14-34 A water pump is used to pump water from one large reservoir to another large…
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Question “14-34 A water pump is used to pump water from one large reservoir to another large…”
14-34 A water pump is used to pump water from one large reservoir to another large reservoir that is at a higher elevation. The free surfaces of both reservoirs are exposed to atmo- spheric pressure, as sketched in Fig. P14-34. The dimen sions and minor loss coefficients are provided in the figure. The pump’s performance is approximated by the expres- sion Havailable H av?, where the shutoff head H 40 m of water column, coefficient a = 0.053 m/Lpm?, available pump hcad H swallable is in units of meter of water column, and capacity Vis in units of liters per minute (Lpm). Estimate the capacity delivered by the pump. Answer: 24.7 Lpm entrance KL. KL. valve Kelb Kit 23-3 -6.7 m (elevation difference) D = 3.0 cm (pipe diameter) 0.50 (pipe entrance) = 2.0 (valve 1) KL valve 2 = 6.8 (valve 2) = 0.34 (cach elbow-there are 3 -1.05 pipe exit = 40 m (total pipe length < =0.0028 em pipe roughness) V20 Reservos 2 Reservoir Valve 2 חוח וויז Vouvet FIGURE P14 34
2. In contrast to polymers and metals, ceramics are not melted for shaping. Explain why (3 marks) 1.
Answer
This question will help us to determine the flow rate
Question 2
You will need to heat ceramic or ceramic material. It is necessary to heat the ceramic material.
Given – zz-2,=6.7m. D = 3.0c.m. – L=40m E = 0.0028 cm, Kuentrance = 0.50, XL vahet Kz valve 2 = 6-8 ke exit = Keelbow = 0.34 (3 nos.) 1.050 Use Bernoulli’s equation at position 1 & 2 Pro +21+ hp 22+ hf URIa 0 0 Kuentrance = 0.50, XL vahet Kz valve 2 = 6-8 ke exit = Keelbow = 0.34 (3 nos.) (k+2) Pipe with Elo = 9.333×104. Consider turbulent F= 0.02 ht= (0.08 0.02×40 0.03+ (0.50 +2+6-8+3×0.34+1.050) 509) 28A2 = 38.0366 v? = 38.0366 – 3880055, 206 U? From equation (1) hp = 6.7 + 38.0366 = $17+ 3880055,206 V2 ag cgiven) Havailable – Ho-av? 40 -0.053 ? hell Y= At what rate in liters per hour convert it to m3/s by multiplying 2.777 X1010 3. Havailable = 40 – 190815265U equaling eano, 6.7+3880055.206V = 40 – 19.15 265 J2 V = = = 4.13565x1024m3/
1 pm = 1.6666818 5 m/s.j – 4.1356555X164 116666 6 X10-5 = 24.08 2pm. Let’s now check for friction factors value and Henate again. To get > 2 10-3 (4.13565×104) 0.5850 M/S (0.033) Re= POD 1000×0.5850×0.03 M 17552.243 x0.03 EUR = 0.00281.m. H-0.028 is obtained from moody Chaut. = 6.7+4968148.9 2g equcate the to given equatim H-6.7+4968148.9 V-40 – 190815265V2 V = 4.1241610-4 Mals ag HP u = 2407 2pm
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