Air required to cool water temperature

[Vig16]

I'm trying to cool 8 ounces of 70 deg C water down to 38 deg C. I want to do this all in 2 minutes, but with as low a cost as possible. I was thinking of using a fan or air pressure of some short to do this. The only question I had is how do I determine the pressure or amount of air that I'll need to obtain the temperature drop?

My initial test was to dispense 3 ounces at 70dC and then have room temperature water dispensed into the 70dC water. In my two attempts at this, I got the temperature in the cup down to 44dC and 47dC, but am still off by a bit. I was thinking that if I introduced air of some sort into the flow of room temperature water, that I'd be able to have the final water temp drop the 5-10dC that I'm looking for.

Does anyone have any insight as to how I'd go about calculating the amount of cooling that I'd need to get to the desired temperature? Even just the equation to start with would be greatly appreciated!

Thanks so much!

 

[edgepflow]

Could you dispense just a little more room temperature water to get to your final temperature?

If not, you cooling heat rate is:

Qrate = m-water X cp-water X (T final - T initial) / elapsed time

You can then determine the velocity of coolant as follows; from Newton's law of cooling:

Qrate = h A (Tsurface - Tfluid)

You can estimate h, the convection heat transfer coefficient, from the Dittus Boelter correlation corrected to your geometry or a simlilar one. The h is a function of velocity.

 

[Vig16]

Could you dispense just a little more room temperature water to get to your final temperature?

If not, you cooling heat rate is:

Qrate = m-water X cp-water X (T final - T initial) / elapsed time

You can then determine the velocity of coolant as follows; from Newton's law of cooling:

Qrate = h A (Tsurface - Tfluid)

You can estimate h, the convection heat transfer coefficient, from the Dittus Boelter correlation corrected to your geometry or a simlilar one. The h is a function of velocity.

I can add a little more room temperature water, but the final solution needs to be 8oz.

Thank you for the equation, I'll plug some numbers into it and see what I come up with!

 

[Vig16]

edgepflow - I did all the math for my cooling rate and I came up with this...

Mass = 8 ounces = 0.237 kg (approx)
Specific heat of water = 4187 J/kgK
Temperature decrease = 32 °C
Time = 2 mins = 120 seconds

Power = 4187 * 0.237 * 32 /120 = 249 W

How would you suggest I determine the velocity if I'm unsure about the Tsurface? Also, is this the fluid temp when it's hot or the ideal temp that I'd like to get it down to?

 

[edgepflow]

There are two ways to approach this.

1. Perform a time dependent solution.

You could apply the simple but powerful "lumped capacity" model if the Biot number is < 0.1. If not, you would need to include spatial effects. Most good heat transfer textbooks present both of these.

2. Assume an average surface temperature.

For a back of the envelope crude estimate, you could assume the surface temperature is constant and the average of the initial and final temperature.

You could figure out the velocity and hence volumetric flow rate and then size a fan or whatever device you may need. You may also consider a "cold air gun" which supplies chilled air and only requires a supply of compressed air.

 

[Unrest]

I'm trying to cool 8 ounces of 70 deg C water down to 38 deg C. I want to do this all in 2 minutes, but with as low a cost as possible. I was thinking of using a

How about a heat exchanger? Just pour the hot water through a room-temperature metal pipe and it'll come out colder at the other end.

If you have to repeat the process quickly you'd need to then cool the pipe, maybe with a fan or just passively.