I How to Operate a Swamp Cooler in Phoenix in June

[rude man]

Guys, this is a real-life problem, especially now in Phoenix in June!
Wolfram Alpha gave me this:

Must be easier than that??

 

[bob012345]

Guys, this is a real-life problem, especially now in Phoenix in June!
Wolfram Alpha gave me this:

Must be easier than that??

Not clear what you are asking here?

 

[Chestermiller]

The problem statement makes sense to me. Is there more to it than that?

 

[bob012345]

http://www.perfect-home-hvac-design.com/swamp-coolers.html

 

[russ_watters]

...and while it says "evaporative cooler", the diagram and process description in the OP doesn't look anything like an evaporative cooler to me.

 

[rude man]

...and while it says "evaporative cooler", the diagram and process description in the OP doesn't look anything like an evaporative cooler to me.

I am looking for the expression for $\rho(t)$ so I can optimally adjust the effluent flow rate.

Basic swamp cooler operation:
Water runs over pads; evaporates; that cools the pads, then air is blown past the pads into the house. If the air is very dry a swamp cooler can produce air almost as cold as an air conditioner.

Details:
You have a tank of impure water of volume V(t) and impurity (solute) density $\rho(t)$. Water at flow rate dV1/dt, controlled by V, and impurity density $\rho_1$ flows into it. Tank water effluent flows out at $dV2/dt$ and $\rho$. Water at $dV3/dt$ and zero impurity density gets evaporated away out of the tank. A float valve controls $dV1/dt$ via tank volume V. Tank water continuously flows over the pads via a pump.

We have a servo controlling $\rho_{~final}$ and $V_{final}$.

The math as I see it is

$dV/dt = dV1/dt - dV2/dt - dV3/dt$ ... (1)
; water rises in tank.

$dV1/dt = k_1(V_r - V)$ ... (2)
; feedback action of an adjustable inlet valve. $k_1$ forms the servo loop gain.

$d/dt~ (\rho V) = \rho_1 ~dV1/dt - \rho ~dV2/dt$ ... (3)
; buid-up of solute in tank

V and $\rho$ are functions of time. V and $\rho$ adjust automatically until equilibrium is reached at $dV/dt = d\rho/dt = 0$.

dV2/dt, dV3/dt and $\rho_1$ are constants.
$V_r$ is a constant reference volume. If the tank is empty (V=0), dV1/dt flow is max. at $k_1V_r$.

The tank fills until dV/dt = 0 set by the servo action, at which point dV1/dt = dV2/dt + dV3/dt.

For that I need an expression for $\rho(t)$.

So just look at it as a math problem with what I've given you, or dispute any of it.. And refer to the attached jpeg.

 

[rude man]

Not clear what you are asking here?

cf. post 6.

 

[rude man]

The problem statement makes sense to me. Is there more to it than that?

Nope, just give me $\rho(t)$.
cf. post 6 and scrutinize the jpeg.

 

[rude man]

...and while it says "evaporative cooler", the diagram and process description in the OP doesn't look anything like an evaporative cooler to me.

Well I tried ...

 

[rude man]

I should add that deriving $\rho_{~final}$ does not require solving for the time function but I also wanted an idea of the time required to reach (close to) equilibrium. Besides, it looked like a good physics challenge.

 

[rude man]

http://www.perfect-home-hvac-design.com/swamp-coolers.html

This cooler would not work - for long. There is no provision for effluent so reservoir salinity would build up until the pads are solid salt! (Inlet water, depending on your area, always contains impurities, particujlarly salts).

 

[russ_watters]

Well I tried ...

I wasn't criticizing your art skills, I was saying you didn't describe an evaporative cooling process. But I think I see it now...

I am looking for the expression for $\rho(t)$ so I can optimally adjust the effluent flow rate.

Basic swamp cooler operation:
Water runs over pads; evaporates; that cools the pads, then air is blown past the pads into the house. If the air is very dry a swamp cooler can produce air almost as cold as an air conditioner.

Why $\rho(t)$ ? If you want a specific effluent impurity density, why not just do a single-point(in time)/steady-state mixing calculation to find the required flow rate for it? $V1/V2 = \rho2/\rho1$
Where $V1 - V3 = V2$

What are your knowns? Do you know or need to calculate the evaporation rate? I don't see how you can calculate the supply water rate without it.

Also, what is the point of the tank? For systems I've seen they just run the water over the pad and what's left goes straight to drain without a tank...or pump. If you are circulating the tank water over the pad and filling/emptying from the tank, then you have high impurity water going over the pad. This isn't critical to your question though...

 

[russ_watters]

I should add that deriving $\rho_{~final}$ does not require solving for the time function but I also wanted an idea of the time required to reach (close to) equilibrium. Besides, it looked like a good physics challenge.

Ah... ok, so it's a side question that you are asking about in addition to or instead of the main question...

For that you can use the differential equation for concentration in mixing/dilution vs time:
https://en.wikipedia.org/wiki/Dilution_(equation)#Dilution_ventilation_equation

As described it's for ventilation, but should work the same for your scenario.

 

[rude man]

Ah... ok, so it's a side question that you are asking about in addition to or instead of the main question...

For that you can use the differential equation for concentration in mixing/dilution vs time:
https://en.wikipedia.org/wiki/Dilution_(equation)#Dilution_ventilation_equation

As described it's for ventilation, but should work the same for your scenario.

My question was 'what is $\rho(t)$.'

The link I don't think applies to my situation: "The equation can only be applied when the purged volume of vapor or gas is replaced with "clean" air or gas."
In my case the impure solvent (water) is replaced by less impure solvent. (Of course I could possibly introduce distilled water into the tank but that would be expensive & have to be pumped. I use an RO setup for my influent.)
Thx anyway.

 

[rude man]

Why $\rho(t)$ ? If you want a specific effluent impurity density, why not just do a single-point(in time)/steady-state mixing calculation to find the required flow rate for it? $V1/V2 = \rho2/\rho1$
Where $V1 - V3 = V2$

I wanted the time to reach equilibrium, not just the final values.

What are your knowns? Do you know or need to calculate the evaporation rate? I don't see how you can calculate the supply water rate without it.

Knowns:
effluent rate dV2/dt
loop gain component $k_1$
influent purity $\rho_1$
Reference volume $V_r$
The fly in the ointment is admittedly the evap rate which I have to estimate. I do have empirical data of sorts.

I never claimed that tank impurity density is independent of evap rate. It's part of the physics prolem I was led to investigate, partly from curiosity.

Also, what is the point of the tank? For systems I've seen they just run the water over the pad and what's left goes straight to drain without a tank...or pump. If you are circulating the tank water over the pad and filling/emptying from the tank, then you have high impurity water going over the pad. This isn't critical to your question though...\quote

Well since you're interested: for the system you describe (1) the water is not cool as it is in a tank (summer water here runs around 90F!); (2) our city water would soon deposit salts on the pads to the extent that they have to be replaced quite frequently. Lots of fun climbing onto the roof when it's 115F+ ! Also, pad replacement is a bear to do.

Operating an R/O (reverse osmosis) system introduces almost pure water to the pads. My pads last for years and years in consequence.

(3) Your system is wasteful of water since the intake flow has to be set well above what is needed for very dry conditions (high evap rate), then as our "monsoon" season starts the humidity goes up, evaporation goes down, more water is wasted. To be fair, an R/O system also typically dumps waste water (to preserve the membrane). My water bill is not significantly impacted in summer though.

(4) Using a pump system allows use of a thermostat in the home just as in air conditioning. Most swamp coolers come with pumps. I use one in my garage, works great.

Etc.