Question

Tank 1: Tank 2: Water h(t) h(t) Q1 (70 pts). Suppose experiments have been performed to determine the exit velocity (u) as a function of water height (h) for two different tanks. In Tank I, a smooth rounded nozzle with a diameter of a 2 cm is used whereas a square gate opening with a 2 cm side is utilized in Tank 2. The measured flow rates at different tank heights are given further below; the initial height for both tanks is 6 m. a) Fit the experimental values below to the equation to determine a and b for each valve. Find the exit velocity when the tank is half-empty b) Find u as a function of h for cach tank using the Bernoulli Equation (i.c. the Mechanical Energy Balance) together with the continuity equation, applied between the points of the surface and the opening of each tank. Assume pseudo steady-state; that is, h varies slowly enough so that you may use the steady-state expression. Make additional reasonable assumptions as necessary Evaluate the exit velocities when the tank becomes half-empty from the expression you obtained in part (b) c) d) Compare the result obtained from the empirical approach (in part a) to those from the Bernoulli Equation (parts b and c). Suggest reasons for the differences

Answer 1

I have considered second table for second tank ( title is written as tank 1). Find below pictures for solution-

I have considered second table for second tank (title is written as tank 1) . Find below pictures for solution- a) u = a . h^b From the experimental data for tank 1, at h = 6m average volumetric flow rate = 3.32 L/s = A times V DeltaD velocity = 3.32 times 10^3 cm^3/s/pi cm^2 {volumetric flow rate = Area times velocity} = 1056.79 cm/s similarly at h = 4m, Average volumetric flow rate = 2.94 L/s velocity = 2.94 times 10^3/pi = 935.83 cm /s So, 1056.79 = a (6) ^b - 1 935.83 = a(4) ^b - 2 dividing 1 & 2 we get 1.12 = 1.5^b taking logarith both sides we get log 1.129 = b log 1.5 So b = 0.3 and a = 0.62 m/s [Notice g have taken a in m/s in cm/s a = 617.28] u = 0.62 h^0.3 u rightarrow m/s similarly for tank 2 at h = 6, vol.flow rate = 2.84 L/s U = 2.84/4 = 0.71 m/s at h = 4, vol.flow rate = 2.46 L/s, U = 2.46/4 = 0.615 m/s 0.71 = a(6) ^b - 1 0.615 = a(4) ^b - 2