Thermodynamics Heat Transfer Heat: Heat transfer: {tis the form of energy that can be The science that deals with finding rates of such transferred from one system to another due energy transfer to difference in temperature SS It is the thermal energy in transit due to temperature Thermodynamics deals with difference End states of the process during which an interaction occurs Thermodynamics does not provide Heat transfer deals with Information about nature of the Systems that lack thermal equilibrium interaction Time rate of occurrence of the interaction Q=.} Mechanical , ) One ae ore ttp://www.eschooltoday, Rourkela (Odisha), india Basic modes of heat transfer Spoon in ht coffee Convection Radiation) (Umino pool a: Transfer of energy from a more energetic particle to its adjacent less energetic particle due to interaction between the particles Energy transfer by conduction can occur in solid, fluid (liquid and gas) ue to combination of lattice vibration and free electrons Jue to collision and diffusion of the molecules during random motion % Soli % Fluid: © (Quiescent fluid i.e. no motion of fluid particles)om fae omsat ice ae http://wwweschooltoday Rourkela (Odisha), India Basic modes of heat transfer ‘Spoon in hot coffee get [Conduction] Convection padiation ete tho ed of he spoon Conduction is the only mode of heat transfer that occur in non-transparent solid |The science of heat conduction is principally concerned with the determination |of the temperature distribution and flow of energy with in solids, Note: Heat transfer cannot be measured directly, it can be derived. Temperature distribution: Variation of temperature with space and time 1D steady: T(x) 2steady: T(x,y) 3Dsteady: T(x,y,z) 1D unsteady: T(x,t) 2D unsteady: T(x,y,t) 3D unsteady: T(x, y,z,t)Fourier’s Law of Heat Conductioi Based on the continuum concept Origin: Experimental observation by Jean-Baptiste Biot, French Physicist J.B.J Fourier, Theorie Analytique de la Chaleur (The analytical theory of heat), 1822 ee Heat flow: QocA ° k= thermal conductivity of the material Thermal conductivity is a measure of the ability of a material to conduct heat a = Temperature gradient (slope of the temperature curve on a T-x diagram at x) + Heat is conducted in the direction of decreasing temperature * Temperature gradient is negative as temperature decreases with increasing x ‘* The negative (-) sign ensures that heat transfer along positive x direction is positive quantityTemperature profile: T(x) dx T=15°C oe T=85°C = T=35°C lo k = 100 w/mk x x —1000 w/m*Problem-1: 1 kW heat conducted through an insulating material as shown. The area normal to the direction of heat conduction is 20 m®. The left surface is at 100°C. The ‘thermal conductivity of the material is 0.5 w/mK. Find the right side surface temperature. If the heat conducted through the insulating material increases five times the previous value, then what should be the new surface temperature on the right side. State your observations about this. Given: Q=1kW=1000W, A= 20m’, dx =L =40 cm=04 m, k=0.5 w/mK av T,-T, = KAS --fA-E ot ° dk & 1000=-05x202=1%_.7,-60°c ® Again: 5000 =-0.5x20 820 =>T, =-100 °CConduction Convection Basic modes of heat transfer Radiation Convection is the mode of energy transfer between a solid surface and the adjacent fluid (liquid/gas) that is in motion. Convection heat transfer mode consists of two mechanisms: “Energy transfer due to random molecular motion (diffusion) ‘ Bulk or macroscopic motion of the fluid Convection: Referring to cumulative transport of energy Advection: Referring to transport due to bulk fluid motion Total heat transfer Physical property of fluidae Typical convection heat transfer coefficient values Process h (W/m?k) Free convection Gases 02-25 Liquid 50 - 1000 Forced convection Gases 25 - 250 Liquids 100- 20,000 Convection with phase change 2500 - 100,000 boiling/ condensation 2 Note: Though convection heat transfer is a separate mode of heat transfer, it will appear in conduction problems very frequently as a condition on the boundary (called Boundary Condition (BC)) While solving problems on conduction, it will be assumed that value of convection heat transfer coefficient (h) is known.Problem-2: Water at 27°C, heated in a steel container using an induction heater which can produce heat in the range of 100 - 1000 W/m?. If the convective heat transfer coefficient is 25 W/m’k, find the minimum. and maximum surface temperature of the steel pan that can be achieved. For what value of convective heat transfer coefficient corresponding to maximum heat flux produced by the induction heater, the surface temperature of the pan can reach 107°C. Giver = 27°C, q’=100 Wim’, h=40 Win?K Q=hA(T,-T,) >q"=h(T,-T.) 100 = 25(T, -300) => T, =304K =31 °C (minimum surface temperature of the steel pan) Again: 1000 = 25(T, -300) => T, =340K =67 °C (maximum surface temperature of the steel pan) Secondly: 1000 = h(380-300) — h =12.5 Wim?K °Problem-3: Hot water at 87°C flowing over a flat plate maintained at 27°C. If the convective heat trang coefficient is 50 W/m’k, find the heat transfer rate into the plate per unit area. Given: T, = 27°C, T, =87°C, h=50 Win?K, A=01 m? Q=hA(T, -T_) =50x01(300 -360)=-3000 W Negative sign indicates heat transfer from hot water in to the plate Heat transfer rate into the plate per unit area Q=3000 W Convective heat transfer from hot solid surface to cold fluid Q=hA(T, -T,,) Convective heat transfer from hot fluid to cold solid surface Q= hA(T, ~T,) °Basic modes of heat transfer Conduction Convection Radiation Radiation is Energy emitted by matter that is at finite temperature. Any surface which is above absolute zero temperature, emits energy in the form of electromagnetic waves (or photons) in all directions and at all wave lengths ( A= 0 - °) * Maxwell's electromagnetic theory: Radiation is treated as electromagnetic waves “ Max Planck’s concept (Theory of quantum): Radiation is treated as photons or quanta of energy Emission of radiation occur from solids as well as fluids (liquids, and gases) Unlike conduction, and convection, radiation does not require any medium. Radiation heat transfer occurs most efficiently in vacuum (no attenuation). Emission or absorption of radiation energy by a body is a bulk processes; ie. * radiation originating from the interior of the body is emitted through the surface. * Radiation incident on the surface of a body penetrates to the depths of the medium where it is attenuated(ME4SI: Heat Transfer” Dr. Manoj Kumar MoharanaMetals are opaque to thermal radiation: * Thermal radiation incident on a metal surface is attenuated within few angstroms from the surface Water and glass are semi transparent to solar radiation: + solar radiation incident on water body is gradually attenuated as the bean penetrates to the depths of water. * solar radiation incident on glass sheet is partially absorbed, partially reflected, and partially transmitted Stefan-Boltzmann Law: Energy radiated into all direction, and over all wave length Incident energy Maximum radiation flux emitted from a surface Qu = 97,’ W/m? reflected energy 9 = Stefan-Boltzmann constant = 5.67x10°° W/m?K* Black body emissive power E, = oT; transmitted energy The radiation flux emitted by a real body at an absolute temperature T is always less than that of the black body emissive power E,, q=6E, =oeT? emissivity (0<¢<1) For all real bodies, e< 1Problem-4: Steam at 200°C is flowing through a pipe of outer diameter 10 cm. The surrounding ambient, 27°C. Neglecting temperature gradient with in the tube wall, find per unit length of the tube, the total energy emitted by radiation from the outer surface of the tube to the ambient. What should be the temperature of the steam if the energy emitted by radiation to the ambient need to be doubled. Given: T, = 200°C, T, =27°C, d=0.1m Q, =0AT} =5.67x10" x mx 0.1x (473)! = 891.974 W Secondly, Q, = 2x891.974 W =1783.94 W=oAT# =5.67x10 x mx0.1xT} => T, = 562.49 K= 289.49 °CConservation of energy: CV: Control volume Heat transfer and Thermodynamics are highly complementary CS: Control surface Heat transfer may be viewed as an extension of Thermodynamics. Conservation of energy for a control volume E,, Thermal and mechanical energy enter the CV through the CS E,, Thermal and mechanical energy leave the CV through the CS E, Energy generation: Creation of thermal energy within the CV due to conversion from other energy forms dE, Rate of change of energy stored with inthe CV (E.,) Inflow and outflow: Surface phenomena L Proportional to surface area E, +E, 6... =6, i i i ia +E, ~Foy = aj On a rate basis on any instant of time t | eration : Volumetric phenomena Eig +E, —E.y = AE.) Over a time interval At Proportional to magnitude of the volume inflow | ; generation Gitflow Storage: Volumetric phenomena Proportional to magnitude of the volumeSurface energy balance: Apply conservation of energy at the surface Note: The control surface include no mass or volume % Energy generation may be occurring in the medium, it will not affect the energy balance at the control surface This conservation equation is true for both steady and unsteady condition. sed = Vow * Ytme — Problem-5: The two surfaces of a wall of 0.25 m thick, and thermal conductivity of 0.5 W/mK are at 2 and 60°C respectively. The surface which is at 60°c is exposed to ambient at 27°C, and heat transfer from this surface to the ambient takes place by means of convection and radiation. Considering this surface to be black body, find the convective heat transfer coefficient under steady state condition. If heat transfer by radiation is neglected, then calculate the convective heat transfer coefficient. Giver T, =100°C, T, = 60°C, T, =27°C, k=100 WimK, L =0.5 m,h=? Tena = Vows + Vat ACT, -T,)+ 0(T-T) 3h +5.67x10"*[ 333* 300° | 0.25 400=33h +2374 =>h=491 Wim’K 2.12 Win?Ka} Department of Mechanical Engineering National Institute of Technology Rourkela Ae Vest oi Heat Transfer by Conduction * Conduction heat transfer is governed by Fourier’s law The manner in which the temperature varies with in the medium is called temperature distribution “ Fourier’s law helps us to determine the heat flux The objective in a conduction analysis is to find the temperature distribution in a medium resulting from conditions imposed on its boundaries. ar Heat flux is a directional quantity (vector quantity) Fourier'slaw q’=—k ix Heat flux in the x direction qf = 1-D Heat conduction: je pS ‘> Linear temperature distribution q! o v Non-linear temperature distribution 4! (x) 2-D Heat conduction: Sa = Isotherm - Constant temperature curve ar dxHeat Transfer by Conduction ar OLA a AS >\- sad ene dx aT/dx dT /dx Specific heat (C,): Energy required to raise the temperature of a unit mass of a substance by one degree + Different materials store heat differently. + The property specific heat indicates measure of a material's ability to store thermal energy. Ge Water can store almost 10 times the energy that iron can store for same mass. 18 kT /keK Cyleg = 0-45 1 / eK (at room temperature) Thermal conductivity (k): It is the measure of a ability of the material to conduct heat K ganar = 0.608 kJ /kgK Kp: = 80.2 KI /kgK (at room temperature) Iron can conduct heat more than 100 times faster than water Thermal conductivity (k): It is defined as the rate of heat transfer through a unit thickness of the material per unit area per unit temperature difference.‘Department of Mechanical Engineering [National Institute of Technology Rourkela Rourkela (Odisha, Ind Thermal conductivity (Transport property) Thermal conductivity depends on physi ruct Kous > Kuga > Kye OT / ox Thermal conductivity of a medium depends on intermolecular spacing. Material k, W/mK Material k, W/mK Diamond 2300 Brick = Silver 430 Water 0.62 Copper 400 Wood 0.18 Gold 315 Helium 0.15 Aluminum 235 Rubber 0.13 tron 80 Glass fiber 0.043 Mercury a5 air 0.026 Glass os Urethane (foam) 0.026 0.01 NONMETALLIC ‘CRYSTALS 0.1 1 100 1000 Thermal conductivity (W/mk) "Di: Manoj Kumar Moharaname Temperature dependence of Thermal conductivity 03 Thermal conductivity of materials vary with temperature on = pe 939 | of wl i . le : : C ie ? z ain] | ‘ ‘eens oe : Tee Soo 400 oo 1650 2000 mass transfer, Bergman et al., John Wiley and Sons Inc., 2011me Thermal diffusivity (material property) It represents how fast thermal energy diffuses through a material Tenia po es aay pC, heat stored Heat storage capacity of a material C, Per unit mass pC, Per unit volume The larger the thermal diffusivity, the faster the propagation of heat in to the medium Smaller thermal diffusivity means + Heat is mostly absorbed by the material + Small amount of heat conducted further Material silver 149 x 10% Gold Copper Aluminum: Iron Mercury Marble ce Concrete Brick Soll Glass Glass wool Water meat Wood 127 10° 113 x 10° 97.5 «10° 22.8 10% 47x 10° 1.2% 106 1.2106 0.75 x 10% 0.52 10¢ 0.52 10* 0.34 10° 0.23 x 10° 0.14« 10% 0.14 10° 0.13 «10%me) eee Measurement of thermal conductivity of material Total heat generated by the heater Q= VI u I] Heat transfer through each sample Y! —j.4 AT It} i tl | te Problem-6: In an experiment to find the thermal conductivity of a brick of size 8x10x18 cm?, a strip heater is used whose electrical resistance across its two ends are found to be 10 ohm. if it draws 2 amp current, then find the thermal conductivity of the brick sample. The difference in temperature between the two thermocouple readings is found to be 10 K. The bead of the thermocouples are placed 5 cm apart. Given: A = 0.08x 0.1m’, R=10 0, 1=2 Amp, AT=5K, L=0.05 m VixL _PRxL — 2?x10x0.05 = 2 Boel *1040.05 5125 Wink 2AxAT 2AxAT ~ 3x0,08x0.1x10 =Hane ai : Initial and Boundary Conditions: Their need? ar Sobitions oF ot GDE describes the generalized phenomena aa 0 Ta)-ca c, T=60°C, x=0, T=25°C, x=5 T(x)=—7x +60 How many ICs and BCs do we need for completely defining a problem? ‘The number of ICs and BCs in the direction of each independent variable of a problem is equal to the order of the highest derivative of the GDE in the same direction. ar Tt (S ot). Tee) —+u—+v—=al Se Time t: Space x: Space y: a Ie: 1 oT BC: 2 an BC: 2 et ox ey" MEASI:HeatTranster ie Maing] Kumar MohanBoundary Conditions: How to know it? The boundary condition can be derived by writing an energy balance equation at the surface of the solid Note that no energy can be stored at an infinitely thin surface 4 = are Ga it lite a oO X=L Classification of Boundary/Initial Conditions: Ahomogeneous condition is one in which all nonzerotermsin T=0 ‘1-0 the expression contain the dependent variable T (x,y,z,t) or its dx derivative.» Department of Mechanical Engineering A national institute of Technology Rourkela Rourkela (Odisha, nd Boundary Condition: Dirichlet BC (BC of 1* kind): Values of the dependent variables (T) are specified at the boundary EZ Ea = _.=Some constant vale Prescribed temperature Phase change occur at | y_ q X=L Special case: T|_,_=0 (Homogeneous BC of the 1* kind) | the surface e.g. boiling or condensation —__. Neumann BC (BC of 2" kind): Prescribed heat flux Derivative of the dependent variable is given as a constant or as a function of the independent variable on one boundary 20 wim? VUILL i.e. the heat flux is prescribed at the boundary surface 5 | 30 w/m? Constant heat flux as - onl, a Thin strip heater | x | Placed over the surface | 5—2y Variable heat flux =f(y) Dr. Manoj Kumar MoharanaBoundary Conditions: Neumann BC (BC of 24 kind): Prescribed heat flux =0 (insulated/adiabatic) ' ar Special case: —| OM lnutce | <0 (Homogeneous BC of the 2" kind) Ineo Cauchy Condition: Both Dirichlet & Numenn BC simultaneously mL OX hoae Boundary Conditions: Robin BC (BC of 3“ kind): Convection from/to surface 1. =, Derivative of the dependent variable is given as a function of the dependent variable on the boundary —} Ty. LT EZ x x X=0 MeL x Ae =. BC of 3" kind —p=0 Special case: Ambient fluid temperature is zero BC of 1° kind BC of 24 kind Tyee (Homogeneous BC of 3° kind) Tet measBoundary Conditions: Interface BCs valance: (ett conduction _( heat transfer ) _ (heat conduction y energy Dalance: | insolid1 | (across the gap) | in solid 2 4, 24 ohn 7) Imperfect h, = Contact conductance for the interface, W/m?K Ee thermal contact Rr=t. =Thermal contact resistance, m?k/W =k Special Case: Perfect thermal contact between the surfaces h, 0 Factors that affect contact conductance: TL =] “ Surface roughness heats = Teheaee % Interface contact pressure and temperature , Oh _ 2S ‘% Thermal conductivity of contacting solid ae ox “Type of fluid in the gapHeat Conduction Equation: Differential formulation The temperature distribution ( or temperature at each and every point) is adjusted such that the principle of conservation of energy is satisfied everywhere. Consider heat conduction in a 3D differential element Assumptions for generalized formulation: + Differential element (or the material) is moving * Internal heat generation per unit volume (q”) E=E,+E, -E,.ourkela( Heat Conduction Equation: Differential formulation in cartesian coordit dz Assumptions: + Constant pressure and density Yd + Negligible change in PE + Uniform velocity of material (element) i anste ——HesecOnduction Equation: Energy generation: E:, = q"dxdydz eqs 3 p+ Seay oe on Energy in: E,, = qfdyde + qfdxdz + qfdxdy + 28 a la (« ee } - Energy out: Ey = Udydz + qidxdz + qfdxdy 1c oar a5, Oat eae (5 Su Change in energy with in the element: E = pC, Senay axdz Principle of conservation of energy E, +E,, a écdyce - = ae - «ay Gd (Z- ot aa 90, 1D =E >E,+E, =E,, +E SA aaydeapc, — axdydz.Heat Conduction Equation: Thermal diffusivity a= msec Pe, Heat conduction equation in generalized form For a moving system eT OT OT eae +ULS+Vl+W. a Wx oy (Fst vt wo)me For stationary material with constant thermal conducti 0 Oo fA Tt yt yt v a) ox | oy | & ey Named after Generalized GDE for heat conduction French mathematician Pierre-Simon Laplace Steady state Unsteady term: unsteady heat conduction steady heat conduction wih | without with without — internal heat generation _ internal heat generation internal heat generation internal heat generation er er erg’ 1a OT eT oT ie eT eT oT aw 30 ae oP "ek aa wy e aa wy & ion oT pas LOT ae OT _10aT eT oT aay ow ok aa oy act oe Oy k 2 2 fo Gal ce eT_la ox1-D Steady State Heat Conduction: Cartesian Coordinate system Problem-6: During a hot summer, the outer surface of an air conditioned house wall is at an ambient temperature of 40°C and the wall inner surface is at room temperature of 20°C. Find the temperature distribution across the wall thickness of 40 cm. Consider the thermal conductivity of the wall to be 1 W/mK. Find the rate of heat penetration to the house through the walls if the room ae is 10x 10x 3m’, Generalized heat conduction equation ST ae t. ¢. yf Governing differential equation (GDE): “4 From BC-II: Boundary conditions (BCs): X=0, T=313K (BC-1) 293 =0.4C, +313 X=0.4,T=293K (BC-II) C,=-50 SGxEG From BC-I: C) = 313 Now, the general solution is T= C,x +313 General solution: ‘Temperature Profile: T=-50x+313 ECDC ° o1 02 03 041-D Steady State Heat Conduction: Cartesian Coordinate system Problem-6: During a hot summer, the outer surface of an air conditioned house wall is at an ambient temperature of 40°C and the wall inner surface is at room temperature of 20°C. Find the temperature distribution across the wall thickness of 40 cm. Consider the thermal conductivity of the wall to be 1 W/mK. Find the rate of heat penetration to the house through the walls if the room size is 10 x 10 x 3m’. ‘Temperature Profile: T=-50x+313 ao er @ o on xm) X= Eo 0 32 o1 0.2 03 0.4 308 303 298 293 x a 40 35 30 25 20_p va 1-D Steady State Heat Conduction: Cartesian Coordinate s) Problem-6: During a hot summer, the outer surface of an air con| at an ambient temperature of 40°C and the wall inner surface is of 20°C. Find the temperature distribution across the wall Consider the thermal conductivity of the wall to be 1 W/mk. penetration to the house through the walls if the room size is 10 Conduction through 39 38 37 36 im 35 34Department of Mechanical Engineering a National Institute of Technology Rourkela al Ne Rourkela (dish, India He , Three dimensional temperature distribution X oc 39 38 7 36 35 34 33 32 31 30 29 28 ie 27 26 25 24 23 22 a (ME4SI: Heat Transfer Dr. Manoj Kumar MoharanaSolution i in dimensionless form Problem-7: During a hot summer, the outer surface of an air conditioned house wall is at an ambient temperature of 40°C and the wall inner surface is at room temperature of 20°C. Find the temperature distribution across the wall thickness of 40 cm. The Consider the thermal conductivity of the wall to be 1 W/mK. Find the total heat | 40°C} penetration to the house through the walls if the room size is 10 x 10 x 3 m’. > coe: ©T_g —Bes:x=0,T=T,=313k (ac) | X_1 oe a 1 x=04,T=T, =293K (BCI et ae yh = Dimensionless parameters: 0-2 x-* Ox’ OX\ EX” LL = 2 a a 2(Zc+00,- Ty 1) GDE and BCs in dimensionless form: Ox! OX\ OX L T=T, +@(1,-T,), x=XL er a ne 1 _@0(-T.) Z a ex 7p oe eT _ afar) a(@ 8 (2 0X) OX 2 Soe 5 ox? 42) 2(2a) aa Ve x rath) a GDE in dimensionless formSolution in dimensionless form Problem-7: During a hot summer, the outer surface of an air conditioned house wall is at an ambient temperature of 40°C and the wall inner surface is at room temperature of 20°C. Find the temperature distribution across the wall thickness of 40 cm. The Consider the thermal conductivity of the wall to be 1 W/mK. Find the total heat | 40°C} penetration to the house through the walls if the room size is 10 x 10 x 3 m?. — General solution: ape G1 2x=0,T=T,=313K (BC mo 2 oe) eeaxic, X= 04,T= T= 293K (BOM | Vinge. c, io Dimensionless parameters: @- 1h x -X O=CX+1 T,-T, L Using BC-II:_ C, =-1 GDE and BCs in dimensionless form: Temperature profile: #0» @=1-X T=T, +O(1,-T) ae T=1, +(I-X\(T,-T,) 0,0=1 (BSolution in dimensionless form Problem-7: During a hot summer, the outer surface of an air conditioned house wall is. at an ambient temperature of 40°C and the wall inner surface is at room temperature of 20°C. Find the temperature distribution across the wall thickness of 40 cm.| T,= Consider the thermal conductivity of the wall to be 1 W/mK. Find the total heat | 40°C} penetration to the house through the walls if the room size is 10 x 10 x 3 m’. ‘Temperature profile: @ =1-X = o t=t,+(1-2\4-n) A " Ee one . ° a Olea 51371 40) = 25 0.75 01 308 35 a | OS Jor ens | Es 67500) (025) (Osn) 298 125) 00. ' ay | © 04 293 20Solution in dimensionless form Problem-7: During a hot summer, the outer surface of an air conditioned house wall is. at an ambient temperature of 40°C and the wall inner surface is at room temperature of 20°C. Find the temperature distribution across the wall thickness of 40 cm.| T,= Consider the thermal conductivity of the wall to be 1 W/mK. Find the total heat | 40°C} penetration to the house through the walls if the room size is 10 x 10 x 3 m’. 35 Temperature profile: @ =1-X x 310 t=t,+(1-2\4-n) =o" ee ESE COO =-kA— oO Q = 0 a 313 40 GT _ %-7, 2 0.75 9a 308 35 (L/KA) 0.5 02 303 30 . 0.25 03 298 25 20 ee 10 0 04 293 20 00 o 02 08 04 im 10x34)"One-dimensional plane wall without heat generation Problem-8: Consider @ plane wall of thickness 25 cm, area 10 m? whose one surface is at constant temperature of 127°C. Heat gets conducted from this surface to the other surface, and from this surface heat gets convected to the ambient at 27°C. If the thermal conductivity of the wall material is 10 W/mK, find (a) temperature distribution with in the wall across its thickness (b) heat transfer across the wall for different values of convective heat transfer coefficient h (10, 20, 30, 40 W/mK). Plot variation of (c) mid wall temperature (d) surface temperature with varying h values. Given: L =0.25 m,A=10 m’, T,=127°C | Using BC: C,=T, T,=27°C, k=10 W/mK T=Cx+T, >1,=CL+1, h=10, 20, 30, 40 W/mK Using BC-II: -kC, =h(C,L+T,-T,) >C, coe: ¢T _9 ox Temperature profile: T — BCs: x=0, T=400K = (BC+) nol, -k2=n(E-7.) ecw) | 9--X_9- T= General solution: T=C,x +C, BiOne-dimensional plane wall without heat generation Problem-8: Consider a plane wall of thickness 25 cm, area 10 m? whose one surface is at constant temperature of 127°C. Heat gets conducted from this surface to the other surface, and from this surface heat gets convected to the ambient at 27°C. If the thermal conductivity of the wall material is 10 W/mK, find (a) temperature distribution with in the wall across its thickness (b) heat transfer across the wall for different values of convective heat transfer coefficient h (10, 20, 30, 40 W/mK). Plot variation of (c) mid wall temperature (d) surface temperature with varying h values. Given: L =0.25m,A=10m’, T,=127°C | Using BC: C,=T, T, =27 °C, k=10 W/mK. T=Ca47, ST-=CLiT h=10, 20, 30, 40 W/mK Using ACHi:-kC, =h(C\L +1, -T,) $C, Temperature profile: (i= Ts), * Lk Temperature profile: [ — h (L=T.) x T-7,-+1 Ss s : ec ee eee ie Zoe a # iaf}; Department of Mechanical Engineering % A wational institute of Technology Rourkela Seve | Rourkela (Odisha), Ini One-dimensional plane wall without heat generation Problem-8: Consider a plane wall of thickness 25 cm, area 10 m? whose one surface is at constant | temperature of 127°C. Heat gets conducted from this surface to the other surface, and from this | surface heat gets convected to the ambient at 27°C. If the thermal conductivity of the wall material | Ty is 10 W/mK, find (a) temperature distribution with in the wall across its thickness (b) heat transfer | across the wall for different values of convective heat transfer coefficient h (10, 20, 30, 40 W/m?k). | Plot variation of (c) mid wall temperature (d) surface temperature with varying h values. | eS [ktm] REYOWineK = 20W/mik |h=30WmK h=AO W/m ° 400 400 400 400 396 393.34 391.42 390 h, T, ch) x o1 392 386.67 382.85 380 0.15 388 380 374,28 370 02 384 373.34 365.71 360 0.25 380 366.67 357.14 350 —(-TL)(skshy x = Dr. Manoj Kumar Moharana—‘Department of Mechanical Engineering [National Institute of Technology Rourkela Rourkela (Odisha} India One-dimensional plane wall without heat generation Problem-8: Consider a plane wall of thickness 25 cm, area 10 m? whose one surface is at constant temperature of 127°C. Heat gets conducted from this surface to the other surface, and from this surface heat gets convected to the ambient at 27°C. If the thermal conductivity of the wall material | Ty is 10 W/mK, find (a) temperature distribution with in the wall across its thickness (b) heat transfer across the wall for different values of convective heat transfer coefficient h (10, 20, 30, 40 W/mK). Plot variation of (c) mid wall temperature (d) surface temperature with varying h values. ° 0 0 0 0 02 0.04 0.0667 0.0875 ot 04 0.08 0.1333 0.1714 0.2 06 0.12 02 0.2571 03 o8 0.16 0.2667 0.3429 0.4 10 02 0.3334 0.4286 os @=X/(1+Bi')One-dimensional plane wall without heat generation Problem-8: Consider a plane wall of thickness 25 cm, area 10 m? whose one surface is at constant temperature of 127°C. Heat gets conducted from this surface to the other surface, and from this surface heat gets convected to the ambient at 27°C. If the thermal conductivity of the wall material | T) is 10 W/mK, find (a) temperature distribution with in the wall across its thickness (b) heat transfer across the wall for different values of convective heat transfer coefficient h (10, 20, 30, 40 W/m?k). Plot variation of (c) mid wall temperature (d) surface temperature with varying h values. jenvenbie pte Cyan etn) 00 10 8000 et per, -boe), 20 13333.34 i : 00 L+— 30 17142.86 g h 3 on 40 20000 = Heat transfer: ‘am ° TT) ot ool oat) Ge) vote &k ood het Ce me bh KA TAOne-dimensional plane wall without heat generation Problem-8: Consider @ plane wall of thickness 25 cm, area 10 m? whose one surface is at constant temperature of 127°C. Heat gets conducted from this surface to the other surface, and from this surface heat gets convected to the ambient at 27°C. If the thermal conductivity of the wall material | Ty is 10 W/mK, find (a) temperature distribution with in the wall across its thickness (b) heat transfer across the wall for different values of convective heat transfer coefficient h (10, 20, 30, 40 W/m?K). Plot variation of (c) mid wall temperature (d) surface temperature with varying h values. Temperature profile: Mid wall temperature 5 i L+ Surface temperature =f Ci ~ mi way [+ surtace1 Probler : Consider heat exchange between two fluids at 127°C and 27 °C separated by a plane wall of width 25 cm, constant thermal conductivity of 5 w/mk. Heat transfer takes place from the hot fluid to the cold fluid by conduction through the separating plane wall. The convective heat transfer coefficient on the hot and the cold side fluids are 30 W/m?k, and 10 W/m’K respectively. Find ne (a) temperature distribution with in the plane wall (b) total heat transfer from the hot fluid to the cold fluid per unit area of the plane wall. Given: T,, 27°C, Ty = 27°C, L 0 25m, A=1m?| Using BCH: h,(T,,-(C,e+C,) k=10 Wink, h, = 30 W/m’K, h, = 10 W/n?K_ aT ae GDE: Be Using BC-II: -kC, = h, (cy Cc, =) bet ec -te) BCs: x=0, h,(T,,-T) ST. k & (BC ao =(CL+C, -T,, at =L, -k==h,(T-T,,) (8c x a ( ) (Bc-HI) General solution: T4C,x+C, eee fea aT/Problem-9: Consider heat exchange between two fluids at 127°C and 27 °C separated by a plane wall of width 25 cm, constant thermal conductivity of 5 w/mk. Heat transfer takes place from the hot fluid to the cold fluid by conduction ‘through the separating plane wall. The convective heat transfer coefficient on the hot and the cold side fluids are 30 W/mK, and 10 W/m’K respectively. Find ne (a) temperature distribution with in the plane wall (b) total heat transfer from the hot fluid to the cold fluid per unit area of the plane wall. k } Using BC-I: h,(T,, -(C,+C,) T=Gx+C, aCxrTy+ eG =T, vofnef 1 Temperature profile: betes -te) Heat transfer:a) ,—+— Problem-9: Consider heat exchange between two fluids at 127°C and 27 °C separated by a plane wall of width 25 cm, constant thermal conductivity of 5 w/mk. Heat transfer takes place from the hot fluid to the cold fluid by conduction through the separating plane wall. The convective heat transfer coefficient on the hot and the cold side fluids are 30 W/m’K, and 10 W/m?K respectively. Find h, (a) temperature distribution with in the plane wall (b) total heat transfer from the hot fluid to the cold fluid per unit area of the plane wall. Temperature profile: T=7,, tts _[, 4 ¥ Cel BD = bane Cra hy hy 0 381.81 108.81 0.05 376.36 103.36 0.1 37090 97.90 #0. 0.15 365.45 92.45 0.2 3600 87.0 : 0.25 354.54 81.54: The inner surface of a furnace wall ing constant heat flux of 200 W/m? while the outer surface is at the ambient temperature of 27°C. Find (a) temperature distribution (b) total heat transfer through furnace wall (c) maximum wall temperature. Given thickness of the wall 25 cm, wall conductivity 10 W/mK, surface area of the furnace wall 10 m?. Given: A=10 m’, k=10 WimK q’ = 200 Wim’, T, =300K Temperature: distribution: 1 T=1,+4 (L-x)=305-208 Using BCI: Heat transfer: dc) ee a =-1afl| T=T,, = Pee ee el ap k k Using BC-I: q” = —kC, 305K_ ‘National institute of Technology Rourkala con Rourkela (Osha), nia Problem 11: Consider the west wall of a house which is 0.2 m thick. The interior of the house is maintained at 20°C using air conditioning. A fan located at the centre of the roof is mixing the air inside the room. Assuming the heat transfer through the wall to be steady and one-dimensional, express the boundary conditions on the outer and the inner surfaces of the wall. Find the wall temperature at the inner and the outer surface. The convection heat transfer coefficients on the inner surface of the wall is 25 W/m*K, respectively. The thermal conductivity of the wall material is 2.5 W/mK. The intensity of sun light is 100 W/m?. Given: | =0.2 m, T, =293 K, h=25 Wim’K k=2.5 WimK, q"=100 W/m? 'T =o GDE: Bes: x=0, q" (ec!) . kL nar, -1,) (6c) General solution: T=C,x+¢, 41 axProblem 11: Consider the west wall of a house which is 0.2 m thick. The interior of the house is maintained at | 20°C using air conditioning. A fan located at the centre of the roof is mixing the air inside the room. Assuming the heat transfer through the wall to be steady and one-dimensional, express the boundary conditions on the outer and the inner surfaces of the wall. Find the wall temperature at the inner and the outer surface. The convection heat transfer coefficients on the inner surface of the wall is 25 W/m’K, respectively. The thermal conductivity of the wall material is 2.5 W/mK. The intensity of sun light is 100 W/m?. Given: | =0.2 m, T, = 293K, h=25 Wim*K | using Bc-: q’= k=2.5 WimK, q"=100 W/m? 1 GbE: 55 =0 ar : x=0, g’=—k— - BCs: x=0, q’=-k (Bc:)) : -K- h¢T, 7.) (BCH) General solution: T=c,x+c,me Problem 11: Consider the west wall of a house which is 0.2 m thick. The interior of the house is maintained.at | 20°C using air conditioning. A fan located at the centre of the roof is mixing the air inside the room. Assuming the heat transfer through the wall to be steady and one-dimensional, express the boundary conditions on the outer and the inner surfaces of the wall. Find the wall temperature at the inner and the outer surface. The convection heat transfer coefficients on the inner surface of the wall is 25 W/m?K, respectively. The thermal conductivity of the wall material is 2.5 W/mK. The intensity of sun light is 100 W/m?. Temperature profile: T=T, ch +Ee =x) Using BC-l: q’= Wall temperature at outer surface T Toon fea fepee ce oa Using BC-II: -kC, = 2934100 , 100 9 9-305k = 32°C eo 1 ee 8 i os kn kk Wall temperature at inner surface ‘Temperature profile: ae on aL + i ane 293410 297K =24°C oo ue Tcme One-dimensional plane wall with heat generation Problem-12: Consider a plane wall of width 30 cm, thermal conductivity 10 W/mK, and internal heat generation 100 W/m? with convective heat transfer to the ambient (at 27°C) from both surfaces. Find (a) temperature distribution, (b) total heat transfer (c) temperature at the solid-fluid interface (d) maximum ‘temperature any where inside the solid. Take h = 25 W/mK, A= 10 m2, BCs: x=0, of) (BC-I) dx eat «Ze nc, ~T,) (BC-II) ion rx jae General solution: t=-4* ,ox4c SG ST he 5 TOG: eS og? To h Using BCI: C,=0 =>T= +C, | Temperature profile: T =T, +q” 2kensional plane wall with heat generation f-e(-(3) in dimensionless form: eo GDE: ox? +1=0 sca: x=0, 2-9 ax BC-Il: X=1, Oe eed aX dT BCs: x=0, —-=0 (BC-I x=0, 5 =0 (BC+) General solution: x=L, 1 (BCI) xe dx O=-F4+GX+G, Dimensionless parameters: 2 ee x hL Using ac: C,-0 >e@=-*~4¢, Poe Gere e2ensional plane wall with heat generation (Fx{-() in dimensionless form: 2 GDE: ae 1=0 ox? Solution in Dimensionless form: do & BCI: X=0, ==0 Using BC-II(—X)|_ -a(-X-c,] as beet BCI: X=1, See 0 x General solution: = s0x+0, : x Temperature profile: Using BCH: C,=0 => O=-7-+C,One-dimensional fae wall with heat generation =(- Xe T=T, sa(E+ Temperature profile: ‘Temperature profile +8105 +-Bi1.0 812.0 he eee 3 ne Temperature profile 16 a h 2k 2kOne-dimensional plane wall with internal heat generation Problem-13: A 40 mm thick plate having thermal conductivity 100 W/mK generating heat internally at a rate of 100x10° W/m?. Its two surfaces are subjected to constant temperature of 127°C and 27°C. Find (a) temperature variation across its thickness (b) location of maximum temperature (c) maximum temperature. Also find heat transfer through the (d) mid plane and its two faces (e) plane at which temperature is maximum. Given: L =0.5 m, k=10 WimK 000 Wim’, T, = 293K, T, = 25K (BC) (BC:) General solition: T 24x +, 2 Using BCI: C,=T,One-dimensional plane wall with internal heat generation Problem-13: A 40 mm thick plate having thermal conductivity 100 W/mK generating heat internally at a rate of 100x10° W/m?. Its two surfaces are subjected to constant temperature of 127°C and 27°C. Find (a) temperature variation across its thickness (b) location of maximum temperature (c) maximum temperature. Also find heat transfer through the (d) mid plane and its two faces (e) plane at which temperature is maximum. Temperature profile: oe 0 wel Tat ce Dee oe AT 40 a : 21000 = Maximum temperature: at = 100 Ses Tis maximum 5 =0 fou ag 1000" 100 x =0.02-0,0025 = 0.0175 x0 Toe: = 553.125 K so ‘om ra om obs o xm) = 280.125 %One-dimensional plane wall with internal heat generation Problem-13: A 40 mm thick plate having thermal conductivity 100 W/mK generating heat internally at a rate of 100x10° W/m?. Its two surfaces are subjected to constant temperature of 127°C and 27°C. Find (a) temperature variation across its thickness (b) location of maximum temperature (c) maximum temperature. Also find heat transfer through the (d) mid plane and its two faces (e) plane at which temperature is maximum. Heat transfer: Q__, 47 at gil 2) Ga) A dx ee Li ue Atleftface: x= _l0ortx0.04 100 0 om 2100 004 =17500 K/m : 00 10017500 =] 1.75 10° Wim? on See aaa ae ay Note: Negative sign indicates heat flows in the negative x direction ain Problem-13: A 40 mm thick plate having thermal conductivity 100 W/mK generating heat internally at a rate of 100x10° W/m?. Its two surfaces are subjected to constant temperature of 127°C and 27°C. Find (a) temperature variation across its thickness (b) location of maximum temperature (c) maximum temperature. Also find heat transfer through the (d) mid plane and its two faces (e) plane at which temperature is maximum. One-dimensional plane wall with internal heat genera Heat transfer: aT age ace) dx ook c BS oe QL _(G-T,)__100*10°*0.04 100 4p eu 2100004 Zao = ~22500 Kim e 0 100x (22500) | * =2.25x10° Win? = a a a xm)One-dimensional plane wall with internal heat generation Problem-13: A 40 mm thick plate having thermal conductivity 100 W/mK generating heat internally at a rate of 100x10° W/m?. Its two surfaces are subjected to constant temperature of 127°C and 27°C. Find (a) temperature variation across its thickness (b) location of maximum temperature (c) maximum temperature. Also find heat transf through the (d) mid plane and its two faces (e) plane at which temperature is maximum. Heat transfer: Q -_, 47 aT _qV(1 2x) ¢ A dx ek Te -T) ™% i At mid plane: x 100 x(-2500) = 0.25%10° W/m* At plane at which temperature is maximum: Lg) at xm)One-dimensional plane wall with variable internal heat generation jt pe Given: q"(x) = aye GDE: ee ae General solution:4 Heat Conduction Equatior Differential formulation in cylindrical coordinate System @Q ar a Ne Q.. Q=al (ride) 9.4 =0, +r = a. : conduction “unsteady heat conduction with internal heat generation. isteady conduction ste: eondiction Do gt lat ley ar [without with | without rece 1) =. linternal | internal internal jheat heat heat Aor fg. éeneration generation] generation, generation Lara a at, rar “at foe pateRourkela (Odisha), India 1-D Steady State Heat Conductior Problem-12: Water at 100°C is flowing through a circular tube of inner and Institute of Technology Rourkela : Conduction through cylindrical wall Radial heat conduction outer radius of 0.25 and 0.5 m respectively. The outer surface of the tube is 30°C the tube wall. Consider the thermal conductivity of the wall to be 1 W/mK. Find found to be at a temperature of 30°C. Find the temperature distribution across oS the rate of heat loss to the ambient through the outer surface if the length of the pipe is 30 m. Heat conduction equation: Using BCI: T, = C,In(;)+C, eens Using BC-II: T, =C, In(r,)+C, eee eee C,=T-C,Ing) Igy) —InG,) »(2) C.=T,-CInG,) 5 Temperature Profile: T=C,In(r)+T,-C,Ing) | T=C,In(r)+T, -C,In(,) T=T, +C,[In(r)-in@)] T=T, +C, [In(r)—In(r,)] T=T, +C,Ine/1) 7 r/t,)1-D Steady State Heat Conduction: Conduction through cylindrical wall Problem-12: Water at 100°C is flowing through a circular tube of inner and outer radius of 0.25 and 0.5 m respectively. The outer surface of the tube is 30°C found to be at a temperature of 30°C. Find the temperature distribution across C2] the tube wall. Consider the thermal conductivity of the wall to be 1 W/mK. Find the rate of heat loss to the ambient through the outer surface if the length of the pipe is 30 m. = 120.25" TT (gg) | UsIMBBCE T=C.Inle) +, 105m In(r, /5) Using BCI: T, =C, In(r,) +C, manyhe!) ee ©, =T,-CIn@) a Inge) —InG,) (5) ,=T,-C,In) ‘Temperature distribution ‘Temperature Profile: T=C,In@)+T,-C,Ing) @ T=C,In@)+T, -C, In) 1a, + BE (ern) oe aes T=,+C,[In@)—in@)] | T=T, +¢,fn(e)—Inte)] In( s/t, i e/8) | T=T,+CIn(e/y) T +C,In(r/1,)Solution in dimensionless form Problem-12: Water at 100°C is flowing through a circular tube of inner and the pipe is 30 m. outer radius of 0.25 and 0.5 m respectively. The outer surface of the tube is 30°C found to be at a température of 30°C. Find the temperature distribution across the tube wall. Consider the thermal conductivity of the wall to be 1 W/mK. Find the rate of heat loss to the ambient through the outer surface if the length of BCs: R=1, @=1 (BC-l) — es Temperature Profile: R=R, ©=0(8CHl) Ia(R ec tan,T=T (CH a ee t General solution: © = C, In(R) +C, In(R) r=n,T=T, (BCI) a From BC+: C; =1 Dimensionless parameters: er Now, the general solution is ©=1+C,In(R) 1 From BC-II: =0.25 r=0.5m.Solution in dimensionless form Problem-12: Water at 100°C is flowing through a circular tube of inner and outer radius of 0.25 and 0.5 m respectively. The outer surface of the tube is found to be at a temperature of 30°C. Find the temperature distribution across the tube wall. Consider the thermal conductivity of the wall to be 1 W/mK. Find the rate of heat loss to the ambient through the outer surface if the length of the pipe is 30 m. ‘Temperature Profile: 30°C r=0.25 1m r=0.5m. 3 373 100 0.736 354.5 81.5 0.514 339.0 66.0 0.321 325.5 52.5 0.152 313.6 40.6 0 303-30me Temperature Contour, Department of Mechanical Engineering National institute of Te Rourkela (Odish Temperature Contour 0.95 09 0.85 08 0.75 07 0.65 06 055 05 0.45 04 0.35 03 0.25 02 0.15 o4 0.05 (ME4SI: Heat Transfer Dr. Manoj Kumar MoharanaGo eeeserereaae wen (onal tte of Teclog ourala / Se) Reactor ha Radial heat conduction i Problem-15: In a shell and tube heat exchanger steam at 127°C is flowing ht. through a circular tube of inner and outer radius of 2.5 and 5 cm respectively. h, T, ‘The cooling water at 27°C surrounds the tube flowing through the shell for which the convective heat transfer coefficient is 50 W/m2K. Find the temperature £ distribution across the tube wall for the thermal conductivity of the tube 8g material to be 1 W/mK. Find the rise in temperature of the cooling water if its flow rate is 1.7 kg/sec. Total length of the tube inside the shell is 25 m. 72 00237~/ | 1=0.05 m Heat conduction equation: Using BC-l: T, =C,In(n)+C, ia a Fae ae = C, =7,-C\lng) 1), T=C,In(r)+T, -C, In() or T=T,+C,In(r/y) »T=T (BC-I) | Using BC-II: aT -k==h(T-T, (8c a ote )(BC-II) ~({&) 4 T Sear, General solution: T =C, In(r) +C, (ME4SI: Heat Transfer ‘Dr: Manoj Kumar Moharanaeee ales concn HR In a shell and tube heat exchanger steam at 127°C is flowing through a circular tube of inner and outer radius of 2.5 and 5 cm respectively. The cooling water at 27°C surrounds the tube flowing through the shell for which the convective heat transfer coefficient is 50 W/m2K. Find the temperature distribution across the tube wall for the thermal conductivity of the tube material to be 1 W/mK. Find the rise in temperature of the cooling water if its flow rate is 1.7 kg/sec. Total length of the tube inside the shell is 25 m. ~Eo=1-1.+¢H(®) Using BC-I: T, =C,In(,)+C, " 5 => C,=T,-C, In) ae T=C,In(r) +T, -C,In(g) +In| & r x T=T,+C,In(r/5) ‘Temperature Profile: Using BC-II: T= Fag) b@LaeeG) hy,ee tai nea canscon RE In a shell and tube heat exchanger steam at 127°C through a circular tube of inner and outer radius of 2.5 and 5 cm respectively. ‘The cooling water at 27°C surrounds the tube flowing through the shell for which the convective heat transfer coefficient is 50 W/m2K. Find the temperature distribution across the tube wall for the thermal conductivity of the tube material to be 1 W/mK. Find the rise in temperature of the cooling water if its flow rate is 1.7 kg/sec. Total length of the tube inside the shell is 25 m. r, } Given: T, =127 °C, T, a nf hr, 5) k=1W/mK, h=50 W/mK a 110 7c. 0 =2.5em, ,=Sem, L=25m tn=1ke/s =” Heat transfer: Q=—kA 4 r % ar fe 80: cl S : 60. 20.25 30 35 40 45 50 55RS Smee aaa ea cancion RR In a shell and tube heat exchanger steam at 127°C through a circular tube of inner and outer radius of 2.5 and 5 cm respectively. ‘The cooling water at 27°C surrounds the tube flowing through the shell for which the convective heat transfer coefficient is 50 W/m2K. Find the temperature distribution across the tube wall for the thermal conductivity of the tube material to be 1 W/mK. Find the rise in temperature of the cooling water if its flow rate is 1.7 kg/sec. Total length of the tube inside the shell is 25 m. () Given: T, =127 °C, T, =27 °C) Heat transfer: 1 )/k=1W/mK, h=50 W/mK 2nx 110x100 oS rah = i 5 =25em, ,=5em,L=25m| 2=—7 =14369.48 W +In(2) th=1ke/s 50x 0.05 Heat transfer: Q= wat Energy balance: Q=m-C, -AT Temperature rise: & ATa_ 2 2 1BOM aocProblem-16: A layer of insulation 0.25 m thick is wrapped around a solid nuclear rod of 0.5 m diameter which generates energy at a rate of 100 W/m?. The outer surface of the insulation layer is at a temperature of 30°C. Find the ‘temperature distribution across the insulation material. Consider the thermal conductivity of the insulation to be 1 W/mK. Find the heat loss to the ambient through the insulation if the length of the nuclear rod is 30 m. By energy balance: From BC-I: q” q’me2L = q’2mL > 4" o GDE: soc ear EA my Iq >me Problem-16: A layer of insulation 0.25 m thick is wrapped around a solid nuclear rod of 0.5 m diameter which generates energy at a rate of 100 W/m?. The outer surface of the insulation layer is at a temperature of 30°C. Find the 30¢ pote temperature distribution across the insulation material. Consider the thermal conductivity of the insulation to be 1 W/mK. Find the heat loss to the q 4 ambient through the insulation if the length of the nuclear rod is 30 m. ‘Temperature Profile: Given: 1, = 0.25 m, r, = 0.5 m, T, = 30°C, q”=100 W/m’, =a 0.25 305.16 32.166 : 03 304.59 31.596 =-#0m(-$5] -avtav 035 304.11 31.114 = 0.4 303.69 30.697 0.45 303.32 30.329 05 303 300 Heat transfer: Q = 7x100%x30x(0.25) = 589.04 WProblem-17: In a nuclear reactor 2 cm diameter cylindrical nuclear fuel rod is cooled by water. Heat is generated uniformly within the nuclear rod (k = 25 W/mK) at a rate ‘of 500x10° W/m, If the outer surface temperature of the rod is at 250°C, find the ‘temperature at the centre of the nuclear fuel rod. Te O0y dt os Sa GDE: 12(-2). =0 From BC-: C,=0 Se aT BCs: r=0, —=0 - “i; T,=-2- Tv a (BC-!) From BC-II: T, ae r=n,T=T, (Bc-il) Temperature Profile: General solution: ee i agen jeneral solution: pe Se ae 4k 4k ak OF aS a@ 2k Centre line temperature: pede T=-T¥+Cn+c, | 7 oa] a EL 23+ Oat xo0y =573K =300°C xProblem-18: A cylinder of diameter 0.2 m, of 1000 w/m? is placed in ambient fluid at a temperature of 30°C. Find the radial temperature distribution considering that it is a very long cylinder compared to its diameter. Given the thermal conductivity of the cylinder material is 1 w/mk, and the convective heat transfer coefficient is 50 w/m’k. Calculate (i) total heat loss to the ambient per unit length of the cylinder (ii) heat flux on the surface of the cylinder. GDE: BCs:Problem-18: A solid cylinder of diameter 0.2 m, of 1000 w/m? is placed in ambient fluid at a temperature of 30°C. Find the radial ‘temperature distribution considering that it is a very long cylinder compared to its diameter. Given the thermal conductivity of the cylinder material is 1 w/mk, and the convective heat transfer coefficient is 50 w/m’k. Calculate (i) total heat loss to the ambient per unit length of the cylinder (ii) heat flux on the surface of the cylinder. ™ ‘Temperature Profile: (- r(m)| T(k) | T(ec) T=T,+ 4 a 3065 33.5 0.025 306.34 33.34 - 0.05 305.87 32.87 0.075 305.09 32.09 0.1 304 31 000 ues 0030 oars: 0100 Fm)Problem-18: A solid cylinder of diameter 0.2 m, of 1000 w/m? is placed in ambient fluid at a temperature of 30°C. Find the radial ‘temperature distribution considering that it is a very long cylinder compared to its diameter. Given the thermal conductivity of the cylinder material is 1 w/mk, and the convective heat transfer coefficient is 50 w/m’k. Calculate (i) total heat loss to the ambient per unit length of the cylinder (ii) heat flux on the surface of the cylinder. ‘Temperature Profile: 4 ba) US | T (°c) a 306.5 33.5 0.025 306.34 33.34 0.05 305.87 32.87 0.075 305.09 32.09 o1 30403 Heat loss to the ambient: (per unit length)maintained between the two ends of an electrical wire of 0.5 m length and 2 mm diameter. The thermal conductivity of the wire material is 15 w/mk, and the resistivity of the wire is 60 4O-cm. If the surface temperature of the wire is 80 °C, calculate the temperature at the center of the electrical wire. Given: V=l2v, 1, =0.1 om, L=50 cm k=15 w/mK, p = 60 1Q-cm, T, =353 K Ree A Centre line temperature: Electrical resistance: p =~ > 2 Power generated: p= vi = ¥— = AV Dia or P Power generated per unit volume: 4” =~ EAVES Ve dF Ee i ; “pL AL pL) 60x10*x(0.5)° wy Me x: =369KProblem-20: A circular rod of infinite length generates energy continuously with and gets convected to the ambient fluid. Prove that for a given heat release rate per unit length, the maximum radial temperature difference in it is independent of its |), diameter. ‘Temperature profile: : Total heat release to ambient fluid: Tate 22 (15 ae = cy hr, Q=-kA Z| “kKomt)(-£E Centre line temperature: . Tok Heat release rate per unit length: T.=TL., =H (1.2) 4k ‘Surface temperature: - Maximum radial temperature difference in the rod: r-1-28 4k - ME@Si:Heat Transfer i Main] Kumar MoharanaSE Hextensctinatonin ce Heat Conduction Equation: Cartesian coordinate system For constant thermal conductivity: 2 fa) e at +2(c2 sarap, a aT q"_ lar ex ox) ay ay) az ao oe Oy ak aa Cylindrical coordinate system 1é oT Lé6f, er é(,6T eee k 12 a). 4 a a) -2/ z).. Coordinate transformation: x =rcos0, y=rsin9, z=z ‘Spherical coordinate system 16 oT LO (oF 1-4 = OT we oO Fal) paargaa tae) Pangan Ent) -P8() For constant thermal conductivity: i a3(r2}- i gt é aT) q” (4) 7 apo + <| sino |+2 =-|— Pal a) rsin’?¢ do? rsingal” "a) k alat = Zenith angle (0° <$ <180°) or Lan) ae a lear "Var ro or) 08 Of ka & x=reosOsing Azimuthal angle (0° <@<360°) yy ~rsinOsing,ner and the outer surface of a sphere is found to be at a| temperature of 100°C and 30°C respectively. Find the temperature distribution across the sphere wall for inner and outer radius of 0.25 and 0.5 m respectively . Consider the thermal conductivity of the wall to be 1 W/mK. | Find the rate of heat loss to the ambient through the outer surface. r2(rf ==|r ror Heat conduction equatio generalized form een Yr &' or (BC-I) Scares T= i tee General solution: Cc, tTProblem-21: The inner and the outer surface of a sphere is found to be at a temperature of 100°C and 30°C respectively. Find the temperature distribution across the sphere wall for inner and outer radius of 0.25 and 0.5 m respectively . Consider the thermal conductivity of the wall to be 1 W/mK. Find the rate of heat loss to the ambient through the outer surface. T-T, eel YE) Temperature distribution toProblem-21: The inner and the outer surface of a sphere is found to be at a temperature of 100°C and 30°C respectively. Find the temperature distribution across the sphere wall for inner and outer radius of 0.25 and 0.5 m respectively . Consider the thermal conductivity of the wall to be 1 W/mK. Find the rate of heat loss to the ambient through the outer surface. 2 ak xa-o-t 3/2) > Temperature distribution Given: r, =0.25 m, r, =0.5 m, T, =100 °C, T, =30 °C, k=1 WimK. Tt (ly Heat transfer: ( ree T-1,PE tana cndconn tO Problem-22: The inner surface of a hollow spherical container is at constant temperature of 127°C while heat transfer by convection occurs from its outer surface to the surrounding fluid at 27°C. The inner and the outer radius of the container are 2.5 and 5 m respectively. The convective heat transfer coefficient is 50 W/mK. Find the temperature distribution across the wall if the thermal conductivity of the wall material is 1 W/mK. Find the heat transfer by convection to the ambient fluid, GDE: 32(e a ror [epee Me (BC-l)Problem-22: The inner surface of a hollow spherical container is at constant ‘temperature of 127°C while heat transfer by convection occurs from its outer surface to the surrounding fluid at 27°C. The inner and the outer radius of the container are 2.5 and 5 m respectively. The convective heat transfer coefficient is 50 W/mK. Find the temperature distribution across the wall if the thermal conductivity of the wall material is 1 W/mK. Find the heat transfer by convection to the ambient fluid. ‘Temperature Profile:me Problem-23: A solid sphere of diameter 0.3 m, with internal heat generation at rate of 30,000 W/m? is placed in ambient fluid at a temperature of 30°C. Find the radial ‘temperature distribution considering that there is no temperature gradient along the angular directions. Given the thermal conductivity of the sphere material is 4.5 W/mK, and the convective heat transfer coefficient is 50 W/mK. Calculate (i) total heat loss to the ambient (ii) heat flux on the surface of the sphere. Also find centerline and surface temperature. /1A( dT), of k General solution:1 Problem-23: A solid sphere of diameter 0.3 m, with internal heat generation at rate of 30,000 W/m? is placed in ambient fluid at a temperature of 30°C. Find the radial ‘temperature distribution considering that there is no temperature gradient along the angular directions. Given the thermal conductivity of the sphere material is 4.5 W/mK, and the convective heat transfer coefficient is 50 W/mK. Calculate (i) total heat loss to the ambient (ii) heat flux on the surface of the sphere. Also find centerline and surface temperature. 2 =! w= 4x0 x(0.15) 30000 _ 9115 md 2 100005015 spa want