ce 323 homework 3

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University Of Arizona *

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323

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Mechanical Engineering

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Apr 3, 2024

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HW 3 - CE 323 Due 2/27/24 – 11:59pm 1. For the pump curve above that is pumping water, a) Read the bep Q and h p for the 9.5-inch impeller bep Q= 490 gpm, hp = ~78 ft b) What is the maximum efficiency? max efficiency= 80% c) What is the pump speed? pump speed= 1750 rpm d) Compute the specific speed (as a practicing engineer would apply) Specific speed n s = nQ 1/2 /h p 3/4 = 1750(490) 1/2 /76 3/4 = 1505
e) Based on the specific speed, f) what type of pump do you expect this to be and Mixed Flow g) Does the maximum efficiency fall in the expected range? No, the maximum efficiency is 75%. h) Compute the brake horsepower (in HP) for the 8.5 inch pump at 340 gpm bhp = (6472/550)=11.767 = 11.8 HP i) How does the value in part F compare to the value you read on the pump curve? The bhp from the pump curve is 13 HP, therefore the value in part F is less than the pump curve HP value. 2. a) Fit power function and quadratic pump curves for the 8.5” diameter impeller in pump curve. b) Compare the fit curve with the actual head and discharge at 500 gpm (don’t use this point in part a). a) n p = =2.8486 ???( 71−38 71−60 ) ???( 625 425 ) k p = = = 𝑐 −ℎ 2 ? 2 ? ? (71−60) 425 2.8486 3. 582 * 10 −7 h p = h c -K p Q np = 71-3.582*10 -7 Q 2.8486
Q (gpm) hp 0 71 50 70.97523653 100 70.82162845 150 70.43384005 200 69.71518787 250 68.57396261 300 66.92194398 350 64.67358303 400 61.74548481 450 58.05605234 500 53.52522728 550 48.07429294 600 41.62571979 650 34.10304137 700 25.43075236 800 4.33963022 hp= 0.0000003582 np= 2.8486 3. An inlet to a pump is a pressurized 20 cm line. The pressure at the pipe inlet is 200 kPa. The flow rates will vary between 0.015 and 0.04 m 3 /s (at an increment of 0.005 m 3 /s). The 3-m long cast iron suction line contains a 90 o bend, a check valve and a gate valve (open). The 2-m cast iron discharge line contains a check valve and a gate valve. The friction factor is equal to 0.031 for the range of flow considered. The manufacturer’s pump curve with h p in m and Q p in m 3 /s is: Compute a modified pump curve for this configuration. h L =h f +h m Q=AV; V=Q/A; V=0.04/((π/4)0.2) 2 )=1.273 m/sec
Friction factor: 0.031; L=2m+3m=5m (For hf) Kvalues: K=0.5 K=0.26 K=2 K=0.07 Q (m^3/s) A(ft^2) V(m/s) hf hm hL hp (Calculated d) hp (manufactur er) 0.015 0.0314 0.4775 0.0090 0.0569 0.0659 10.2596 23.515 0.02 0.0314 0.6366 0.0160 0.1012 0.1172 10.3109 22.36 0.025 0.0314 0.7958 0.0250 0.1582 0.1832 10.3768 20.875 0.03 0.0314 0.9549 0.0360 0.2277 0.2638 10.4574 19.06 0.035 0.0314 1.1141 0.0490 0.3100 0.3590 10.5527 16.915 0.04 0.0314 1.2732 0.0640 0.4049 0.4689 10.6626 14.44 4. The inlet pressure for the pump configuration in Problem 3 is 100 kPa (gage) for all flow rates. The eye of the pump’s impeller is located 3.5 m above the inlet. The NPSH R increases from 1.5 m at a flow of 0.015 m 3 /s to 4.0 m at a flow rate of 0.04 m 3 /s. Determine if cavitation will occur with this configuration by computing the NPSH A and for the range of flows (0.015 to 0.04 m 3 /s at an 0.005 m 3 /s increment). The water temperature is 20 o C and standard atmospheric conditions hold. If a safety factor of 1.35 is applied, will cavitation occur at any flow rates? a. ? 𝑎?? γ + ? 1 = ? ??? γ + ? ??? + 𝑣 ??? 2 2? + ℎ ? + ℎ ? ? ??? γ = ? 𝑣 γ + ???𝐻 𝐴
? 𝑎?? γ + ? 1 = ? 𝑣 γ + ???𝐻 𝐴 + ? ??? + 𝑣 ??? 2 2? + ℎ ? + ℎ ? ???𝐻 𝐴 = ? 𝑎?? γ ? 𝑣 γ − ? ??? − ℎ ? − ℎ ? flow rate (cms) z_eye (m) hf hm p_atm/y (m) p_v/y (m) NPSHA (m) 0.015 3.5 0.0090 0.0569 10.19367992 0 6.6277 0.02 3.5 0.0160 0.1012 10.19367992 0 6.5765 0.025 3.5 0.0250 0.1582 10.19367992 0 6.5105 0.03 3.5 0.0360 0.2277 10.19367992 0 6.4299 0.035 3.5 0.0490 0.3100 10.19367992 0 6.3347 0.04 3.5 0.0640 0.4049 10.19367992 0 6.2248 No, cavitation will not occur because the available NPSH is greater than the required NPSH. b. ???𝐻 𝐴 = ? ? * ???𝐻 ? flow rate (cms) NPSHR (m) safety factor NPSHA, given safety factor 0.015 1.5 1.5 2.25 0.02 2 1.5 3 0.025 2.5 1.5 3.75 0.03 3 1.5 4.5 0.035 3.5 1.5 5.25 0.04 4 1.5 6 Required NPSH is still less than available NPSH so cavitation won’t occur at any flow rates. Show all hand calculations for Q = 0.015 m 3 /s. 5. Pump curve coefficients for two pumps are given below. a) Compute the equivalent pump curve for three pumps in series configuration (two of pump A and one of pump B). Use feet and gpm units. Pump A Pump B h c 77.0 81.0 K p 2.8 x 10 -7 3.3 x 10 -7 n 2.10 2.005
b) Develop the pump curve table of the one pump 1 and one pump 2 in a parallel configuration. Use meters and cubic meters/hour. Pump 1 Pump 2 h c 18.43 15.63 K p 6.69 x 10 -8 1.32 x 10 -7 n 2.10 2.005 6. Over time a pump wears and alters its head-discharge relationship. a) Assuming the system curve does not change over time, sketch a system curve on an h p vs Q axis and the new and old pump curves.
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