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- 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. 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. Conduction Metal rod fire t1 t2 Hotter body Colder body
- 4. Convection 1• 2 • surface t1 t2 Fluid
- 5. Radiation Hot body Cold body
- 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. 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. 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. Thermal Circuit Diagram R1 R2 R3 41 3 2 Q L1 R1 = k1A L2 R2 = k2A L3 R3 = k 3A
- 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. Shell and Tube Type th1 mh 1 mc tc2 B A t c1 2 mh mc th2
- 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. 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. Log Mean Temperature Difference (LMTD) θ2 − θ1 LMTD = θ2 ln θ1Where: θ1 – small terminal temperature difference, °C θ2 – large terminal temperature difference, °C
- 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. 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. 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. 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. 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)

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