#### Transcript Reading Materials: Chapter 9 LAST LECTURE 1

Reading Materials: Chapter 9 Heat transfer results from a temperature difference. LAST LECTURE 1 CONVECTION HEAT TRANSFER Modes forced flow induced by external agency e.g. pump eg. forced-draught air cooler, evaporators natural fluid motion caused by temperature-induced density gradients within fluid Examples air flow over hot steam pipe, fireplace circulation, cooling electronic devices 2 CONVECTION HEAT TRANSFER Warm (lighter) air rises Figure: Natural convection flow over a heated steam pipe Cool (more dense) air falls to replace warm rising air 3 Modelling Convection Forced convection generally most-effective transport of energy from solid to fluid. .Q Engineer's prime concern Solid Ts Flowing fluid Tb rate of convection enables sizing of equipment 4 Modelling Convection Experimentally found that: Q A Ts Tb Q hA Ts Tb h - convective heat transfer coefficient. Main problem predict h value for: • variety fluids & flow rates • range of shapes 5 Resistance Concept Rate equation • Q Written in same form as Ohm’s Law: surface Ts fluid Tb 1 hA Potential Difference V Current flow I = Resistance R Q hA Ts Tb Ts Tb Ts Tb 1/ hA Ts • Q Tb R (Ts-Tb)= driving force (1/hA) – thermal resistance (R) for convection heat transfer. 6 TYPICAL UNITS FOR h -2 -1 -1 -2 -1 S.I.: W m K or J s m K British: Btu hr ft (F deg) Conversion: 1 W m K = 0.176 Btu hr ft (F deg) -1 -2 -2 -1 -1 -1 -2 -1 Typical Values free convection (air) 5 - 60 forced convection (air) 25 - 300 forced convection (water) 200 - 10,000 boiling water 2,000 - 25,000 condensing steam 4,000 - 110,000 7 Illustration 27.1 Air at 20°C is blown over an electrical resistor to keep it cool. The resistor is rated at 40,000 ohm and has a potential difference of 200 volts applied across it. The expected mean heat transfer coefficient between the resistor surface and the air is 50 W m-2K-1. What will be the surface temperature of the resistor, which has a surface area of 2 cm2? Air 20°C h = 50 W m K R = 40,000 Potential = 200 V 8 SOLUTION Energy Balance Generation = heat loss by convection Rate of heat generation PQ 2 V V P VI V R R V2 2002 Q 1 watt 4 R 4x10 Convective loss Q hA(Ts Tb ) 1 (50)(2x10 4 )(Ts 20) Ts 120o C 9 Determining the size (H/T area) of the exchanger Q h1A T1 T2 T2 T3 kA h x 2A T3 T4 (10.25) Q T1 T4 1 x 1 h1A kA h2 A overall driving force resis tances (10.26) Figure 1. Heat transfer between two flowing fluids separated by a rectangular 10 Determining the size (H/T area) of the exchanger Q Q Figure 2: Heat transfer between two flowing fluids separated by a cylindrical wall Ti To r ln o ri 1 1 hi A i 2kL ho A o Ti To R1 R2 R3 (10.27) 11 Overall heat-transfer coefficient As a short-hand method of describing heatexchanger performance, we use the overall heattransfer coefficient, Q hA Ts Tb Qduty Uo AT Qduty T T 1 R Uo A (10.28) where Uo overall heat-transfer coefficient W / m 2K 12 Determining the size (H/T area) of the exchanger 13 Illustration 27.2 Consider the kettle below. For the conditions given, find the flame temperature for the following values of the heat transfer coefficients: hi (boiling) = 4000 W m-2K-1 ho (gas flame) = 40 W m-2K-1 • Q = 1.88 kW Water hi Flame ho FLAME 1 ho T3 T2 T1 T0 ? .Q T2 x k T1 To 1 hi T3 = 100°C BOILING WATER 14 Solution Plane slab- area constant, eliminate A: 1 1 x 1 Uo A ho A kA h1A 1 1 1.2x10 3 1 Uo 40 204 4000 Uo 39.6 W m-2K -1 15 Solution Q 1883 W (as before) A = 3.14x10-2 m2 (as before) Qduty Uo A To T3 Qduty To T3 AUo 1883 o 1514K 1514 C To T3 2 3.14x10 39.6 Water hi Flame ho .Q T3 T2 T1 To To 1514 100 1614o C 16 Determining the size (H/T area) of the exchanger (10.28) Qduty Uo ATavg • The term Tavg in equation 10.28 represents the temperature difference between the hot and cold streams averaged. • For single-pass exchangers, the appropriate form of Tavg is the log-mean temperature difference, Tlog mean (often abbreviated LMTD), defined as Log mean temperature difference Tlog mean Tlog mean T1 T2 ln T1 T 2 (10.29) 17 Determining the size (H/T area) of the exchanger Qduty Uo ATavg A Qduty Uo Tlog mean (10. 30) For shell-and-tube exchangers, the inside area (Ai) of the tubes is smaller than the outside area (Ao). However, the differences between Ai and Ao will be neglected. 18 Example Saturated Steam, Hot 280oF, mstream Cold Oil, 110oF, 960 lbm/min Saturated water, 280oF, mstream Oil, 35oF, 960 lbm/min Balance on cold stream: mCp Tout Tin cold Qduty 10.24b lbm Btu o 960 0.74 110 35 F Qduty o min lbm F cold Btu Qduty 53,280 min 19 Example How much area is required for the counter-current heat exchanger in Example 10.5? 1 Saturated Steam, 280oF, mstream 2 Saturated water, Hot 280oF, mstream Cold Oil, 35oF, 960 lbm/min Oil, 110oF, 960 lbm/min T1 280 35 245o F T2 280 110 170o F Tlog mean 245 170 205o F ln 245 170 20 Example From table 10.5 o Uo 150 Btu / hr ft 2 F A 53,280 Btu/min 150 Btu hr ft 2 oF 60min 2 104 ft o 1hr 205 F 21 Example How much area is required for the co-current heat exchanger in Example 10.5? 1 Saturated Steam, 280oF, mstream Oil, 35oF, 2 Saturated water, Hot 280oF, mstream Cold 960 lbm/min Oil, 110oF, 960 lbm/min T1 280 110 170o F T2 280 35 245o F Tlog mean 170 245 205o F ln 170 245 22 Example From table 10.5 o Uo 150 Btu / hr ft 2 F A 53,280 Btu/min 150 Btu hr ft 2 oF 60min 2 104 ft o 1hr 205 F 23 Summary 24