MET 214 Module 2

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MET 214 Module 2

  1. 1. MODULE-2HEAT EXCHANGERS
  2. 2. ONE DIMENSIONAL STEADY CONDUCTION• The simplest example of steady state conduction in one dimension is the transfer of heat through a single plane slab.• Example:x-axis single plane slab• Many simple problems such as conduction through the wall of a building ,approximate to this.
  3. 3. T2 < T1 dT q kT1 T2 dx q’’ Conduction through a solid
  4. 4. CONDUCTION IN PLANE SLABS: FOURIER Law applied to conduction through slabQ=-kA dT/dxQ- Rate of heat conduction through the wall,Wk - Thermal conductivity of the wall material,W/m-K.A- the area of heat flow taken at right angles to the direction of flow of heat,m2.-(dT/-dx) ---- temperature gradient, K/m.Q(X2-X1)=-KA(T2-T1)=KA(T1-T2)Q=KA[(T1-T2)/(x2-x1)]=kA/L ∆T
  5. 5. Where ∆ T =(T1-T2) =Temperature driving force,k.L= (x2-x1)= thickness of the wall,m.Q= ∆ T/RWhere R=L/kA =thermal resistance of the wall k/W.
  6. 6. 1.One face of a copper plate 3cm thick is maintained at 400 0 C,and the other face is maintained at 100 0 C.How much heat is transfers through the plate? Thermal conductivity of copper=370W/m 0 C and the surface area of the plate is 1m2.• SOLUTION:Q=KA( ∆ T)/L=370*1(400-100)/0.03Q=3.7W
  7. 7. CONDUCTION THROUGH SERIES RESISTANCES• Conduction through a system of plane slabs of different material has often to be considered.• A furnace wall consisting of a layer of firebrick and a layer of insulating brick is a typical example,such slabs are called composite walls.• ∆ T= ∆ T1+ ∆ T2+ ∆ T3---------------(1), ∆ T1=Qa*R1,• ∆ T2=Qb*R2, ∆ T3=Qc*R3------(2)• Qa=Qb=Qc=Q---------------(3)• Adding equations (1 )and (2)∆ T1+ ∆ T2+ ∆ T3= ∆ T=Q(R1+R2+R3)Q=(∆ T1+ ∆ T2+ ∆ T3)/(R1+R2+R3) = ∆ T total/R total.In steady heat flow , all the heat that passes through the first resistance must pass through the second and in turn through third (Qa=Qb=Qc)
  8. 8. q Ts1 Ts2 T 2,h 2 T 1 T 2 q Ts3 R Ts4T 1,h 1 KA KB KC x x=L 1 LA LB LC 1 h 1A kA A kB A kC A h 2 A
  9. 9. Where Rtotal=total thermal resistance of the composite wallQ= ∆ T/R Q = ∆ T1/R 1 =∆ T2/R 2 = ∆ T3/R3
  10. 10. RADIAL CONDUCTION IN HOLLOW CYLINDRICAL LAYERS• Conduction through thick walled pipes is a common heat transfer problem, and may be treated one- dimensionally if surface temperatures are uniform.• To find an expression Q in the radial direction consider the cross section of a hollow pipe of length(L).• Q=-kA dT/dr (1)• A=2πrL (2)• Substituting (2) in (1) and integrating• Q={2 πkL/ ln(r0 /ri )}*(Ti-To)= ∆ T/R (note:ln=loge)R= ln(r0 /ri )/2 πkL= thermal Resistance
  11. 11. 1 ln( r2 / r1 ) ln( r3 / r2 ) ln( r4 / r3 ) 1h1 2 r1 L 2 k L 2 k L 2 kC L h2 2 r4 L A B
  12. 12. Cold fluidh 2 ,T 2 r1 Ts2 1 d dT kr r dr dr r2 L Ts1
  13. 13. • Q= kA L ∆ T /(r0 /ri )• A L=2 πL (r0 /ri )/ ln(r0 /ri ) = A0- Ai/ ln(A0/Ai)rL=ro-ri/ln(ro/ri)= logarithmic mean radius.
  14. 14. Multilayer cylindrical system:1.Athick walled tube of stainless steel [18% Cr, 8%Ni, k=19W . 0 C ]with a 2-cm inner diameter(ID) and 4-cm outer diameter (OD) is covered with a 3-cm layer of asbestos insulation [k=0.2 W/m . 0 C] If the inside wall temperature of the pipe is maintained at 600 0 C, and the outside wall temperature is maintained at 100 0 C, calculate the heat loss per meter of length.
  15. 15. RADIAL CONDUCTION IN HOLLOW SPHERICAL LAYERS• Conduction will be in the radial direction if the temperatures of the inner and outer spherical surfaces are uniform Q= -kA dT/drA=4πr2Q=4πkr1r2(T1-T2)/(r2-r1) = ∆ T/R
  16. 16. PURPOSE OF INSULATION• The insulation is defined as a material which retards the heat flow with reasonable effectiveness.• The purpose of insulation is two fold (a) –to prevent the flow of heat from the system to the surroundings as in the case of steam and the hot water pipes which are used for air-conditioning in winter (b) – to prevent the flow of heat from the surroundings to the system as in the case of brine pipes which are used for air-conditioning in summer.
  17. 17. T2 < T1q insulator x A
  18. 18. • The insulations are commonly used for the following industrial purposes1. Air-conditioning systems2. Refrigerators and food preserving stores3. Preservation of liquid gases4. Boilers and steam pipes5. Insulating bricks in furnaces
  19. 19. • Factors affecting the thermal conductivity: The thermal conductivity of insulating materials is one of the most important physical property ,its low value is required for reduction in heat flow rate.
  20. 20. CRITICAL THICKNESS OF INSULATION• The addition of small amount of insulation to small diameter wires or tubes frequently increases the rate of heat flow through the tube to the ambient air.• An experiment showed that the rate of heat loss increased by the addition of asbestos sheet.
  21. 21. • Critical thickness of insulation for cylinder q= T1Ta/[{ln(r2/r1)/2πK} + {1/2πr2 ht } ] 1/k * 1/r2 -1/(r22 *ht ) =0 r2 = K/htThe thickness up to which heat flow increases and after which heat flow decreases is termed as critical thickness
  22. 22. Asbestos string wound on small diameter glass tubes or wires generally increases the rate of heat lossRubber covered wires transmit more heat radially outward than bare wires if the bare wire has the same emissivity as rubber.This means that a rubber covered wire can carry more current than a bare wire for the same temperature rise in the wire.
  23. 23. • If we have relatively good conductors such as concerete with a K value of 1 kcl/m-hr 0 C .• The prime purpose of insulation is to provide protection from electrical hazard, but by using the proper thickness, the ability of the insulated wire is to dissipate heat may be greater than that of bare wire.• r2 – r1 =r1[K /ht r1 –1]• Critical thickness of insulation for spheres:• q = T1-Ta/ {(r2 –r1/4πr1r2K ) + (1/ 4πr2 2 ht)}• r2 =2K/ht
  24. 24. MODULE 3HEAT EXCHANGERS
  25. 25. INTRODUCTION• When a fluid flows past a stationary solid surface ,a thin film of fluid is postulated as existing between the flowing fluid and the stationary surface• Diagram• It is also assumed that all the resistance to transmission of heat between the flowing fluid and the body containing the fluid is due to the film at the stationary surface.
  26. 26. • The amount of heat transferred Q across this film is given by the convection equationWhere h: film co-efficient of convective heat transfer,W/m2KA: area of heat transfer parallel to the direction of fluid flow, m2.T1:solid surface temperature, 0C or KT2: flowing fluid temperature, 0C or K∆t: temperature difference ,K
  27. 27. • Laminar flow of the fluid is encountered at Re<2100.Turbulent flow is normally at Re>4000.Sometimes when Re>2100 the fluid flow regime is considered to be turbulent• Reynolds number=• Prandtl number=• Nusselt number=• Peclet number=
  28. 28. • Grashof number=• Where in SI system• D: pipe diameter,m• V :fluid velocity,m/s• :fluid density,kg/m3• μ :fluid dynamic viscosity N.s/m2 or kg/m.s• √:fluid kinematic viscosity, m2/s
  29. 29. • K:fluid thermal conductivity,W/mK• h: convective heat transfer coefficient,W/m2.K• Cp:fluid specific heat transfer,J/Kg.K• g:acceleration of gravity m/s2• ϐ:cubical coefficient of expansion of fluid=• ∆t:temperature difference between surface and fluid ,K
  30. 30. Functional Relation Between Dimensionless Groups in Convective Heat Transfer• For fluids flowing without a change of phase(i.e without boiling or condensation),it has been found that Nusselt number (Nu) is a function of Prandtl number(Pr) and Reynolds number(Re) or Grashof number(Gr).• And for forced convection
  31. 31. Emperical relationships for Force Convection• Laminar Flow in tubes:• Turbulent Flow in Tubes:For fluids with a Prandtl number near unity ,Dittus and Boelter recommend:• Turbulent Flow among flat plates:
  32. 32. Empirical Relationships for natural convection• Where a and b are constants Laminar and turbulent flow regimes have been observed in natural convection,GrPr<109 Wdepending on the geometry.• Horizontal Cylinders: when 10

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