The balance tank in our milk processing plant in our building holds about 10 gallons of milk. The inlet pipe is 1.5 inches in diameter abd the outlet is 3/4 inches diameter. If milk flows into the tank at .5 ft/sec how fast must it depart to keep from overflowing?

Use the mass continuity equation...

Area1*velocity1=Area2*velocity2

I will be happy to critique your work.

To find how fast the milk must depart the tank to prevent overflowing, we will apply the mass continuity equation, which states that the product of the area and velocity at one point is equal to the product of the area and velocity at another point in a system.

Let's first determine the area and velocity at the inlet and outlet pipes.

The area of a pipe can be calculated using the formula:
Area = π * (diameter/2)^2

For the inlet pipe:
Inlet diameter = 1.5 inches

Converting inches to feet:
Inlet diameter = 1.5 inches * (1/12) feet/inch = 0.125 feet

Inlet area = π * (0.125/2)^2 = 0.01227 ft^2

Given inlet velocity = 0.5 ft/sec, we have:
Inlet mass flow rate = Inlet area * Inlet velocity = 0.01227 ft^2 * 0.5 ft/sec = 0.006135 ft^3/sec

Now, let's calculate the area of the outlet pipe:
Outlet diameter = 3/4 inches

Converting inches to feet:
Outlet diameter = (3/4) inches * (1/12) feet/inch = 0.0625 feet

Outlet area = π * (0.0625/2)^2 = 0.003069 ft^2

To prevent overflowing, the outlet mass flow rate must be equal to the inlet mass flow rate:
Outlet area * Outlet velocity = Inlet area * Inlet velocity

Rearranging the equation to solve for Outlet velocity:
Outlet velocity = (Inlet area * Inlet velocity) / Outlet area

Outlet velocity = (0.006135 ft^3/sec) / (0.003069 ft^2)
Outlet velocity ≈ 2 ft/sec

Therefore, the milk must depart the tank at a velocity of approximately 2 ft/sec to prevent overflowing.