# Determine conditions of water tank and pump

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1. Nov 2, 2015

### Carbaro

1. The problem statement, all variables and given/know
In rural communities where landowners have their own water wells, large tanks are used to store water, which is pressurized by a sealed “air cushion”. Water is pumped intermittently from the well to restore the tank supply when the level is low. This problem considers the work associated with one charging event. The water storage tank is 0.618 [m] in diameter and 1.667 [m] tall. At the beginning of a charging event, the tank contains 140 litres of water with the remaining space filled with air; the tank contents (air and water) are at 20 [oC] and 200 [kPa]. At the end of a charging event, the tank contains 340 litres of water pressurized by the air cushion. During filling, the compression of air is slow enough that the pressurization process can be considered isothermal. Water is pumped into the tank using a submerged pump in a drilled well that has a water level 60 [m] below the water surface of the tank; inefficiencies in the pump can be ignored. Determine the following:

1. The total volume of the tank, and the volumes occupied by water and air at the beginning of the charging event;

2. (ii) The mass of air in the tank, and the specific volumes of air at the beginning and end of the charging event;

3. (iii) The pressure at the end of the charging event;

4. (iv) The work required to compress the air cushion, and the associated heat transfer (hint:

treat the tank as a closed piston-cylinder device where the rising water level is analogous

to a moving piston);

5. (v) The pump work required in the charging event (hint: find the work required to raise the

water charge 60m in the absence of an air cushion, and then add the additional work of

compressing the air cushion).

6. (vi) If the submerged pump was 0.75 hp (560 W), how long would the charging event
require?

2. Relevant equations

First Law of Open Systems: dE = Q - W +Mass in(h + 1/2v2 + gz) - Mass out(h + 1/2v2 +gz)

3. The attempt at a solution

Couldn't get past the first part, really not understanding where do start and how to solve the problem

I know that the total volume of the tank is the volume of the water + the volume of the air

I know that once i have the correct mass, i can use the volume of the air to find specific volume

Mainly I know kinda what route I need to be taking, but im not sure how to construct those routes

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2. Nov 2, 2015

### SteamKing

Staff Emeritus
You should be able to calculate the volume of the tank given its dimensions from the problem statement. Based on that volume, you know that initially there are 140 liters of water in the tank with the rest of the volume containing air at 20° C and a pressure of 200 kPa.

After the tank is filled, it now contains 340 liters of water, with the rest of the volume of the tank occupied by the original mass of air in the tank before pumping started.

Can you calculate what the new pressure is in the tank, after pumping is completed?

3. Nov 2, 2015

### Carbaro

Okay yea I missed the dimensions, so the volume is the volume of a cylinder using diameter and height, thanks!

With the 340 liters of water and the mass, I can find specific volume and then since its isothermal I have 20 celsius. I can then use those two to look up the pressure in the steam tables? And I can find the state of the air by using 20celsius and pressure of 200kPa from the initial condition?

4. Nov 2, 2015

### SteamKing

Staff Emeritus
I don't think you need steam tables for this problem. Most of the work comes from pumping the water up the well and compressing the air cushion in the tank.

The initial condition of the air cushion in the tank is 20° C at a pressure of 200 kPa. Knowing that there are 140 liters of water in the tank at this time will allow you to calculate the mass of air in the cushion. After pumping has stopped, there are now 340 liters of water in the tank, but with the same mass of air in the cushion at 20° C. What's the pressure of the air in this smaller volume inside the tank?

5. Nov 2, 2015

### Carbaro

So, would you use Pv=RT, where you have temperature, where you know P, V, R, T so you can just use that to find m where m = V/v. V=volume, v = specific volume?
Now, you know the new V, you have T, you have mass, you have R, so you can solve for P? P = RT/v = RT/(V/m)

Then the work required to compress the air is the work required to compress from 200kPa at initial volume to the pressure that we just found and 340 litres of water

And since it is an isothermal process, Q=W, with W = integral of( PdV)? But since P is not constant, you treat it as a const throughout the integration so how do you get work?

By the way thanks! The question is making a lot more sense now, it really helped to break it up

6. Nov 2, 2015

### SteamKing

Staff Emeritus
How did you come to that conclusion, to treat P = const, even though it clearly is not constant?

The process of filling the tank occurs slowly enough that you can assume that temperature remains constant in the air cushion, even though pressure is increasing.
This article on isothermal processes shows how to find the integral pf P dV:

https://en.wikipedia.org/wiki/Isothermal_process

7. Nov 2, 2015

### Carbaro

I don't think I said P = const, I just didn't know how to analyze the work now after finding the final pressure but the link helped, I remember the process now.
I said that because its isothermal, you can say Q = W because all heat transfer is in the form of work since there is no change in temperature (no change in internal energy)

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