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The bathroom plumbing of a building consists of 1.5-cm-diameter copper pipes with threaded connectors, as shown in Fig. 8-49. (a) If the gage pres-
The bathroom plumbing of a building consists of 1.5-cm-diameter copper pipes with threaded connectors, as shown in Fig. 8-49. (a) If the gage pres- sure at the inlet of the system is 200 kPa during a shower and the toilet reservoir is full (no flow in that branch), determine the flow rate of water through the shower head. (b) Determine the effect of flushing of the toilet on the flow rate through the shower head. Take the loss coefficients of the shower head and the reservoir to be 12 and 14, respectively. Shower head K = 12 Cold water Toilet reservoir with float KL=14 1 m K = 10 K = 2 5 m K = 0.9 4 m 2 m H Globe valve, fully open K = 10 Properties The properties of water at 20C are p = 998 kg/m, - 1.002 x 10-3 kg/m s, and v = lp = 1.004 x 10-6 m/s. The roughness of copper pipes is a = 1.5 x 10-6 m. . Analysis This is a problem of the second lype since il involves the delermi- nation of the flow rate for a specified pipe diameter and pressure drop. The solution involves an iterative approach since the flow rate (and thus the flow velocity) is not known. (a) The piping system of the shower alone involves 11 m of piping, a tee with line flow (K-0.9), two standard elbows (K-0.9 cach), a fully open globe valve (K = 10), and a shower head (K = 12). Therefore, K = 0.9 + 2 x 0.910 + 12 = 24.7. Noting that the shower head is open to the atmosphere, and the velocity heads are negligible, the energy equation for a control volume between points 1 and 2 simplifies to P P pg +a +2+hpump, u = 2g pg +2 2g +22+ h turbine, e + h Pl.gage pg = (22-21)+hL = Therefore, the head loss is h = Also, 200,000 N/m (998 kg/m)(9.81 m/s) - 2 m = 18.4 m + -> 18.4 = 11 m 0.015 m +24.7 V2 2(9.81 m/s) 28 h = ( +
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