Friction Losses and Pump Horsepower [PDF]

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16 1. Momentum Transfer. Hot water in an open storage tank at 82.2°C is being pumped at the rate of 0.379m3/min from this storage tank. The line from the storage tank to the pump suction is 6.1m of 2-in. schedule 40 steel pipe and it contains1 three elbows. The discharge line after the pump is 61 m of 2-in. pipe and contains two elbows. The water discharges to the atmosphere at a height of 6.1 m above the water level in storage tank. a. Calculate all frictional losses ƩF. b. Make a mechanical-energy balance and calculate Ws of the pump in J/kg. c. What is the kW power of the pump if its efficiency is 75%? Given: f= 0.0048 Ff=



2 4 fρLV 22 4 (0.0048)(67.91)(2.918) = (0.0525)(2) 2D



= 105.74 J/kg Total loss ƩF= 2.333+ 15.965+ 105.74= 122.77 J/kg a.) ƩF= 122.77 J/kg



V 22 + g Z 2 + ƩF +Ws=0 2 2.9182 9.81 m 122.77 J + (6.1 m)+ +Ws=0 2 2 kg s b.) Ws = -186.85 J/kg mdot = (6.317x10-3 m3/s) (970.4 kg/m3) = 6.130 kg/s Ws= -ɦWp -186.9 J/kg = -0.75 Wp c.) Wp= 249.13



6.130 = 1.527 kW 1000



2. Heat and Mass Transfer. The liquid metal bismuth at a flow rate of 2.00 kg/s enters a tube having an inside diameter of 35 mm at 425oC and is heated to 430oC in the tube. The tube wall is maintained at temperature of 25oC above the liquid bulk temperature. Calculate the tube length required. The physical properties are as follows (H1): k = 15.6 W/m•K, cp = 149 J/kg•K, μ = 1.34 x 10-3 Pa•s. Given:



𝐷 = 35/1000 = 0.035 𝑚 𝑚 = 2.0 𝑘𝑔/𝑠



𝑇1 = 425 𝑜𝐶 𝑇2 = 430 𝑜𝐶 𝑘 = 15.6 𝑊/ 𝑚 ∙ 𝐾



𝑐𝑝 = 149 𝑗/ 𝑘𝑔 ∙ 𝐾 𝜇 = 1.34𝑥10−3 𝑃𝑎 ∙ 𝑠



Solution: 𝐴 = 𝜋D2/ 4 = (𝜋/4) (0.035)2 = 9.621𝑥10−4 𝑚2 𝐺 = 𝑚/𝐴 = 2.0/9.63𝑥10−4 = 2.079𝑥103 𝑘𝑔/𝑚2∙𝑠 𝑁𝑅𝑒 = 𝐷𝐺/ 𝜇 = 0.035(2.079𝑥103) 1.34𝑥10−3 = 5.430𝑥104 𝑁𝑃𝑟 = 𝑐𝑝𝜇/ 𝑘 = 149(1.34𝑥10−3) 15.6 = 0.01280 ℎ𝐿 = (𝑘/𝐷) (0.625) (𝑁𝑃𝑒)0.40 𝑁𝑃𝑒 = 𝑁𝑅𝑒𝑁𝑃𝑟 = (5.430x104) (0.01280) h𝐿 = 15.6/ 0.035 (0.625) ((5.430𝑥104) (0.01280))0.40 𝑞 = 𝑚𝑐𝑝∆𝑇 = 2.0 (149) (430 − 425) = 1490 𝑊 𝑞/ 𝐴 = 1490/ 𝐴 = ℎ𝐿 (𝑇𝑤 − 𝑇) = (3817) (25) 𝐴 = 0.01561 𝑚2 = 𝜋𝐷𝐿 = 𝜋 (0.035) (𝐿) 𝑳 = 𝟎. 𝟏𝟒𝟐𝟎 𝒎



3. Transport Phenomena. A flat plug 30 mm thick having an area of 4.0 x 10-4 m2 and made of vulcanized rubber used for closing an opening in a container. The gas CO2 at 25 oC and 2.0 atm pressure is inside the container. Calculate the total leakage or diffusion of CO2 through the plug to the outside in kgmol/CO2/s at steady state. Assume that the partial pressure of CO2. The diffusivity is 0.11 x10-9 m2 /s Given: 𝐿=



30 = 0.030𝑚 1000



𝐴 = 4.0𝑥10−4 𝑚2 𝑇 = 25𝑜𝐶 𝑝𝐴1 = 2.0 𝑎𝑡𝑚 𝑎𝑏𝑠 𝑆 = 0.90



𝐷𝐴𝐵 = 0.11𝑥10−9



m2 s



Solution: 𝑐𝐴1 =



Sp A 1 0.90(2.0) kgmol = = 8.031𝑥10−2 3 22.44 22.44 m rubber



𝑁𝐴 =



N´ A D AB (c A 1−c A 2) = z2− z1 A



kgmol (4 x 10−4)( 0.11 x 10−9)(8.031 x 10−2) = 𝟏. 𝟏𝟕𝟖𝒙𝟏𝟎−𝟏𝟑 N´ A = s 0.030 4. Separation Processes. It is proposed to recover material A from an aqueous effluent by washing it with a solvent S and separating the resulting two phases. The light product phase will contain A and the solvent S and the heavy phase will contain A and water. Show that the most economical solvent rate, W (kg/s) is given by: W = [(F2 ax0)/mb)]0.5 − F/m where the feedrate is F kg/s water containing x0 kg A/kg water, the value of A in the solvent product phase = £a/kg A, the cost of solvent S = £b/kg S and the equilibrium data are given by: (kg A/kg S)product phase = m (kg A/kg water)water phase where a, b and m are constants Solution: The feed consists of F kg/s water containing F x0 kg/s A, and this is mixed with W kg/s of solvent S. The product consists of a heavy phase containing F kg/s water and, say, F x kg/s of A, and a heavy phase containing W kg/s of S and Wy kg/s A, where y is the concentration of A in S, that is kg A/kg S. The equilibrium relation is of the form: y = mx Making a balance in terms of the solute A gives: F x0 = Wy + F x or: F x0 = (Wmx + Fx) = x (mW + F) and x = F x0/ (mW + F) The amount received of A recovered = F (x0 − x) kg/s which has a value of: F a(x0 − x)£/s. Substituting for x, the value of A recovered = F ax0 − F2ax0/ (mW + F )£/s The cost involved is that of the solvent used, W b £/s.



Taking the profit P as the value of A recovered less the cost of solvent, all other costs being equal, then: P = F ax0[1 − F/ (mW + F )] − W b £/s Differentiating: dP /dW = F2 ax0m/(mW + F )2 − b Putting the differential equal to zero for maximum profit, then: F2 ax0m = b (mW + F)2 and: W = (F2ax0/mb)0.5 − F/m



5. Chemical Reaction Engineering. For a gas reaction at 400 K the rate is reported as:



−dp A 2 =3.66 P A dt



atm/ hr



(a) What are the units of the rate constant? (b) What is the value of the rate constant for this reaction if the rate equation is expressed as:



−r A =



−1 d N A 2 =k C A V dt



mol/m3•s



Solution: a) We are given:



−dp A −dp A 2 2 =k p P A  =3.66 P A dt dt



Balancing dimensions we find:



k p=3.66(atm−1)¿) b) For an ideal gas P A V =n A RT or P A =C A RT



−dp A −d (C A RT ) 2 =3.66 ¿ =3.66 P A becomes dt dt −dC A −1 −1 2 or =( 3.66 ( atm ) hr ) RT C A dt −1 −1 ( 3.66 ( atm ) hr ) RT = new rate constant = k’ Thus



Where: k’ = 3.66



−1 ( 1 atm ) ( 22.4 L ) 1 • •400 K = 120(hr −1)( mol ) atm• hr (1 mol )( 273 K ) L