Time to Reach equilibrium temperature

[spokora]

if anyone could provide guidance, it would be appreciated.
here's the problem:

a hot gas of known temperature is exhausting at a known rate into a room. the room is being ventilated at a known rate. i need to estimate the time it takes for the room to reach an equilibrium temperature 20 deg higher than the original temperature. any basic formula/method for estimating this?

 

[MintJulep]

Equilibrium may or may not (more likely not) be 20 degrees above the original temperature.

Basic method would be to perform an energy balance.

 

[magicme]

if you can assume that the air exiting the room is all at the initial room temperature, this would be a basic heat balance solution as MintJulep says.

but if the gas mixes and the exiting gas is at some partial mixture temperature, that's one heck of a tough problem.... maybe impossible to calculate.

you have to make some assumptions about the temperature of the exiting gas as a function of time. its not reasonable to say that the exiting gas is always the gas that was initially in the room.

not much help, i know.

magicme



------------------------------------
"not all that glitters is gold"

 

[MintJulep]

Easier to make some assumptions about the composition of the exiting air.

The exhausting air would be made up of:

The hot gas
The ventilating air
Air that was already in the room

A few initial combinations should suffice to bound the problem and get an idea of the solution's sensitivity to the assumptions.

 

[spokora]

i hear you magicme...i realize it is near impossible to figure out the problem when you factor in mixtures, but i didn't know if there was some sort of idealized diffEQ for these types of problems... it is not really my background

 

[prex]

If by ventilating you mean air extraction only (no fresh air inlet), then I'm afraid the equilibrium temperature will be that of the hot gas, unless you account for convective heat losses through room walls.
That said, the time to equilibrium will be be anyway influenced by the heat capacity of anything that's in the room (machinery, presumably 50% of wall thickness or 100% if they are externally insulated).
As you see you should better specify your goals and your constraints before going to any equation set up.

prex

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[25362]

Google around for sweep-through purging and you'll probably learn of the problems involved.

 

[BronYrAur]

As spokora suggests, this reminds me of a differential equation, and a fairly simple linear one as I recall. That's been 17 yrs ago, so I will have to pull out the book. I'm sure that you will have to assume a constant mixing of the room (which may not be reality) and you will probably have to assume a constant density air for both temps (which also may not be aceptable). That way, is should be a simple energy balance with known mass flows. You probably only know a volumetric flow (CFM,etc.), so we will have to use density to get mass. I'll try to look it up.

 

[MortenA]

A really rough assumption would off course assume perfect blending and the you could do a time step "simulation" in excell?

Best regards

Morten

 

[BronYrAur]

I have been trying to work out the equation. Please explain the question further. When you say, "a hot gas of known temperature is exhausting at a known rate into a room." I assume you mean that is hot air is entering the room. The word exhaust is throwing.

Then when you say the the room is ventilated, do you mean that you are introducing a cold air stream into the room and exhausting that air back out?

When I read this last week, I thought you were just dumping hot air into the room and exhausting out a mixture of that hot air and the room air. Now I'm not so sure what you are asking???

Also, you need to know the volume of the room and if the air flows are different, you will have a pressure gradient that will cause in(ex)filtration. I am going to assume that the air flows are constant, but please provide more detail.

 

[vpl]

BronYrAur

It could be a situation where a piece of equipment is giving off an exhaust gas into the room (hopefully nothing dangerous) and there is also an exhaust fan pulling the air from the room (giving rise to a ventilation rate.) Or maybe it's just the heat from a large piece of equipment.

I ran across a situation not too long ago where the ventilation setup was intake air dampers in small room with large diesel engine (combustion exhaust air vented to outside). Small room connected to another room via some wall openings. Second room had exhaust fans of (supposedly) known ventilation flow rate. So from the view of the larger room, there would be hot air "exhausting" into the larger room. However, in that case, what was being questioned was the equilibrium value, not the time it took to get to equilibrium.

Don't know if this helps your calculation any...

Patricia Lougheed

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[magicme]

how about this....

mdot is the mass flow rate thru the room (kg / s)
MASS is the constant mass (kg) of the gas in the room
T is the absolue temp (K) in the room (changing with time)
Tinlet is the temp of the entering gas
Texit = T
time = time ins seconds

net heat into the room is
Qdot = mdot * Cp * (Tinlet - T)

In terms of the MASS in the room, that net heat into the room also equals (MASS * T) / time

so....

mdot * (Tinlet - T ) = (MASS * T) / time

and the room temperature is ...

T = (mdot * Tinlet) /(MASS/time + mdot)

???

I'll have to make some numbers before I really believe this.

regards,

magicme





------------------------------------
there's no place like gnome.

 

[magicme]

oops..... correction....

Tzero = initial room temperature, then...

T = (mdot * Tinlet + {MASS/time}*Tzero}) /(MASS/time + mdot)

or alternately....

T = { (time/MASS)*mdot*Tinlet +Tzero } / { 1 + mdot *time / MASS }


i think that's correct...

magicme




------------------------------------
there's no place like gnome.

 

[BronYrAur]

I have to put this down now and get back to work on a big project. Sorry I can't be more helpful, but I was trying to put together an energy balance by using a diff Eq example from my book. In that example, they are finding the concentration of salt in a tank at any time t. It would seem that the same could be done with heat. Here is a summary of the salt example:

A tank has 10 gal of brine in which 5 Lb of salt are dissolved. Brine consisting of 3 Lbs of salt per gallon enters the tank at 2 GPM and a well-stirred mixture leaved at the same rate.

Let A be the number of Lbs of salt in the tank at time t.
dA/dt = rate gained - rate lost

Since 2 GPM of brine at 3 Lbs salt per gallon enter, we have 6 Lbs per Minute entering. This is the rate gained.

Since there are always 10 gallons in the tank and since there are A Lbs of salt at time t, the rate lost is:
(A Lb/10 gal)*(2 GPM) = (A Lbs/5 Min)

So, dA/dt = 6 - (A/5)

Initial conditions are A=5 at t=0

Solving the Diff Eq, you get:

-ln(30-A) = (t/5) + c

plugging in the initial conditions, c = -ln(25)

So, the solution is:

A = 30 - 25e^(-t/5)

Hope this helps point you in the right direction.