P1≈0(表压) u,=V/A= =162m/s 3600×-×0033 P2=0.×9.807×10=9807N/m2 ∑h=30J/kg therefore we can get that 6229807 2a5o+30)/981=-437m the liquid level of header tanker should be 4. 37m higher than the orifice for raw stuff. 11. The liquid level of the elevated tank is &m higher than the floor. The water flows out of the pipeline((108x4mm) The exit of the pipeline is 2m higher than the floor. In the given condition, the energy loss of the water flowing through system (the energy loss of the exit is not included )can be calculated by 2hf=6.5m2, where u is water velocity (in m/s).Try to calculate 1)The velocity of the water at the"A--A "cross section; 2)The flow rate of water(in m/h) Solution: 1)Suppose the liquid level of header tanker as the upper reaches, and the inner side of the pipe's exit as the lower reaches. Suppose the ground as basic level. Then we can get the quation g g ∑b in the equation Z1=8m Z2 0 P1=P2 ∑h=6.52=65h2 based on the above, we can getu V A m s p u s 1.62 / 0.033 4 3600 5 / 0( 0 2 2 1 1 =   = =    表压） 4 2 2 p = 0.19.80710 = 9807N / m hf = 30J / kg therefore we can get that Z 30)/ 9.81 4.37m 850 9807 2 1.62 ( 2 2 = − + + = − the liquid level of header tanker should be 4.37m higher than the orifice for raw stuff. 11. The liquid level of the elevated tank is 8m higher than the floor. The water flows out of the pipeline(Φ1084mm).The exit of the pipeline is 2m higher than the floor. In the given condition, the energy loss of the water flowing through system (the energy loss of the exit is not included )can be calculated byΣhf=6.5m²,where u is water velocity(in m/s).Try to calculate : 1) The velocity of the water at the “A--A” cross section; 2) The flow rate of water(in m²/h) Solution: 1) Suppose the liquid level of header tanker as the upper reaches, and the inner side of the pipe’s exit as the lower reaches. Suppose the ground as basic level. Then we can get the equation + + = + + +hf u p gZ u p gZ   2 2 2 2 1 2 1 1 2 2 in the equation Z1 = 8m Z2 = 2m 2 2 2 1 2 1 6.5 6.5 0 h u u p p u f = = =   based on the above，we can get u 9.81 6 / 7 2.9m /s 2 =  =