Heat TransferFitra Dani, Dwi Laura Pramita, Indah Zuliarti & Yohana Siregar Kelompok 8 Kelas C Labtek II 4.5-5. Cooling and Overall U. Oil flowing at the rate of 7258 kg/h with a cpm 2.01 kJ/kg K is cooled from 394.3 K to 338.9 K in a counterflow heat exchanger by water entering at 294.3 K and leaving at 305.4 K. Calculate the flow rate of the water and the overall Ui if the Ai is 5.11 m2. Solution Assume cpm water is 4.187 kJ/kg K. Heat balance Qoil = Qwater (m cpm ΔT)oil = (m cpm ΔT)water 7258 . 2.01 . (394.3-338.9) = m . 4.187 . (305.4-294.3) m = 17389 kg/h LMTD Hot fluid (K) 394.3 338.9 Higher temperature Lower temperature 88.9 44.6 64.22 ln (88.9/44.6) Cold fluid (K) 305.4 294.3 Difference 88.9 44.6 ΔTm = Q = Ui Ai ΔTm 808207.332 . 1000 = Ui 5.11 . 64.22 3600 Ui = 684 W/m2 K 4.5-6. Laminar Flow and Heating of Oil. A hydrocarbon oil having the same physical properties as the oil in Example 4.5-5 enters at 1750F inside a pipe having an inside diameter of 0.0303 ft and a length of 15 ft. The inside pipe (4. The viscosity of oil varies with temperature as follows: 1500F.0303 1/3 11. m = 84.05 cp. 5. 15 . 1.95 cp. 116620 = = 308. The viscosity at 212.14 L μw k ha 0. How many lbm/h oil can be heated? Solution From Example 4. 2.6 .4191 = 11. 3500F.50F.surface temperature is constant at 3250F.0303) 2 0. 0.2 . .7375 cp (with interpolation) and at 3250F is 2.000722 ft2 4 4 m 84. so assume is correct.80 cp.82 cp.385 . π .50F is 4.46 NRe = NPr = 0.2 = = 116620 lbm/ft2 h A 0.46 0.385 cp (with interpolation).86 (308.000722 Dvρ DG 0. The bulk mean temperature of the oil is (175+250)/2 or 212.5 btu/h ΔTa = (Tw-Tbi)+(Tw-Tbo) (325-175)+(325-250) = = 112. 6. 112.6 btu/h ft2 0F Heat balance Q = m cpm ΔT = 84. 69 ) ( ) 0.5 btu/lbm 0F and km 0.5-5.34 μ μ 11.46 lbm/ft h μ w = 2.77 ha = 19. Assume flow rate 84.0303.7375 . 3.083 k Since Reynold number below 2100. 2000F.5-4) will be used.34 .42 btu/h Both Q are same. μ b = 4.5 = 3148.50 cp.86 (NRe NPr )1/3 ( ) 0.5 . 2.14 = 1. so Eq. 2.2 lbm/h.77 lbm/ft h The cross section area of the pipe A is A G= πD 2 π(0.4191 = 5. (250-175) = 3157. 2500F. D μb ha D = 1. 0.2 lbm/h.5 11.0303 0. properties of the oil are cpm 0. The oil is to be heated to 2500F in the pipe.083 btu/h ft 0F.0303 .46 cp = = 69 0.5 2 2 Q = ha A ΔTa = 19.083 15 5. 3000F. Heat Loss Through Thermopane Double Window. The thermal conductivity of the glass is 0.869 W/m K and that of air is 0.914 .35. Solution ΔT = 27.146 k A 0. For a temperature drop of 27. Repeat Problem 4.66 W 0. 1.35 mm thick separated by a 6. calculate the heat loss for a window 0. However. include a convective coefficient of h 11. 1. 1. Solution From Problem 4. R will added by R4 and R5 R4= 1 1 = 0.00635 = 0.3-3 for heat loss in the double window.83 m.83 . each of the glass layers is 6.914 .004368 k A 0.1547 Then.004368 k A 0.05267 h A 11.4.914 .869 .3-10.3-3 ΣR = 0. 0.00635 = 0.8 K over the system. In a given window. 1.83 Δx 0.83 q= 27. Also calculate the overall U.3-3.155 4.026 .35 on the other outside surface.83 R1= R2= R3= Δx 0.8 K q= T R1 R 2 R3 Δx 0. A double window called thermopane is one in which two layers of glass are used separated by a layer of dry stagnant air.914 m x 1.00635 = 0.914 .026 over the temperature range used.8 =179.35 mm space of stagnant air. 0. 0. Effect of Convective Coefficients on Heat Loss in Double Window. 0.35 W/m2 K on the one outside surface of one side of the window and an h 11.869 . average temperature of 449.R5= 1 1 = 0.4 m measured to the center line of pipe.83 New of ΣR = 0.24 W 4. 27.914 .9 = 2.914. It is buried horizontally in the ground at a depth of 0. A water pipe whose wall temperature is 300 K has diameter of 150 mm and a length of 10 m.85 W/m K.075) q = 0. The inside wall temperature is held constant at 477.05267 = 0.29 W/m2 K A ΔT 0. Calculate the loss of heat from the pipe.85 . Heating Air by Condensing Steam.000025 kg/m s .1547+0.54 ln (2H/r1) ln (2 . 1.2 1. Calculate the heat transfer coefficient for a long tube and the heat transfer flux. The ground surface temperature is 280 K and k 0.4-3.4-1 q = kS (T-T0) S= 2 L 2 10 = =26.35.05267+0.1.4/0.9 W 0.26 q= 27.83.71 m/s. 0. Heat Loss from a Buried Pipe. 26.9 K and pressure of 138 kPa.5-1.414 101.33 449.97 1 138 273.9 From Appendix A. Solution ρ=28. Air is flowing through a tube having an inside diameter of 38.6 K by steam condensing outside the tube wall.26 q=U A T U= q 106.068 kg/m3 22.05267 h A 11. (300-280) = 451.54 .3 μ b = 0. Solution From Table 4. 0.1 mm at a velocity of 6.8 4.8 =106. 61 m/s at 101.687 NRe = 10921 (Turbulen) 1 hD b =0.14 k w h = 39.32 kPa pressure and 316.6-2.5 NPr 3 k h = 6.0225 W/m K NPr = 0.027 NRe 0.03721 W/m K NPr = 0. The sides and bottom of this rectangular slab of meat are insulated and the top surface is 254 mm by 254 mm square. . predict the average heat transfer coefficient to the surface.664 NRe0. Heat Transfer to an Apple. Solution T average = -17.05 W/m2 K 4.5 K.000026 kg/m s k = 0.80C From Appendix A. Assume that it is sphere.3 k = 0.379 kg/m3 NRe = 13189 (Turbulen) 1 hL =0.6 K and its average diameter is 114 mm.6-2) can be used.6-3.μ w = 0.90C and 1 atm is recirculated at a velocity of 0. (4. If the surface of the meat is at -6.0000162 kg/m s ρ = 1.35 W/m2 K 4. It is desired to predict the heat transfer coefficient for air being blown by an apple lying on a screen with large openings.61 m/s over the exposed top flat surface of a piece of frozen meat.8 NPr 3 ( ) 0. The surface of the apple is at 277. The air velocity is 0.70C.72 μ b = 0. Chilling Frozen Meat. Cold water at -28. As an approximation assume that either Eq. 5 =0.02512 W/m K NPr = 0.Solution T average = 297.05 K From Appendix A.201 kg/m3 NRe = 5860 (Turbulen) 1 hD 0. Jawab: = 0 0125 =02 =2 =2 =0 010416666667 f =0 0166666666 f =2 3 14) 1) 0 010416) =2 3 14) 1) 0 0166) ⁄ K = 7 75+7 78. ⁄ .664 NRe NPr 3 k h = 9.988 W/m2 K 4.2-2 (heat removal of a cooling coil) Diket : = 40 = 80 = 0 25 = 0 40 K = 7 75+7 78.709 μ b = 0.000018 kg/m s ρ = 1.3 k = 0. 4.2-5.3-7. Pipa 2ln schedule 40 = 2 067 = 2 357 Tebal isolasi= 51mm = 2. Convection. Temperature Distribution in Hollow Sphere. Conduction and Overall U.007874 inchi = 450 K=350 33 = 300 K=80 33 ⁄ . =4 ∫ ∫ ⁄ ⁄ 4. Jawab: = = =2 =2 =2 =2 3 14) 3 280839895) =2 3 14) 3 280839895) =2 3 14) 3 280839895) ⁄ ⁄ ∑ ∑ . 4.5-8 Heat Transfer with a Liquid Metal Diket: ⁄ =1490 W A= m . 4.2 0K.20 K K = = = 0. Calculate the heat loss per m2 of surface area for a temporary insulating wall of a food cold storage room where the outside temperature is 299. The wall is composed of 25.1 W/m2. Calculate the thermal conductivity in Btu/h ft 0F and in W/mK.4 and 303.9 W 4.025 m T1 T2 jawab : = 318.4 0K = 303.1-1 Insulation in a cold Room.0577 W/m K .1-2 Determination of Thermal Conductivity.50 K. The heat flux was measured as 35. In determining the thermal conductivity of an insulating material.9 0K = 276.5 0K X2-X1 = 0. Penyelesaian : Dik : X2-X1 = 0.4 mm of corkboard having k of 0.0433 W/mK Penyelesaian : Dik : T2 T1 = 299. the temperatures were measured on both sides of a flat slab of 15 mm of the material and were 318.90 K and the inside temperature 276.0254 m K = 0.0433 w/m K Dit : q/A = ? = = -39. 762 W/m K. The total inside surface area of the roomto use in the calculation is approximately 39 m2.4 0C = 0.0433 = 0.762 = 586 W = 39 m2 T1 T4 kA kB kC q A Jawab : RA = RC = Q = = = 3.8 0C and the outside surface temperature is 29. cork. The inside wall surface temperature is-17. At the outer concrete surface. and concrete. What thickness of cork board is needed to keep the heat loss to 586 W??? Penyelesaian : Dik : = -17.8 mm of concrete. 0. The mean conductivities are for pine. A food cold storage room is to be constructed of an inner layer of 19.1 mm of pine wood.151 = 0.0433 X 39) = 0.07559333835 (0.0433. 0. a middle layer of cork board.7049401709 X -RA + RB + RC = = 0.07559333835 XB = 0.3-1 Insulation needed for Food Cold Storage Room.8 0C = 29.243335031 X = 1. 0.08054607509 RB = 0.151 .1276544705 m . and an outer layer of 50.4 0 C.4. The liquid metal bismuth at a flow rate of 2. Calculated the thickness of insulation required.3-2 Insulation of a Furnace. The tube wall is maintained at a temperature of 25 0C above the liquid bulk temperature.4.244 m thick is constructed of material having a thermal conductivity of 1.1876923077 RA + RB = 0.7043715847 RB XB = 0. Calculated the tube length require. The physical properties are as follows (H1) .346 q T1 T2 = 1830 = 1588 = 299 Dit : x = ?? Jawab : RA = Q = = = 0. cp = 149 J/kg K. so the heat loss from the furnace will be equal to or less than 1830 W/m2. The inner surface temperature is 1588 K and the outer 299 K.5-8 Heat transfer with a Liquid metal. Peneyelesaian : Dik : m = 2 kg/s ID = 35 mm .1787710298 m 4.516679277 = 0. The wall will be insulated on the outside wiht the material having an average k of 0.00 kg/s enters a tube having an inside diameter of 35 mm at 425 0C and is heated to 430 0C in the tube. k = 15.6 W/m K. A wall of furnace 0.3 kB = 0. Penyelesaian : Dik : kA = 1.346 W/m K.30 W/m K. K. 4691 = 0.05 m/s. Calculated the heat transfer coeffisient assuming laminar flow.0156116247 = L = 0.012798795 = 9.K = 15. smooth falt plate at 3.6-1 Heat transfer from a Flat palte.2 K.3 kpa and a temperature of 188. .305 m and is at 333.6999 A = A = 0.625) )/D = 54323.142053 m 4.667996 (25) = 95441.6 W/m K Cp = 149 J/kg K T1 = 4250C T2 = 430 0C Tw = 25 0C Dit : L Jawab : A = G = Nre = Npr = hL = (k ( 0.8 k is flowing over a thin.667996 W/m2 K Q = m cp = 2 (149) (5) = 1490 W = 3817. The plate length in teh direction of flow is 0.61625 x 10-4 m2 = ? = 3817. Air at a pressure of 101. 3 K. the equivalent L to use for the top flat surface it 0.2 K L = 0. Use the simplified equation of table 4.34331032 W/m2 = 0.9 times the diameter. At its surface.027 W/m K Npr = 0. It losses heat by natural convection to air at 294.7-2 Losses by Natural Convection from a Cylinder. Heat is lossses from the cylindrical side and the flat circular and at the top. Calculate the heat loss neglecting radiation losses.1 0K.305 m V = 3.05 m/s Tf = = 311 K = 37. A vertikal cylinder 76.85 0C K = 0.Penyelesaian : Dik : Tw = 288. Penyelesaian : Dik : L = 76.0048 kj/kg K Dit : h = ? Jawab Nre x L = N = h = 12.11842 4.7-2 and those equation for the lowest range of NGr Npr.9 mm high is maintained at 397.8 K Tb = 333.705 Cp = 1.664 ( = = 55668.2 mm .2 mm in diameter and 121. 799545 W 4. NpR 3.44 m .8) = 74.1 0K Tb = 294.3 0K T Tf = K = 0.37 ( ) = ( ) = (5255325707) (0.14 (0.9)3 (102.6 x 10 9 L3 H Maka : A = ( ) + ( ) + 3. A current of 250 A is passing through a stainless steel wire having a diameter of 5.1219)(0.0297232852 W/mK Npr = 0.08 mm. The wire is 2.9 times the diameter = 345.7000072202) = 3678765939 = = 0.0762) (0.9412 = 1.7 0K = 0.7000072202 Jawab : NGR = = = 5255325707 NGR.9 mm Tw = 397.04083156305 m2 Q = hA (Tw-Tb) = 18.D = 121. Temperature Rise in Heating Wire.3-13. 01283 = 0. 1 . 104 h = 2.24 .44 q = 3.6 K. 1 . 0. 108 W/m3 T = 3.0843 = q π 0. Ri = 2. 108 . 2.24 m2 q = h A (T–T0) q = 2. 0.5) = 12.086125 = 0. The thermal conductivity is k 22.14 . 0.3-8.001075 R = 0. 0.54 ft2 A1 = 2 .6 = 451.01283/(26 .423 .086125+0.4 K NGr = 3.62/0.4 = 0.005625 = 26665 btu/h . Heat Transfer in Steam Heater. (394.002542 / 90 + 427.005625 q = (220-70)/0.086125 ft Thickness = 0. Solution I2R = q π r2L 2502 .58 ln (0. From Example 4.099 = 0.0843 ohm. 12) = 0.01283 ft R1 = 0.6 . 3.154 in = 0.14 . 3.35 .0037 R1 = 0.7-5.35 .53 W 4.002542 . 0. 0.54 0.62 ft2 AAlm = 0.58) = 0.01 . 0.7-3 T average = 380.54) = 0. Natural Convection on Plate Spaces.long and has resistance of 0. 0.067 / (2 .62 0.5 W/m K. The outer surface is held constant at 427.01 W/m2 K Area = 0.099 ft Ai = 2 .6 K 4. Calculate the center-line temperature at steady state.3-366.62) = 0.54) Ri = 1/hi Ai = 1/(500 .00085 Ro = 1/ho A1 = 1/(1500 . 0. Ui = 26665/(0.62 .4 btu/h ft2 0F . (220-70)) = 286.54 . (220-70)) = 329.2 btu/h ft2 0F Uo = 26665/(0.