A 10-m diameter cylindrical tank has two holes in it through which water drains out. One is a 13-cm diameter, sharp-edged circular orifice whose center is at 1.35 above the bottom of the tank, and the second is a50-cm diameter sharp-edged circular orifice at 4.52m. If the water level is above the middle of the orifice, assume the flowrate of water through each orifice can be approximated as 0.61A2gh, where A is the cross-sectional area of the orifice, g is the acceleration due to gravity, h is the height (head) of the liquid height above the center of the orifice, and 0.61 is a dimensionless coefficient for flow out of a sharp-edged orifice. If the water level drops below the middle of the orifice, assume the flow rate of water through that hole is zero. Assume the tank is initially filled to h = 20.0 m. The overall goal is to plot the water height vs. time over 3 hours of draining the tank. Complete the following elements to solve this problem:
B. Write an equation for water height as a function of time, h(t). C. Differentiate your h(t) equation with respect to time, to get dh/dt as a function of dv/dt. D. Write an equation for the change in water volume of the tank, recognizing that dv/dt is a volumetric flow rate (equal to the total water flow out of the tank). Your equation may have several parts (that is, you may need a different form of the equation for different circumstances).
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