Heat transfer

2,103 views

Published on

0 Comments
1 Like
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total views
2,103
On SlideShare
0
From Embeds
0
Number of Embeds
3
Actions
Shares
0
Downloads
24
Comments
0
Likes
1
Embeds 0
No embeds

No notes for slide
  • Modes of Heat Transfer
  • Radiation heat transfer
  • Heat transfer

    1. 1. ENGR. YURI G. MELLIZA Heat TransferIt is that area of mechanical engineering that dealswith the different principles and mechanisms involvedin transferring heat from one point to another.
    2. 2. Modes of Heat Transfer1. Conduction: Is the transfer of heat from one point to another point within a body or from one body to another body when they are physical contact with each other.2. Convection: Is the transfer of heat from one point to another within a fluid. a. Natural or Free convection – motion of the fluid is due to the difference in density because of a difference in temperature. b. Force Convection – motion of fluid is accomplished by mechanical means, such as a fan or a blower.3. Radiation: It is the flow of heat from one body to another body separated by a distance due to electromagnetic waves.
    3. 3. Conduction Metal rod fire t1 t2 Hotter body Colder body
    4. 4. Convection 1• 2 • surface t1 t2 Fluid
    5. 5. Radiation Hot body Cold body
    6. 6. Conduction From Fourier s Law t1 L 1 − kA(t2 − t1 ) kA(t1 − t2 )A Q= = k 2t L L A ( t1 − t 2 ) ( t 1 − t 2 ) 2 Q Q= = L L k kA Where: L – thickness, meters A – surface area, m2 k – thermal conductivity, mW°C or mWK − - Q – conductive heat flow, Watts
    7. 7. Thermal Circuit Diagram R 1 2 Q −(t2 − t1 ) − Δt (t1 − t2 ) Q= = = L R R kA °C K R −resistance, or W W − ∆t − temperatur e potential, °C or K L R= kA
    8. 8. Conduction through a Composite Plane Wall L1 L2 L3 1 A 2 3 k1 k2 k3 4 Q ( t1 −t 4 ) ( t1 −t 4 ) Q= = L1 L L R1 +R2 +R 3 + 2 + 3 k1 A k2 A k 3 A (t1 −t 4 ) A ( t1 −t 4 ) Q= = 1 L 1 L 2 L 3  L 1 L 2 L 3  k +k +k  k +k +k  A 1 2 3  1 2 3 Q (t1 −t 4 ) = A L 1 L 2 L 3  k +k +k   1 2 3
    9. 9. Thermal Circuit Diagram R1 R2 R3 41 3 2 Q L1 R1 = k1A L2 R2 = k2A L3 R3 = k 3A
    10. 10. A furnace is constructed with 20 cm of firebrick, k = 1.36 W/m-°K, 10 cm of insulatingbrick, k = 0.26 W/m-°K, and 20 cm of building brick, k = 0.69 W/m-°K. The insidesurface temperature is 650°C. The heat loss from the furnace wall is 56 W/m2.Determine a. the interface temperature and the outside wall temperature, °C b. the total resistance Rt, for 1 m2 L1 L2 L3 R1 R2 R3 1 A 2 4 3 1 2 3 Q 4 Q Given: L1 =0.20 m ; L2 = 0.10 m ; L3 = 0.20 m k1 = 1.46 ; k2 = 0.26 ; k3 = 0.69 t1 = 650°C Q/A = 56 W/m2
    11. 11. At 1 to 2 At 1 to 3Q ( t1 − t 2 ) Q ( t1 − t 3 ) = =A L1 A L1 + L 2 k1 k1 k2 Q  L 1  Q  L 1 L 2 t2 = t1 −    t3 = t1 −   +    A  k 1      A  k1 k 2    0.20    0.20 0.10 t2 = 650 − 56  = 641.7°C t3 = 650 − 56 +  = 620.2°C   1.36    1.36 0.26  At 1 to 4 Q ( t1 − t 4 ) = A L1 + L2 + L 3 k1 k 2 k 3 Q  L L L  t 4 = t1 −   1 + 2 + 3     A  k 1 k 2 k 3    0.20 0.10 0.20  t 4 = 650 − 56 + +  = 604°C   1.36 0.26 0.69 
    12. 12. Convection AFluid If t1 > t2 Q = hA ( t1 − t2 ) Watts 1• t1 If t1 < t2 (heat flows in opposite direction) 2• t2 Q = hA ( t2 − t1 ) Watts h ( t 2 − t1 ) Q Q= Watts 1 hA Where: Q – convective heat flow, Watts A – surface area in contact with the fluid, m2 h – convective coefficient, W/m2-°C or W/m2-K t1, t2 – temperature, °C
    13. 13. Conduction from Fluid to Fluid separated by a composite plane wall L1 L2 L3 1 A 2 3 o• ho, i• to hi ti k1 k2 k3 4 Q ( ti − t o ) (ti − to ) Q= = 1 L L L 1 R i + R1 + R 2 + R 3 + R o + 1 + 2 + 3 + hi A k 1 A k 2 A k 3 A ho A ( ti − t o ) A ( ti − t o ) Q= = 1  1 L1 L 2 L 3 1   1 L1 L 2 L 3 1  h + k + k + k + h  h + k + k + k + h  A i 1 2 3 o  i 1 2 3 o Q ( ti − t o ) = A  1 L1 L 2 L 3 1  h + k + k + k + h   i 1 2 3 o
    14. 14. Thermal Circuit Diagram Ri R1 R2 R3 Ro 1 2 3 4 i o Q 1 L1 Ri = R1 = hi A k1A 1 L2 Ro = R2 = ho A k2A L3 R3 = k3A
    15. 15. Overall Coefficient of Heat Transfer (ti − t o ) Q= 1 L1 L2 L3 1 + + + + hi A k 1 A k 2 A k 3 A ho A A (ti − t o ) Q=  1 L1 L2 L 3 1  h + k + k + k + h   i 1 2 3 o Q = UA (−∆t) 1 U=  1 L1 L2 L 3 1  h + k + k + k + h   i 1 2 3 o Where: U – overall coefficient of heat transfer, W/m2-°C or W/m2-K
    16. 16. CONDUCTION THROUGH CYLINDRICAL COORDINATES Q ( t1 − t 2 ) Q= ln r2 1 2 r1 Where: 2πkL r1 – inside radius, m − (∆t) r2 – outside radius, m Q= R L – length of pipe, m t1 − (∆t) = (t1 − t2 ) k – thermal conductivity of r1 material, W/m-°C k ln r2 r1 r2 t2 R= 2πkL
    17. 17. For composite cylindrical pipes (Insulated pipe) ( t1 − t 3 ) 2πL (t1 − t 3 ) Q= = Q r r ln r2 ln 3 ln r2 ln 3 r1 r2 r1 r2 + + 1 2 3 2πk 1L 2πk 2L k1 k2 − (∆t) Q= R1 + R 2 k1 t1 − (∆t) = (t1 − t3 ) r1 r2 r3 ln ln k2 r2 t2 r1 r2 R1 = ; R2 = r3 t3 2πk 1L 2πk 2L
    18. 18. Heat Flow from fluid to fluid separated by a composite cylindricalwall (ti − to ) Q= Q r r ln 2 ln 3 1 r1 r2 1 o + + + hiA i 2π k1L 2π k 2L ho A o i ti 1 2 3 to − ( ∆ t) hi ho Q= R i + R1 + R 2 + R o − ( ∆ t) = ( t i − t o ) t1 r3 k1 r1 ln r2 ln 1 r1 r2 1 Ri = ; R1 = ; R2 = ; Ro = r2 t2 hi A i 2π k1L 2π k 2L ho A o k2 t3 A i = 2π r1L ; A o = 2π r3L r3
    19. 19. Overall Coefficient of Heat Transfer Q = Uo A o (−∆t) Q =Ui A i (−∆t) 1 Uo A o =Ui A i = r2 r3 1 ln r1 ln r2 1 + + + hi A i 2πk 1L 2πk 2L ho A o where : Uo -coefficie nt of heat transfer based on outside area Ui - coefficien t of heat transfer based on inside area
    20. 20. RadiationFrom Stefan - Boltzmann Law:The radiant heat flow Q for a blackbody is proportionalto the surface area A, times the absolute surfacetemperature to the 4th power. A blackbody, or blacksurface is one that absorb all the radiation incidentupon it. Q = δAT4 Watts →1 where: δ = 5.67x 10-8 W/m2-°K4 δ -Stefan-Boltzmann constant A- surface area,m2 T - absolute temperature, -°K
    21. 21. Body # 1 T1 A1 Body # 2The net radiant heat transfer T2between two bodies or surfaces is; →2 Q = δA1(T14 - T24)Real bodies, surfaces, are not perfect radiators andabsorbers,but emit, for the same surface temperature, a fraction of theblackbody radiation. This fraction is called the “emittance” ∈
    22. 22. Emittance(∈) Actual surface radiation at T ∈= Black surface radiation at T Actual bodies or surfaces are called “Gray Bodies” or “Gray surfaces” . Thus, the net rate of heat transfer between gray surface at temperature T1 to a surrounding black surface at temperature T2 is; Q = δ ∈ 1 A1(T14 - T24) Watts The enclosure being total and a black surface may be modified by the modulus F1- 2, which accounts for the relative geometries of the surfaces(not all radiation leaving 1 reaches 2) and the surface emmitances, thus the equation becomes, Q = δ F 1-2 A1(T14 - T24) Watts
    23. 23. Radiant heat transfer frequently occurs with other modes ofheat transfer, and the use of a radiative resistance R is veryhelpful. Let us now define Q to also be, T -T 1 2 Q= R T -T 1 2 R= QWhere T2’ is an arbitrary reference temperature.
    24. 24. Heat ExchangersTypes of Heat Exchangers 1. Direct Contact Type: The same fluid at different states are mixed. 2. Shell and Tube Type: One fluid flows inside the tubes and the other fluid on the outside.Direct Contact m1, h1 m2, h2 m3, h3
    25. 25. Applying First Law for an OPEN SYSTEMMass Balancem1 +m2 = m3 → Eq. 1Energy balance, ∆KE and ∆PE negligiblem1h1 +m2h2 = m3h3 → Eq. 2m1h1 +m2h2 = m1h3 + m 2 h 3m1 (h1 − h 3 ) = m 2 ( h 3 − h 2 ) →Eq. 3
    26. 26. Shell and Tube Type th1 mh 1 mc tc2 B A t c1 2 mh mc th2
    27. 27. By energy balance Heat rejected by the hot fluid = Heat absorbed by the cold fluid Qh = Qc Q h = m h C ph (t h1 − t h 2 ) →Eq .1 Q c = m c C pc (t c 2 − t c1 ) →Eq .2Where: mc – mass flow rate of cold fluid, kg/sec mh – mass flow rate of hot fluid, kg/sec h – enthalpy, kj/kg t – temperature,°C Cpc – specific heat of the cold fluid, KJ/kg-°C Cph – specific heat of the hot fluid, KJ/kg-°C Q – heat transfer, KW h, c – refers to hot and cold, respectively 1, 2 – refers to entering and leaving conditions of hot fluid A, B – refers to entering and leaving conditions of cold fluid
    28. 28. Heat Transfer in terms of OVERALL COEFFICIENT Of HEAT TRANSFER U UA (LMTD )Q= KW 1000where : WU - overall coefficient of heat transfer, 2 or m - °C W m2 - KA - total heat transfer surface area, m2LMTD −log mean temperatur e difference
    29. 29. Log Mean Temperature Difference (LMTD) θ2 − θ1 LMTD = θ2 ln θ1Where: θ1 – small terminal temperature difference, °C θ2 – large terminal temperature difference, °C
    30. 30. Terminal Temperature Difference 1. For Counter Flow 2. For Parallel FlowT th1 T th1 Ho tF θ2 Ho tF luid th2 luid tc2 th2 θ2 θ1 Co ld θ1 Flu id tc2 Flu ld id Co tc1 tc1 A A θ1 = t h 2 − t c1 θ1 = t h 2 − t c 2 θ 2 = t h1 − t c 2 θ 2 = t h1 − t c 2
    31. 31. 3. For Constant temperature HEATING 4. For Constant temperature COOLING T T th1 t Hot Fluid t θ1 Ho tF θ2 luid θ2 tc2 th2 id Flu Co l d θ1 t Cold Fluid t tc1 A A θ1 = t − t c 2 θ1 = t h 2 − t θ2 = t − t c1 θ2 = t h1 − t
    32. 32. Correction Factor FUA(LMTD )Q= KW 1000where : WU - overall coefficient of heat transfer, 2 or m - °C W m2 - KA - total heat transfer surface area, m2LMTD −log mean temperatur e differenceF - Correction Factor
    33. 33. Exhaust gases flowing through a tubular heat exchanger at the rate of 0.3 kg/sec arecooled from 400 to 120°C by water initially at 10°C. The specific heat capacities ofexhaust gases and water may be taken as 1.13 and 4.19 KJ/kg-°K respectively, andthe overall heat transfer coefficient from gases to water is 140 W/m2-°K. Calculate thesurface area required when the cooling water flow is 0.4 kg/sec; a. for parallel flow (4.01 m2) b. for counter flow (3.37 m2) Given θ2 - θ1 390 − 53.3 mc = 0.4 kg/sec ; Cpc = 4.19 KJ/kg - C LMTD = = = 169.2°C θ2 390 ln ln mh= 0.3 kg/sec ; Cph = 1.13 KJ/kg - C θ1 53.3 th1 = 400°C ; th2 = 120°C For Counter Flow tc1 = 10°C θ2 = th 1 - tc2 = 400 − 66.7 = 333.3°C Qh = Qc θ1 = th2 − tc1 = 120 −10 = 110°C 0.3(1.13)(400 −120) = 0.4(4.19)(t c2 −10) 333.3 −110 LMTD = = 201.43°C 333.3 tc2 = 66.7°C ln 110 For Parallel Flow Q = UA(LMTD) θ2 = th1 - t c1 = 400 - 10 = 390°C Q A= θ1 = th2 - tc2 = 120 - 66.7 = 53.3°C U(LMTD)
    34. 34. Q = 0.3(1.13)(400 −120) = 94.92 KWQ = 94920 Watts 94 ,920A= = 4.01 m2 140(169.2) 94 ,920A= = 3.4 m2 140(201.43)

    ×