The first part of the question has been worked on, where the heat transfer was calculated to be 6.770 kilowatt, but the problem also asks to calculate the temperature of the outer most surface. What is the temperature of the outer most surface? Consider a thick cylindrical wall with insulation at both the innerand outer walls. The inner and outer radii of the thick cylinder are12 = 40 cm and 13 = 80 cm, respectively. The length of cylinderis 1 meter. The thermal conductivity of the wall is k₂ = 120-The thickness of outer insulation is 5 cm and its thermalconductivity is k3 = 12 . The thickness of the inner insulationWmkWmkWThe heatmk=is also 5 cm, but the thermal conductivity is k₁ = 8dissipates through the surface of outer sides of the insulationthrough convention with h, 50- and the ambienttemperature outside the pipe is Tsurface is maintained at 80°C. Compute the heat transfer through the wall and the temperature ofthe outer most surface.Wm²K= 25°C. The inner mostUse resistance models to solve this problem.wallk2 = 120 W/mKr3= 80 cmr2 = 40 cmr1 = 35 cminner insulationk1 = 8 W/mKr4 = 85 cmouter insulationk3 = 12 W/mK બબતો/K₂= /201=350m = 0.35mN₂ == 400m = 0.4muV3 = 80cm = 0.8mVy=85cm = 0.85m1 W/mklos (only sli)82lufInnerInsulationK₁=8 W/mkryк тыwwwTi In (²4) In (13)27 K₂L2741Total ResisternaRtotal = pthi + R+42 +R+h3 + RoomRotolIn [/2) + In (141)Dim tr₂/ +27K1127/24Y327K3Lho=ambient temperature outside pipe Too = 2.5°CInner most surfde Tou perature Ti = 80°Cconvection Heat transfer coefficient of outside surpree ho= 50wm² kLength of cylinder= IMRT42kth 3Heat dispitale through surfreeouterInsulation50 W=1²15k3= 12W/mktRconvmm Tooㅗho (27 XL)110:45) + In (0:50) + In (0.65)0.827x8x127 x (20x) 27X12X127 KL ho(2784L)f.Rolal = 2.6565X10 3 +9.193x10-4+8.04659 × 10 ++3.744x10-3Rotol = 8.1238 X1b-3 k/wHeat dissipates through surface50x (27x 0.85X1,(T₁ - Too) waltR Total(80-25)8.1238 X103= 6770.18 watt-6.770 kilowattAr

Given: Tin= 80°CT∞=25°CRth1 = 2.6565×10-3 °CWRth2 = 9.193×10-3 °CWRth3 = 2.6565×10-3 °CWRconv =…

Consider a thick cylindrical wall with insulation at both the inner and outer walls. The inner and outer radii of the thick cylinder are 1₂ = 40 cm and 13 = 80 cm, respectively. The length of cylinder is 1 meter. The thermal conductivity of the wall is k₂ = 120- W mk The thickness of outer insulation is 5 cm and its thermal conductivity is k3 = 12 The thickness of the inner insulation W mk W mk wall k2 = 120 W/mK . The heat r3=80 cm 2 = 40 cm is also 5 cm, but the thermal conductivity is k₁ = 8 dissipates through the surface of outer sides of the insulation W through convention with h, = 50 and the ambient m²K temperature outside the pipe is T = 25°C. The inner most surface is maintained at 80°C. Compute the heat transfer through the wall and the temperature of the outer most surface. Use resistance models to solve this problem. r1 = 35 cm r4 = 85 cm inner insulation k1 = 8 W/mK outer insulation k3 = 12 W/mK

Consider a thick cylindrical wall with insulation at both the inner and outer walls. The inner and outer radii of the thick cylinder are 1₂ = 40 cm and 13 = 80 cm, respectively. The length of cylinder is 1 meter. The thermal conductivity of the wall is k₂ = 120- W mk The thickness of outer insulation is 5 cm and its thermal conductivity is k3 = 12 The thickness of the inner insulation W mk W mk wall k2 = 120 W/mK . The heat r3=80 cm 2 = 40 cm is also 5 cm, but the thermal conductivity is k₁ = 8 dissipates through the surface of outer sides of the insulation W through convention with h, = 50 and the ambient m²K temperature outside the pipe is T = 25°C. The inner most surface is maintained at 80°C. Compute the heat transfer through the wall and the temperature of the outer most surface. Use resistance models to solve this problem. r1 = 35 cm r4 = 85 cm inner insulation k1 = 8 W/mK outer insulation k3 = 12 W/mK

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)

8th Edition

ISBN:9781305387102

Author:Kreith, Frank; Manglik, Raj M.

Publisher:Kreith, Frank; Manglik, Raj M.

Chapter1: Basic Modes Of Heat Transfer

Section: Chapter Questions

Problem 1.37P: 1.37 Mild steel nails were driven through a solid wood wall consisting of two layers, each 2.5-cm...

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2.38 The addition of aluminum fins has been suggested to increase the rate of heat dissipation from one side of an electronic device 1 m wide and 1 m tall. The fins are to be rectangular in cross section, 2.5 cm long and 0.25 cm thick, as shown in the figure. There are to be 100 fins per meter. The convection heat transfer coefficient, both for the wall and the fins, is estimated to be K. With this information determine the percent increase in the rate of heat transfer of the finned wall compared to the bare wall.
1.37 Mild steel nails were driven through a solid wood wall consisting of two layers, each 2.5-cm thick, for reinforcement. If the total cross-sectional area of the nails is 0.5% of the wall area, determine the unit thermal conductance of the composite wall and the percent of the total heat flow that passes through the nails when the temperature difference across the wall is 25°C. Neglect contact resistance between the wood layers.
2.30 An electrical heater capable of generating 10,000 W is to be designed. The heating element is to be a stainless steel wire having an electrical resistivity of ohm-centimeter. The operating temperature of the stainless steel is to be no more than 1260°C. The heat transfer coefficient at the outer surface is expected to be no less than in a medium whose maximum temperature is 93°C. A transformer capable of delivering current at 9 and 12 V is available. Determine a suitable size for the wire, the current required, and discuss what effect a reduction in the heat transfer coefficient would have. (Hint: Demonstrate first that the temperature drop between the center and the surface of the wire is independent of the wire diameter, and determine its value.)
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