pump. The liquid level of the reservoir keeps constant, as shown in the figure. The diameter of the pipeline is (762.5mm. In the operating condition, the reading of the vacuum gauge in the pump entrance is 185mmHg The energy loss of the suction tube (The entrance of the pipe is not included )and the discharge tube can be calculated according to formulachf, 1=2u2 and Ehf,2=10u?, where u is water velocity in the suction and discharged tubes. The gauge pressure of the joint of the discharged tube and the spray nozzle is lkgf/cm2. Try to calculate the effective power of the pump Solution: Suppose liquid level of storing tank as section 1---l, the connection of vacuum meter as section 2---2. And suppose section 1---I as basic level. Then the Bernoulli equation Z 82+Y3 +2+∑h in the equation Z=0 Z2 1.5m PI ∑h/:=2 2s、l85 ×10133×105=-247×104N/ From the above, we can get the velocity of water in the pipe 2.47×10 9.81×1.5)2.5=2m/s 1000 w,=uAp=2××0.0712×1000=7.92kg/s 2) Suppose the liquid level of storing tank as upper reaches section 1--l; suppose the connection point of drain pipe and blow head as lower reaches section 3---3; Suppose section I---1 as basic level Then we get the equation pI +w=gZ +∑b+∑ in the equationpump. The liquid level of the reservoir keeps constant, as shown in the figure. The diameter of the pipeline is Φ762.5mm.In the operating condition ,the reading of the vacuum gauge in the pump entrance is 185mmHg.The energy loss of the suction tube (The entrance of the pipe is not included)and the discharge tube can be calculated according to formulaΣhf，1=2u² and Σhf‚2=10u²,where u is water velocity in the suction and discharged tubes. The gauge pressure of the joint of the discharged tube and the spray nozzle is 1kgf/cm2 . Try to calculate the effective power of the pump. Solution: Suppose liquid level of storing tank as section 1---1, the connection of vacuum meter as section 2---2. And suppose section 1---1 as basic level. Then the Bernoulli equation is + + = + + + ,1 2 2 2 2 1 2 1 1 2 2 hf u p gZ u p gZ   in the equation Z1 = 0 Z2 =1.5m 0 2 0 2 2 ,1 1  = =  u h u p f 5 4 2 2 1.0133 10 2.47 10 / 760 185 p = −   = −  N m From the above，we can get the velocity of water in the pipe u 9.81 1.5)2.5 2m/s 1000 2.47 10 ( 4 −  =  = the water mass flow rate w uA k g s s 0.071 1000 7.92 / 4 2 2 = =    =   2）pump’s effective power Suppose the liquid level of storing tank as upper reaches section 1---1; suppose the connection point of drain pipe and blow head as lower reaches section 3---3; Suppose section 1---1 as basic level. Then we get the equation + + + = + + + ,1 + ,2 2 2 2 2 1 2 1 1 2 2 e hf hf u p W gZ u p gZ   in the equation