Z1=022=14m P1=0p2=9.807×104N/m ∑h+∑ha2=2n2+10 from the data above, we get W=9,81×14+807×10 +12.5×22=2854J/k 1000 he pump's effective power is N=W,=2854×7.92=2260W=226W 14. As shown in the figure. the inside diameter d of the reservoir is 2m and its bottom is connected with a steel pipe whose inside diameter do is 32mm. There is no liquid supply to the reservoir, and the heigh of the liquid level h2 is 2m. The total energy loss of the liquid flowing through the pipe can be calculated by the formula zhf-20u, where u is the velocity of liquid (in m/s). Try to calculate the required time when the liquid level of the reservoir drops Im? Solution: The problem attributes to the instability flow, the time when the surface of the reservoir descend Im can be calculated from material balance Suppose F is instant feed into system, D'is instant discharge out of system, dA is the accumulated quantity in the de time, then in the de time, the material balance equation is Fde-D'de=dA And then suppose in the de time, the surface of reservoir descend dh, the instant flow rate of liquid in the tube is uo In the equation F’=0 dA'=-D-dh Then the equation reduces to D、2ah In order to work out the relation between the height of instant surface h take the cente of drainpipe as benchmark and the instant velocity u in equation (a), we can use the quation of Bernoulli0 0 0 1 1 1 =  = p u Z 4 2 2 2 2 9.807 10 / 2 / 14 p N m u m s Z m =  = = 2 2 2 hf ,1 +hf ,2 = 2u +10u =12u from the data above, we get We 12.5 2 285.4J / k g 1000 9.807 10 9.81 14 2 4 +  =  =  + the pump’s effective power is Ne = Wews = 285.47.92 = 2260W = 2.26kW 14. As shown in the figure, the inside diameter D of the reservoir is 2m and its bottom is connected with a steel pipe whose inside diameter do is 32mm.There is no liquid supply to the reservoir, and the heigh of the liquid level h2 is 2m.The total energy loss of the liquid flowing through the pipe can be calculated by the formula Σhf=20u²,where u is the velocity of liquid (in m/s).Try to calculate the required time when the liquid level of the reservoir drops 1m? Solution：The problem attributes to the instability flow，the time when the surface of the reservoir descend 1m can be calculated from material balance. Suppose F’ is instant feed into system，D’ is instant discharge out of system，dA’ is the accumulated quantity in the dθ time，then in the dθ time, the material balance equation is F’dθ―D’dθ=dA’ And then suppose in the dθ time，the surface of reservoir descend dh，the instant flow rate of liquid in the tube is u。 In the equation F’=0 D’= d u 2 0 4  dA D dh 2 4   = Then the equation reduces to d ud D dh 2 0 4 4    − = or u dh d D d 2 0  = −（ ） In order to work out the relation between the height of instant surface h（take the center line of drainpipe as benchmark）and the instant velocity u in equation（a），we can use the instant equation of Bernoulli