Heat loss when combining two water streams........

[esimp2k]

Given: 20gpm @ T=180F flowing into a tank at T=-ambient.

Unknown: volume of tank such that T(effluent) = 120F.

Incoming fluid and tank fluid are water. Thanks!

 

[25362]

Using simple algebra, after making some assumptions, such as:

[•] the only temperature changes are those resulting
from mixing, meaning no external heat exchange
[•] complete and thorough mixing before the tank
starts overflowing
[•] water specific heat is constant at all temperature
levels under consideration
[•] original temperature of water in the tank = T
[•] filling time = t, minutes
[•] tank volume = V, gallons

V = [30 + 10[÷](120-T)][×]t​

Correct me if I'm wrong.

 

[katmar]

To 25362 : I cannot follow your calculation but there are probably 2 reasons for this. The first is that I am not at home in your units. I guess the number 10 is the specific heat of water in BTU/gallon.F, but I don't know what the 30 represents. The second, and more important reason, is that I am not sure that I understand esimp2k's question.

The title of the thread hints that there are two streams entering the tank, but the wording of the actual query makes it look like the combining they are talking of is the single entering stream plus what is already in the tank.

If we were talking of 2 separate streams entering the tank and creating an average temperature of 120F then the volume of the tank would be irrelevant. All that is required is that the two streams are well mixed. As the question specifically asks for the volume of the tank I must conclude that we are not talking about combining two entering streams.

If we are looking at a single stream entering a tank at 180F and the overflow (effluent) from the tank leaving at 120F then the heat can only go to one place, i.e. that it is lost through the tank walls to the atmosphere. I presume also that we are looking at a steady state question i.e. we are not trying to calculate how long it takes for the tank to get to 120F.

What we really need to calculate then is how much heat is entering the tank (in the 180F water) and how much wall area we need to be able to get this heat out of the tank. The tank has to be at 120F (given) so the heat entering the tank relative to this is simply (mass flow x specific heat x temperature change)

Assuming the 20 gpm are US gallons the flowrate is 10,000 lb/h. The specific heat of water is 1 btu/lb.F and the temperature change is 60 F (i.e. 180 - 120). The heat entering the tank is therefore 10,000 x 1 x 60 = 600,000 btu/h.

The heat transfer from the tank walls will depend on the tank material of construction, the air velocity around the tank etc, so it is impossible to say anything other than that the heat transfer coefficient will be low, probably less than 2 btu/h.ft2.F. Assuming an ambient temperature of 70F gives a temperature difference of 120-70=50F. This gives an area of 600,000 / (2 x 50) = 6,000 ft2. Assuming a cube shaped tank and no heat transfer out of the top or bottom faces, the area of each face is 1,500 ft2. This gives an edge of virtually 40 ft.

If the assumption of a single stream flowing in at 180 F is correct then a cubic tank 40x40x40 ft would be required. The volume is therefore about 480,000 US gallons, but will depend on the actual heat transfer conditions around the tank. Putting a cooling coil in the tank, or a heat exchanger on the effluent will be a more practical solution.

Harvey

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[25362]

To katmar, as I understood it, the problem, and esimp2k is asked to confirm or deny, was:

1. The tank contains water at ambient temperature which I called T, it receives a constant stream of 20 gpm water at 180 F.

2. When the tank is full and starts to overflow, the effluent is at 120 F.

I assumed complete mixing and equilibrium up to the moment the tank starts to overflow. Other assumptions were that the specific heats at all the considered temperatures are equal, and there was no heat exchange with the surroundings.

The tank's volume is then obtained by making a mass and enthalpy balance.

 

[katmar]

25362, that is indeed another plausible interpretation, and one that I had not seen. Putting together a good problem statement is a large part of the way towards getting the right solution.

esimp2k, I'm sure you have a clear picture in your mind of what is happening but it is not obvious from your description. You will have to give us more information if you want a clearer answer.


Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[Zapster]

Is this a homework problem?

 

[BronYrAur]

I'm not sure the question is being asked clearly. Something is missing. As others have stated, assuming no heat loss, etc., there is a point when the mixed temperature (and thus effluent) will be 120 deg. However, this is only a single point in time, and that time is dependent on the volume of the tank. As you keep introducing 180 deg water, your mixed temperature will creep up.

For example, if you introdice 180 deg water for 1 minute and you have a tank with 70 deg ambient water, an initial tank volume of 24 gal (44 once you add the 20 gpm for 1 minute) will give you a mixed temp of 120 deg.

If you wait 2 minutes, an initial volume of 48 gal (88 after 2 min of 20 gpm is added) will be at 120 deg.

There are an infinite number of possibilities.

 

[esimp2k]

Just got back in, thanks for the responses, and yes, it was poorly worded. To hopefully clarify, it is one stream entering a fixed volume (tank). The stream enters the tank and water overflows via gravity out of the tank. I think katmar got it. I need to assume that worse case the stream flow is constant in which case the max. temp in the tank is 120F, so the only way to dissipate the approx. 600,000btuh is via losses through the tank walls or surface. Unfortunately it's not a homework problem per say.