Question

Consider a hemi-spherical tank with radius R = 16 see figure that is initially entirely filled with a fluid. At time t=0, the fluid begins to drain through an opening in the bottom of the tank see figure] until the tank is completely empty at t = tend- t= 0 te (0, tend) (a) At any time t, consider the maximum depth of fluid in the tank, h = h(t), and the corresponding radius of the surface of the fluid, r = r(t). For R = 16, use the tank geometry given in the figure to determine a relation between r and h. (b) Use your relation in (a) to isolate p2, and write p2 as a function of h. From physics, it is known that the rate of discharge of a fluid flowing through an opening at the bottom of a container is given by dV de = -k Av(t) where V(t) is the volume of fluid remaining in the container at time t, k is a constant that depends on fluid viscosity, shape of the opening, and the fact that the cross-sectional area of fluid pouring out of the opening is slightly less than the area of the opening itself, A, and v(t) is the velocity of the discharge of fluid through the opening. Geometrically, it can be determined that the volume of the fluid draining from the tank is related to the depth of the fluid h(t) and the radius at that depth, r(t) see figure, as At any t, the work done by gravity g in moving the maximum depth of fluid in the tank downward a distance h(t) through the opening at the bottom must equal the change in the kinetic energy. It follows that v(t) = 2gh(t). (c) Combine the given information and your result from (b) to obtain a separable, first-order ODE for h=h(t). (d) Suppose the fluid initially fills the entire tank. Set up the corresponding initial value problem (IVP). (e) Let g = 32, the area of the opening A = 1e, and k = 0.8. Determine the implicit solution for h = h(t) that satisfies the IVP derived in (d). (f) At what time tend , does the tank completely drain?

Answer 1

R-17 " / from the figure R² = (R-h)² +12 (b) 912=R?_(R-6) 2 = C2R-6) (h) 92 = 2Rh-h² since R=16 » 92= 32h-h² r2 = tfh +32)] (i) C 192 dh - dt. - KAU (A) => dan AVCHD uts given that v6+) = Naghct => eo -KA Jag het Tr2 -kalig dt Jh(+) 7912 12 h (82-h) we get. h2 (32-h) dh =-KA 2g dt

(d) if initially the tank wan filled then t= R at tao. (e) we have the differential equation V2(82-1)dh = -KA Ja det integrating both sides we get. 62 - - - - - - - - etish ( 2 ) = -KA Jog +c. now putting the initial value at t=0; h=r. → (= 2 R12 (32- $) = 2x(14)" (12 - 13 c = 2764 (7,966) C = 955,73 => ) 2bem os meses - Home Jeg + 955,73% how putting K=0,B, A = 7/16, 9= 32 we get. 23% 0,8 12 3 [ 984 +955.73 240V (921) = -0,44+ 95543 (fl at t= tend h=0. hence 0.44 = 955,73 → /* = 2389,825)
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