InterviewSolution
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InterviewSolution
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A side wall of a wide open tank is provided with a narrowing tube (figure) through which water flows out. The cross-sectional area of the tube decreases from S=3.0cm^2 to s=1.0cm^2. The water level in the tank is h=4.6m higher than that in the tube. Neglecting the viscosity of the water, find the horizontal component of the force tending to pull the tube out of the tank. |
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Answer» SOLUTION :Suppose the radius at A is R and it decreases uniformly to r at B where `S=piR^2` and `s=piR^2`. Assume also that the semi vectical angle at 0 is `alpha`. Then `(R)/(L_2)=(r)/(L_1)=y/x` So `y=r+(R-r)/(L_2-L_1)(x-L_1)` where y is the radius at the point P distant x from the vertex O. Suppose the velocity with which the liquid flows out is `V` at A, v at B and u at P. Then by the equation of continuity `piR^2V=pir^2v=piy^2u` The velocity v of efflux is given by `v=sqrt(2gh)` and BERNOULLI's theorem gives `p_p+1/2rhou^2=p_0+1/2rhov^2` where `p_p` is the pressure at P and `p_0` is the atmospheric pressure which is the pressure just outside of B. The force on the nozzle tending to pull it out is then `F=int(p_(rho)-p_0)sintheta2piyds` We have subtracted `p_0` which is the force due to atmospheric pressure the factor `sintheta` gives horizontal component of the force and `ds` is the length of the element of nozzle SURFACE, `ds=dxsectheta` and `tantheta=(R-r)/(L_2-L_1)` Then `F=underset(L_1)overset(L_2)int 1/2(v^2-u^2)rho2piy(R-r)/(L_2-L_1)dx` `=pirhounderset(r)overset(R)intv^2(1-(r^4)/(y^4))ydy` `=pirhov^2 1/2(R^2-r^2+(r^4)/(R^2)-r^2)=rhogh((pi(R^2-r^2)^2)/(R^2))` `=rhogh(S-s)^2//S=6*02N` on putting the values. Note: If we try to calculate F from the momentum change of the liquid flowing out will be wrong even as REGARDS the sign of the force. There is of course the effect of pressure at S and s but quantitative derivation of F from Newton's law is DIFFICULT. 
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