A mixing tank initially contains 1800 lb of liquid water. The tank is fitted with two inlet pipes, one delivering hot water at a mass flow rate of 0.7 lb/s and the other delivering cold water at a mass flow rate of 1.2 lb/s. Water exits through a single exit pipe at a mass flow rate of 2.6 lb/s. Determine the amount of water, in lb, in the tank after 30 minutes.

The amount of water in the tank after 30 minutes is 540lb

Explanation:

In this model, the control volume encloses the water in the tank and has two inlets and one exit.

Also, the entering and exiting mass flow rates each remain constant.

Now, to solve the question,

Let's apply a mass rate balance from the initial state when the tank contains 1800 lb of water until the

final state after 30 minutes and solve for the mass of water in the tank at the final state. This is started by;

dmcv/dt = Σm (inlet) - Σm (exit)

Thus, dmcv/dt = (m1 + m2) - m3

Where Σm (inlet) is the sum of the mass flow rate of hot and cold water while, Σm (exit) is the sum of mass flow rate through the exit.

Now, integrating dmcv/dt, we get;

M (final) - M (initial) = [ (m1 + m2) - m3] Δt

M (final) is the final amount of water in tank after 30 minutes while M (initial) is the initial quantity of water in the tank.

So M (final) = M (initial) + [ (m1 + m2) - m3]Δt

Δt is in minutes from the question, so let's convert to seconds.

Thus 30 minutes = 30 x 60 = 1800 seconds

Plugging in the values given in the question;

M (final) = 1800 + [ (0.7 + 1.2) - 2.6]1800

M (final) = 1800 + [1.9 - 2.6]1800

M (final) = 1800 - (0.7 x 1800)

M (final) = 1800 - 1260 = 540lb