Question

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

Answer 1

In this question we have to find th flow rate

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 Apply Bernoulli's equation at position ① and ② Pro +21+ hp P₂7 22+ hf УРЯ 0 0 V + V2 + +22+ Pg ag - <I> hp= 22-2, the = 6.7+ht here has head loss = 0? (k+2) For pipe having Elo = 9.333 x104 and consider turbulent How the friction factor from moody chart is f= 0.02 ht= (0.08 0.02x40 0.03 + (0.50 +2+6-8+3x0.34 +1.050) 509) ů? 28A2 = 38.0366 v? = 38.0366 - 3880055, 206 Ú? from equation (1) hp - 6.7 + 38.0366 ? = $17+ 3880055,206 V2 ag cgiven) Havailable - Ho-av? 40 -0.053 ? hell Ý= How rate in liters per minute convert it into m3/s by dividing 2.777 X1010 3 Havailable = 40 - 190815265Ú? equalting eano and 6.7+3880055.206 V = 40 - 1908.15 265 j2 V = = 4.13565 x1024 m3/

1 pm = 1.6666 818 5 m/s j - 4.1356555 X164 116666 X10-5 = 24.8 2pm Let's check for friction factors value of flow rate and Henate again to get > 2 10-3 4.13565x104 0.5850 M/S (0.03) Re= POD 1000 x 0.5850 x0.03 M 17552.243 and € = 0.00281.m. from moody chaut are get H-0.028 so ht 02 48.7033 = 607 + 48.7033 U? = 6.7+ 4968148.9 2 2g equcate this to given equatim 6.7+ 4968148.9 V - 40 - 190815265 V2 V = 4.1241610-4 mals ag hp ů = 2407 2pm

Question 2

We can melt ceramic or ceramic material but need sufficient temp. To melt the ceramic material because ceramic materials have property to withstand at high temperature without any deformation also it is known as refractory material.Ceramics by virtue of their strong bonds require high temperatures for melting not only this the process require for melting have various complication.The melting range of ceramic material 800–1500 degree Celsius.