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www.webassign.net/web/Student/Assignment-Responses/last?dep- 16328398801 -/1 points 11 1.3.014. My Notes Ask Your The right-circular conical tank shown in the figure below loses water out of a ciroular hole of area An at its bottom. When water leaks through a hole, friction and contraction of the stream near the hole reduce the volume of water leaving the tank per second to CAO 2S , where e o·ce i) is an empnal constant Oetemne 0.6. (Assume a differential equation for the height of the water h at time t o. The radius of the hole is 3 in..g -32 ftu/s?, and the friction/contraction factor is c the removed apex of the cone is of negligible height and volume.) dh dt ft/s 8 ft 20 ft circular hole Sabmit AnswnBave Progress Practice Another VersionExplanation / Answer
SOLUTION:
In this case, R/h is always equal to (8ft/20ft)= 2/5,
so R(h) = 2h/5 and:
V(h) = (?/3)h(2h/5)^2 = (4?/75)*h^3
Now, differentiate this equation to get:
dV(h)/dt = (4?/15)*(h^2)*dh/dt
The volumetric rate at which the fluid in the tank drains from the tank is given by the speed of the flow
through the hole multiplied by the area of the hole.The speed of the flow is given by Torricelli's Law,
which states that:
s = sqrt(2*g*h),
where s is the speed of the fflow,
h is the height of the fluid above a hole in a tank, and
g is the acceleration due to gravity.
In reality the flow rate is lower than this due to friction between the fluid and the sides of the hole,
so this law is often written with a constant that acts as a "fudge factor" to account for the effect of the geometry of the hole, the surface tension between the fluid and the material of the tank, etc.:
s = b*sqrt(2*g*h). In this case, we will assume b = 1 (the ideal case).
The area of the hole is ?*r^2, where r= 2in is the radius of the hole, so the flux of water out of the tank is given by:
f = (?*4in^2)*sqrt(2*g*h)
But the flux out of the tank is simply the rate at which the volume of water in the tank decreases, so:
dV(h)/dt = (4?/15)*(h^2)*dh/dt = -(?*4in^2)*sqrt(2*g*h)
dh/dt = -(15in^2)*sqrt(2g)*h^(-3/2)
This is the differential equation describing how the height of the water changes with time. It is a separable equation:
h^(3/2) dh = -(15in^2)*sqrt(2g) dt
(2/5)*[h^(5/2) - ho^(5/2)] = -(15in^2)*sqrt(2g)*t
where ho is the height of the water at t = 0.
h^(5/2) = ho^(5/2) - (5/2)*(15in^2)*sqrt(2g)*t
h(t) = [ho^(5/2) - ((75/2)in^2)*sqrt(2g)*t]^(2/5)
This gives the height of the water as a function of time.In this case, ho = 20 ft = 240 in.
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