# Solution Geankoplis Chapter 4

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### Heat Transfer

Fitra 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) Cold fluid (K) Difference

394.3 Higher temperature 305.4 88.9 338.9 Lower temperature 294.3 44.6 ΔTm = 88.9 44.6 64.22 ln (88.9/44.6)   Q = Ui Ai ΔTm 808207.332 . 1000 3600 = Ui 5.11 . 64.22 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

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surface temperature is constant at 3250F. The oil is to be heated to 2500F in the pipe. How many lbm/h oil can be heated?

Solution

From Example 4.5-5, properties of the oil are cpm 0.5 btu/lbm 0F and km 0.083 btu/h ft 0F. The viscosity of oil varies with temperature as follows: 1500F, 6.50 cp; 2000F, 5.05 cp; 2500F, 3.80 cp; 3000F, 2.82 cp; 3500F, 1.95 cp.

The bulk mean temperature of the oil is (175+250)/2 or 212.50F. The viscosity at 212.50F is 4.7375 cp (with interpolation) and at 3250F is 2.385 cp (with interpolation). Assume flow rate 84.2 lbm/h.

μb = 4.7375 . 2.4191 = 11.46 lbm/ft h μw = 2.385 . 2.4191 = 5.77 lbm/ft h The cross section area of the pipe A is

A 2 2 π π(0.0303) 0.000722 4 4 D    ft2 G = m A= 84.2 0.000722 = 116620 lbm/ft 2 h NRe = Dvρ μ = DG μ = 0.0303. 116620 308.34 11.46  NPr = p c k  = 0.5 11.46= 69 0.083

Since Reynold number below 2100, so Eq. (4.5-4) will be used. a h D k = b 1/3 0.14 Re Pr w D μ 1.86 (N N ) ( ) L μ a h 0.0303 0.083 = 1/3 0.14 0.0303 11.46 1.86 (308.34 . 69 ) ( ) 15 5.77 ha = 19.6 btu/h ft20F Heat balance Q = m cpm ΔT = 84.2 . 0.5 . (250-175) = 3157.5 btu/h ΔTa = w bi w bo (T -T )+(T -T ) 2 = (325-175)+(325-250) 2 = 112.5 Q = ha A ΔTa = 19.6 . π . 0.0303 . 15 . 112.5 = 3148.42 btu/h Both Q are same, so assume is correct, m = 84.2 lbm/h.

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4.3-3. Heat Loss Through Thermopane Double Window. A double window

called thermopane is one in which two layers of glass are used separated by a layer of dry stagnant air. In a given window, each of the glass layers is 6.35 mm thick separated by a 6.35 mm space of stagnant air. The thermal conductivity of the glass is 0.869 W/m K and that of air is 0.026 over the temperature range used. For a temperature drop of 27.8 K over the system, calculate the heat loss for a window 0.914 m x 1.83 m. Solution ΔT = 27.8 K 1 2 3 T q= R R R    1 Δx 0.00635 R = = 0.004368 k A 0.869 . 0.914 . 1.83 2 Δx 0.00635 R = = 0.146 k A 0.026 . 0.914 . 1.83 3 Δx 0.00635 R = = 0.004368 k A 0.869 . 0.914 . 1.83 27.8 q= =179.66 0.155 W

4.3-10. Effect of Convective Coefficients on Heat Loss in Double Window.

Repeat Problem 4.3-3 for heat loss in the double window. However, include a convective coefficient of h 11.35 W/m2 K on the one outside surface of one side of the window and an h 11.35 on the other outside surface. Also calculate the overall U.

Solution

From Problem 4.3-3 ΣR = 0.1547

Then, R will added by R4 and R5

4

1 1

R = = 0.05267

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5 1 1 R = = 0.05267 h A 11.35. 0.914 . 1.83 New of ΣR = 0.1547+0.05267+0.05267 = 0.26 27.8 q= =106.9 0.26 W q=U A T q 106.9 U= = 2.29 A ΔT 0.914.1.83. 27.8 W/m 2 K

4.4-3. Heat Loss from a Buried Pipe. A water pipe whose wall temperature is

300 K has diameter of 150 mm and a length of 10 m. It is buried horizontally in the ground at a depth of 0.4 m measured to the center line of pipe. The ground surface temperature is 280 K and k 0.85 W/m K. Calculate the loss of heat from the pipe. Solution From Table 4.4-1 q = kS (T-T0) 1 2 L S= ln (2H/r )  2 10 = =26.54 ln (2 . 0.4/0.075)  q = 0.85 . 26.54 . (300-280) = 451.24 W

4.5-1. Heating Air by Condensing Steam. Air is flowing through a tube having

an inside diameter of 38.1 mm at a velocity of 6.71 m/s, average temperature of 449.9 K and pressure of 138 kPa. The inside wall temperature is held constant at 477.6 K by steam condensing outside the tube wall. Calculate the heat transfer coefficient for a long tube and the heat transfer flux.

Solution

1 138 273.2

ρ=28.97 1.068

22.414 101.33 449.9 kg/m 3

From Appendix A.3 μb = 0.000025 kg/m s

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μw = 0.000026 kg/m s k = 0.03721 W/m K NPr = 0.687 NRe = 10921 (Turbulen) 1 0.8 3 0.14 h D =0.027 NRe NPr ( ) k b w   h = 39.35 W/m2 K

4.6-2. Chilling Frozen Meat. Cold water at -28.90C and 1 atm is recirculated at a velocity of 0.61 m/s over the exposed top flat surface of a piece of frozen meat. The sides and bottom of this rectangular slab of meat are insulated and the top surface is 254 mm by 254 mm square. If the surface of the meat is at -6.70C, predict the average heat transfer coefficient to the surface. As an approximation assume that either Eq. (4.6-2) can be used.

Solution

T average = -17.80C From Appendix A.3 k = 0.0225 W/m K NPr = 0.72 μb =0.0000162 kg/m s ρ = 1.379 kg/m3 NRe = 13189 (Turbulen) 1 0.5 3 h L =0.664 NRe NPr k h = 6.05 W/m2 K

4.6-3. Heat Transfer to an Apple. It is desired to predict the heat transfer

coefficient for air being blown by an apple lying on a screen with large openings. The air velocity is 0.61 m/s at 101.32 kPa pressure and 316.5 K. The surface of the apple is at 277.6 K and its average diameter is 114 mm. Assume that it is sphere.

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Solution

T average = 297.05 K From Appendix A.3 k = 0.02512 W/m K NPr = 0.709 μb =0.000018 kg/m s ρ = 1.201 kg/m3 NRe = 5860 (Turbulen) 1 0.5 3 h D =0.664 NRe NPr 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. Jawab: = 0 0125 =0 010416666667 f = 0 2 =0 0166666666 f = 2 =2 3 14) 1) 0 010416) = 2 =2 3 14) 1) 0 0166) ⁄ K = 7 75+7 78.

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4.2-5. Temperature Distribution in Hollow Sphere.

= 4 ∫ ∫ ⁄ ⁄

4.3-7. Convection, Conduction and Overall U.

Pipa 2ln schedule 40 = 2 067 = 2 357

Tebal isolasi= 51mm = 2,007874 inchi = 450 K=350 33 = 300 K=80 33 ⁄

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Jawab: = = = 2 =2 3 14) 3 280839895) = 2 =2 3 14) 3 280839895) = 2 =2 3 14) 3 280839895) ⁄ ⁄

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4.5-8 Heat Transfer with a Liquid Metal Diket: ⁄ =1490 W A= m

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4.1-1 Insulation in a cold Room. 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 299,90 K and the inside temperature 276,50 K. The wall is composed of 25,4 mm of corkboard having k of 0,0433 W/mK Penyelesaian : Dik : T2 = 299,9 0K T1 = 276,5 0K X2-X1 = 0,0254 m K = 0,0433 w/m K Dit : q/A = ? = = -39,9 W

4.1-2 Determination of Thermal Conductivity. In determining the thermal

conductivity of an insulating material, the temperatures were measured on both sides of a flat slab of 15 mm of the material and were 318,4 and 303,2 0K. The heat flux was measured as 35,1 W/m2. Calculate the thermal conductivity in Btu/h ft 0F and in W/mK.

Penyelesaian : Dik : X2-X1 = 0,025 m T1 = 318,4 0K T2 = 303,20 K jawab : K = = = 0,0577 W/m K

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4.3-1 Insulation needed for Food Cold Storage Room. A food cold storage

room is to be constructed of an inner layer of 19,1 mm of pine wood, a middle layer of cork board, and an outer layer of 50,8 mm of concrete. The inside wall surface temperature is-17,8 0C and the outside surface temperature is 29,4 0C. At the outer concrete surface. The mean conductivities are for pine, 0,151 ; cork, 0,0433; and concrete, 0,762 W/m K. The total inside surface area of the roomto use in the calculation is approximately 39 m2. What thickness of cork board is needed to keep the heat loss to 586 W??? Penyelesaian : Dik : T1 = -17,8 0C T4 = 29,4 0C kA = 0,151 kB = 0,0433 kC = 0,762 q = 586 W A = 39 m2 Jawab : RA = = = 3,243335031 X RC = = 1,7049401709 X Q = -RA + RB + RC = = 0,08054607509 RB = 0,07559333835 XB = 0,07559333835 (0,0433 X 39) = 0,1276544705 m

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4.3-2 Insulation of a Furnace. A wall of furnace 0,244 m thick is constructed of

material having a thermal conductivity of 1,30 W/m K. The wall will be insulated on the outside wiht the material having an average k of 0,346 W/m 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. Calculated the thickness of insulation required.

Penyelesaian : Dik : kA = 1,3 kB = 0,346 q = 1830 T1 = 1588 T2 = 299 Dit : x = ?? Jawab : RA = = = 0,1876923077 Q = RA + RB = 0,7043715847 RB = 0,516679277 XB = 0,1787710298 m

4.5-8 Heat transfer with a Liquid metal. The liquid metal bismuth at a flow

rate of 2.00 kg/s enters a tube having an inside diameter of 35 mm at 425 0C and is heated to 430 0C in the tube. The tube wall is maintained at a temperature of 25 0C above the liquid bulk temperature. Calculated the tube length require. The physical properties are as follows (H1) ; k = 15,6 W/m K. K, cp = 149 J/kg K,

Peneyelesaian :

Dik : m = 2 kg/s ID = 35 mm

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K = 15,6 W/m K Cp = 149 J/kg K T1 = 4250C T2 = 430 0C Tw = 25 0C Dit : L = ? Jawab : A = = 9,61625 x 10-4 m2 G = Nre = = 54323,4691 Npr = = 0,012798795 hL = (k ( 0,625) )/D = 3817,667996 W/m2 K Q = m cp = 2 (149) (5) = 1490 W = 3817,667996 (25) = 95441,6999 A = A = 0,0156116247 = L = 0,142053 m

4.6-1 Heat transfer from a Flat palte. Air at a pressure of 101,3 kpa and a

temperature of 188,8 k is flowing over a thin, smooth falt plate at 3,05 m/s. The plate length in teh direction of flow is 0,305 m and is at 333,2 K. Calculated the heat transfer coeffisient assuming laminar flow.

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Penyelesaian : Dik : Tw = 288,8 K Tb = 333,2 K L = 0,305 m V = 3,05 m/s Tf = = 311 K = 37,85 0C K = 0,027 W/m K Npr = 0,705 Cp = 1,0048 kj/kg K Dit : h = ? Jawab Nre x L = = = 55668,11842 N = = 0,664 ( h = 12,34331032 W/m2

4.7-2 Losses by Natural Convection from a Cylinder. A vertikal cylinder 76,2

mm in diameter and 121,9 mm high is maintained at 397,1 0K. At its surface. It losses heat by natural convection to air at 294,3 K. Heat is lossses from the cylindrical side and the flat circular and at the top. Calculate the heat loss neglecting radiation losses. Use the simplified equation of table 4.7-2 and those equation for the lowest range of NGr Npr, the equivalent L to use for the top flat surface it 0,9 times the diameter.

Penyelesaian :

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D = 121,9 mm Tw = 397,1 0K Tb = 294,3 0K

T = 0,9 times the diameter

Tf = = 345,7 0K K = 0,0297232852 W/mK Npr = 0,7000072202 Jawab : NGR = = = 5255325707 NGR. NpR = (5255325707) (0,7000072202) = 3678765939 = 3,6 x 10 9 L3 (0,9)3 (102,8) = 74,9412 H = 1,37 ( ) = ( ) Maka : A = ( ) + ( ) + 3,14 (0,1219)(0,0762) = 0,04083156305 m2 Q = hA (Tw-Tb) = 18,799545 W

4.3-13. Temperature Rise in Heating Wire. A current of 250 A is passing

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long and has resistance of 0.0843 ohm. The outer surface is held constant at 427.6 K. The thermal conductivity is k 22.5 W/m K. Calculate the center-line temperature at steady state.

Solution

I2R = q πr2L

2502 . 0.0843 = qπ0.002542 . 2.44 q = 3.35 . 108 W/m3

T = 3.35 . 108 . 0.002542 / 90 + 427.6 = 451.6 K

4.7-5. Natural Convection on Plate Spaces.

From Example 4.7-3 T average = 380.4 K NGr = 3.423 . 104 h = 2.01 W/m2 K Area = 0.6 . 0.4 = 0.24 m2 q = h A (T–T0) q = 2.01 . 0.24 . (394.3-366.5) = 12.53 W

4.3-8. Heat Transfer in Steam Heater.

Ri = 2.067 / (2 . 12) = 0.086125 ft Thickness = 0.154 in = 0.01283 ft R1 = 0.086125+0.01283 = 0.099 ft Ai = 2 . 3.14 . 1 . 0.086125 = 0.54 ft2 A1 = 2 . 3.14 . 1 . 0.099 = 0.62 ft2 AAlm = 0.62 0.54 0.58 ln (0.62/0.54)  Ri = 1/hi Ai = 1/(500 . 0.54) = 0.0037 R1 = 0.01283/(26 . 0.58) = 0.00085 Ro = 1/ho A1 = 1/(1500 . 0.62) = 0.001075 R  = 0.005625 q = (220-70)/0.005625 = 26665 btu/h

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Ui = 26665/(0.54 . (220-70)) = 329.2 btu/h ft20F Uo = 26665/(0.62 . (220-70)) = 286.4 btu/h ft20F

References

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