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THERMO Spoken Here! ~ J. Pohl © ~ 2017 | www.ThermoSpokenHere.com (B7200-109) |

Each tank is square in cross-section and has a valve for filling with water (or venting of air) at its top. The tanks are connected at their bottoms by a pipe with a valve.

**Two cases are to be studied.** The sketch shows the setup for both: all valves are closed, the depth of water in tank **A** is 5 meters and the water depth in tank **B** is zero. Events initiate when valves are opened. For each event, calculate the final depth of water in each tank. Since the events take a long time, temperature is constant.

**Case I:** If all valves are opened, what will be the final depths of water in the tanks?

♦ We take the system to be water "above" the zero level of both tanks. Since the density of water is constant, the mass of water initially above **H = 0** will equal the mass above **H = 0** finally.

**
Δm** = 0 or **m**_{water,2} - **m**_{water,1} = 0

The mass equation is expanded to account for the two tanks:

[**m**_{A,2} + **m**_{B,2}] - [**m**_{A,1} + **m**_{B,1}] = 0

We expect the final depths of water in the tanks to be equal at **H**_{2}. So we write each mass as:
**ρ**_{water}**A**_{tank}**H**. The tank areas are known:

**Area A** = 9m^{2}, **Area B**
= 4m^{2} and Also **H**_{A,1} = 5m.

Place these facts into the mass equation:

[**ρ**_{water}(9m^{2}) **H**_{2} +
**ρ**_{water} (4m^{2}) **H**_{2}] -
[**ρ**_{water} (9m^{2})(5m) + 0]= 0.

The above is a mass equation and each of its terms has the dimension mass, [m]. However the equation
has been reduced to the point of containing a single unknown - the second height we seek. Thus now is the
time to manipulate the equation algebraically for a solution. Reduce by division by ρ_{water}:

9m^{2}**H**_{2} + 4m^{2}**H**_{2} -
9m^{2}(5m) = 0

Which gives us:

**H**_{2} = (45/13) meters.

**Case II:**Suppose from the initial condition, only valves **2)** and **3)** are opened,
what then will be the second depth?

Valve **1)** is left closed and air is trapped in tank **B**. This will influence the final depths.
We will study two systems: i) all of the water and ii) the air in tank B. For the water, by reasoning as above
but with the second depths, we have:

9m^{2}**H**_{A,2} +
4m^{2}**H**_{B,2} -
9m^{2}(5m) + [**m**_{air,2} - **m**_{air,1}] = 0

The mass equation for the air is:

(1) 1 |

The air is compressed, apply the Ideal Gas Equation to it:

(2) 2 |

The original depth of air in **Tank B** from the top was 6 meters. As water flows into tank B,
the depth decreases to **L**_{2} which equals 6m - **H**_{B,2}. The
second time is a long time after the valves were opened. Consequently
**T**_{air,2} = **T**_{air,1}. These facts are applied
to obtain a final equation for the air. Just below that equation is written the equation for the water:

[**p**_{air,2}(6m - **H**_{B,2}) - **p**_{atm}(6m)] = 0

[9m^{2}**H**_{A,2} +
4m^{2}**H**_{B,2} -
9m^{2}(5m)] = 0

These are two equations and we see there are three unknowns: **p**_{air,2},
**H**_{A,2} and **H**_{B,2}. The added equation we need is the hydrostatic
equation written from the water surface in **Tank A** through water to the water surface in **Tank B**.
We expect **H**_{A,2} will be greater than **H**_{B,2}. With that,
the equation becomes:

**p**_{atm} + **ρ**_{water}g_{o}(**H**_{A,2} -
**H**_{B,2}) = **p**_{air,2}

The event is physical, it has an answer and only one answer. The three equations above are based upon sound principles; by a double-check, they are correct. There being three unknown quantities, three independent equations are sufficient for solution. Solve the set and you'll get the answer.