HW 3 KEY 2024.docx

PSE 360 HW 3 KEY Due February 2, 2024 For all problems, assume the entrance is inward projecting (K factor = 0.78) See Cameron 3-130 1. (10 points) Oil with a kinematic viscosity of 162.3 is to be pumped using a pump which can deliver 150 ft lb f /lb m of head. The length of pipe (4-inch Sch 40) is 400 ft and the change is height between the tanks is 90 ft (Tank B is higher) and there are two elbows and one globe valve in the line. Both tanks are open to atmosphere. Calculate the flow in gpm. ****Should be 14.7 psia**** ∆ P ρ + g g c ∆z + ∆ v 2 2 g c + h f = ηW p 90 + h f = 150 h f = 60 ft h fs = X 1000 ft (h fs is over 1000 ft for viscous fluids) h ff for 4 inch Globe valve: 114 ft 2 90° Elbows: 2 ∗ 10.1 = 20.2 ft Entrance + exit: kd f = 1.78 ∗ 0.3355 0.017 = 35.1 ft Total: 20.2 + 114 + 35.1 + 400 = 569.3 ft To receive credit, you must show ALL work, draw a diagram where applicable, box in final answers, and complete each problem on a new page. This homework will be graded out of a total of 50 points. 90 ft

PSE 360 HW 3 KEY Due February 2, 2024 h fs 1000 ( 569.3 ft ) = h f = 60 ft h fs = 105.39 1000 ft Use kinematic viscosity to match column and h fs value 260 gpm = 101 280 gpm = 109 Flow is about 270 gpm

PSE 360 HW 3 KEY Due February 2, 2024 2. (10 points) Bleached sulfite pulp at 3.5% OD consistency must be delivered from Tank A to Tank B at a flow rate of 600 gpm through a 6-inch Sch. 40 pipe. Based on the diagram below, calculate the HP of the pump with an efficiency of 72%. ∆ P ρ + g g c ∆z + ∆ v 2 2 g c + h f = ηW p Atmospheric pressure: 14.7 PSI = 29.92 in Hg P A = 29.92 − 20 = 9.92 ∈ .Hg P B = 29.92 ∈ .Hg ρ = 8.45 l b m gal @ 3.5% OD ( Cameron 3 − 107 ) 8.45 l b m gal → 63.2 lb m f t 3 ∆ P ρ = ( 29.92 − 9.92 ) ∈ Hg ∗ 0.4912 psi ¿ Hg ∗ 144 in 2 f t 2 ∗ f t 3 63.2 lb = 22.38 ftl b f l b m 400 ft 50 ft 8 ft 4 ft P = 20” Hg vac 10 ft

PSE 360 HW 3 KEY Due February 2, 2024 ∆ z = 50 − 10 − 4 = 36 ft h f = h fs + h ff %K AD = 0.9 %K OD→ 3.5 %K OD = 3.9 %K AD h fs = 31 ft l b f / lb m 100 ft ∗ 0.9 ∗ ( 8 + 50 + 400 + 10 ) = 130.57 ftl b f l b m h ff : EL Entrance and Exit: 1.78 ∗ 0.5054 0.015 = 59.97 ft 3* 90° elbows: 3 ∗ 15.2 = 45.6 ft Total: 59.97 + 45.6 = 105.57 ft 130.57 + 29.45 = 160 ftlb f lb m 22.38 + 36 + 160 = 0.72 W p W p = 301 ft l b f lb m 301 ftl b f l b m ∗ 600 gal min ∗ 8.45 l b m gal ∗ HP 33,000 ftl b f / min = 46.2 HP

PSE 360 HW 3 KEY Due February 2, 2024 3. (10 points) It is desired to pump 4% OD consistency Southern Kraft pulp to from Tank A to Tank B which is 80 feet above the pump centerline at a flow rate of 800 gpm. Both tanks are open to the atmosphere. You have available a used centrifugal pump that will deliver a total head of 250 feet. It is estimated that 1000 feet of piping will be required with 7 - 90 degree elbows. Calculate the pipe size you would recommend (in). ∆ P ρ + g g c ∆z + ∆ v 2 2 g c + h f = ηW p g g c ∆ z + h f = ηW p 80 ft lb f l b m + h f = 250 ft l b f l b m h f = 170 ft l b f lb m We know we have 1000 ft of piping – so the minimum h fs is 17 (x/100*1000=170) Let’s find the smallest pipe size where 4.4%K has head loss is below 17. Smallest pipe size possible is 10” – h fs is equal to about 15, but we have to multiply it by 0.9 to account for southern kraft pulp 80 ft

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