Heat Transfer Study Material PDF

Heat and Mass Transfer Chapter:01 INTRODUCTION CONCEPTS, DEFINITIONS AND CONDUCTION 1. What is heat transfer? Briefly explain three modes of heat transfer. Heat transfer is the transmission of energy from one...

Heat and Mass Transfer Chapter:01 INTRODUCTION CONCEPTS, DEFINITIONS AND CONDUCTION 1. What is heat transfer? Briefly explain three modes of heat transfer. Heat transfer is the transmission of energy from one region to another region as a result of temperature gradient There are three modes of heat transfer 1. Conduction 2. Convection 3. Radiation 1.1 Conduction: Conduction is transfer of heat from one part of the substance to the other part of the same substance. Or Conduction is transfer of heat from one substance to other in physical contract with it. Examples: i. Hold a metal rod in a Bunsen burner flame. After some time the end you’re hold will heat up. This is because of conduction of heat by collision of molecules due to increase in K.E of the molecules within the metal. ii. Cooking: - while cooking food on the stove by heat conduction from container to the food indirectly. 1.2 Convection: It is mode of heat transfer between the solid surface and adjacent liquid or gas that is in motion and it involves the combined effect of conduction and fluid motion. There are two types of convection Forced convection: If the mixing motion induces by some external means such as Fan, Pump or blower is said to be forced convection. Free or Natural convection : If the mixing motion takes places due to density gradient or buoyancy effects caused by temperature gradient is said to be Free or Natural convection 1.3 Radiation: radiation is the transfer of heat through a space or matter by means other than conduction and convection. Properties  It does not require the presence of material medium for its transmission. RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 1  Heat and Mass Transfer  Radiant heat can be reflected from the surface and obeys the law of reflection  It travels with velocity of light  Like light, it shows interference, diffraction and polarization. 2. State and derive the Fourier Law of heat conduction. Identify the different terms of Fourier’s equation of heat conduction. State the assumptions made on which the law is based. It is states that “The rate of heat through a simple homogeneous solid is directly proportional to the area of cross section (A) at right angle to the direction of heat flow and temperature gradient (dt). Inversely proportional to thickness (dx) of the body in the direction of heat flow” T1 T1 A T2 T2 dx x Mathematically, it can be represented by 1 Q α A ; Qαdt ; Qα dx dt Qα A(dx) dt Q = −k A(dx) Where ; Q= Rate of heat flow through a body / unit time, in W A = Surface area of heat flow, in m2 dt = Temperature differences of the faces of block , in 0C or 0K dx= Thickness of the body in the direction of heat flow , in m k = Constant of proportionality and is known as Thermal conductivity of a body RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 2  Heat and Mass Transfer Where: - ve sign indicates decrease in the temperature along the direction of 𝑑𝑡 increase in the thickness and the term ( 𝑑𝑥 ) is always along + ve X-direction. Hence the value of Q become + ve. Assumptions 1. Conduction of heat takes places under steady state conditions 2. Heat flow is unidirectional 3. There is no internal heat generation 4. The material is homogeneous and isotropic (i.e: The value of thermal conductivity is constant in all direction) 5. The temperature gradient is constant and temperature profile is linear. 3. State and derive the law of heat convection. Consider hot plate with surface temperature Tw and cold fluid temperature Tf flowing parallel to the plate as shown above. At the interface of surface and fluid, heat is transported only by conduction. Due to viscous force the velocity of fluid will be zero at the wall and will be increase to umax as shown above fig. Velocity Profile Temperature Profile Fluid flow Tf Hot Plate Q Tw The rate of equation for convection heat transfer was given by Newton and is referred to as the Newton’s law of cooling. It states that “when a fluid at a temperature Tf is contact with solid surface at a 𝑄 different temperature T w , the heat flux ( 𝐴 ) from the surface to the fluid is proportional to the temperature difference between the surface and the fluid.” 𝑄 (𝐴 ) (Tw - Tf) 𝑄 (𝐴) = h(Tw - Tf) Q = Ah(Tw - Tf) RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 3  Heat and Mass Transfer Q ℎ= in W/m2oC A(Tw −Tf ) Where: h = constant of proportionality is called heat transfer coefficient W/m2oC Q= Rate of heat flow through a body / unit time, in W A = Surface area of heat flow, in m2 Tw = surface temperature in oC or K Tf = Fluid temperature in oC or K 4. Define the law of radiation. 4.1 Stefan-Boltzmann Law: It states that “The emissive power of a black body is directly proportional to fourth power of its absolute temperature”. Mathematically Eb 4 in W Where : Eb= Q = energy emitted by a perfect or block body -Boltzmann constant =5.67x 10-8(W/m2 K4) A = Surface area, m2 T = absolute surface temperature of the body (K) 4.2 Wein’s displacement Law: It states that “The wave length of 𝑚 corresponding to the maximum energy is inversely proportional to the absolute temperature of the body.” 1 m = T 4.3 Kirchhoff’s law : It states that “The emissivity of the body at a particular temperature is numerically equal to its absorptivity for radiant energy from hot body at the same temperature ” 5. Define the following. 5.1 Thermal conductivity (k): it is a property of a material and is defined as the ability of the material to conduct heat energy through it. SI unit : W/moK  Material having high thermal conductivity are good conductor of heat, whereas material with low thermal conductivity they are good insulator.  Conduction of heat occurs most readily in pure metal less so in alloy & much so readily in non metals. RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 4  Heat and Mass Transfer  Thermal conductivity depends on the following factors o Material structure o Moisture content o Density of material o Pressure and temperature OR 𝑑𝑡 𝑑𝑡 We know that Q = k A (𝑑𝑥) , the value of k=1, when Q=1 , A=1, and (𝑑𝑥)= 1. Thus the thermal conductivity can be defined as. “The amount of energy conduced through a body of unit area, unit thickness, in unit time when the differences in temperature between the faces causing heat flow is unit temperature difference” 5.2 Briefly explain conduction thermal resistance or Thermal Resistance. 1.1. When two physical systems are described by similar equation and have similar boundary condition, these are said to be analogous. The heat transfer process may be compared by analog with flow of electricity in electric resistance. As the flow of electric current (I) in the electrical resistance is directly proportional to potential difference (dv). T1 Q T2 Rth = (dx/ka) Similarly the heat flow (Q) is directly proportional to the temperature differences (dt), the driving force for heat conduction through a medium. As per Ohm’s law (in electric- circuit theory) Potential difference(dV) Current (I) = ( Electrical resistance (R) ) __________(1) By analogy, the heat floe equation (Fourier’s equation) may be written as Tempearture difference(dt) Heat Flow rate (Q) = ( 𝑑𝑥 ) _______(2) ( ) 𝑘𝐴 By comparing equation (1) & (2) , we find that I is analogous to Q , dV RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 5  Heat and Mass Transfer 𝑑𝑥 𝑑𝑥 analogous to dt and R is analogous to (𝑘𝐴). The quantity (𝑘𝐴) is called thermal conduction resistance (Rth)cond 𝑑𝑥 (Rth)cond = (𝑘𝐴) The reciprocal of the thermal resistance is called thermal conductance. 5.3 Convective heat transfer coefficient (h) : It is defined as the ability of the fluid carries away the heat from the surface. This in turns depends upon the velocity and other thermal properties of the fluids. It expressed in W/m2K or W/m2oC 5.4 Heat flux (q): It is defined as quantity of heat transferred per unit time per unit area. 𝑄 𝑞=𝐴 It is expressed as W/m2 6. What do you mean by boundary condition of First, Second and Third Kind. 6.1 Boundary condition of First Kind: This is also called Specified temperature boundary condition. There are many situations, where the temperature of the boundary surface is known or the distribution of temperature at the boundary surface may be known as the function of time. Let L= length of the plate be maintained at a uniform temperature T1 and T2 at x=0 and x=L, respectively. As shown in fig T(x,t)x=0= T1 T(x,t)x=L= T2 L x=0 x=L One dimensional conduction heat transfer T(x,t)x=0= T(0,t)= T1 T(x,t)x=L= T(L,t)= T2 As the temperatures are known at the boundary surfaces, the boundary conditions are known as those of first kind. 6.2 Boundary condition of Second Kind: There are many situations where the RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 6  Heat and Mass Transfer heat flux at the surface is known. Fig shows the boundary conditions of the second kind qr0 = -k [dt/dr]r=r0 Heat `Supply Conduction q0 = -k [dt/dx]x=0 Heat qri = -k [dt/dr]r=ri r `Supply ri r0 Heat qL = -k [dt/dx]x=L `Supply X X=0 X=L (1) Rectangular Coordinate (2) Cylindrical or Spherical coordinates 𝑑𝑡 Let; q0= -k ]x=0 = heat supply through the boundary surface at x=0 𝑑𝑥 𝑑𝑡 qL = -k ]x=L = heat supply through the boundary surface at x=L 𝑑𝑥 In the above equations, q0 and qL are positive which means that heat flows into the medium, where negative valves of q0 and qL means that heat flows from the medium. Similarly, we can write the boundary conditions for hollow cylinder and hollow spheres. 𝑑𝑡 qri= -k ]r=ri = heat supply through the boundary surface at r=ri 𝑑𝑟 𝑑𝑡 qr0 = -k ]r=r0 = heat supply through the boundary surface at r=r0 𝑑𝑟 6.3 Boundary condition of third Kind: Convection boundary condition is the most common boundary condition since most heat transfer is exposed to an environment at a specified temperature. The convection boundary condition is based on the surface energy balances. As shown in fig h2[T2 - T(r.t)]r=r0 = -k [dt(r,t)] r=r0 dr Heat `Supply Convection Convection Convection T1,h1 Conduction T2,h2 h1[T1 - T(r.t)]r=ri = -k [dt(r,t)] r=ri Heat dr r h1[T1 - T(x.t)]x=0 = -k [dt(x,t)] x=0 `Supply dx ri r0 T1,h1 Heat T2,h2 Convection h2[T2 - T(x.t)]x=L = -k [dt(x,t)] `Supply dx x=L X X=0 X=L (1) Rectangular Coordinate (2) Cylindrical or Spherical coordinates RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 7  Heat and Mass Transfer 1. Derive three dimensional heat conduction equations in Cartesian coordinates or rectangular coordinates Assumption: the following assumption are considered for deducing three dimensional conduction heat transfer equation  Material is considered as isotropic  Density of the material romance same and does not change with time  Specific heat of the material remains same  Temperature is the function of all the three dimensions as well as time Derivation Considered an infinitesimal rectangular parallelepiped (Volume Element) of sides dx, dy & dz parallel, respectively to the three axes (X,Y,& Z) Let : Qx, Qy & Qz are the rate of heat flow along X,Y&Z direction respectively kx, ky & kz are the thermal conductivity along X,Y & Z direction respectively. Applying Energy balance equation at inlet and outlet of the element ⌊(𝑹𝒂𝒕𝒆 𝒐𝒇 𝒉𝒆𝒂𝒕 𝒄𝒐𝒏𝒅𝒖𝒄𝒕𝒆𝒅 𝒊𝒏𝒕𝒐 𝒕𝒉𝒆 𝑬𝒍𝒆𝒎𝒆𝒏𝒕 𝒊𝒏 𝒕𝒉𝒆 𝑿 − 𝒅𝒊𝒓𝒆𝒄𝒕𝒊𝒐𝒏 𝒕𝒉𝒓𝒐𝒖𝒈𝒉 𝒍𝒆𝒇𝒕𝒇𝒂𝒄𝒆 𝒑𝒆𝒓 𝒖𝒏𝒊𝒕 𝒂𝒓𝒆𝒂 𝒊. 𝒆 𝑨𝑩𝑪𝑫 (𝑸𝑿 ) ⌋ + [(𝑅𝑎𝑡𝑒 𝑜𝑓 ℎ𝑒𝑎𝑡 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 )] = 𝑑𝐸 ⌈(𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑡𝑖𝑚𝑒 ( 𝑑𝑡 ))⌉ + [𝑬𝒏𝒆𝒓𝒈𝒚 𝒄𝒐𝒏𝒅𝒖𝒄𝒕𝒆𝒅 𝒐𝒖𝒕 𝒐𝒇 𝒕𝒉𝒆 𝒆𝒍𝒆𝒎𝒆𝒏𝒕 𝒊𝒏 𝒕𝒉𝒆 𝑿 − 𝒅𝒊𝒓𝒆𝒄𝒕𝒊𝒐𝒏 𝒕𝒉𝒓𝒐𝒖𝒈𝒉 𝒓𝒊𝒈𝒉𝒕𝒇𝒂𝒄𝒆 𝒑𝒆𝒓 𝒖𝒏𝒊𝒕 𝒂𝒓𝒆𝒂 𝒊. 𝒆 𝑬𝑭𝑮𝑯 (𝑸𝑿+𝒅𝒙 )⌉ _____(A) Rate of Heat Conducted into the Element in the X-direction through the left Face per unit area i.e ABCD (Qx) 𝜕𝑇 𝑄𝑋 = −𝑘𝑋 𝜕𝑋 RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 8  Heat and Mass Transfer Energy conducted out of the element in the X-direction through right face per unit area i.e EFGH (Qx+dx) 𝜕𝑇 𝜕 𝜕𝑇 𝑄𝑋+𝑑𝑋 = [−𝑘𝑋 + (−𝑘𝑋 )𝑑𝑋] 𝜕𝑋 𝜕𝑥 𝜕𝑋 Total heat change in the X-direction per unit area 𝜕𝑇 𝜕𝑇 𝜕 𝜕𝑇 𝑄𝑋 − 𝑄𝑋+𝑑𝑋 = −𝑘𝑋 − (−𝑘𝑋 + (−𝑘𝑋 )𝑑𝑋) 𝜕𝑋 𝜕𝑋 𝜕𝑥 𝜕𝑋 𝜕 𝜕𝑇 𝑄𝑋 − 𝑄𝑋+𝑑𝑋 = (𝑘𝑋 )𝑑𝑋 𝜕𝑥 𝜕𝑋 As The area heat stored along X-direction is dy and dz 𝜕2 𝑡 𝑄𝑋 − 𝑄𝑋+𝑑𝑋 = (𝑘𝑋 𝜕𝑋 2 ) 𝑑𝑋 𝑑𝑌 𝑑𝑍 _______(1) Similarly 𝜕2 𝑡 𝑄𝑌 − 𝑄𝑌+𝑑𝑌 = (𝑘𝑌 𝜕𝑌 2 ) 𝑑𝑋 𝑑𝑌 𝑑𝑍 _______(2) 𝜕2 𝑡 𝑄𝑍 − 𝑄𝑍+𝑑𝑍 = (𝑘𝑍 𝜕𝑍 2 ) 𝑑𝑋 𝑑𝑌 𝑑𝑍 _______(3) Rate of Heat Generated in the material per unit volume (or area) = 𝒒. 𝑑𝑋 𝑑𝑌 𝑑𝑍 ________(4) Change in the internal energy per unit time (dE/dt) =mass of the element X Specific Heat X change in the temperature of the element in the time dt 𝜕𝑻 = 𝝆 𝑪𝒑 𝑑𝑋 𝑑𝑌 𝑑𝑍 ________(5) 𝜕𝒕 Now Equation (1+2+3+4) = (5) ∂2 t ∂2 t ∂2 t ∂𝐓 (k X 2 ) dX dY dZ + (k Y 2 ) dX dY dZ + (k Z 2 ) dX dY dZ + 𝐪. dX dY dZ = 𝛒 𝐂𝐩 dX dY dZ ______(6) ∂X ∂Y ∂Z ∂𝐭 Dividing throughout by 𝑑𝑋 𝑑𝑌 𝑑𝑍 & Since the material isotropic then kx = ky= kz = k then 𝜕2 𝑡 𝜕2 𝑡 𝜕2 𝑡 𝒒. 𝝆 𝑪𝒑 𝜕𝑻 (𝜕𝑋 2 ) + (𝜕𝑌 2 ) + (𝜕𝑍 2) + 𝒌 = ___(7) 𝒌 𝜕𝒕 𝒌 ∝= = Thermal diffusivity, equation (7)becomes 𝝆 𝑪𝒑 RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 9  Heat and Mass Transfer 𝜕 2𝑡 𝜕 2𝑡 𝜕2𝑡 𝒒. 𝟏 𝜕𝑻 ( 2) + ( 2) + ( 2) + = _____(8) 𝜕𝑋 𝜕𝑌 𝜕𝑍 𝒌 ∝ 𝜕𝒕 The above equation is called three dimensional heat transfer equation in Cartesian or rectangular coordinates 𝜕𝑻 Note: (i) For Steady State; = 0 ; 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (8) 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 𝜕𝒕 𝜕2 𝑡 𝜕2 𝑡 𝜕2 𝑡 𝒒. (𝜕𝑋 2 ) + (𝜕𝑌 2 ) + (𝜕𝑍 2) + 𝒌 = 𝟎 this equation is called as Poisson’s Equation (ii) For Steady state, with no heat generation q’=0 𝜕 2𝑡 𝜕 2𝑡 𝜕2𝑡 ( 2) + ( 2) + ( 2) = 𝟎 𝜕𝑋 𝜕𝑌 𝜕𝑍 (iii) ∇2 T= 0 Where 𝜕 2𝑡 𝜕 2𝑡 𝜕 2𝑡 𝛁 = ( 2) + ( 2) + ( 2) 𝜕𝑋 𝜕𝑌 𝜕𝑍 (iv) For Steady State, 1-D Heat Transfer, with heat generation 𝜕 2𝑡 𝒒. ( 2) + = 𝟎 𝜕𝑋 𝒌 (v) For Steady State, 1-D Heat Transfer, without heat generation 𝜕 2𝑡 ( 2) = 𝟎 𝜕𝑋 (vi) For Steady State,2-D Heat Transfer, with heat generation 𝜕2𝑡 𝜕 2𝑡 𝒒. ( 2) + ( 2) + = 𝟎 𝜕𝑋 𝜕𝑌 𝒌  Thermal diffusivity: thermal diffusivity is the properties of a material that represents how fast heat propagates through the material. Denoted by 𝐤 𝝆 𝐂𝐩 RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 10  Heat and Mass Transfer 02 Heat conduction through plane walls {Without Heat Generation} Considerer a simple plane wall of homogeneous material having constant thermal conductivity, and with each face held at a constant uniform temperature and without heat generation. Let: dx = L = Thickness of plane wall (m) A = Cross-sectional Area perpendicular to the rate of heat transfer (m2) K = thermal conductivity of the material (W/m-k) T1, T2 = constant uniform temperature at x=0 , x=L , respectively (0k or 0C) We know that general heat conduction equation in Cartesian co ordinates , i.e 𝒅𝟐 𝑻 𝒅𝟐 𝑻 𝒅𝟐 𝑻 𝒒′ 𝟏 + + + = 𝒅𝒙𝟐 𝒅𝒚𝟐 𝒅𝒛𝟐 𝒌 𝛂 Assume 1-D, steady state and No heat generation then above equation becomes 𝒅𝟐 𝑻 = 𝟎__________(𝟏) 𝒅𝒙𝟐 Integrating above equation Twice, We get 𝒅𝑻 = 𝑪𝟏 𝒅𝒙 𝑇(𝑥) = 𝐶1𝑥 + 𝐶2 _________(2) Where C1 and C2 are arbitrary constants The valves of these constant can be obtained from the boundary conditions. At x=0, T(x) = T1; At x=L, T(x) = T2 Substituting these conditions in equations in equation (2) T1 = 0 +C2; Therefore C2= T1 RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 11  Heat and Mass Transfer T2 = C1L + C2 T2= C1L + T1; (𝑇2 − 𝑇1) ∴ 𝐶1 = 𝐿 Substitute C1 and C2 in equation (2) (𝑇2− 𝑇1) T(x) = [ ] x + T1 ______(3) 𝐿 The above equation is temperature distribution equation for one dimensional steady state, no heat generation in Cartesian coordinates Differentiate equation (3) with respect x , we get 𝒅𝑻 𝒅 (𝑇2− 𝑇1) 𝐓𝟐 −𝐓𝟏 = [ ] x + 𝐓𝟏 ] = ____________(4) 𝒅𝒙 𝒅𝒙 𝐿 𝐋 Substitute equation (4) in Fourier Equation then 𝑑𝑇 −kA(T2 − T1 ) 𝑄 = −𝐾𝐴 = 𝑑𝑋 L (𝐓𝟏 −𝐓𝟐 ) (𝐓𝟏 −𝐓𝟐 ) 𝐋 Q= 𝐋 = R= = Thermal resistance to the heat flow 𝐑 𝐤𝐀 𝐤𝐀 *** Data Hand Book “C.P.Kothandaraman” page No ‘___’ *** RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 12  Heat and Mass Transfer 03 Heat conduction through composite walls {Without Heat Generation} Composite wall is composed of several different layers, each having a different thermal conductivity. Consider a composite wall made up of three parallel layers in perfect thermal contact Let: L1,L2 and L3 are thickness of layer 1,2 and 3 respectively K1,K2 and K3 are the thermal conductivity of the material Heat transfer through layer (1) (𝐓𝟏 −𝐓𝟐 ) (𝐓𝟏 −𝐓𝟐 ) 𝑄1 = 𝐋𝟏 = ∴ (𝑇1 − 𝐓𝟐 ) = 𝐑 𝟏 𝑄1 𝐑𝟏 𝐤𝟏 𝐀𝟏 Heat transfer through layer (2) (𝐓𝟐 −𝐓𝟑 ) (𝐓𝟐 −𝐓𝟑 ) 𝑄2 = 𝐋𝟐 = ∴ (𝑇2 − 𝐓𝟑 ) = 𝐑 𝟐 𝑄2 𝐑𝟐 𝐤𝟐 𝐀𝟐 Heat transfer through layer (3) (𝐓𝟑 −𝐓𝟒 ) (𝐓𝟑 −𝐓𝟒 ) 𝑄3 = 𝐋𝟑 = ∴ T3 − T4 = R 3 Q3 𝐑𝟑 𝐤𝟑 𝐀𝟑 Now (T1-T4) = (T1-T2)+ (T2-T3)+ (T3-T4) = Q1R1 + Q2R2+ Q3R3 Under steady conditions, heat flow does not vary across the wall i.e Q1 =Q2=Q3= Q (T1-T4) = Q (R1+R2+R3) (𝐓𝟏 −𝐓𝟒 ) (𝐓𝟏 −𝐓𝟒 ) (𝐓𝟏 −𝐓𝟒 ) 𝑄=𝐑 = 𝐋𝟏 𝐋 𝐋 = 𝟏 +𝐑 𝟐 +𝐑 𝟑 + 𝟐 + 𝟑 𝚺𝐑 𝐤𝟏 𝐀𝟏 𝐤𝟐 𝐀𝟐 𝐤𝟑 𝐀𝟑 For n-layers (𝐓𝟏 − 𝐓𝐧+𝟏 ) 𝑄= 𝚺𝐑 RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 13  Heat and Mass Transfer 04 The overall heat transfer coefficient In most of the time science and engineering application, fluid flow on the both side of the composite walls. In such situation the overall heat transfer coefficient is used. Denoted by ‘U’ The following fig shows composite slab with fluid flow on both side of the slab and equivalent thermal resistance k1 A1 (T1 − T2 ) k 2 A2 (T2 − T3 ) k 2 A3 (T3 − T4 ) Q = ha A(Ta − T1 ) + = = = Ahb k1 A1 (T4 − Tb ) L1 L2 L3 Above equation can be written as Q (Ta − T1) = = QRa ______(1) ha A QL1 (T1 − T2 ) = = QR1 _______(2) K1 A QL2 (T2 − T3 ) = = QR 2 _______(3) K2 A QL3 (T3 − T4 ) = = QR 3 _______(4) K3 A Q (T4 − Tb ) = = QR b _______(5) hbA Where Ra and Rb are the thermal resistances of convection. Adding all equations we get (𝑇𝑎 − 𝑇𝑏 ) = Q (Ra+R1+R2+R3+Rb) (Ta – Tb ) 𝑄= (R a + R1 + R 2 + R 3 + R b ) RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 14  Heat and Mass Transfer 1 𝐿1 𝐿2 𝐿3 1 (Ta – Tb ) = Q ( + + + + ) ℎ𝑎 𝐴 𝐾1 𝐴 𝐾2 𝐴 𝐾3𝐴 ℎ𝑏𝐴 𝑄 1 𝐿1 𝐿2 𝐿3 1 (𝑇𝑎 − 𝑇𝑏 ) = ( + + + + ) 𝐴 ℎ𝑎 𝐾1 𝐾2 𝐾3 ℎ𝑏 𝐴(𝑇𝑎 − 𝑇𝑏 ) ∴ 𝑄= 1 𝐿 𝐿 𝐿 1 ( + 𝐾1 + 𝐾2 + 𝐾3 + ) ℎ𝑎 1 2 3 ℎ𝑏 A(Ta – Tb ) = UA(Ta − Tb ) 1 1 L ( + +Σ i ) ha hb ki Where; U is overall heat transfer coefficient 𝟏 𝐔= 𝟏 𝐋 𝐋 𝐋 𝟏 ( + 𝐊𝟏 + 𝐊𝟐 + 𝐊𝟑 + ) 𝐡𝐚 𝟏 𝟐 𝟑 𝐡𝐛 *** Data Hand Book “C.P.Kothandaraman” page No ‘___’ *** 05 Heat conduction through Hollow Cylinder Consider a hollow cylinder made of a material having constant thermal conductivity (k) and insulated at both ends. Let: r1and r2 = inner and outer radii respectively. T1and T2= inner and outer surface temperature, L= length of the cylinder RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 15  Heat and Mass Transfer Consider an element at radius ‘r’ and thickness ‘dr’ for a length of the hallow cylinder through which heat is transmitted. Let dT be the temperature drop over the element. We know the Fourier law of heat conduction We know the Fourier law of heat conduction 𝑑𝑇 𝑑𝑇 𝑄 = −𝐾𝐴 ( ) = −𝐾. 2𝜋𝑟. 𝐿 ( ) 𝑑𝑟 𝑑𝑟 or 𝑑𝑟 𝑄 = 𝐾. 2𝜋L dt 𝑟 Integrating both side 𝑟2 𝑇2 𝑑𝑟 𝑄∫ = k. 2. L ∫ 𝑑𝑡 𝑟1 𝑟 𝑇1 𝑟2 𝑄. ln( ) = 𝐾. 2𝜋. 𝐿 (𝑇2 − 𝑇1 ) 𝑟1 k. 2π. L [T2 − T1] [T2 − T1] ∆𝑇𝑜𝑣𝑒𝑟𝑎𝑙𝑙 𝑄= = = ln (r2/r1)] ln (r2/r1) 𝑅 k. 2π. L 06 Heat conduction through composite Cylinders Consider the rate of heat transfer through a composite cylinder and their equivalent thermal resistance as shown in fig Let : T1,T2 and T3 = temperature at inlet surface, between first and second surface , respectively L = length of the cylinder RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 16  Heat and Mass Transfer ha,hb = convective heat transfer coefficient at inside and outside the composite cylinder respectively. Ta,Tb = temperature of the flowing fluid inside and outside the composite cylinder. k1,k2 = thermal conductivity of first and second material respectively. The rate of heat transfer equation is given by k1. 2π. L(T1 − T2 ) k 2. 2π. L (T2 − T3 ) 𝑄 = ha. 2πr1 L(Ta − T1 ) = = = hb. 2πr3 L(T3 − Tb ) ln (r2 /r1 ) ln (r3 /r2 ) Rearranging above equation Q (Ta − T1 ) = = QR a ________________(1) ha. 2πr1 L Q (T1 − T2 ) = = QR1 ________________(2) k1. 2π. L ⌊ r ⌋ ln (r2 ) 1 Q (T2 − T3 ) = = QR 2 ________________(3) k 2. 2π. L ⌊ r ⌋ ln (r3 ) 2 Q (T3 − Tb ) = = QR b ________________(4) hb. 2πr3 L Adding equation 1,2,3 & 4 , we get (Ta − Tb ) = Q(R a + R1 + R 2 + R b ) (Ta − Tb ) 𝑄= (R a + R1 + R 2 + R b ) 2πL(Ta − Tb ) Q= r2 r3 ________________(5) ln ( ) ln ( ) 1 r1 r2 1 ( + + + ) ha r1 k1 k2 hb r3 If equation (5) Divided and multiplied by r1, then we have 2πr1 L(Ta − Tb ) Q= r2 r3 1 r1. ln (r1 ) r1. ln (r2 ) r ( + + + 1 ) ha k1 k2 hb r3 Q = Ui Ai (Ta − Tb ) Where; Ui= overall heat transfer coefficient at inside the surface of the composite cylinder RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 17  Heat and Mass Transfer 1 Ui = r2 r3 1 r1. ln (r1 ) r1. ln (r2 ) r ( + + + 1 ) ha k1 k2 hb r3 If equation (5) Divided and multiplied by r3, then we have 2πr3 L(Ta − Tb ) Q= r r r3. ln (r2 ) r3. ln (r3 ) 1 r3 1 2 ( + + + ) ha r1 k1 k2 hb Q = U0 A0 (Ta − Tb ) Where; U0= overall heat transfer coefficient at outside the surface of the composite cylinder and A0= outside surface area = 2𝜋r3L 1 U0 = r r r3. ln (r2 ) r3. ln (r3 ) 1 r3 1 2 ( + + + ) ha r1 k1 k2 hb 07 Heat flow through the Hollow sphere Consider the hollow sphere made of a material having constant thermal conductivity. Let r1,r2 = inner and outer radii T1,T2= Temperature of inner and outer surface and K = thermal conductivity of the material with the given temperature range Consider a small elemental of thickness dr at any radii r Area through which the heat is transmitted, A = 4𝜋r2 We know Fourier law of heat conduction 𝑑𝑡 𝑑𝑡 𝑄 = −k A ( ) = −k. 4 πr 2. ( ) 𝑑𝑟 𝑑𝑟 𝑑𝑟 𝑄. 2 = −k. 4 π dt r RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 18  Heat and Mass Transfer Interacting above equation, we get 𝑟2 𝑇2 𝑑𝑟 𝑄∫ = − k. 4 ∫ 𝑑𝑡 𝑟1 𝑟2 𝑇1 𝑟 −2+1 𝑄[ ] = −4 𝑘 π [𝑡2 − 𝑡1] −2 + 1 1 −𝑄[ ] = −4 𝑘π [𝑡2 − 𝑡1] 𝑟2 − 𝑟1 𝑟 −𝑟 -𝑄 ( 𝑟2 𝑟 1 ) = 4 𝑘π [𝑡1 − 𝑡2] 1 2 4πk𝑟1 𝑟2 [t1 − t2] [t1 − t2] 𝑄=( )= 𝑟 −𝑟 𝑟2 − 𝑟1 2 1 4πk𝑟1 𝑟2 08 Heat Transfer Through a Composite Sphere Consider the rate of heat transfer through a composite sphere and their equivalent thermal resistance as shown in fig r3 Ta,ha r2 r1 T1 T2 Tb,hb T3 Ta T1 T2 T3 Tb Q Q Ra R1 R2 Rb Let : T1,T2 and T3 = temperature at inlet surface, between first and second surface , respectively ha,hb = convective heat transfer coefficient at inside and outside the composite cylinder respectively. Ta,Tb = temperature of the flowing fluid inside and outside the composite sphere. k1,k2 = thermal conductivity of first and second material respectively. The rate of heat transfer equation is given by 4π.k1.𝑟1 𝑟2 [T1−T2] 4π.k2.𝑟2 𝑟3 [T2−T3] Q= ha 𝑟12.(Ta-T1) = (r2−r1)] = (r3−r2)] = hb 𝑟32 (T3-Tb) Rearranging above equation RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 19  Heat and Mass Transfer 𝑄 (𝑇𝑎 − 𝑇1 ) = = 𝑄𝑅𝑎 − − − − − (1) ℎ𝑎 4𝜋𝑟12 𝑄(𝑟2 − 𝑟1 ) (𝑇1 − 𝑇2 ) = = 𝑄𝑅1 − − − − − (2) 4𝜋𝐾1 𝑟1 𝑟2 𝑄(𝑟3 − 𝑟2 ) (𝑇2 − 𝑇3 ) = = 𝑄𝑅2 − − − − − (3) 4𝜋𝐾2 𝑟2 𝑟3 𝑄 (𝑇3 − 𝑇𝑏 ) = = 𝑄𝑅𝑏 − − − − − (4) ℎ𝑏 4𝜋𝑟32 Adding equation 1, 2, 3 & 4, we get 𝑄 𝟏 (r2 − r1) (r3 − r2) [ 2 + + ] = (𝑇𝑎 − 𝑇𝑏 ) 4𝜋 ha 𝑟1 k1. 𝑟1 𝑟2 k 2. 𝑟2 𝑟3 𝑄 (𝑇𝑎 − 𝑇𝑏 ) = (Ra + R1 + R2 + Rb) 4𝜋 4π(Ta − Tb) (Ta − Tb) 𝑄= = (Ra + R1 + R2 + Rb) 𝟏 (r2 − r1) (r3 − r2) 𝟏 2 + 4π. k. 𝑟 𝑟 + 4π. k. 𝑟 𝑟 + ha 𝑟1 1 1 2 2 2 3 hb4π𝑟32 4π(Ta − Tb) 𝑄= 𝟏 𝟏 2 + 𝚺𝐑 + ha 𝑟1 hb𝑟32 RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 20  Heat and Mass Transfer 09 Critical Thickness of Insulation “The thickness up to which heat flow increases and after which heat flow decreases is termed as critical thickness. In case cylinder and sphere it is called critical radius. ” Critical Thickness Of Insulation For Cylinder Consider a solid cylinder of radius r1 insulated with an insulation of thickness (r2-r1) as shown in figure s The rate heat transfer through the solid cylinder to the surrounding is given by 2π. L [T2 − T1] 𝑄= ln (r2/r1) 1 [ + ] k hb r2 r2 ln( ) From above equation it’s evident that as r2 increases, the factor r1 increases, but k 1 ln (r2/r1) the factor decreases. Thus Q becomes maximum when the denominator [ + ha r2 k 1 ] becomes minimum. The required condition is hb r2 r2 d ln( ) 1 r1 [ + ]=0 dr2 k hb r2 1 1 1 1. + (− r2 ) = 0 k r2 hb 2 1 1 + =0 k r2. hb h0.r2= k r2 (rc)= (k/hb) Analysis of critical thickness  If r1rc , the heat loss decreases with addition of insulation  If r1=rc , the heat loss maximum. Heat loss decreases with the addition of insulation. RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 21  Heat and Mass Transfer CRITICAL THICKNESS OF INSULATION FOR SPHERES Fluid film hb,Ta T2 Insulation T1 K Solid Sphere r2 - r1 r1 r2 The rate of heat transfer through the sphere is given by [T1 − T2] 𝑄= 𝑟2 − 𝑟1 1 + 2 4πk𝑟1 𝑟2 4π𝑟2 h0 Adopting the same procedure as that of cylinder, we have d 𝑟2 − 𝑟1 1 [ + ]=0 dr2 4πk𝑟1 𝑟2 4π𝑟22 h0 d 1 1 1 [ − + 2 ]=0 dr2 𝑘𝑟1 𝑘𝑟2 𝑟2 h0 1 2 2− 3 =0 𝑘𝑟2 𝑟2 h0 𝑟23 h0 = 2 𝑘𝑟22 r2(rc)= (2k/h0) RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 22  Heat and Mass Transfer PROBLEMS RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 23  Heat and Mass Transfer RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 24  Heat and Mass Transfer RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 25  Heat and Mass Transfer RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 26  Heat and Mass Transfer RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 27  Heat and Mass Transfer RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 28  Heat and Mass Transfer RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 29  Heat and Mass Transfer RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 30  Heat and Mass Transfer RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 31  Heat and Mass Transfer RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 32  Heat and Mass Transfer Chapter : 02 FINS OR EXTENDED SURFACES 01 What is a fin? What are the different types of fins? Where is fin used? A fin is extended surfaces used the transfer heat OR Heat transfer by the connection between a surface and the fluid surrounding it can be increased by attaching to the surface thin strips of metals called Fins. Types of Fins  Uniform straight fin  Tapered straight fin  Annular fin  Pin fins  Splines Uses of Fins  Economizer for steam power plant  Convectors for steam and hot water heating systems  Radiators of Automobiles  Air-Cooled Engine Cylinder heads  Cooling coils and condenser coils in refrigerators and air-conditioning  Small capacity compressor  Electric motor bodies  Transformer and electronic equipment 02 Derive an general equation for the Heat flow through rectangular fin of uniform cross section Let us consider an element of fin of thickness ‘dX’ at a distance X from the surface. The heat enters and leaves the element by conduction along the length and heat convected around the surface of the element. Assumptions  Steady state heat conduction along X-direction  No heat generation within the element  Uniform heat transfer coefficient (h)over the entire surface of the fin  The material is homogeneous and isotropic  Negligible radiation heat transfer RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 33  Heat and Mass Transfer Heat conducted into the element is given by dT Qx = −KA dx _____(1) Heat leaving the element is given by d(Qx ) Q (x+dx) = Qx + dx_____(2) dx The heat leaving by convection is given by QConv = hA(T(X)− T∞ ) _____(3) By energy balancing Heat enter into 𝐻𝑒𝑎𝑡 𝐶𝑜𝑛𝑣𝑒𝑐𝑡𝑒𝑑 [ ] = [𝐻𝑒𝑎𝑡 𝐿𝑒𝑎𝑣𝑖𝑛𝑔 𝑡ℎ𝑒 𝐸𝑙𝑒𝑚𝑒𝑛𝑡] + [ ] the element 𝐹𝑟𝑜𝑚 𝑡ℎ𝑒 𝐸𝑙𝑒𝑚𝑒𝑛𝑡 d(QX ) Qx = Qx + dx + hA(T(X)− T∞ ) dx d(QX ) dx + hA(T(X)− T∞ ) = 0 dx dT d(−KA ) dx dx + hA(T (X)− T∞ ) = 0 dx RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 34  Heat and Mass Transfer d2 T −KA dx + h(P. dx)(T(X)− T∞ ) = 0 dx 2 Dividing throughout by “- KAdx”, will get, d2 T hp + ( ) (T(X)− T∞ ) = 0 dx 2 dx d2 T + m2 θ = 0 _____(4) dx2 Where ℎ𝑝 𝑚2 = & 𝜃 = (𝑇(𝑋)− 𝑇∞ ) 𝑑𝑥 𝜃 = (𝑇(𝑋)− 𝑇∞ ) 𝑑𝜃 𝑑𝑇 = 𝑑𝑥 𝑑𝑥 Differentiating again 𝑑2 𝜃 𝑑2 𝑇 = _____(5) 𝑑𝑥 2 𝑑𝑥 2 From equation (4) and (5) 𝑑2𝜃 + 𝑚2 𝜃 = 0 𝑑𝑥 2 This equation is describes the temperature as a faction of ‘x’ And ‘m’. It is second order linear differential equation and solution for above equation is given by θ = (𝑇(𝑋)− 𝑇∞ ) = 𝐶1 𝑒 −𝑚𝑥 + 𝐶2 𝑒 𝑚𝑥 _____(6) Where C1 and C2 are arbitrary constant required boundary conditions to get the values We know that 𝑒 𝑚𝑥 − 𝑒 −𝑚𝑥 𝑒 𝑚𝑥 + 𝑒 −𝑚𝑥 𝑠𝑖𝑛ℎ𝑚𝑥 = & 𝑐𝑜𝑠ℎ𝑚𝑥 = 2 2 𝑒 𝑚𝑥 + 𝑒 −𝑚𝑥 𝑒 𝑚𝑥 − 𝑒 −𝑚𝑥 𝑐𝑜𝑠ℎ𝑚𝑥 − 𝑠𝑖𝑛ℎ𝑚𝑥 = − 2 2 𝑚𝑥 −𝑚𝑥 𝑒 +𝑒 − 𝑒 + 𝑒 −𝑚𝑥 𝑚𝑥 2𝑒 −𝑚𝑥 𝑐𝑜𝑠ℎ𝑚𝑥 − 𝑠𝑖𝑛ℎ𝑚𝑥 = = = 𝑒 −𝑚𝑥 2 2 RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 35  Heat and Mass Transfer From equation (6) θ = C1 (coshmx − sinhmx) + C2 (coshmx + sinhmx) θ = coshmx(C1 + C2 ) + sinhmx(C1 − C2 ) θ = Acoshmx + Bsinhmx Where; A and B are constant Boundary conditions I) Fin is sufficiently long (x=∞) OR Derive the temperature distribution equation and equation for the heat transfer through a fin of infinite length Applying boundary conditions in the general solution equation (𝑇(𝑋)− 𝑇∞ ) = 𝐶1 𝑒 −𝑚𝑥 + 𝐶2 𝑒 𝑚𝑥 ______(1) (i) At x=0 ; T(x)= T0 (𝑇0− 𝑇∞ ) = 𝐶1 𝑒 −𝑚0 + 𝐶2 𝑒 𝑚0 𝜃0 = 𝐶1 + 𝐶2 ___________(2) (ii) At x=∞ ; T(x)= T(∞) (𝑇∞ − 𝑇∞ ) = 𝐶1 𝑒 −𝑚∞ + 𝐶2 𝑒 𝑚∞ 𝐶2 = 0 ___________(3) from (2) and (3) 𝐶1 = 𝜃0 Substituting the values of C1 and C2 in equation (1) will get, 𝜃 = 𝜃0 𝑒 −𝑚𝑥 + 0𝑒 𝑚𝑥 𝜃 = 𝜃0 𝑒 −𝑚𝑥 RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 36  Heat and Mass Transfer 𝜃 = 𝑒 −𝑚𝑥 𝜃0 (𝑇(𝑋)− 𝑇∞ ) = 𝑒 −𝑚𝑥 (𝑇0− 𝑇∞ ) This is required equation for temperature distribution The Rate of Heat flow We know that dT dθ Q = [−KA ] at x = 0 = [−KA ] at x = 0 ______(4) dx dx Now θ = θ0 e−mx Differentiating wrt ‘x’ dθ = θ0. −m. e−mx + 0 at x = 0 dx dθ = −m. θ0 _____(5) dx From equation (4)& (5) dθ Q = [−KA ] dx Q = [−KA. (−m. θ0 ) ] Q = [KA. m. θ0 ] Substituting the value of ‘m’ in the above equation will get hp 𝑄 = KA. √.θ KA 0 Q = √hpKA. θ0 Q = √hpKA (T0− T∞ ) RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 37  Heat and Mass Transfer II) Long fin (of length L) (X=𝐋) Derive the temperature distribution equation and equation for the heat transfer through a fin of finite length or long fin Applying boundary conditions in the general solution equation 𝜃 = (𝑇(𝑋)− 𝑇∞ ) = 𝐴𝑐𝑜𝑠ℎ𝑚𝑥 + 𝐵𝑠𝑖𝑛ℎ𝑚𝑥 _____(1) When X=0; T(x) = T0 from equation (1) 𝜃0 = 𝐴________(2) When X=L; T(x) = 𝑇∞ from equation (1) 0 = AcoshmL + BsinhmL −BsinhmL = 𝜃0. coshmL −𝜃0.coshmL 𝐵= ________(3) sinhmL Substitute the value of A and B in (1), will get 𝜃0. coshmL θ = 𝜃0 coshmx − sinhmx sinhmL 𝜃0 sinhmLcoshmx − 𝜃0. coshmL sinhmx 𝜃0 (sinhmLcoshmx − coshmL sinhmx) θ= = sinhmL sinhmL RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 38  Heat and Mass Transfer θ (𝑇(𝑋)− 𝑇∞ ) = 𝜃0 (𝑇0− 𝑇∞ ) θ sinhm(L − X) = [ ∵ sinh(𝐴 − 𝐵)] 𝜃0 sinhmL This is required equation for temperature distribution Heat flow We know that 𝑑𝜃 𝑄 = [−𝐾𝐴 ] ________(4) 𝑑𝑥 𝑥=0 From equation (1) θ = Acoshmx + Bsinhmx 𝑑𝜃 = 𝐴𝑚𝑠𝑖𝑛ℎ𝑚𝑥 + 𝐵𝑚𝑐𝑜𝑠ℎ𝑚𝑥 𝑑𝑥 𝑑𝜃 [ ] = 𝐴𝑚𝑠𝑖𝑛ℎ𝑚0 + 𝐵𝑚𝑐𝑜𝑠ℎ𝑚0 𝑑𝑥 𝑥=0 𝑑𝜃 [ ] = 𝐵𝑚 𝑑𝑥 𝑥=0 From equation (4) 𝑄 = −𝐾𝐴 𝐵𝑚 𝑠𝑢𝑏 𝐵 & 𝑚 𝑣𝑎𝑙𝑢𝑒 𝑤𝑖𝑙𝑙 𝑔𝑒𝑡 ℎ𝑝 −𝜃0. coshmL 𝑄 = −𝐾𝐴 √ 𝐾𝐴 sinhmL Q = √hpKA. (T0− T∞ ). cot h mL III) End of fin Insulated OR Derive the temperature distribution equation and equation for the heat transfer through a fin which is insulated at the tip Applying boundary conditions in the general solution equation RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 39  Heat and Mass Transfer θ = (T(X)− T∞ ) = Acoshmx + Bsinhmx _____(1) When X=0; T(x) = T0 from equation (1) 𝜃0 = 𝐴________(2) 𝑑𝜃 When X=L; = 0 from equation (1) 𝑑𝑥 θ = Acoshmx + Bsinhmx dθ = mAcoshmL + mBsinhmL = 0 dx sinhmL B = −θ0 = −θ0 tanhmL coshmL B = − θ0. tanhmL _______(3) Substitute the value of A and B in (1), will get sinhmL θ = 𝜃0 coshmx − 𝜃0 sinhmx coshmL 𝜃0 [(coshmL. coshmx) − (sinhmL. sinhmx)] θ= coshmL RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 40  Heat and Mass Transfer 𝜃0 cosh(𝐿 − 𝑋) θ= [ ∵ Cosh(𝐴 − 𝐵)] coshmL θ (𝑇(𝑋)− 𝑇∞ ) = 𝜃0 (𝑇0− 𝑇∞ ) θ cosh(𝐿 − 𝑋) = This is required equation for temperature distribution 𝜃0 coshmL Heat flow We know that dθ Q = [−KA ] ________(4) dx x=0 From equation (1) θ = Acoshmx + Bsinhmx dθ = Amsinhmx + Bmcoshmx dx dθ [ ] = Amsinhm0 + Bmcoshm0 dx x=0 dθ [ ] = Bm dx x=0 From equation (4) 𝑄 = −𝐾𝐴 𝐵𝑚 𝑠𝑢𝑏 𝐵 & 𝑚 𝑣𝑎𝑙𝑢𝑒 𝑤𝑒 𝑤𝑖𝑙𝑙 𝑔𝑒𝑡 ℎ𝑝 𝑄 = 𝐾𝐴 𝜃0. √. tanhmL 𝐾𝐴 𝑄 = √ℎ𝑝𝐾𝐴. (𝑇0− 𝑇∞ ). tan ℎ 𝑚𝐿 IV Convective Boundary Conditions (Short fin) OR Derive the temperature RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 41  Heat and Mass Transfer ) distribution equation and equation for the heat transfer through a fin which is of finite length and losses at the TIP or by Convection. Applying boundary conditions in the general solution equation θ = (T(X)− T∞ ) = Acoshmx + Bsinhmx _____(1) When X=0; T(x) = T0 from equation (1) θ0 = A________(2) dθ dθ When X=L; −k. A dx = h. A. θ 𝐎𝐑 − k dx = h. θ from equation (1) dθ k = h. θ dx −k[Amsinhmx + Bmcoshmx] = h(Acoshmx + Bsinhmx ) Substitute the value of A & x=L in above equation, will get, −k[θ0 msinhmL + BmcoshmL] = h(θ0 coshmL + BsinhmL ) −km [θ0 sinhmL + BcoshmL] = h(θ0 coshmL + BsinhmL ) h − [θ0 sinhmL + BcoshmL] = (θ coshmL + BsinhmL ) km 0 h sinhmL +coshmL B = −θ0 km _________(3) h coshmL + sinhmL km Divide both Numerator and denominator by coshmL h tanhmL + B = −θ0 km h 1+ tanhmL km Substitute the value of A and B in (1), we will get, h tanhmL + θ = θ0 coshmx − θ0 km sinhmx h 1+ tanhmL km RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 42  Heat and Mass Transfer h tanhmL + θ = θ0 coshmx − θ0 km sinhmx h 1+ tanhmL km h tanhmL + θ = θ0 [coshmx − ( km ) sinhmx] h 1+ tanhmL km h θ (T(X)− T∞ ) tanhmL + = = [coshmx − ( km ) sinhmx] θ0 (T0− T∞ ) 1+ h tanhmL km This is required equation for temperature distribution Heat flow We know that dθ Q = [−KA ] ________(4) dx x=0 From equation (1) θ = Acoshmx + Bsinhmx dθ = Amsinhmx + Bmcoshmx dx dθ [ ] = Amsinhm0 + Bmcoshm0 dx x=0 dθ [ ] = Bm dx x=0 From equation (4) Q = −KA Bm sub B & 𝑚 𝑣𝑎𝑙𝑢𝑒 𝑤𝑒 𝑤𝑖𝑙𝑙 𝑔𝑒𝑡 h hp sinhmL + Q = KA θ0. √.θ km KA 0 1 + h tanhmL km h tanhmL + Q = √hpKA. (T0− T∞ ). km h 1+ tanhmL km (V) Heat Transfer from a Bar Connected to the Two Heat Sources at Different RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 43  Heat and Mass Transfer Temperatures Applying boundary conditions in the general solution equation θ = (𝑇(𝑋)− 𝑇∞ ) = Acoshmx + Bsinhmx _____(1) When X=0; T(x) = T1 from equation (1) 𝜃1 = 𝐴________(2) When X=L; T(x) = T2 from equation (1) 𝜃2 = AcoshmL + BsinhmL Substitute the value of A in above equation 𝜃2 = 𝜃1 coshmL + BsinhmL 𝜃2 − 𝜃1 coshmL 𝐵= _________(3) sinhmL Substitute the value of A & B in equation (1) 𝜃2 − 𝜃1 coshmL θ = (𝑇(𝑋)− 𝑇∞ ) = 𝜃1 (coshmx) + ( ) sinhmx sinhmL This is required equation for temperature distribution Heat flow We know that 𝑑𝜃 𝑄 = [−𝐾𝐴 ] ________(4) 𝑑𝑥 𝑥=0 From equation (1) θ = Acoshmx + Bsinhmx 𝑑𝜃 = 𝐴𝑚𝑠𝑖𝑛ℎ𝑚𝑥 + 𝐵𝑚𝑐𝑜𝑠ℎ𝑚𝑥 𝑑𝑥 𝑑𝜃 [ ] = 𝐴𝑚𝑠𝑖𝑛ℎ𝑚0 + 𝐵𝑚𝑐𝑜𝑠ℎ𝑚0 𝑑𝑥 𝑥=0 𝑑𝜃 [ ] = 𝐵𝑚 𝑑𝑥 𝑥=0 From equation (4) 𝑄 = −𝐾𝐴 𝐵𝑚 𝑠𝑢𝑏 𝐵 & 𝑚 𝑣𝑎𝑙𝑢𝑒 𝑤𝑒 𝑤𝑖𝑙𝑙 𝑔𝑒𝑡 ℎ𝑝 𝜃2 − 𝜃1 coshmL 𝑄 = − 𝐾𝐴. √.( ) 𝐾𝐴 sinhmL √ℎ𝑝𝐾𝐴. (𝜃1 coshmL − 𝜃2 ) 𝑄= sinh mL RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 44  Heat and Mass Transfer 03 Define Fin Efficiency (𝛈): Of The Fin? Explain. The performance of the fin is expressed in terms of the fin efficiency. It is defined as the ratio of the actual heat transfer from the fin to the maximum possible heat transfer from the fin. The maximum heat would occur if the temperature of the entire extended surface were equal to the base temperature Tb or (T0) at all points Actual heat transfer from the fin θa η= = Ideal heat transfer from the fin if the entire surafce θI ( ) were maintained at the base temperature √𝐩𝐡𝐚 𝐊𝐀 (𝐓𝐛 −𝐓𝐚 ) 𝐊𝐀 𝟏  For an infinitely long fin, 𝛈 = = √𝐡 𝟐 = 𝐦𝐋 𝐡𝐚 𝐩(𝐓𝐛 −𝐓𝐚 ) 𝐚 𝐏𝐋  For an end-insulated fin of finite length, √pha KA (Tb −Ta )tanh(mL) tanh(mL) η= = ha pL(Tb −Ta ) mL 04 Define Fin effectiveness (𝛆) of fin and what is the significance? it is defined as the ratio heat lost with fin to the without fin.  If the fin end is insulated, then Heat transfer with fin √hpKA. (T0− T∞ ). tan h mL 𝛆= = Heat transfer without fin hA(T0− T∞ ) √hpKA. tan h mL 𝛆= hA  If the fin is infinitely long , then √hpKA(T0− T∞ ) √hpKA pK 𝛆= = =√ hA(T0− T∞ ) hA hA Significance pK  Fin effectiveness √h should be greater than unity if the rate of heat transfer aA from the primary surface is to be improved, it has been observed that use fins pK on surface is justified only if >5 hA  If the ratio of P(Perimeter)and A (Cross Section Area) is increased , the effectiveness of the fin is improved. Due to this reason thin and closely spaced fins are preferred. RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 45  Heat and Mass Transfer  Use of fins only justified where ha is small. if the value of ha is larger, the fins will produce reduction in heat transfer.  The use of fins will be more effective with materials of high thermal conductivity. RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 46  Heat and Mass Transfer RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 47  Heat and Mass Transfer RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 48  Heat and Mass Transfer RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 49  Heat and Mass Transfer RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 50  Heat and Mass Transfer RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 51  Heat and Mass Transfer RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 52  Heat and Mass Transfer Chapter : 03 ONE DIMENSIONAL TRANSIENT HEAT CONDUCTION 1. What do you mean by transient heat conduction or Unsteady state heat transfer? Give examples for this. Transient heat conduction is the condition where in the heat flow and temperature distribution at any point of the system varies continuously with time. Examples  Cooling of IC Engine  Heating and cooling of metal Bodies  Heat transfer of metals by quenching  Cooling and freezing of food. 2. What are the types of unsteady state changes in temperature of a body? A. Non-Periodic Variations : In a non-periodic transient state , the temperature at any point within the system varies Non-linearly with time Ex: 1.Heating of ingots in a furnace 2. Cooling of bars blanks and metal billets in steel works. 3.Various heat treatment process B. Periodic Variation: In a non-periodic transient state , the temperature undergo periodic changes which are either regular or irregular but definitely Cyclic. RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 53  Heat and Mass Transfer Ex: 1.Cylinder of IC Engine 2. Surface of earth during 24 hours 3. Heat processing of regenerators 3. What is lumped system? Derive an expression for the temperature distribution in case of a lumped system in terms of dimensionless parameters. All solids have a finite thermal conductivity and there will be always a temperature gradient inside the solid whenever heat is added or removed. The internal resistance 𝑳 𝟏 (𝑲𝑨) can be assumed to be small or negligible with convective resistance (𝒉𝑨) at the surface. The process in which the internal resistance is assumed negligible in comparison with its surface resistance is called the Newtonian Heating or Cooling process. The temperature in this process is considered to be uniform at a given time. Such an analysis is called Lumped parameter analysis and such system is called lumped system. Lets us consider a body as shown above Let : V= Volume of the solid A= Surface area of the solid K = Thermal Conductivity of the solid 𝜌 =density of the solid C= Specific heat of the solid T = Temperature of the solid at the given time interval T0= uniform initial temperature If the solid suddenly immersed in a well stirred fluid kept at uniform temperature 𝑇∞ at RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 54  Heat and Mass Transfer time. Assuming temperature is a function of time ‘t’ only, 𝐑𝐚𝐭𝐞 𝐨𝐟 𝐡𝐞𝐚𝐭 𝐟𝐥𝐨𝐰 𝐢𝐧𝐭𝐨 𝐭𝐡𝐞 𝐈𝐧𝐜𝐫𝐞𝐚𝐬𝐞 𝐢𝐧 𝐢𝐧𝐭𝐞𝐫𝐧𝐚𝐥 𝐞𝐧𝐞𝐫𝐠𝐲 𝐢𝐧 𝐭𝐡𝐞 [ ]=[ ] 𝐬𝐨𝐥𝐢𝐝 𝐨𝐟 𝐯𝐨𝐥𝐮𝐦𝐞 𝐕 𝐬𝐨𝐥𝐢𝐝 𝐨𝐟 𝐭𝐡𝐞 𝐯𝐨𝐥𝐮𝐦𝐞 𝐕 dT hA(T∞ − T) = mC dt dT hA(T∞ − T) = ρVC ∵ 𝑚 = 𝜌𝑉 dt dT hA = (T − T) dt ρVC ∞ dT −hA = dt (T − T∞ ) ρVC Integrating the above expression 𝐝𝐓 −𝐡𝐀 ∫ (𝐓−𝐓 ) = 𝛒𝐕𝐂 ∫ 𝐝𝐭 ∞ −hA ln(T − T∞ ) = t + C1 − − − − − (1) ρVC Applying initial boundary conditions i.e at t=0 ; T=Ti −hA ln(Ti − T∞ ) = (0) + C1 ρVC C1 = ln(T0 − T∞ ) Substitute the value of C1 in the equation (1) −hA ln(T − T∞ ) = t + ln(T0 − T∞ ) ρVC −hA ln(T − T∞ ) − ln(T0 − T∞ ) = t ρVC T − T∞ −hA ln ( )= t Ti − T∞ ρVC T − T∞ 𝜃 −hA t ( )= = 𝑒 ρVC − − − − − (2) T0 − T∞ 𝜃0 This is the temperature variation expression for 1-Diemensional transient heat conduction −𝐡𝐀 From equation (2) the quantity [ 𝛒𝐕𝐂 𝐭] is expressed as follows RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 55  Heat and Mass Transfer −hA h V A2 k hLc ∝ t=. t=..t ρVC kA ρCV 2 k L2c Where Lc=Characteristics length V Volume of the body Lc = = A Surface Area hA t = Bi. F0 ρVC hLc Bi = Biot Number = k ∝t F0 = Fourier Number = L2c 𝐁𝐢𝐨𝐭 𝐍𝐮𝐦𝐛𝐞𝐫:. = its the ratio of internal conduction resistance to surface convective resistance hLc Bi = k Significance :  When the Bi is small, it indicates that the internal resistance has a small resistance, i.e., relativelly small temperature gradient or the existence of practical uniform temperature within the system.  If Bi < 0.1, the lumped heat capacity approach can be used to advantage with simple shapes such as plates, Cylinder and cubes Fourier Number : 𝐅𝟎 : fourier number signifies the degree of penetration of heating or cooling effect through a solid. ∝t F0 = L2c hLc Bi = Biot Number = k ∝t F0 = Fourier Number = L2c RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 56  Heat and Mass Transfer 𝐁𝐢𝐨𝐭 𝐍𝐮𝐦𝐛𝐞𝐫:. = its the ratio of internal conduction resistance to surface convective resistance hLc Bi = k Significance :  When the Bi is small, it indicates that the internal resistance has a small resistance, i.e., relativelly small temperature gradient or the existence of practical uniform temperature within the system.  If Bi < 0.1, the lumped heat capacity approach can be used to advantage with simple shapes such as plates, Cylinder and cubes Fourier Number : 𝐅𝟎 : fourier number signifies the degree of penetration of heating or cooling effect through a solid. ∝t F0 = L2c To find heat flow The heat ‘Q’ at any time ‘t’ dT Q = ρCV − − − − − (1) dt Consider temperature distribution (𝑇 − 𝑇∞ ) −ℎ𝐴 ( )𝑡 = 𝑒 𝜌𝑉𝐶 (𝑇0 − 𝑇∞ ) −ℎ𝐴 ( )𝑡 𝑇 = 𝑇∞ + (𝑇0 − 𝑇∞ )𝑒 𝜌𝑉𝐶 𝑑𝑇 −ℎ𝐴 (−ℎ𝐴 )𝑡 =. 𝑒 𝜌𝑉𝐶 (𝑇0 − 𝑇∞ ) − − − − − (2) 𝑑𝑡 𝜌𝑉𝐶 𝐹𝑟𝑜𝑚 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (1)& (2) −ℎ𝐴 (−ℎ𝐴 )𝑡 Q = ρCV. 𝑒 𝜌𝑉𝐶 (𝑇0 − 𝑇∞ ) 𝜌𝑉𝐶 −ℎ𝐴 ( )𝑡 Q = −ℎ𝐴. 𝑒 𝜌𝑉𝐶 (𝑇 − 𝑇∞ ) 0 The total quantity of heat ‘QT’ given-off during a time interval (0,t) is given by RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 57  Heat and Mass Transfer 𝑡 𝑡 −ℎ𝐴 ( )𝑡 𝑄𝑇 = ∫ 𝑄. 𝑑𝑡 = ∫ −ℎ𝐴(𝑇0 − 𝑇∞ )𝑒 𝜌𝑉𝐶 𝑑𝑡 0 0 −ℎ𝐴 ( )𝑡 𝑒 𝜌𝑉𝐶 𝑄𝑇 = −ℎ𝐴(𝑇0 − 𝑇∞ ) ⌈ ⌉ −ℎ𝐴 𝜌𝑉𝐶 𝜌𝑉𝐶 (−ℎ𝐴 )𝑡 𝑄𝑇 = −ℎ𝐴 (𝑇0 − 𝑇∞ ) ⌈𝑒 𝜌𝑉𝐶 ⌉ −ℎ𝐴 −ℎ𝐴 −ℎ𝐴 ( )𝑡 ( )0 𝑄𝑇 = (𝑇0 − 𝑇∞ )𝜌𝑉𝐶 ⌈𝑒 𝜌𝑉𝐶 −𝑒 𝜌𝑉𝐶 ⌉ −ℎ𝐴 ( )𝑡 𝑄𝑇 = (𝑇0 − 𝑇∞ )𝜌𝑉𝐶 ⌈𝑒 𝜌𝑉𝐶 − 1⌉ −ℎ𝐴 ( )𝑡 𝑄𝑇 = 𝜃0 𝜌𝑉𝐶 ⌈𝑒 𝜌𝑉𝐶 − 1⌉ 𝑄𝑇 = 𝜃0 𝜌𝑉𝐶[ 𝑒 −𝐵𝑖 𝐹0 − 1] 4. Explain Heisler’s chart. Consider a plane wall of thickness 2L and extending to infinity in Y and Z-directions Let : T0= initial temperature of the wall T∞ = Ambient temperature The governing differential equation 𝑑2𝑇 1 𝑑𝑇 2 = 𝑑𝑥 ∝ 𝑑𝑡 The boundary conditions are i. When t=0; T=Ti 𝑑𝑇 ii. At x=0 ; =0 𝑑𝑡 𝑑𝑇 iii. At x =±𝐿, 𝑘𝐴 𝑑𝑥 = ℎ𝐴(𝑇 − 𝑇∞ ) i.e., Conduction heat transfer = Convective heat transfer from the wall The mathematically analysis yield following (𝑇 − 𝑇∞ ) 𝑥 ℎ𝐿 ∝ 𝑡 = 𝑓 ( , , 2 ) − − − − − (1) (𝑇0 − 𝑇∞ ) 𝐿 𝑘 𝐿 Thus the temperature history is a function of ℎ𝐿 i. Biot Number 𝑘 RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 58  Heat and Mass Transfer ∝𝑡 ii. Fourier Number 𝐿2 iii. Dimensionless number 𝑥⁄𝐿 for plane wall and 𝑟⁄𝑅 for cylinder and spheres From equation (1) a number of graphical charts are obtained and these are called (𝑇−𝑇∞ ) “Heisler Charts”. These charts are drawn with verses Fn(Fourier Number) for (𝑇0 −𝑇∞ ) varies values of Bn for slabs, Cylinders and spheres etc. 5. What is a semi-infinite solid? Give examples A solid which extends itself infinitely in all direction of space is termed as an infinite solid. If an infinite solid is split in the middle by a plane, each half is known as semi- infinite solid. Examples: 1. The heating of an ingot in a furnace 2. The heat flow in an IC engine 3. The periodic heat flow in a building between day and night RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 59  Heat and Mass Transfer RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 60  Heat and Mass Transfer RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 61  Heat and Mass Transfer RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 62  Heat and Mass Transfer RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 63  Heat and Mass Transfer RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 64  Heat and Mass Transfer RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 65  Heat and Mass Transfer RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 66  Heat and Mass Transfer Chapter: 04 HEAT EXCHANGERS 01 Define heat exchanger? What are its applications? A heat exchanger is a device which transfers the energy from a hot fluid to cold fluid, with maximum rate & minimum investment and running cost. The applications of heat exchangers are:  Condenser and boiler in steam plant  Automobile radiator  Intercooler & pre heater  Regenerator  Milk industry 02 How the heat exchangers are classified? Give examples. I. Based on nature of heat exchange process RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 67  Heat and Mass Transfer Non-Condensable gases Hot Fluid Hot Water Wall Cold Water Steam In Cold Fluid (i) Direct Contact Type (ii) Indirect Contact Type i. Direct contact or open type heat exchangers : In this type of heat exchanger , the exchange of heat takes places by direct mixing of hot and cold fluids : Ex : (i) cooling towers , (ii) Direct contact feed heater ii. Indirect contact type heat exchangers: In this type of heat exchanger, the hot fluid and cold fluids are separated by a wall. This is also known as surface heat exchanger Ex : (i) I.C engine & Gas turbine , (ii) Air heater of blast furnace II. Based on Relative direction of fluid flow i. Parallel flow: In this type of heat exchanger, the hot and cold fluid travels at the same direction. The two fluids enter at one end and leave at other end : EX : (i) Oil cooler , (ii) Oil Heater , (iii) Water heater , etc ii. Counter flow: In this type of heat exchanger, the hot and cold fluid travels in the opposite direction. Hot Fluid In T1 Hot Fluid T dT T2 T1 Cold Fluid Out Cold Fluid In Hot Fluid q1= (T1 - t1) q dA dQ q2= (T2 - t2) T T2 q1= (T1 - t2) T dT dt t2 T q t2 dA dQ t t dt t1 Cold Fluid t1 Cold Fluid L L Hot Fluid Out Parallel Flow Counter Flow Cross Flow RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 68  Heat and Mass Transfer iii. Cross flow: In this type of heat exchanger, the hot and cold fluids flows at right angles each other. III. Based on design & construction : i. Concentric type: In this type of heat exchanger, two concentric tubes are used each carrying one of the fluids. The direction of fluid may be parallel or counter. ii. Shell and tube type: In this type of heat exchanger, one of the fluids flows through a bundle of tubes enclosed by a shell and other fluid is forced through the shell and it flows over the outside surface of the tubes. IV. According to physical state of the fluids i. Condenser: In this type of heat exchanger, the condensing fluid remains at constant temperature throughout heat exchanger while the temperature of the cooler fluid gradually increases from inlet to outlet. ii. Evaporator: In this type of heat exchanger, the cold fluid temperature remains at constant temperature while the temperature of hot fluid gradually decreases from inlet to outlet. RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 69  Heat and Mass Transfer Temperature Temperature T1 Hot Fluid T2 T1 Hot Fluid t2 T2 Cold Fluid t1 t1 Cold Fluid t2 Length Length (1) Condenser (2) Evaporators 03 What do you mean by LMTD? What are the assumptions made during the derivation of LMTD? LMTD means Logarithmic Mean Temperature Difference and is defined as that temperature difference which, if constant, would give the same rate of heat transfer actually occurs under variable conditions of temperature differences The assumptions made during the derivation of LMTD i. The overall heat transfer Co-efficient U is constant ii. The flow conditions are steady iii. The specific heat and mass flow rates of both fluid are remains constant iv. There is no loss of heat to the surrounding, since heat exchanger is insulated v. There is no change of phases of both fluids during heat transfer vi. The change in the KE and PE are negligible vii. Axial condition along the tubes of heat exchanger is negligible Ste Derivation An Expression For LMTD For Derivation An Expression For LMTD For RAVICHANDRA KOTI, Asst.Proff, Department of MED, SVIT Bangalore - 560 064 Page 70  Heat and Mass Transfer p PARALLEL Flow Heat Exchanger COUNTER Flow Heat Exchanger 01 T1 T1 Hot Fluid COLD FLUID Hot Fluid T COLD FLUID dT T2 T T2 HOT FLUID q 1= (T1 - t2) dT q 1= (T1 - t1) q dA dQ q 2 = (T2 - t2) T HOT FLUID T q dA q 2= (T2 - t1) t2 dQ dt t2 COLD FLUID dt t t COLD FLUID t1 t1 Cold Fluid Cold Fluid L L 02 Let m