slurry pump impeller breakdown

A pump motor is used to deliver water at a certain rate from a given

A pump motor is used to deliver water at a certain rate from a given pipe.

To obtain `n` times water from the same pipe same time, the factor by which the power of the motor should be increased is: a. n² b. n³ c. n4 d. n½

Arun Kumar

power = F.v v has to be increased by n oppossing force is proportional to v^2 totally n^3

ravi yadav

mass of water flowing volume density per second==(volume X density)/time =(area x lenght x density)/time =area x density x velocity =a z x v Therefore,Rate of increase of k.e=mv2/2xtime =(A x z x V^3)/2 Mass m, flowing out per second, can be increased to m1 by increasing to V to V1, then power increases from P to P1. therefore,p1/p=(A x z x V1^3)/(A x z x V^3) =(v1/v)^3 so, m1/m=(A x z x v1)/(a x z x v)=v1/v But given m1 = 2m, v1 = 2v therefore,p1/p=2^3=8 so,p1=8p..

Siddhant vikram awasthi

There is increase in the rate of water flowing through the pipe , sodmdt=Avρwhere A is area−crossection, v is the velocity through which water is flowing out of the pipe , ρ is the density of the materialdm`dt=ndmdtAv`ρ=nAvρ , so v` = nv(a) Force=dpdt=dmvdt=vdmdt=F`F=v`dm`dtvdmdt=n2 , F` = n2 F(b) Power = F.v =P`P =v`dm`dtv`vdmdtv= n2 .n=n3 ,P`=n3P

Kushagra Madhukar

Dear student, Please find the attached to your problem. We know, discharge, Q = Area(A) x Velocity(v) To increase Q by n times we need to either increase A or v or both. Since Area of cross section of pipe is same, velocity must be increased by n times. Now, the force by which water comes out and is exerted on pipe is, F = d(mv)/dt = v*dm/dt = v * ρAv = ρAv 2 Hence power delivered is, P = F.v = ρAv 2 * v = ρAv 3 Hence, P is directly proportional to cube of velocity. Thus on increasing v, n times, the power becomes, P’ = ρA(nv)3 = n 3 * P Hence the correct option is b) n 3 Hope it helps. Thanks and regards, Kushagra