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Figure 1(1) Qin di = 0.075 m I lout Figure 1(IT): Qout 2 10 m Qout 1 -dz d= 0.075 m = = kin = 0.18 m3/s. N 0) A cylind

Posted: Mon May 16, 2022 6:28 am
by answerhappygod
Figure 1 1 Qin Di 0 075 M I Lout Figure 1 It Qout 2 10 M Qout 1 Dz D 0 075 M Kin 0 18 M3 S N 0 A Cylind 1
Figure 1 1 Qin Di 0 075 M I Lout Figure 1 It Qout 2 10 M Qout 1 Dz D 0 075 M Kin 0 18 M3 S N 0 A Cylind 1 (24.38 KiB) Viewed 53 times
Figure 1 1 Qin Di 0 075 M I Lout Figure 1 It Qout 2 10 M Qout 1 Dz D 0 075 M Kin 0 18 M3 S N 0 A Cylind 2
Figure 1 1 Qin Di 0 075 M I Lout Figure 1 It Qout 2 10 M Qout 1 Dz D 0 075 M Kin 0 18 M3 S N 0 A Cylind 2 (28.8 KiB) Viewed 53 times
Figure 1(1) Qin di = 0.075 m I lout Figure 1(IT): Qout 2 10 m Qout 1 -dz d= 0.075 m =
= kin = 0.18 m3/s. N 0) A cylindrical tank is open at the top, with water flow rate of Qin discharged into the tank. Water is flow out from the tank through circular outlet with diameter of dı = 0.075 m as shows in Figure 1(1). Design the tank with all necessary dimension, such that: Water will not overflow for given Qin, and Given Qin can fill half of the tank in 100 s, with maximum volume of water. = (ii) After the flow in (1) reached steady state (maximum possible height), the inflow is stopped (Qin = 0), and another opening is created at 10 m from the bottom of the tank, as shows in Figure 1(1) below. Water is let discharged from the tank through two outlets. Predict the minimum value of diameter da of the new outlet so that the water can be fully discharged from the tank in time T s. Note that water discharge from second outlet (Qout,2) will be 0 once the water level is below 10 m. Take T = 188 s,