ce 323 homework 3

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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 = = = ℎ 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

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● 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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b) Explain what happens to the operating point over time The operating point is when the difference in head losses equates to zero. At the beginning, as the discharge/operating point increases, the difference in the length of the head losses gets increasingly smaller, until the difference of the two head losses becomes zero. When the difference of the head loss is zero, that is the operating point that showcases the capacity that is expected from the pump. After the operating point, as the discharge continues to increase, the difference in head loss will get increasingly larger and further away from each other. 7. a) Given the data below, determine the optimal pipeline diameter considering initial and operation costs for a pump-pipeline system. Pipe costs as a function of diameter are given below. Pipe Length (ft) 5280 Pump efficiency 0.78 Pumping time/day 16

Pipe roughness (CHW) 100 Motor efficiency 0.92 Interest rate 4% Elevation change (ft) 60 Cost/Pump HP ($/HP) 12000 Design Life (years) 30 Flow rate (cfs) 2.4 Unit Energy cost ($/kw-hr) 0.13 USPW 17.29 diameter (in) unit pipe cost ($/ft) pipe cost ($) friction loss (ft) [hazen williams] n=1.85 total lift (ft) [system curve] horsepower to water (HP) pump capital cost ($) HP to operate pumps (HP) energy cost/year ($/yr) present worth of energy cost for N years ($) total PW cost ($) cheapest 14 54.56 218240 11.88 71.88 19.57255773 234870.6927 27.275024 7 15447.5702 7 $267,119.90 $720,230.5 9 b) How and WHY will the optimal pipe size and overall cost change if the unit energy cost is $0.05/kw-hr, rather than $0.13/kw-hr The optimal pipe size stays constant with the cheapest overall cost being a pipe that has a diameter of 14 inches. However, the overall cost decreases $164,381.48, with the cost change occurring primarily in the energy cost of the pipe. When the unit energy cost increases, the overall total PW cost increases in a similar manner, with the present worth of energy cost for N years increasing at the same rate. diameter (in) unit pipe cost ($/ft) pipe cost ($) friction loss (ft) [hazen williams] n=1.85 total lift (ft) [system curve] horsepower to water (HP) pump capital cost ($) HP to operate pumps (HP) energy cost/year ($/yr) present worth of energy cost for N years ($) total PW cost ($) cheapest 14 54.56 218240 11.88 71.88 19.57255773 234870.6927 27.275024 7 5941.37318 1 $102,738.4 2 $555,849.12

c) How and WHY does the optimal pipe size and overall cost change if the desired flow is 1.8 cfs vs 2.4 cfs for the same pumping duration and $0.09/kw-hr) The optimal pipe size has a pipe with a diameter of 12 inches when the desired flow is 1.8 cfs, while the optimal pipe size has a pipe with a diameter of 14 inches when the desired flow is 2.4 cfs. The most drastic change with the increase of the desired flow is in regards to the pump capital cost and the energy cost of the pipe. When the flow increased, the friction loss and horsepower to water both decreased, leading to the total PW cost being incredibly more expensive. Flow=1.8 & $0.09/kw-hr: diameter (in) unit pipe cost ($/ft) pipe cost ($) friction loss (ft) n=1.85 total lift (ft) horsepower to water (HP) pump capital cost ($) HP to operate (HP) energy cost/year ($/yr) PW of energy cost/ N years ($) total PW cost ($) cheapest 12 42.64 170560 14.78 74.78 15.27195529 183263.4635 21.2819889 7 8344.6168 $144,295.3 9 $498,118.86 Flow=2.4 & $0.09/kw-hr: diameter (in) unit pipe cost ($/ft) pipe cost ($) friction loss (ft) [hazen williams] n=1.85 total lift (ft) [system curve] horsepower to water (HP) pump capital cost ($) HP to operate pumps (HP) energy cost/year ($/yr) PW of energy cost for N years ($) total PW cost ($) cheapest 14 54.56 218240 11.88 71.88 19.57255773 234870.6927 27.2750247 10694.47173 $184,929.1 6 $638,039.85 Diameter [in.] Cost ($/ft) 1 0.80 2 2.43 3 4.64 4 7.35 6 14.06 8 22.29

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10 31.85 12 42.64 14 54.56 16 67.56 18 81.57 20 96.55 22 112.45 24 129.25 8. The NY City tunnel problem is a classic WDS problem to expand the system from the system in place in the 1960’s (figure this page) to meet future demands. You can learn more about the current system and the actual expansions with a quick web search. The best design is show in the EPANET. I’ve modified the elevations here to give reasonable pressures at the surface. Tunnels are 100 or more feet below the surface and tunnels are drilled through the granite below Manhattan. A) What are the unknowns and how many of each type need to be solved for in the 1960 system (this page figure)? The unknowns are pressure and head for all 20 nodes and the 1 reservoir. B) What are the equations and how many of each type are solved in the 1960 system? ? 1 γ + 𝑣 1 2 2? + ? 1 = ? 2 γ + 𝑣 2 2 2? + ? 2 C) Set up and solve for the pressures using EPANET for the (1) existing and (2) expanded network (next page)? The expanded system includes pipes 22-27. You can set up two data files or set up for the expanded system then use the status option to close the new pipes. a.

Elevati on Base Demand ID ft CFS Junc 2 155 92.4 Junc 3 155 92.4 Junc 4 155 88.2 Junc 5 155 88.2 Junc 6 155 88.2 Junc 7 155 88.2 Junc 8 155 88.2 Junc 9 155 170 Junc 10 155 1 Junc 11 155 170

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Junc 12 155 117.1 Junc 13 155 117.1 Junc 14 155 92.4 Junc 15 155 92.4 Junc 16 160 170 Junc 17 172.8 57.5 Junc 18 155 117.1 Junc 19 155 117.1 Junc 20 155 170 Resvr 1 300 #N/A

Length Diameter Link ID ft in Pipe 1 11600 180 Pipe 2 19800 180 Pipe 3 7300 180 Pipe 4 8300 180 Pipe 5 8600 180 Pipe 6 19100 180 Pipe 7 9600 132 Pipe 8 12500 132 Pipe 9 9600 180 Pipe 10 11200 204 Pipe 11 14500 204 Pipe 12 12200 204 Pipe 13 24100 204 Pipe 14 21100 204 Pipe 15 15500 204 Pipe 16 26400 72 Pipe 17 31200 72 Pipe 18 24000 60 Pipe 19 14400 60

Pipe 20 38400 60 Pipe 21 26400 72 Pipe 22 15500 120 Pipe 23 26400 84 Pipe 24 31200 96 Pipe 25 24000 84 Pipe 26 14400 72 Pipe 27 26400 72

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9. For the system below compute: a. the pressure in psi at node 90 → 𝐻 90 = ? 90 γ + 𝑣 90 2 2? + ? 90 ? 10 γ = 𝐻 90 − 𝑣 90 2 2? − ? 90 ? 90 62.4 1728 = 267. 02(12)𝑖? − 0 − 50(12) 𝑖? ? 90 γ = 94. 04 ??𝑖 b. the flow rate in pipe 10 ? 10 = 𝑉𝐴 = 1. 84 * ( 1 4 π( 12 12 ) 2 ) ? ?−10 = 1. 45 ?? 3 /? c. the velocity in pipe 32 → ? = 𝑉𝐴 𝑉 32 = ? 32 𝐴 32 = 237.26 448.8 1 4 π( 8 12 ) 2 𝑉 ?−32 = 1. 51 ??/? d. the head at node 110 e. the demand at node 60 ? ?−14 + ? ?−30 − ? ?−18 − ? ?−16 = ? 60 ? 60 = 719. 04 + 248. 27 − 314. 48 − 197. 83 ? 60 = 455 ??? f. pipe roughness in pipe 32 → 𝐻 = 4.73?? 1.85 ? 1.85 ? 4.87 ? = ( 4.73?? 1.85 𝐻? 4.87 ) 1/1.85 = ( 4.73(600)( 237.26 448.8 ) 1.85 (2.51)( 8 12 ) 4.87 ) 1/1.85 ? ?−32 = 69

Elevation Demand Head Pressure Node ID ft GPM ft psi Junc 20 20 455 272.64 109.47 Junc 30 268.73 94.78 Junc 40 267.23 94.12 Junc 50 182 267.12 94.08 Junc 60 267.6 94.29 Junc 70 94.77 Junc 80 65 87.54 Junc 90 50 910 267.02 Junc 100 455 94.37 Junc 110 95.38 Junc 150 120 182 267.18 63.77 Resvr 10 10 -4368 10 0 Length Diameter Roughne ss Flow Velocity Unit Head Loss Link ID ft in GPM fps ft/Kft Pipe 2 1120 16 110 2144.2 3.42 3.49 Pipe 4 1250 12 110 947.4 2.69 3.12 Pipe 6 Pipe 8 950 12 110 -34.12 0.1 0.01 Pipe 10 600 12 110 1.84 1.55 Pipe 12 600 10 70 354.42 1.45 2.84 Pipe 14 12 110 719.04 2.04 1.87 Pipe 16 10 70 -197.83 0.81 0.96 Pipe 18 12 70 314.48 0.89 0.94 Pipe 20 10 70 -29.38 0.12 0.03 Pipe 22 10 -97.39 0.4 0.26 Pipe 24 0.94 1.28 Pipe 26 600 330.29 0.94 1.03 Pipe 28 70 50.89 0.32 0.23

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Pipe 30 110 -248.27 1.58 1.89 Pipe 32 600 8 237.26 2.51 Pipe 34 245.75 1.57 2.68 Pipe 36 55.26 0.35 0.17 Pipe 38 10 120 119.01 0.49 0.14 Pipe 48 600 8 110 -366.4 2.34 3.88 Pump 82 #N/A #N/A #N/A 4368 0 -262.64