HW 3 KEY 2024

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 ∆ 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

PSE 360 HW 3 KEY Due February 2, 2024 4. (10 points + used for ABET) Make up a problem with water where you have to calculate the HP of the pump or the flow rate in a system. For full credit, solve the problem and include the following: o The problem should require the application of two or more basic concepts of fluid flow as taught in 360  (e.g. required flow for a target production; estimate friction loss; size pumps or valves; calculate required motor horsepower) o The problem should be solvable o The problem should be straightforward but not considered a “gimme” o The problem should involve some creativity  o The solution should be accurate and clear o The solution should include a sketch 

PSE 360 HW 3 KEY Due February 2, 2024 5. (10 points) The figure below shows two tanks. Draw the system curve for the system shown below. If we use a MT3x4-10 A70 pump with a 7” impeller. See pump curves on the last page. (2.5 points each) Calculate: a. What rpm motor would you use and what will be the flow rate with the pump? b. Calculate the NPSHA for the pump using a 7” impeller and compare with the NPSHR required. Do you think the suction is designed properly? c. What will be the flow rate if the impeller size is increased to 8”? d. Use the affinity equations and the GPM with a 7” impeller to find the GPM with an 8” impeller and compare with your answer in Part C. Part A: ∆ P ρ + g g c ∆z + ∆ v 2 2 g c + h f = ηW p g g c ∆ z + h f = ηW p 150 ft + h f = nW p We need to plot multiple flowrates (GPM) to find the head the pump needs to provide At 0 GPM (no flow): 150 ft = nW p → h sys = 150 ft We need to use 3560 RPM motor, since the minimum head provided by the pump must be 150 ft. In order to find h f at other flowrates, we need to find the EL of the valves and exit + entrance: System is at 70°F 150 ft Suction pipe: 20 Discharge pipe: 1000 ft 5-inch Sch. 40 pipe Tanks: open to atmosphere

PSE 360 HW 3 KEY Due February 2, 2024 Anywhere between 400-500 GPM works. Part B NPS H A = P A ρ + g g c z A − h f − h V P A @ 70 ° F→V P H 2 O = 0.36292 l b f i n 2 Suction: h ff gate valve :3.36 ft entrance : 0.78 ( 0.4206 ) 0.016 = 20.5 ft

PSE 360 HW 3 KEY Due February 2, 2024 ∑ EL = 23.86 ft Total = 43.9 ft h f = 4.15 100 ( 43.9 ) = 1.82 NPSHA = 14.7 l b f i n 2 ∗ 144 i n 2 f t 2 62.43 l b m f t 3 + 20 ft − 1.82 − 0.36292 l b f i n 2 ∗ 144 in 2 f t 2 62.43 lb m f t 3 = 51 ft NPSHR = ¿ therange of 12 − 15 ft ( see pumpcurve ) Therefore, the suction is designed properly since NPSHA > NPSHR Part C The pump curve crosses the 8” impeller at 600 gpm Part D Q 2 = Q 1 ( D 2 D 1 ) = 450 GPM ( 8 7 ) = 514 GPM The affinity equations are just approximations. It is close “enough” to the pump curve.

PSE 360 HW 3 KEY Due February 2, 2024

PSE 360 HW 3 KEY Due February 2, 2024