- The Valve Position In Part B Is Changed Reducing The Mass Flow Rate To 25 Kg S And The Power Output Changes To 3 7 Kw S 1 (84.29 KiB) Viewed 38 times
WITH CORRECT ANSWERS IS REQUIRED. IF DONE SO WELL, DEFINATELY A
THUMBS UP AND PERFECT FEEDBACK WILL BE GIVEN.
The valve position in part B is changed, reducing the mass flow rate to 25 kg/s and the power output changes to 3.7 kW's. Determine, the new operating conditions (velocity, friction factor, hydraulic head, etc.). Then determine, to the nearest quarter, the new valve position of the gate valve. The following equations are given, as well as table 1:1 for the equivalent lengths of pipe and a moody diagram overleaf with the friction factor equation. Thus head loss for turbulent flow is: Exit/Entry losses are: foLU2 kv2 Ah 2gD hexit = 2g Hydraulic output: Reynold's number: pvd Re = и Powerout = pg(Eh)Q Table 1:1 Equivalent lengths of pipe for selected fittings 100mm diameter Gate valve fully open 0.8 m Gate valve 1/4 closed 4.7 m Gate valve half closed 22 m Gate valve 3/4 closed 92 m T-piece - flow through run 2.1 m T-piece-flow through branch 8.5 m long sweeping elbow 2.1 m Mitre bend 5.2 m
0.1 0.09 0.08 Wholly turbulent flow 1.325 [In{(8/3.7D) + (5.74/ Re" 9)]]> 0.07 0.05 0.04 0.03 0.06 0.05 0.02 0.015 0.04 0.01 0.008 0.006 0.004 f 0.03 E D 0.025 0.002 0.02 Laminar flow 0.015 0.001 0.0008 0.0006 0.0004 0.0002 0.0001 0.00005 Smooth Transition range Laminar flow, f = 64/Reo 0.01 0.009 0.008 103 2(103) 4 6 8 2(104) 4 681 2(105) 4 68 2(106) 4 681 21107 4 6 8 0.00001 104 105 106 107 108 Rep = p.V.D/u