KConduction = = - -- -1.11.3 Its unit is Watt / 0K If it is K per unit area then unit is Watt /m2 0KElectrical analogy between flow of heat and Electricity:Electrical resistance Re , resists the flow of Current i , across Potential difference V.Similarly the Thermal resistance, Rth , resists the flow of heat, q , across the temperature difference, (t1 t2). L

T1

T2

Rth = ()Q

a)Temperature profile in slab i E1 E2 Re i = b) Electrial circuit.

T1T2 =

= where = c) Equivalent Thermal circuit & Conduction resistance. Fig: 1.5. Electrical Analogy of Conduction The Ohms law for electricity is given as; i = = ----------------------------------------------( i)where i = current in amperes

V = is potential difference in volts, & Re = electrical resistance in ohms

The Fouriers law of conduction gives;

Q = = ( where T1 > T2)This can be written as: Q = ----------------(ii) Rth = = is the thermal resistanceComparing the above equations I & ii, the following quantities are analogous to each other while considering the electrical circuit in equivalent to a thermal circuit. Electrical circuit. Thermal circuit.1) Current , i, in amperes 1) Heat flow rate , q , in watts2)Voltage , V= (E2 E1) in volts 2) Temperature difference, t = (T1 T2) (Electrical Potential difference) (Thermal potential difference) 3) Electrical Resistance, Re , in Ohms 3)Thermal resistance, Rth = L/KA, in 0C/watt 4) Charge, C, in coulomb 4) Heat flow , Q , in Joules

Problem-1.11-1: The outer surface of 10m x 3mt x o.2mt thick concrete wall is kept at temp of 50C, while the inner surface is at 300C. The thermal conductivity of concrete is 1.2 w/m K. Determine rate of heat loss from room and thermal resistance of wall.Answer:Thermal resistance of conduction , Rth = / KA = 0.2 / 1.2 (10 x 3) = 0.00556 0c / watt Rate of heat loss = q = (t1 t2) / Rth = (30-5) / 0.00558 = 4500 wattProblem-1.11-2:The insulation of 7.5cm thick is fixed on side wall of a oven. The inside of wall is at 420K. The electric coils dissipate the heat of 36.5 watt to make up the heat loss through the walls. Calculate the wall surface area, so that the temperature on other side of wall does not exceed 310 K. The thermal conductivity of insulation is 0.04 w/mK.Ans:Q = - KA (t2 t1) / 36.5 = - 0.04 A (310-420) / 0.75 = 5.87 AHence A = 36.5 / 5.87 = 6.218 m2Problem-1.11-3 One surface of a copper plate thickness of 3cm is maintained at 4000C and other surface at 1000C. How much heat is transferred through the plate per square mt. The thermal conductivity of copper is 370 w/m K Answerq/A = -K (t2 t1) / = -370 (100-400) / 3 x 10-2 =3.7 MW / m2

Problem-1.11-4. A plane wall of 10cm thick and 3sq.mt area is made of a material whose conductivity is 8.5w / m K. The temperature of wall surfaces are steady at 1000C and 300C respectively.Find temperature gradient and heat flow across wall.AnswerTemperature gradient in the direction of heat flow is: dt/dx = (t2 t1) / = (30-100) / 0.1 = -700 0C/mHeat flow across the wall is by applying Fourier equation. q = -KAdt/dx = - 8.4 x 3 x (-700) = 17850 w or 17.85 kw.Problem 1.11-5 Two metal plates of 2.5cm and 15cm thick are bonded by an epoxy and heat source of 43.5 kw/m2 is applied uniformly and steadily, through the thinner plate by a radial heat source. The bonding epoxy must be held at 320K for a short time.A thermocouple installed on the side of thinner plate next to heat source indicates a temperature of 345K. Calculate the temperature gradient for heat conduction through the thinner plate and the thermal conductivity of its material.Answer q =43.5 kw/ m2 , L1 =2.5cm =0.025m ; L2= 15cm =o.15m Temperature gradient , dt/dx =( t2 t1) / L = (320- 345) / 0.025 = - 1000 0C/mWe have q/A = -K ( t2 t1) / L ie: 43.5 x 103 = - K (-1000) K = 43.5 w/mKProblem 1.11-6: A cold storage consists of a cubical chamber of dimension 2m x 2m x 2m, maintained at 10C inside temperature. The outside wall temperature is 55C. The top and side walls are covered by a low conducting insulation with thermal conductivity k = 0.06 W/mK. There is no heat loss from the bottom. If heat loss through the top and side walls is to be restricted to 500W, what is the minimum thickness of insulation required?Solution:To find: Thickness of insulation needed to maintain heat loss below 500W.Assumptions: (1) perfectly insulted bottom, (2) one dimensional conduction through five walls each of area A=4m2, (3) steady state conditionsAnalysis: Using Fouriers law, the heat rate is given byq = -K Atotal (t2 t1)/LSolving for L and recognizing that A total =5 x AL= -K Atotal (t2 t1) / q = - 0.06 x 5 x 4 x (10-55) / 500 =0.108m =108mmProblem-1.11-7. A square silicon chip is of width W=5mm on a side and of thickness t=1mm. The chip is mounted in a substrate such that there is no heat loss from its side and back surfaces. The top surface is exposed to a coolant. The thermal conductivity of the chip is 200W/m.K. If 5W are being dissipated by the chip, what is the temperature difference between its back and front surfaces?AnswerAssumptions: (1) steadystate conditions, (2) constant properties, (3) uniform dissipation,(4) negligible heat loss from back and sides, (5) onedimensional conduction in chip.Analysis: All of the electrical power dissipated at the back surface of the chip is transferred by conduction through the chip. Hence, Fouriers law,P=q= K A (t2-t1) / L ; (t2-t1) = L q/ K AL = t= 1mm =0.001m; A =0.005 x 0.005 = 25 x 10-6t2-t1) = 0.001 x 5/ 200 x 25 x 10-6 = 1.00CComments: for fixed P, the temperature drop across the chip decreases with increasing k and width, as well as with decreasing thickness, t.Problem-1.11-8. Sheets of brass and steel, each of thickness 1cm, are placed in contact. The outer surface of brass is kept at 100C and the outer surface of steel is kept at 00C. What is the temperature of inter-face? The ratio of K of brass and steel is 2:1.Answer t1 =100C; t2 = interface temp; t3 = 0C; Lb =Ls = L =1cm =0.01m Heat flow is same ie q brass = q steel ; - Kb A (t2 100)/L = - Ks A ( 0 - t2) / L Hence Kb / Ks = - t2 / (t2 100) : ie. 2 (t2 100) = -t2 or 3t2 = 100 ; t2 = 100/3 =66.70CProblem-1.11-9How long will it take to form 4cm thick slab of ice on the surface of a lake when the air temperature is -60C? K of ice is 1.675 w/mK, Density is 920 kg/ m3. Take the latent heat of fusion of ice as 335 kj/kg.(Lfusion) AnswerLet d is the differential time to form ice of thickness dyThen the energy balance gives A dy Lfu = -K A (t2 t1) / y . d = y dy Integrating; = [ ] = = = 24533.33 sec =6.815 hours = 6 hours 49minProblem-1.11-10.A oven wall of thickness 7.5 cm is made of insulation of thermal conductivity 0.04 w/mK. The temperature of wall on oven side is 4200 C. The electrical coils in the oven dissipates 36.5 watts of electrical energy to make up for the heat loss through wall. Calculate the wall surface area perpendicular to heat flow, so that the outside temperature of wall does not exceed 3100K. The rate of heat dissipation = Q = 36.5 watts Q = ie: 36.5 = = 5.87 A or A = = 6.218 m21.11-11) Calculate the temperature gradient for heat flow and the thermal conductivity of material plate of thickness 2.5 cm for a steady state heat flow of 43.5 kw/m2 . The temperatures on both sides of the plate are 3200K and 3450K. Answer: T1 = 345 K and T2 = 320 K, L = 2.5 cm = 0.025 m. Q/A = 43.5 Kw / m2 =43.5 x 103 w / m2 Temperature gradient = = = = - 1000 0C/m From Fourier law: = - K 43.5 x 103 = - K x (-1000) Hence K = 43.5 W/ m 0K1.11-12)Find the thermal conductivity of a rod of diameter 30mm and 200mm long when it is maintained at 1000C at one end and 100C at other end. The surface of the rod is completely insulated such that the heat flow is maintained at 6 watts.Answer: The Fouriers law is; Q = A = area perpendicular to heat flow = = = 0.71 x 10-3 6 = = 0.318 KHence K = = 18.87 W/ m-0K1.11-13) The guarded hot plate method is used to find the thermal conductivity of the material. Two similar 1cm thick specimens receive heat from 6.5cm x 6.5 cm guard heater. The wattmeter shows 15 watt and thermocouples inserted in hot and cold surfaces indicate 325 K and 300 K respectively. Calculate the thermal conductivity of the specimen material.Answer: We have Q = hence K = Q = 15 watt, A = 0.065 x 0.065 = 0.00423 m2, L = 0.01 mt K = = 0.71 W/ m KExercises of Conduction Principles.Ex-1) The heat flow rate through a 30mm thick wood board for a temperature difference of 300 C between the two surfaces is 120 w/m2. Calculate the thermal conductivity of wood.Ex-2) If 3 kw is conducted through a section of insulating material of 0.6m2 cross section and 2.5 cm thick and the thermal conductivity may be taken as 0.2 w/moC, compute the temperature difference across the material.Ex-3) A temperature difference of 85 C is maintained across a fibre layer of 13cm thickness. The thermal conductivity of fiberglass is 0.035 w/m C. Compute the heat transferred through the material per hour per unit area.Ex-4) The temperatures of the faces of a plane wall of 15 cm thick are 375 C and 85 C. The wall is constructed of a special glass of properties K = 0.78 w/m C, = 2700 kg/m3 , Cp = 0.84 KJ/kgC. What is the heat flow through the wall under steady state conditions.

Convection:Newtons Law of Cooling for convection heat transfer = As (- ) The heat flow rate is proportional to product of area perpendicular to heat flow and the temperature difference. This equation is called as Newtons law of cooling. where = Convective heat flow from the surface in watt or J/sec. It is scalar quantity. As = Surface area from which convection occurs in m2 = convective heat transfer coefficient in w/ m2 0K.It is not thermodynamic property of material, but may depend on geometry of surface , flow characteristics, thermodynamic properties of fluid etc. It is the proportionality constant in above equation. (- ) = t = temperature difference between the hot surface and fluid in 0K Convective heat transfer Resistance and equivalent electrical circuit.The Newtons equation for Convection heat transfer is; Q = h A (t ) = h A( T1 T2 ) = = = Kconvection (t1 t2 )Comparing with the Ohms law,Thermal resistance in convection = ( Rth)convection = 1/hA 0C/watt --------- 1.15.4Thermal Conductance in convection =Kconvection = hA Watt/ 0C ------------1.15.5 Circuit is given below It is seen that the thermal resistances in combined conduction and convection are shown in series in the electrical circuit. (However the resistances in combined convection and radiation are shown as parallel in the circuit) Fluid filmi = Qc

T1T1T2E1E2 T2 Rconvection = Rc = i = Qc = = Qc Fig (a): Circuit of convection heat flow through stationary fluid film

L TA

TATB T1T2T1 T2 Qc R1 = , R2 = , R3= h1 h2TB Rtotal = R1 + R2 + R3Qc = = Fig (b): Circuit of Combined convection and conduction heat flow through stationary fluid film and a wall separating the two fluids Fig Circuit of convection heat flow. through stationary fluid film

Boundary layerProf.Prandtl suggested that the fluid of flow can be divided in to two regions; a thin layer next to the wall, which is called as Velocity boundary layer where the shear stress is confined, and the regions outside this layer, where the fluid is ideal ie, non-viscous and in-compressible. The Velocity boundary layer thickness, , is defined as the distance from the wall where velocity, u = 0.99 ; ie the fluid velocity (u) is 99% of the stream velocity ( . As a result of viscous forces at the wall (on the surface of wall) the velocity of the fluid is zero and increases to the (velocity of fluid) as shown. Vertical hot plate. Boundary layeryFlow yTemperature profilevelocity profile Velocity profile Temp.profile Boundary layer uyTy

Laminar flow y y=0, u =0 Tw a) Natural convection on vertical plateb) Forced convection on horizontal plate Tw = temperature of hot plate; = temperature of fluid stream Fig:1.6: Velocity and Temperature profile during convection A thermal boundary layer thickness, is defined as the thickness from the wall where the temperature difference (Tw T ) = 0.99 (Tw - In thermal boundary layer, the temperature varies from Tw to In general is not equal to . The thermal boundary layer is regarded as consisting of stationary fluid film as shown in fig 1.6, above, through which heat is conducted and then transported by fluid motion. Thermal conductivity of fluid film is taken as and rate of heat transfer is calculated.Problem-1.15-1Air at 200C blows over a hot plate of 50 x 75cm maintained at 250 C. The convection heat transfer coefficient. is 25 w/m2. Calculate the heat transfer.AnswerFrom Newtons law q = hA (ts - t) = 25 x (0.5 x 0.75) (250-20) = 2.156 kwProblem -1.15-2An electric current is passed through a wire of 1mm diameter and 10cm long. The wire is submerged in liquid water at atmosphere pressure and current is increased till water boils. h= 5000 w/m2 0c and water temperature is 1000C. How much electric power is supplied to maintain the wire temperature at 1140C.Answerq= hA (tw - t ) A = d L = (1 X 10-3) (10 X 10-2) = 3.142 x 10-4 m2 q = 5000 x 3.142 x 10-4 x (114-100) = 21.99 watt 21.99 watts is supplied to maintain the temperature at 1140CProblem-1.15-3A 120 watt heater has been employed to maintain a plate of 0.25 m2 area at temperature of 600C when its surroundings are at temperature 200C. What fraction of heat supplied is lost by natural convection. The convection coefficient conforms the relation h = 2.5 ( T)0.25 w/m2 K . What happens to the rest of the heat? AnswerConvective heat transfer coefficient h = 2.5 ( T)0.25 = 2.5 (60 20 )0.25 = 6.287 w/m2 KHeat lost by convection = q = h A ( T) = 6.287 x 0.25 x (60 -20) = 62.87 wHeat lost by convection as fraction of heat supplied of 120 watt = 62.87/ 120 = 0.5239 = 52.39%

Rest of the heat is lost by radiation = 120- 62.87 = 57.i3 w =100-52.39 =47.61 wProblem-1.15-4Air flows over a rectangular plate having dimensions 0.5 m x 0.25 m. The free stream temperature of the air is 300C. At steady state, the plate temperature is 40C. If the convective heat transfer coefficient is 250 W/m2.K, determine the heat transfer rate from the air to one side of the plate.AnswerKnown: air flow over a plate with prescribed air and surface temperature and convection heat transfer coefficient.Find: heat transfer rate from the air to the plateAssumptions: (1) temperature is uniform over plate area, (2) heat transfer coefficient is uniform over plate areaAnalysis: the heat transfer coefficient rate by convection from the airstreams to the plate can be determined from Newtons law of cooling written inthe formq = hA (tair-tplate) = 250 x (0.25 x 0.5) (300-40) =8125 watts. Comments: recognize that Newtowns law of cooling implies a direction for the convection heat transfer rate. Written in the form above, the heat rate is from the air to plate.Problem-1.15-5An electric heater of exposed area of 0.09 m2 and output of 600watt is designed to operate fully submerged in water. Calculate the surface temperatureof the heater when the water temperature is 370C and the surface heat transfer coefficient is 285.3 w/m2-deg. How is it effected if the heater is operated in air by mistake, assuming the air is at temperature of 370C and the surface coefficient is 8.5 w/m2-deg. Comment on result.Answer: a) Heater operating in water;q = h A (ts ta) Hence ts = (q / hA) + ta = {600 / (283.5 x 0.09) } + 37 = 60.50C b) When operating in air; ts = {600 / (8.5 x 0.09) } + 37 = 8210C Coment:-The temperature of surface of heater is too high, so that the heater element may melt.Problem-1.15-6A container with out side surface area of 0.36m2 and outside temperature of 00C contain ice at 00C. The container is placed in ambient air at 240C and the surface coefficient of heat transfer between the container surface and surroundings air is estimated as 6.25W/m2-deg. Cakculate rate at which ice would be changed in to liquid water. Take latent heat of fusion of ice as 340 J / g .Answer. Heat flow from air to Ice, q = hA (t2 t1) = 6.25 x 0.36 (24-0) = 54 w = 54 J/sec Heat utalized in melting the ice in container, q = m hfg 54 = m x 340Hence the mass rate of change of ice in to water is; m = 54/ 340 = 0.1588 g / s =0.1588 x 3600/ 1000 = 0.572 kgProblem-1.15-7A lake surface is covered with 8cm thick layer of ice (K = 8 kj/m-hr-deg) when the ambient air temperature is -12.5 C. A thermocouple on the upper surface of the layer indicate temperature of - 5C. Assuming steady state condition of ice and no liquid sub cooling at the bottom surface of the ice layer, find the heat transfer coef. At upper surface. Also work out heat loss per square kilometer area.Answer. Ta = -12.50C T2 = -50 C

T1 = 00C Ice Layer 8 cm thick

(Liquid) LAKE.

Since the water at bottom of the ice layer is liquid (no sub cooling) the minimum temperature at surface is 0 C,Area A = 1 sq km = (1000x 1000) = 106 m2Fouriers law q = K A (t2 t1) / =8 x 106 x { 0- (-5)} / 0.08 = 5x 108 kj/hr Steady flow means Heat flow across the ice slab = convective heat transfer at top of slab and surrounding air. q = hA (t2 t1); Hence 5 x 108 = h x 106 x{ -5- (-12.5)}or h = 5 x 108 / 106 x 7.5 = 66.67 m2 hr-degProblem-1.15-8Hot gases at 980C flow past the upper surface of the blade of a gas turbine and the lower surface is cooled by air bled off the compressor. The convective heat transfer at upper and lower surfaces are estimated as 2830 and 1415 w/m2-deg. Respectively.The blade material has the thermal conductivity of 11.6 w/m-deg. If metallurgical considerations limits the blade temperature at 870 C, work out the temperature of cooling air. Consider the blade as a flat plate of 0.115 thick and presume the steady state conditions have been reached.Answer Hot gasesh1tg = 9800CUpper surface, tu = 8700C

= 0.115 cmGas turbine blade.

h2 Cool Air, tc Fig Lower surface , tL AnswerConsider per unit area ie: A = 1m2 , h1 =heat transfer coef. at upper surface = 2830 w/m2-deg h2 =heat transfer coef. at lower surface = 1415 w/m2-deg K = thermal conductivity of material =11.6 w/m-deg tg temperature of hot gasses = 980 C tu - temperature at upper surface = 870 C tl temperature at lower surface of blade. tc temperature of cooling airHeat flow from hot gasses to the upper surface of blade q = h1 A (tg tu ) = 2830 x 1 x (980 - 870) = 311300 W / m2This heat is conducted through the blade, hence q = KA ( tu - tl ) / 311300 = 11.6 x 1 x (870 tl ) / 0.00115 = 10087 (870 tl)Hence, tl = 870 311300/ 10087 =839 CThe heat transferred across the blade lower surfce is transferred to cooling air. 311300 = 1415 x 1 x (839- tc) or tc = 839 311300 /1415 = 619 CProblem:1.15-9:The oven of an electric stove, of total out side surface area 2.9 m2 dissipates electric energy at the rate of 600 watt. The surrounding air is at 20 C and the surface coefficient of heat transfer between room air and surface of oven is 11.35 w/m2 0C. Determine the average steady state temperature of the outside surface of stove. What would be the inside surface temperature if wall thickness of stove is 3.8cm and thermal conductivity of the material is 0.069 W/ m-0C. Answer:The electrical energy is dissipated as convective heat flow from surface of heater to the ambient air. Hence Q = h A (to ta) 600 = 11.35 x 2.9 x (to 20) So; to = + 20 = 38.220 C Ambient air , ta = 20 C Out side surface (to) ho

L = 3.8 cm, Stove wall. to

tiInside surface (ti) Q = 600 WThe electrical energy is first conducted across the wall of the oven. Hence from the Fouriers equation, Q = 600 = = 5.26 (ti 38.22) ti = + 38.22 = 152.290CInside surface temperature of the stove wall = 152.29 C Examples of topics from 1.12 to 1.15.(CONVECTION)Ex-1: A hot plate is maintained at a temperature of 1200C dissipates heat at the rate of 7500 w/ m2 to the ambient air at 30 C. Calculate the gheat transfer coefficient for conversion between the plate and the air. (8.3 W / m2 K.)Ex-2: The inside surface of the insulating layer is at 2700C and the outside surface is dissipating heat by convection in to air at 20 C. The insulation layer is 40mm thick and has a thermal conductivity of 1.2 W/ mK. What is the minimum value of the heat transfer coefficient at the outside surface if the outside surface temperature should not exceed 70 C.(120 w/m2 K).

Absorptivity, Reflectivity and Transmissivity. If Q is the total radiant energy incident on the surface of a body:- i) Some part,QA will be absorbed, ii) Some part, QR will be reflected and iii) Some part, QTr will be transmitted. By energy balance; QA + QR + QTr = Q or + + = 1 or + + = 1 i) = ; called as absorptivity. -------------1.17.1 Absorptivity,, is Fraction of incident energy which is absorbed ii) = ; called as reflectivity ------------ 1.17.2 Reflectivity, , is the fraction of incident energy which is reflected. iii) = ; called as transmissivity. or Transmittance.------------ 1.17.3 Transmissivity, , is the fraction of incident energy which is transmitted Transmissivity,,is fraction of incident energy which is transmitted through body.a) For an opaque body, = 0. (ie: Transmissivity is Zero For Opaque Body), hence + = 1. Most of the solids do not transmit any radiation, hence are opaque. Hence if is reduced, increases. b)The reflectivity, , depends on the character of surface. By increasing the surface polishing, the reflectivity, is increased and the absorptivity, decreases. Hence the absorptivity, , for a opaque body can be increased or decreased by the appropriate surface treatment. -Angle made by incident ray with the normal is angle of incidence Angle made by the reflected ray with the normal is angle of reflection

Normal

Incident Radiation Reflected Radiation Q QR Angle of incidence Angle of radiation

QA Radiation absorbed

QTr Transmitted radiation Fig:- 1.10, Radiation incident on a bodySpecular reflection: When the surface is highly polished, the angle of incidence ( is equal to the angle of reflection ( and this type of reflection is called as Specular reflection.Diffuse reflection: When the surface is rough, the incident radiation is distributed in all directions, and the reflection is said to be diffuse reflection.

Black BodyA body absorbs all incident radiation is called as black body. ie: = 0 and = 0 whch means = 1. A black body is also best emitter ie:emissivity of black body, =1. There is no perfect black body (having = 1 ) in nature. The term black is used, since most black colored surfaces normally shows high values of absorptivity, and they also absorb all visible light rays, because of which they appear black to our eyes. There are some surfaces which absorb nearly all incident radiation, yet do not appear black, Ice, Snow, white-washed walls have absorptivities greater than 0.95. Emissive Power: (E) and ( ) The total emissive power(E), of a body is defined as the total radiant energy emitted by the hot body at a certain temperature per unit time and unit surface area at all wave lengths The monochromatic emissive power(), of a body is defined as the radiant energy emitted by the hot body at a certain temperature per unit time and unit surface area at particular wave length (.Emissivity: ( ); Emissivity is a ratio given by the ratio of radiation emitted by a gray body surface to the radiation emitted by a black body at same temperature and is denoted by and its value varies from 0 to 1.

= The emissivity of a surface indicates how efficient the gray body surface emits the radiation compared to an ideal black body radiation. The value of emissivity is a radiation property of the surface. Its value depends on the surface characteristics. The emissivity of gray body surface is less than unity. ie: < 1 for gray body and = 1 for the black body.STEFAN BOLTZMAN LAWThe law states that total emissive power ( or total radiation energy emitted) of a black body is proportional to the fourth power of the absolute temperature. ie: Eb . Eb = b A Where, Eb = Maximum rate of heat Emission by a black body at given temperature T0 Kelvin of surface in Watts (or Emissive power of black body) b = Stefan- Boltzman constant = 5.67 x 10-8 w/m2 K4 (or in MKS units = 487.6 x 10-10 Kcal / hr m2K4)A = Area of the surface emitting radiation in m2Ts = Absolute temperature of the surface in 0K = ( t0C + 273) Emission by a Real surface ( Gray body) The radiant energy emitted by real surface = ER = b A T4Radiation between two bodiesShape factor is defined as the fraction of radient energy emitted from one surface of a body and received by other body directly and is denoted by . This is also known as Configuration factor or angle factor or view factor or geometric factor. means fraction of energy leaving surface -1 and falling on surface -2. = = Like wise is also defined and = is called as reciprocity theorem.Case-i:- Radiation heat transfer between two black bodiesRadiation leaving 1 and falling on-2 ia ; = Radiation falling on 1 out of energy emitted by 2 is ; = Hence net exchange between two faces is; = = ( ( since = ) or = (( Case-2:- Radiation Heat transfer from real surface to black surface Consider a real surface-1 at temperature T1 surrounded by a black surface-2 at temperature T2 then the heat transfer from real surface to black surface is given by; (F12 = 1 ) Q = 1 b A1 (T14 T24)Case-3)The net exchange of heat between two radiating surfaces = Q1-2 = F1-2 1 b A1 (T14 T24).Radiation Heat transfer from gray surface Reflections and equivalent electrical circuit. We have to consider Surface Resistance and Space resistance.This gives the surface resistance to Radiation in forming the electrical circuit

GJ = Eb + G

= (J G ) Eb J Q = A (J - G) Q = Rth = Rsurface = Current, i = Fig: 1.11- Radiation over a surface of gray body.Two New terms are defined for finding heat exchange between non-black surfaces. i) Irradiation G; It is total radiation incident upon a surface per unit time and per unit area; ie: Watt / sec m2 . ii) Radiosity- J ; It is total radiation which leaves the surface per unit time and unit area Radiosity is the sum of energy emitted and energy reflected when transmitted energy is zero. ie: J = Eb + G -----------------1.20.5Since transmissivity is zero, = 0, then, = 1 or = 1 or = 1 from Kirchoffs identity, we have = Substituting value in 1.17-5, we get J = Eb + (1 G Hence G = Surface ResistanceThe net energy leaving the surface per unit area is = (J G ) Substituting G from above G = = [ J - ] Q = A [J - ] - = = = ie Q = Equation above is useful for conceiving the electrical circuit equivalent to thermal circuit equivalent to electrical circuit as shown .The quantity in denominator, ie: is called as surface resistance in the radiation heat transfer electrical circuit.The quantity in Numerator, , is considered as potential differenceSpace ResistanceNow consider the exchange of radiant energy between two gray surfaces A1 and A2, when both are viewing each other ; = A1 A A

J1 J2

A2Rth = or = . ie: Current, i = Fig- 1.12: Radiatin between two surfaces and electrical circuit with space resistance.The radiation that leaves the surface 1 and reaches the surface 2 = The radiation that leaves the surface 2 and reaches the surface 1 = The net exchange between two surfaces is, = Since We have = ( or = ( ie: = or = . The quantity or is called as Space resistance in constructing the net work.The net work circuit is shown in fig 1.12.

Combined Net Work of Surface and Space Resistances for two surfaces

Rsurface1 = Rspace = Rsurface2 =

Fig. 1-13:- Radiation net work for two surfaces which see each pther and nothing else. Qnet = - Here Total resistance Rr = Above equation can also be written as in terms of temperatures by using Eb1 = b A1 and Eb2 = b A2 with Surface and Space resistances. Q12 = (Fg )1-2 b A1 (T14 T24) where (Fg )1-2 is called as Gray body Factor and is given by; (Fg )1-2 = = Also we can write reciprocity theorem A1F1-2 = A2F2-1 (this will be proved latter)Problem -1A radiator in a domestic heating system operates at a surface temperature of 600CCalculate heat flux at the surface of the radiator if it behaves as a black body.SolutionHeat flux = q = Q/ A = b A T14 / A = b T14 = 5.67 x 10-8 (60+273)4 =5.67 x 1.23 x 102 =697.2 w/ m2

Problem-2 Two infinite black plates at 8000C and 3000C exchange heat by radiation. Calculate heat transfer per unit area. Answerq/A = (T14 T24) = 5.67 x 10-8 (10734 5734) = 69.03 kw/m2Problem-3A cylindrical rod of 1.5cm long and 2cm in diameter, positioned in a vacuum furnace is electrically heated such that its surface of rod is maintained at 1000 K. The furnace interior is at 800 K. Calculate the power supplied to the rod if the emissivity of the surface of rod is 0.9. AnswerFor steady state conditions, the power supplied to rod equal to radiant heat loss from it. Since the rod is completely surrounded by furnace walls the total radiant energy is intercepted by the walls of furnace. q = A ( Ts4 Tsur4 ) ; A = d l = x 0.02 x 1.5 = 0.09424q = 0.09424 x 5.67 x 10-8 (10004 8004) = 2838 w Problem 4A spherical shaped transistor of 2cm diameter is kept in an evacuated case of black walls at 300C. The heat loss from transistor is only by radiation. If the transistor dissipates 300 mW, what will be the temperature of transistor if it is (i) Bright aluminium of =0.035 and (ii) black anodized aluminium of =0.80.Answerq = 300mW = 0.3 Watt. T2 = 273 + 30 = 303 KA = d2 = x 0.022 = 1.26 x 10-3q = A ( Ts4 Tsur4 ) =1.26 x 10-3 x 5.67 x 10-8 x (T14 3034)0.3 = 7.162 x 10-11 (T14 3034).i) When =0.0350.3/7.162 x10-11 x 0.035 = (T14 3034).1.197 x 1011 =(T14 8.43 x 109).T14 = (1.197 + 0.0843) 1011 = 1.2813 x 1011 Hnce T1= 598.29 K ii)When =0.8 T1 = 342K Problem-5A sphere of diameter 10 mm and emissivity 0.9 is maintained at 80C inside an oven with a wall temperature of 400C. What is the net transfer rate from the oven walls to the object?ANSWERKnown: spherical object maintained at a prescribed temperature within a oven.Find: heat transfer rate from the oven walls to the objectAssumptions: (1) oven walls completely surround spherical object, (2) steadystate condition, (3) uniform temperature for areas of sphere and oven walls, (4) oven enclosure is evacuated and large compared to sphere.Analysis: heat transfer rate will be only due to radiation mode. The rate equation isq = A ( Ts4 Tsur4 ) A = d2 for sphere. T1 =400+273 = 673K, T2=80+273= 353K q = . (10 x 10-3)2 . (5.67 x 10-8) . 0.9 . { 6734 - 3534} = 3.04 wDiscussion:This rate equation is applicable when we are calculating the net heat exchange between a small object and larger surface that completely surrounds the smaller one. When performing radiant heat transfer calculations, it is always necessary to have temperatures in Kelvin (K) unitProblem-6A surface of area 0.5m2, emissivity 0.8 and temperature 1500C is placed in a large, evacuated chamber whose walls are maintained at 25 0C. Find the rate at which radiation is emitted by the surface? What is the net rate of radiation exchange between the surface and the chamber walls?AnswerKnown: Area, emissivity and temperature of a surface placed in a large, evacuated chamber of prescribed temperature.Find: (a) rate of surface radiation emission, (b) net rate of radiation exchange between the surface and chamber walls.Assumptions: (1) area of the enclosed surface is much less than that of chamber walls.Analysis (a) the rate at which radiation is emitted by the surface is emittedqemit = A Ts4 = 0.5 . 5.67 x 10-8 . 0.8 . { 150 +273}4 = 726 w(b) The net rate at which radiation is transferred from the surface to the chamber walls isqt = A (Ts4 Tsur4) Ts =150+273 = 423K Tsur = 25 +273 = 298 Kq = 0.5 . 5.67 x 10-8 . 0.8 { 4234 2984} = 547 WProblem-7The quantity of radiation received by earth from sum is 1.4 kw/m2 (solar constant). Assuming that sun is an ideal radiator, calculate the surface temperature of the sun. The ratio of radii of earths orbit to the sun is 216.Ans:Total radiation from sun Qr = (1.4x 103) * 4 R2 where R is the radius of earths orbit.Total radiation emitted by Sun = Qr = A b Ts4 = 4 r2 * Ts4 where r is radius of Sun & b =1 For Sun considered as black body Hence 4 r2 * Ts4 = 1.4x 103 * 4 R2 Ts4 = (216)2 *1.4 x 103 / 5.67 x 10-8 =0.1152 x 1016 Ts = 5826 KProblem -8:

A solid aluminium sphere of emissivity , initially at a high temperature, is cooled by convection and radiation in a chamber having walls at a lower temperature. Convective cooling is achieved with a gas passing through the chamber. Write a differential equation to predict the variation of sphere temperature with time during the coolingprocess.AnswerKnown: Initial temperature, diameter and surface emissivity of a solid aluminium sphere placed in a chamber whose walls are maintained at lower temperature. Temperature and convection coefficient associated with gas flow over the sphere.Find: equation which could be used to determine the aluminium temperature as a function of time during the cooling process.Assumptions: (1) at any time t, the temperature T of the sphere is uniform, (2) constant properties (3) chamber walls are large relative to sphere.Analysis: applying an energy balance at an instant of time to a control volume about the sphere, it follows that energy balanceEstored = - EoutHeat rates out of Control volume due to convection and radiationd/d ( V C T) = - (qconv + qrad) (C is specific heatdT/d = -{1/( V CT)} [h A (t- t) + A (T4 Tsur4)] A = d2 and V = d3/6 Hence A/V = 6/ddT/d = (6/ C d ) [h (t- t) + (T4 Tsur4)] is required equationProblem:-9 An electronic package dissipating 1 kW has a surface area 1m2. The package is mounted on a space craft, such that the heat generated is transferred from the exposed surface by radiation into space. The surface emissivity of the package is 1.0. Calculate the steady state temperature of the package surface for the following two conditions:(a) the surface is not exposed to the sun

(b) The surface is exposed to a solar flux of 750W/m2 and its absorptivity to solar radiation is 0.25? AnswerKnown: surface area of electronic package and power dissipation by the electronics. Surface emissivityand absorptivity to solar radiation. Solar flux.Find: surface temperature without and with incident solar radiation.Applying conservation of energy to the control volume:Ein Eout + Eg = 0s As qs - As qemit + P =0s As qs - As Tsur4 + P = 0Tsur = [(s As qs + P) / As ]1/4 In the shade qs =0Tsur = [ 1000/ 1 x5.67 x 10-8]1/4 = 364 KIn the SunTsur = [0.25 x 1x 750 + 1000 / 1 x 1 x 5.67 x 10-8]1/4 =380KProblem :-10A horizontal steel pipe having a diameter of 5cm is maintained at temperature of 500C in a large room where the air and wall temperature are at 20C. The surface emissivity of the steel may be taken as 0.8. Using h as 6.5 w/m2C, calculate the total heat loss by the pipe per unit length. Coment on the result. AnswerTotal heat loss = loss by convection + loss by radiation. Surface area for convection is taken as d l. Here l = 1mt aske per unit length of pipe. (q/L)convection = d l h (t- t) = x 0.05x 6.5 x (50-20) =30.63 w/mThe pipe is considered as body surrounded by large enclosure so that the radiation heat transfer can be calculated by equation (q/L)radiation = A (T1 4 T2 4) A = d = x 0.05 = 5.669 x 10-8 * 0.8 * x 0.05 (3234 2934 ) = 25.04 w/m(q/L)total = (q/L)convection + (q/L)radiation = 30.63 + 25.04 = 55.67 w/mComment on result is:In this type of problem both convection and radiation are approx. equal and hence we can not neglect any one of them.

Problem:-11A surface at 475 K convects and radiates heat to the surroundings at 335K. If the surface conducts this heat through a solid plate of thermal conductivity 12.5 w/mK, determine the temperature gradient at surface of solid.Take h value as 80w/m2K & radiation factor 0.9.Ans:Under steady state conditions,Heat conducted through plate = convection heat loss + radiation heat loss.- KA dt/ dx = hA (t2 tf) + A (T2 4 Tf 4) Taking unit area A=1m2,- 12.5 dt/ dx = 80(475 335) + 5.67 x 10-8 (4754 3354 ) = 13155 Hence dt/dx = - 13155/ 12.5 = - 1052.4 C/ mProblem-12A small hot surface at temperature 425 K having an emissivity 0.85 discipates heat by radiation in to surrounding area at 400 K Calcolate the radiation heat transfer coefficient.AnswerQ = 1 A1 (T1 4 T2 4) = 1 A1 (T12 + T22 ) (T12 -- T22 ) = 1 A1 (T12 + T22 ) (T1 + T2 ) (T1 -- T2 ) = 1 A1 4 (Tm)3 (T1 -- T2 ) where Tm = (T1 + T2 ) /2 = hr A1 (T1 -- T2 )hr = 4 1 (Tm)3 Tm = (425 + 400) / 2 =412.5C hr = 4 x 5.67 x 10-8 x 0.85 x (412.5)3 = 13.6 W /m2 KConduction through a composite wall , Series Resistances (Electrical analogy)The corresponding electrical circuit is also shown below.

Layer-3K3L 3Layer-2K2L 2 T3 Layer-1 K1L 1 T2

Q T1 T1 > T2 > T3 > T4

Area A T4

T1 T4 T2 T3 = = = Fig 1.16. Conduction through three resistance. Under steady state conditions, the heat flow does not vary across the wall , ie; it is same at each and every layer. Hence q = = = Thus the temperature drop in each layer is indicated as; (T1 T2) = q ; (T2 T3) = q ; (T3 T4) = q Adding all the terms, we get temperature difference across the composite wall. (T1 T4 ) = q [ + + ] or q = q = = - ( like i = ) where =Where R1 , R2 , R3 are conductive thermal resistances and in series circuit the total resistance is equal to sum of all resistances in series.It appear in the same way as the electrical resistors are connected in series.The above analysis can be extended to n-layers of composite wall as below q =

Problem 1A kiln exterior wall is of 10cm layer of common brick with K= 0.75 w/mK. It is followed with a layer of 4cm gypsum plaster of K= 0.5 w/mK. What thickness of loosely packed rockwool insulation of K= 0.065 w/mK should be aded to reduce the heat loss or gain through the wall by 75%.Answer Consider the multi layer wall and we calculate q per unit area, ie A=1m2 :1st layer Brick (inside):- Thickness, 1 = 10cm = 0.1m. K1 = 0.75 w/mK2nd Layer gypsum plaster :- Thickness, 2 = 4cm =0.04m , K2 = 0.5 w/mK3rd layer Rock wool, 3 = ?. K = 0.065 w/mKResistance of brick work R1= 1 / K1 A = 0.1 / 0.75 x 1 = 0.133 deg/ wResistance of gypsum plaster R2= 0.04 / 0.5 x 1 = 0.08 deg/wResistance of Rock wool insulation = 3 / 0.065 x 1 = 0.1538 3Heat flow with out insulation being added = t / R1 + R2 =t / 0.133+ 0.08 = 4.695 t.Heat loss with addition of rock wool = (1- 0.75) 4.695 t = 1.174 t.This must be equal to the heat flow when rock wheel is also considered. 1.174 t = t / ( 0.133 + 0.08 + 0.1538 3 )Hence from this 3 = thickness of rock wool = 0.04153m = 4.153cmProblem2A storage chamber of interior dimentions 10m x 8m x 2.5m height is maintained at temperature of -200C while the out side is at 25 C. The walls and ceiling has 3 layers made of 1)60mm thick board of K = 0.2 w/mK on the inside, 2) 90mm thick insulation of K = 0.04 w/ mK in the middle 3) 240mm thick concrete of K = 1.8w/mK. On out side.

Neglecting flow of heat through the floor, determine the rate at which heat flows in to the chamber.AnswerNeglecting corners and edges and floor, = Area of sides on 10mt wide + Area of side on 8mt wide + area of topThe area of heat flow = A = (2 x 10 x 2.5) + (2 x 8 x 2.5) + (10 x 8) = 170m2Total thermal resistance Rt = (1/A) ((650- 125)1 / K1 + 2 / K2 + 3 / K3 ) = 1/170 ( 0.06/0.2 + 0.09 / 0.04 + 2.4/1.8 ) = 0.01578 deg/wHeat flow rate =Q = t / Rt = 25- (-20) /0.01578 = 2851.7 WProblem 3A furnace wall is made up of a steel plate of 10mm thick, K = 62.8 kj/m-hr-deg, lined inside with silica bricks 150mm thick, K = 7.32 kj/m-hr-deg, and on outer side with magnesia bricks of 200mm thick, K= 18.84 kj/m-hr-deg,. The inside and out side surfaces of the wall are at temperatures of 650C and 125C respectively. Calculate the heat loss per unit area of the wall. It is required to reduce the heat loss to 10MJ / hour by means of air gap between the steel and magnesia bricks. Estimate the necessary width of air gap if the thermal conductivity of air is 0.126KJ/m hr deg.Answer (The units of K can be same through out, ie: KW/ m-deg or KJ / m-hr-deg.)R1 = Resistance of silica bricks = 0.15 / 7.32 x 1 = 0.000159 deg-hr/ KJ (inside of composite wall)R2 = Resistance of steel plate = 0.01/ 62.8 x 1 = 0.000159 deg hr / KJ (middle of composite wall)R3 = Resistance of magnesia bricks = 0.20 /18.84 x 1 = 0.01061 deg hr / KJ (out side of composite wall)Total resistance of composite wall, = Rt = R1 + R2 + R3 = 0.03126 deg hr / KJ.Heat loss from wall = (t2 t1) / Rt = (650- 125) / 0.03126 = 16795 KJ/ hr.The total resistance if heat loss to be restricted to 10 MJ / hr, ie: 10 x 103 kJ / hr. = (650- 125) / 10 x 103 = 0.0525 deg hr / KJHence the resistance to be for air gap Ra = 0.0525 0.03126 = 0.02124 deg-hr / KJ SO, Thickness of air gap = = Ra x K x A = 0.02124 x 0.126 x 1 = 2.676mm Problem 4 A furnace wall comprises three layers: 13.5 thick inside layer of fire brick, 7.5 cm thick middle layer of insulating brick and 11.5cm thick outside layer of red brick. The furnace operates at 870C and out side is maintained at 40C by circulating the air. The wall measures 5m x 2m and thermal conductivities of materials are:K1 (Fire brick) = 1.2 w/m-deg , K2 (Insulating brick) = 0.14 w/m-deg K3 (red brick) = 0.85 w/m-deg Find the rate of heat loss from the furnace and wall interface temperature.

Answer: The wall area 5 x 2 = 10m2 is constant for heat transfer and same for all layers. Resistance of fire brick = R1 = 1/ K1 A = 0.135 / 1.2 x 10 = 0.01125 deg/ wResistance of Insulating brick = R2 = 2 / K2 A = 0.075 / 0.14 x 10 = 0.05357 deg/ wResistance of Red brick = R3 = 3 / K3 A = 0.115 / 0.85 x 10 = 0.01353 deg/ w Total resistance of composite wall = Rt = R1 + R2 + R3 = 0.07835 deg / w Heat flows from inside of furnace at temp 870C to out side at temp 40C. Heat loss = (870 40) / 0.07835 = 10593.5 Wb) Let t1 = 870 C , t2 = interface temperature between fire brick and insulation brickt3 = interface temp. between insulation brick and red brick. t4 =out side temp=40CSince the total heat loss through each layer is assumed to be constant as10593.5 Wthen for the inside layer, fire brick only, 10593.5 = (870 t2) / 0.01125 t2 =750.82 CSimilarly 10593.5 = (750.82 t3) / 0.05357 t3 = 183.33 CProblem 5:A 30 cm wall of a reactor is made up of an inner layer of fire brick , K = 0.85 w/ mKCovered with a layer of insulation, K = 0.15 w/mK. The reactor operates at temperature 1600K while the ambient temp is 295 K.Calculate the thickness of which gives minimum heat loss. Also work out the heat loss presuming the insulating material has maximum temperature of 1475K. If the calculated heat loss is unacceptable, would the addition of another layer of insulation be a satis factory solution? Comment on result.Solution Under steady state conditions, the heat transfer is constant through out the wall and is same for each layer. Assuming per unit area of wall, and ti is interface temperature,Q = (t1 t2) / (1/ K1 + 2 / K2 ) = (t1 ti) / (1/ K1 ) = (ti t2) / (2 / K2 ) (composite wall) (fire brick) (insulation)We have 2 = (0.3- 1), then taking the first two equations, (1600 295) / [ (1/ 0.85) + (0.3- 1)/0.15] = (1600-1475)/ (1/ 0.85) From this equation, 1 = 0.1125 = 11.25cm = Thickness of fire brick Thickness of insulation = 2 = 30 11.25 = 18.75cm Heat flow per unit area of the wall = q = (t1 ti) / 1/ K1 = (1600 1475) / 0.1125/ 0.85 = 944.45 w/ m2The heat loss from the wall will get reduced if further insulating layer is added. This reduce the temperature drop across the fire brick lining and subsequently increase the interface temperature, ti . How ever, this value of interface temperature is set at the maximum possible value (by fixing temperature of 1475 K), the addition of insulation of further layer will increase the interface temperature and hence is not a satisfactory solution.

Conduction through a composite wall , Parallel Resistances (Electrical analogy)The total heat transfer, because A and B are in parallel, is Q = QA + QB T1 L T2

Layer-1 K1 A1L1 = L2 = L Q

Layer-2 K2 A2

=

T1 T2 = = (Like the addition of electrical resistances which are parallel) = or = Fig-1.17. Conduction through two resistances in parallel.

Hence Q = [ K1 A1 ] + [ K2 A2 ] = + = + = [ [ + ] = [ [ (since parallel resistances are added as = = ----------------1.22-1 or Q =

Problem-1:Two slabs each 100mm thick and made of materials with thermal conductivities of 16 W /m-seg and 200 W/m-deg, are placed in contact which is not perfect. Due to roughness of surfaces, only 40% of area is in contact on either side of contact and air fills 0.02mm thick gap in the remaining area. If the extreme surfaces of the arrangement are at temperatures of 2500 C and 300C , determine the heat flow through the composite system, the contact resistance and temperature drop at contact. Take the thermal conductivity of air as 0.032 W/m deg. and assume that half of the contact (of the total contact) is due to either material.SOLUTION:-B100mm0.02100mm mm

CAirQQAE

D

RA RB REQ RC Q

RDIf Area ,AA , is taken as 1 sq.mt, then AB = 0.2 sqmt (ie 20%) = AD and AC = 0.6 (60%)Various thermal resistances are calculated as below. Take perpendicular to flow is 1 mt2 RA = = = 0.00625 deg/ W RB = = = 0.00000625 deg / W RC = = = 0.001042 deg / W RD = = = 0.0000005 deg / W RE = = = 0.0005 deg / W The resistances RB , RC , RD are in series , Hence = + = + = 160000 + 959.7 + 2000000 = 2160959.7 W/ DEG Req = = 0.462 x 10-6 deg/W Req , ie equivalent resistance is in series with RA and RE Hence , Rtotal = RA + Req + RE= 0.00625 +0.462 x 10-6 + 0.0005 = 0.00675 deg/W Hence heat transfer rate through the system = Q = = = 32592 watts b) Contact Resistance = 0.462 x 10-6 deg/ w Temperature drop = T = Q x Contact resistance = 32592 x 0.462 x 10-6 = 0.01505 CThermal Resistance of slab including convection resistance:-Since the energy will flow first through block from fluid A and then through fluid B, we say that the resistances are thermally in series arrangement. Surface films on both sides. Fluid A

L TA

TATB T12 T2 T1 T2 Qc TB R1 = , R2 = , R3= h1 h2 Rtotal = R1 + R2 + R3Fluid-B Qc = =

Fluid films on both sides Fig (1.19): Circuit of Combined convection and conduction heat flow through stationary fluid film and a wall separating the two fluids

(Note: Temp. Distribution in surface film is a curve, It is convex at entry and concave at exit.Temperature distribution in solid is a straight line. Temperature distribution in fluid is a st. line.) Rtotal = R1 + R2 + R3 +R4 +R5 (for a series circuit, the total resistance is sum of all resistances and the resistances of conduction and convection are treated as series circuit))The steady state heat flow rate through the walls is given by: Q = ProblemThe wall in a furnace consists of 125mm thick refractory bricks and 125mm thick insulating bricks separated by an air gap of 12mm. A 12mm thick plaster covers the outer surface.The inner surface is at 1100 C and ambient temperature is 25 C. the heat transfer coefficient of outside wall to air is 17 w/m2 K. and the resistance by air gap to heat flow is 0.16 k/w. The thermal conductivities of refractory brick brick, insulating brick and plaster are 1.6, 0.3, and 0.14 w/ m-K respectively. Calculate a) The rate of heat loss per unit area of wall surface, b) The interface temperature through the wall and c) the temperature of the outside surface of the wall. d)Coment on the result of outer temperature.

Answer..This inner film resistance is neglectedThick plaster Out side film t1 = 11000CR1R2

R4R5Insulating brick R3 Air gapRefractoryBrick

t2 = 250C

125mm 12mm 125mm 12mm Fig Consider A =1m2Assume the resistance due to convection and radiation at inner wall is neglected and temperature at inner wall surface is 1100 K. Hence no drop in temperature. ie (t1 =t2)R1 = Resistance of refractory brick = x1/ K1A = 0.125 / 1.6 x 1 = 0.0781 K/ wR2 = Resistance of air gap = 0.16 K/w (given)R3 =Resistance of insulating brick = 0.125 / 0.3 x1 = 0.417 K/WR4 = Resistance of plaster = 0.012 /0.14 x1 = 0.0857 K/WR5 = Resistance of air film out side = 1/hA =1/17 x1 = 0.0588 K/WRT = total resistance = 0.0781 + 0.16 +0.417 + 0.0857 +0.0588 = 0.7996 0.8 K/wRate of heat transfer per unit area =q = (t1 to) /0.8 = (1100 25)/ 0.8 =1344 w =1.344 kw Let the interface temperatures are t3, t4 , t5 , and out side temperature as t6 , Applying the electrical analogy to each surface, q = ie; 1344 = . Solving, t3 = 995 C q = ie; 1344 = . Solving, t4 = 780 C q = ie; 1344 = . Solving, t5 = 220 C q = ie; 1344 = . Solving, t6 = 104.1 C

Comment: A good furnace design will load to an outer wall temperature of 60 C or below as required by factory act, to prevent injury to personnel when contacted with wall and safe working conditions.1.22) Combined Convection and RadiationQR = b A1 (T14 T24) (refer to topic-1.20; fig-1.14) = b A1 (T12 + T22) (T12 - T22) = b A1 (T12 + T22) (T1 + T2) (T1- T2) = b A1 4 (Tm)3 (T1- T2) ( where Tm = (T1 + T2) / 2 )or QR = hr A1 (T1- T2) ------------------------------------------ = = We have hr = b (T12 + T22) (T1 + T2) It is a good approximation to the radiation heat transfer which resembles the convective heat transfer rate. Where hr = 4 b (Tm)3 where Tm = (T1 + T2)/2 & hr = 6 . at atmospheric conditionsNote that the radiation resistance is drawn as parallel resistance when combined with convective resistances =

T1 T2 =

= = = or = Fig-b:- Electrical circuit, (Thermal circuit)Example:-A steam pipe ( of 0.4 diameter has a surface temperature of 500 K. The pipe is located in a room at 27 C and the convection heat transfer coefficient is 25 W/m2K. Calculate the combined heat transfer coefficient and the rate of heat loss per unit length of pipe.Answer: Radiation heat transfer Q1-2 = hr A1 (T1- T2) ------------------------------------------ 1.20.12where hr = b (T12 + T22) (T1 + T2) = 0.9 x 5.67 x 10-8 (5002 + 3002) (500+300) = 13.88 W/m2K h = hc + hr = 25 + 13.88 = 38.88 w/m2KRate of heat loss = Q1-2 = hr A1 (T1- T2) = 38.88 x D L (500 -300) = 38.88 x x 0.4 x 1 x 200 = 9771.6 W = 9.77 KW.

Combined heat transfer Coefficient or Overall heat transfer coefficient. (U)

L1 L2 L3

Layer-3K3 T4 Layer-2K2 T3 Layer-1 K1 T2 TH Q T1 TH > T1 > T2 > T3 > T4 > TC

Area A TC

TH TC T1 T2 T3 T4 = = = RH = RC = Fig: 1.21; Heat transfer through three layer composite wall with convection on both exterior surfaces showing the temperature profile and thermal circuit Q = = =

Usually the heat flow through composite structure is written in the form Q = UA ( ) = Where U represents over all heat transfer coefficient. = Hence = =RH + R1 + R2 + R3 + RC or U = overall heat transfer coefficient is the reciprocal of over all thermal resistance to the heat flow. The over all surface coefficient has its numerical value same whether the heat flows from either side incase of a slab or multi layer slab. It is because the area considered perpendicular to heat flow is same on both sides. Hence UO AO = Ui AiHow ever in case of heat flow in cylinders the values of U are different for the different directions as area is different.Problem.The walls of a lavishly furnished room cabin consists of two layers of wood (K = 0.10 W/mK) each of 2cm thick, sand-witching 5cm of fiberglass (k=0.038W/m K) insulation. The cabin interior is maintained at 20 C when the ambient air temperature is 20C. The interior and exterior convective heat transfer coefficients are 3 and 6 w/m2K respectively. Exterior is coated with a white acrylic paint ( = 0.9). Estimate the heat flux through the wall.Answer:The heat flux through the wall is, Q = UA (Ti T0) = + + + + The exterior radiation coefficient is given by hR o = 4 Tm3 where Tm = = = = = 284 hR o = 4 x 5.67 x 10-8 x 0.9 x 2843 = 4.2 W/ m2K = + + + + = 2.15 (w/m2K)-1 U = 0.466 w/m2Kq = Q/A = U (Ti T0) = 0.466(20-2) = 8.38 W/ m2HEAT TRANSFER WITH INTERFACE RESISTANCE (or: Contact Resistance)Due to this apparent decrease in the heat flow area and also due to presence of air voids, there occurs a large resistance to heat flow at the interface. The resistance is referred as thermal contact resistance and it causes the drop in temperature between two materials at the interface. Fig. 1.22 (a) shows the temperature distribution with out contact resistance and fig: 1.22 (b) shows the temperature distribution with drop of temperature because of contact resistance (T2A T2B) =temp drop at interfaceT2 is interface temperature.

K2 K1 T2A T2 B

T2T1 T1

T3 T3

L1 L2 LgFig: 1.22 (a)- No temperature drop ; Fig: 1.22 (a)- Temperature drop at interface, shows temp. distribution. at interface; shows temp.distribution Ac = contact area, Av = voids area, Lg = thickness of void space at interface,Kf = thermal conductivity of fluid in voids, hc = contact coefficient, A = Total cross- sectional area normal to heat flow. and taking the half of the length of void , ie Lg/2, on the side of layer- A and other half Lg/2 on to the side of layer-B, then Q = + = Hence, = = [ + ] = [ + ] = [ + ] = [ + ] ------------1.23.1 In most cases the air is filled in voids and value of Kf is very small compared to KA and KB. The problem to find contact coefficient from above equation is the measurement of quantities Ac , Av , and Lg .Problem-1Two large aluminium plates , K =240 w/mK, each 2cm thick, with 10 m surface roughness are placed in contact under 105 N / m2 pressure in air. The temperature at the out side surface s are 390C and 406C. Calculate a)heat flux b) temperature drop due to contact resistance c) contact temperatures. Thermal contact resistance with air as interfacial fluid for 10 m roughnes at 105 N/m2 (or 1bar).Answera)Rate of heat flow = q = (t1 t2) / R1 + R2 + R3 . consider area A=1 sq.mt.R1 = Resistance of plate-1 = L/ KA = 0.02 / 240 x 1 = 8.34 x 10-5 m2 K/WR2 = Resistance of air gap = 2.75 x 10-4 m2 K/WR3 = R1RT = Total resistance = 8.34 x 10-5 + 2.75 x 10-4 +8.34 x 10-5 = 4.418 x 10-4 m2 K/W q = (406-390) / 4.418 x 10-4 = 3.62 x 104 w/m2b) The temperature drop in each section is proportional to the resistance. Hene the fraction of the contact resistance is R2 / RT = 2.75 / 4.418 = 0.622The temperature drop = (R2 / RT) x the total temp.drop = 0.622 x (406-390) = 0.622x 16 = 9.95 CThe temperature drop in both aluninium plates = 16 - 9.95 = 6.05CThe temp.drop in each plate = 6.05/2 = 3.025 Cc) Contact surface Temperature of plate-1 =406 temp. drop in plate-1 = 4063.025 = 402.975C Contact surface temperature of plate-2 = 402.975 9.95 = 393.025 CProblem-2Two nos of 304 S.S steel bars of 3cm diameter and 10cm long have the ground surfaces with surface roughness of about 1 m. If the surfaces are pressed together with apressure of 50 atm. and the combination of this two bar is exposed to a temperature difference of 100 C calculate the axial heat flow and temperature drop across the contact surface.Thermal conductivity of steel is 16.3 w/m deg and the thermal contact coefficient is given by 1/ hc = 5.28 x 10-4 m2 0C/ W.Answer.Area of flow of heat = d2/4 = x (0.03)2 /4 = 7.069 x 10-4 sq mt.L =10cm =0.1 mt ; R 1 = Resistance of bar-1 = L / KA = 0.1 / 16.3 x 7.069 x 10-4 =8.679 C/wR2 = Contact resistance = 1/hc A = 5.28 x 10-4 / 7.069 x 10-4 = 0.747 C/WR3 = R1 : RT = Total resistance = 8.679 + 0.747 + 8.679 = 18.105Over all heat flow q= 100/ 18.105 =5.52 wThe temperature drop at contact surface = Fraction of contact resistance in total resistance x total temperature drop = (R2 / RT) x (T) = (0.747 / 18.105) x 100 = 4.13CContact resistance is 4% of total resistance.HEAT TRANSFER THROUGH A HALLOW CYLINDER: (Heat flow in radial direction)The rate of heat transfer, Q = - KA by Fouriers Law.---1.24-1 = -- K (2 r L) Q = -- (2 L K) dTIntegrating with boundary conditions :At r = r1 , T = T1 and at r = r2 , T = T2 ; Q = - 2 LK Q ln ( ) = -- (2 L K) (T2 T1) (here ln means the natural logaritham)Hence Q = ----------(1.24.2 ) ( T1 > T2 and r2 > r1) or Q = ---------(1.24-3)Where Rth = Thermal resistance = --------(1.24.4)The equivalent thermal circuit is also shown in fig 1.15 dr

Q r2

r r1

Fig-a: Physical system of hallow cylinder Q T1 T2 Rth = (b)- Electrical circuit

Heat flow equation of hallow cylinder in the form written for a plane wall, takingwall thickness.Some times it is convenient to write the heat flow equation of hallow cylinder in the form written for a plane wall, taking wall thickness; Then the wall thickness is (r2- r1) and area is Am. This is given below. We have Q = Multiplying and dividing Eq.1-24-1, with (r2 r1) , and also multiplying and dividingthe inside term of ln. term with 2 L, as we get PERIMETER AREA of cylinder is Q = = = = - K Am [ This equation resembles the heat transfer equation of plane wall. where Alm = Log- Mean area = -------- ---------(1.24.6) A1 = Inside surface area = 2 r1 L A2 = Out side surface area = 2 r2 L ( = Wall Thickness of hollow cylinder.Concept of Log-Mean radius:- We have Log- Mean area = = = 2 rm ------------ (1.24-7)Obviously logarithemic mean radius of the cylinder tube is given as: rm = ------------------(1.28.7) When = 1, then the value of arithmetic mean radius ie differs from the value of logarithmetic radius ie rm = ln ( ) by about 4%. If = 1.5 , then the deviation is by 1.3%. Hence when the value < 2, the logarithametic mean radius can be avoided and in place of it we may use arithmetic mean with out appreciable error.However the log mean radius has important applications in designing the lagging of steam pipes and insulation of electrical cables.Temperature distribution across the cylindrical wall; derived from fundamental Fourier equation.Temperature across the cylindrical wall varies logarithamically with the radius.Equation of temperature variation in cylindrical wall is derived from the heat flow equation.We have equation for heat flow as Q = K . 2 r L . ---------------- 1.24.1 dt = = C1 Integrating,t = C1 ln (r) + C2. ----------------------- (i)where the constants C1 and C2 are evaluated from conditions when r = r1 ; t = t1 and r = r2 ; t = t2 t1 = C1 ln (r1) + C2. ----------------------- (ii) t2 = C1 ln (r2) + C2. ------------------------ (iii)(iii) (ii) gives that (t2 t1) = C1 ln (r2) C1 ln (r1) = C1 ln (r2 / r1)Hence C1 = -------------------------(iv)Substituting C1 in (ii). t1 = + C2 C2 = t1 -------------------- (v) Substituting C1 and C2 from (iv) and (v) in equation (i) and simplifying, we getthe temperature distribution in dimensionless form as;t= t1 + --------1.24-8 = This shows that the temperature varies logarithamically with radius of cylinder.Hence the profile is curve (like a hyperbola) in the cylindrical solid wall, (it is not a line as in case of plane wall). It is independent of the value of K, ie type of material, steel, wood or Rubber etc. Isotherms (or temperature profile of constant temperature) are concentric circles lying between the inner and outer cylinder boundaries.This temperature profile is nearly linear for values of = 1, but non linear for larger values of .

Heat conduction through multi cylindrical wall with internal and external convective coefficient.

hi

K1 T0 K2

Ti r1 h ho T3 T2 Layer-1 (Metal thickness of pipe) (R1) T1 Layer-2 (layer of lagging) (R2)

Ti To r3 Q Ti r2 T1T2 T3 To Ri R1 R2 R0Fig:1.24.One dimentional heat flow through multi cylindrical wall ; electrical analogy.

a) Total Resistance considering the inside and outside convective coefficients. Multi cylindrical walls are frequently employed to reduce heat losses such as in Steam pipes lagging or increase the heat conduction such as in electrical cables. The above fig. 1.24shows the pipe with a layer of Insulation. Consider a length, L of the cylinder. Ri = Resistance of inside fluid film ( in Convection). = R1 = Resistance of the first layer (in conduction) = R2 = Resistance of the 2nd layer layer (in conduction) = Ro = Resistance of out side fluid film ( in Convection). = Rtotal = Ri (convection) + R1 (conduction) + R2 (conduction0 + Ro (convection0 = + + + --------1.25-1 Q = or Q =

Over all heat transfer coefficient. (Ui and Uo )Q = Ui Ai (Ti To ) = Uo Ao (Ti To ) Q = Ui 2 r1 L (Ti To ) = = where Ui is overall heat transfer coefficient on in side area. Similarly over all heat transfer coefficient based on out side area, Uo is given as; = { + r3 ( + r3 ( + Problem-1. A cylindrical cement tube of radii 0.05cm and 1.0 cm has a wire along the axis. To maintain a steady temperature difference of 1200 C between the inner and outer surfaces, current of 5 amp. is made to flow in the wire. Resistance of wire is 0.1 ohm per cm length of wire. Calculate the thermal conductivity of the cement pipe material.Answer: Resistance of wire per meter lengrth = 0.1 x 100 = 10 per mt lengthHeat generated = I2 R = 52 x 10 = 250 Watts/ mtUnder steady state conditions; Heat generated = Heat conducted through the cylinder. Q = 2 K L (t1 t2) L = 1mt ln (r2 / r1)250 = (2 K) x 1 x ( 120 ) ln (1.0 / 0.05) K = 250 x ln (1.0 / 0.05) / 2 120 = 0.994 W / m-degProblem-2:- A stainles steel tube with inner diameter 12mm, thickness 0.2mm and length 50cm is heated with nicrome wire fitted along the axis. The entire 15 kw energy generated in tube is transferred through its outer surface. Find the intensity of current flow and temperature drop across the wall of the tube. The tube material has thermal conductivity of 18.5 W / m-deg and specific resistance of 0.85 -mm2 / m.Answer:Power generated = I2Re = 15 kw = 1500 watts.Electrical resistance = Re = L/ ASpecific resistance = = 0.85 -mm2 / m. = 0.85 (cm/100) 2 / 100 cm. = 0.85 x 10-4 -cm L = 50 cm A = (r22 r12) r1 = 6mm = 0.6cm, r2 = 6 + 0.2 = 6.2mm = 0.62cm = (0.622 0.602) = 0.077 cm2 Hence Re = 0.85 x 10-4` x 50 / 0.077 = 551.9 x 10-4 Power generated = I2Re = 15 KW =15000 w Intensity of current flow =I = (15000 / 551.9 x 10-4 )1/2 = 5.213 x 102 = 521.3 amp Under steady state conditions, the heat generated equal to the heat transfer through the cylindrical tube. Q = 2 K L (t1 t2) ln (r2 / r1) 15000 = 2 K (18.5) x 0.5 x (t1 t2) / ln (6.2/ 6) = 1771.58 (t1 t2) (t1 t2) = 15000/ 1771.58 = 8.467 C.

Problem-3: A jet plane is considered as a tube of 3mt diameter and 20mt length. It is lined with 3cm thick insulation of thermal conductivity of 0. 042 w/m K and is maintained at 20 C for the comfort of passangers travelling in the jet. The average out side temperature is -30C at the operating height. What is the rate of heating needed inside the jet compartment. Neglect the end effects.Answer:Heat supplied in compartment = Heat leaked out of the compartment.Heat leaked out = Q = 2 K L (t1 t2) = ln (r2 / r1) = 2 x 0.042 x 20 x {20- (-30) } / ln (300/294) = 13060 watt = 13.06 kw.Heat supplied = 13.06 kwProblem-4: A wire of 0.5mm diameter is stretched along the axis of a cylinder of 50mm diameter and 250mm in length. The wire is maintained at a temperature of 750 K by passing current through it, while the cylinder is kept at 250 K by the gas inside the tube whose K is 0.0251 w/m K. Find the rate at which the heat is dissipated both by conduction and radiation, if the wire is perfectly black.Answer: Heat dissipated by conduction = QC = 2 K L (t1 t2) ln (r2 / r1) = 2 x 0.0251 x 0.25 x (750 250) ln ( 25mm / 0.25mm) = 4.28 w. Heat dissipated by radiation QR = A1 (T14 T24) = 5.67 X 10-8 x 2 x 0.25 x 10-3 x 0.25 x ( 750 250) =6.958 Watts

Problem-5: A steam pipe is covered with two layers of insulation. The inner layer of 30mm thick has K = 0.17 w/ m K and outer layer of 50mm thick of K = 0,023 w/m K. The pipe is made of steel (k= 58 w/mK and has inner and outer diameters of 160mm and 170mm respectively. The temperature of saturated steam is 300 C and of ambient air is 50 C. If the inside and out side heat transfer coefficients are 30 and 5.8 w/ sq.mt. K resp. Calculate the rate of heat loss per mt length of pipe.

Answer:

85 80115 165

L = 1mt Inner radius of pipe = r1 = 80mm, Outer radius of pipe = r2 = 85mm, outer radius with 1st layer of insulation=r3 = 115mm , Outer radius with 2nd layer of insulation = r4 = 165mm , ti = 300 C , to = 50 CQ = [(ti to) / { + ( + ( + = 2 (300 50)/ ++ + = 82.94 w/ mtPROBLEM-6: An aluminium pipe of K= 185 w/m K carrying steam is maintained at 110 C and is kept in a room whose ambient temperature is 30 C. The pipe has inner dia 10cm and outer dia 12cm. The convective heat transfer coefficient is 15 w/ m2 K . a)Determine the heat transfer rate per unit length. b)To reduce the heat loss from pipe, it is covered with a 5cm thick layer of insulation (k= 0.2 w/m k) Determine the rate of heat loss per unit length and the percentage reduction in heat loss by providing insulation. Neglect the convective resistance on steam side. Answer: Before insulation: Q = [(ti to) / ( + = (110 -30) / ( ) + = 80 / 1.57 x 10-4 + 0.177 = 452 w/mtAfter Insulation: Q = [(ti to) / ( + ( + = ( 110 30) / ( + ( + = 80 / (1.57 x 10-4) + 0.482 + 0.096 = 138 w/ mReduction of heat loss from pipe by insulation = (452- 138)/ 452 = 0.695 or 69.5 %

Problem-7:Steam at 350 C is flowing through a pipe (k= 80 w/m k) of 5cm inner diameter and 5.6 cm outer dia. covered by 3cm thick insulation of , K = 0.05 w/m K. Heat is lost to the surroundings at 5 C by natural convection and radiation. The combined h is 20 w/m K. Taking the heat transfer coefficient inside the pipe as 60 w/mK determine a) The rate of heat loss from the pipe per unit length of pipe. b) The temperature drop across the pipe and insulation.Answer:Fig below gives the thermal resistance in series.Ri = 1/ h1A1 ; h1 = 60 w/m-K; A1 = 2 r1 L = 2 x 0.025 x 1 = 0.157 m2Ri = 1 / 60 x 0.157 = 0.106 K/W

Fig:

R1 = = = 0.00023 K/WR2 = = = 2.318 K/W

Ro = 1/ h2 A3 = 1/20 x 0.364 = 0.137 K/WRtotal = Ri + R1 + R2 + Ro = 0.106 + 0.00023 + 2.318 +0.137 = 2.56123 K/wRate of heat transfer = Q = (T1 T2) / Rtotal = (350 5)/ 2.56123 = 134.7 Wb) T pipe = Q x R1 = 134.7 x 0.00023 = 0.03 C Tinsulation = 134.7 x 2.318 = 312.2 C

Problem-8: A steam pipe of 10cm outside diameter is covered with two layers of insulation, each having thickness of 2.5 cm. The average thermal conductivity of one layer is 3 times that of other and the surface temperature of the insulated pipe is fixed. Examin the position of better insulating layer relative to the steam pipe if heat dissipation from steam is to be minimum. What percentage saving in heat dissipation results from the arreangement.Answer:r1 = 5cm, r2 = 7.5cm, r3 = 10cm. Let K and 3K are thermal conductivities of two insulating materials.i) Better insulator (material with low Thermal conductivity0 is placed inside ie next to steam pipe.Thermal resisstance = Rt = + = + = 0.2506 / Heat dissipation Q1 = T/ Rt = T. / 0.2506 = 3.99 T. .

ii) Better insulation is out side. Rt = + = 0.2114 /Heat dissipated = Q2 = T/ Rt = T / 0.2114 = 4.7303 T. Q1/ Q2 3.99 / 4.7303 = 0.843. Obviously the heat dissipation is small when the material with low thermal conductivity is placed next to the pipe. Saving in heat dissipation = (4.7303- 3.99) / 3.99 = 0.1855 = 18.55 %Problem -9 : A steam main 0f 75mm inside diameter and 90mm outside diameter is logged with two layers of insulation. The layer in contact with pipe is 38mm asbestos ond the other is 25mm thick magnesia layer. the surface coefficients of inside and out side surfaces are 227 w/m2 k and 6.8 w/m2 k respectively. Steam temperature is 375 C and ambiant temperature is 35 C. Calculate the steady state heat loss from steam for 60m length of pipe. Also work out the overall coefficient of heat transfer based on inside and outside surfaces of lagged steam main pipe. Comment on the result. Thermal conductivities of pipe material is 45 w/mk, Asbestos: 0.14 w/m-k and magnesia insulation is 0.07 w/m K .Answer: r1 = 75/2 = 37.5 mm = 0.0375 mt. A1 = 2 r1 L = 2 0.0375 x 60r2 = 90/2 mm = 45mm = 0.045 mt. = 14.1372r3 = 45 mm + 38mm = 83mm = 0.083mt , r4 = 83+25 = 108mm = 0.108mt A0 =2 ro L = 2 0.108 x 60 = 40.715.i) Inner steam film resistance = 1/ hi A1 = 1/ 14.1372 x 227 = 3.117 x 10-4 0K/ Wii) Pipe resistance = = = 0.1075 x 10-4 0K/ Wiii) Asbestos resistance = = = 116.048 x 10-4 0K/ Wiv) Magnesia insulation resistance = = =99.822 x 10-4v) Out side film resistance = 1/ ho A0 = 1/ 6.8 x 40.715 = 36.137 x 10-4Total resistance in series Rt = 255.2315 x 10-4 0K/ WHeat loss through pipe = (T2 T1)/ Rt = (375- 35) / 255.2315 x 10-4 = 13321 W b) Heat lost can be expressed as,Q = Uo Ao (T2 T1) = Ui Ai (T2 T1) Where Uo and Ui are overall coefficients of heat transfer based on out side area A0 and inside area Ai respectively.Uo = Q / Ao (T2 T1) = 13321 / 40.715 x (375- 35) = 0.962 w/ m2 KUi = 13321 / 14.1372 x 340 = 62.19 w/m2 KImportant conclusions:1) Resistances to heat flow is mainly due to laggings and not by metal pipe. 2)Total resistance is mostly and strongly controlled by film with lowest coefficient. Also little gain would be obtained if the steam side film coefficient is increased, but a large gain would result by an increase of air side film coefficient. 3) Resistance of pipe material and resistance due to inner film could have neglected with out any appreciable error.

Heat conduction through a Spherical shell Q = KA = K . 4 r2 . (where A is area of spherical surface at radius r normal to heat flow.)

r r1 r2 T1 Inside temperatureT2 Out side temperature.

T1 T2 R = Fig: 1,26-i: Steady state heat conduction through Sphere and electrical analogy.

Separating the variables of equation 1-26-1 and Integrating at given boundary conditions; = = ( ) Q = Hence Q = Resistance in case of Spherical shell = Rth = Quite often it is convenient to write the heat flow equation through a spherein the same way as that equation written for a plane wall considering thickness, = (r2 r1) and equivalent area as Am . Hence Q = ---------------- 1.26.4We have derived the equation 1.26-1 as Q = Comparing both equations above, we write; Agm = 4 r1 r2. ----------------------------1.26.5 where, Agm = Geometric Mean area , = = . =4 r1 r2Hence Q = -------------------------1.26.6

We can also specify geometric mean area as, Agm = 4 rgm2 , where rgm = geometric mean radius. = -------1.30.5Thermal resistance:.From equation1.26.6, we have Q = = where, Thermal resistance = Rth = ) ---------------1.26.7Different expressions for Thermal resistances are given below.A) For Slab: Rslab = / K A. A = area of surface Temperature distribution in wall : = B) Cylinder: Rcylinder = Rth = Thermal resistance = = where Alm = Logarithamic mean area = or Alm = 2 rlm. L where, logarithmic mean radius = rlm = when the ratio r2 /r1 < 2 then we can use arithmetic mean radius ra = (r1 + r2) / 2 instead of log mean radius, rlm. Arithmetic mean area, Aa = 2 ra. L Temperature distribution in cylinder is: = C) Sphere: Rsphere = where Agm = Geometric mean area = = 4 r1 r2. or Agm = 4 rgm2 where rgm =

Temperature distribution in spherical shell. The equation for temperature profile will be derived in the next chapter from the general conduction equation in spherical coordinates.However the final equation for the temp. profile is given below for information. = -----------------------1.26.8I

MPORTANT RESULTThe equation, above, indicates that the temperature distribution associated with radial conduction through a sphere is represented by a hyperbola.

Note: If Spherical shell with two layers of insulation and inside out side film coefficients

Q =

Conduction through Hollw Spherical shell walls. Problem-1: A hollow sphere of inner radius 30 mm and outer radius 50mm is electrically heated at inner surface at a rate of 105 watt/m2 . At the outer