Effectively cooling molding sands.
Assume that we are trying to cool one ton of sand per minute. The sand is at an average temperature of 302F (150C), and contains 0.6% moisture. We want to cool it to 122F (50C), with a 2% moisture content for use in the foundry.
While there are a number of ways to solve the problem, one of the easiest to understand uses a heat balance:
Heat gained by water and air = Heat lost by sand
Start by mixing water with the sand to cool it to 200F.
Because we are interested in the residual moisture content of the sand after it cools to 200F, we have to take into account that part of the water mixed with the sand will evaporate. If we assume the water is initially at 70F (21C), and that all of the water we add to bring the sand temperature down to 212F (100C) evaporates, we can break the problem into two parts:
A. How much water do we use to cool the sand to 212F (100C)?
B. How much water do we use to cool the sand from 212F to 200F (93.3C)?
To calculate the answer to Question A, use the following published data:
Average specific heat of sand is 0.2 Btu/lb F = 0.2 kcal/kg |degrees~ C
Average specific heat of water is 1.0 Btu/lb F = 1.0 kcal/kg |degrees~ C
Latent heat of vaporization of water is 974 Btu/lb = 540 kcal/kg
Now apply a heat balance to the problem.
(Heat required to raise W lb of water/minute from 70F to 212F) + (heat required to vaporize W lb of water/minute) = (heat required to cool one ton of sand/minute from 302F (150C) to 212F (100C)
(W lb water/min) (212-70) |degrees~ F (1.0 Btu/lb water |degrees~ F) + (W lb water/win) (974 Btu/lb water) = (2000 lb sand/min) (0.2 Btu/lb sand |degrees~ F) (302-212) |degrees~ F
W = 32.2 lb of water/min (14.6 kg of water/min)
Note that this water all vaporized and is not part of the moisture content of the sand in the system.
To calculate the answer to Question B, reducing the temperature of the sand from 212F (100C) to 200F (93.3C), we will need to raise the temperature of Y lb of water/minute from 70F (21C) to 200F (93.3C).
Note that the heat balance looks like this:
(Y lb water/min) (200-70) |degrees~ F (1.0 Btu/lb water |degrees~ F) = (2000 lb sand/min (212-200) |degrees~ F (0.2 Btu/lb sand |degrees~ F)
Y = 36.9 lb/min (16.8 kg/min)
Total water added is 32.2 lb/min + 36.9 lb/min = 69.1 lb/min (31.4 kg/min)
We can now calculate the moisture content of the sand exiting this part of the cooler:
The water we added (36.9 lb/min) is 36.9/2000 = 1.85%.
Added to the 0.6% moisture already in the sand, the total moisture is 2.45%.
To cool the sand to 122F, mix it with air. Remember that every pound of water that evaporates required 974 Btu (or every kg requires 540 kilocalories). Furthermore, water can only evaporate up to the limit that air can absorb it at saturation. This means that only some of the water will evaporate from the sand. So, we need to know not only the original temperature of the air, but also the original water content of the air that we are mixing with the sand and how much water the air can hold.
Let's assume that we start with absolutely dry air at 32F (0C), and also that when we finish mixing the air with the sand the air will be at 122F (50C) and will be completely saturated. To do this calculation, we need to know that:
1 lb of air will hold 0.08 lb of water at saturation
(1 kg of air will hold 0.08 kg of water at saturation)
Average specific heat of air = 0.25 Btu/lb |degrees~ F = 0.25 kcal/kg |degrees~ C
The heat balance now looks like this:
(heat required to raise A lb of air/min from 32F (0C) to 122F (50C) + (heat required to evaporate the amount of water that A lb of air will hold at saturation) = (heat lost by the sand in cooling to 122F (50C). Or:
(A lb air/min) (0.25 Btu/lb air |degrees~ F) (122-32) |degrees~ F + (A lb air/min) (0.08 lb water/lb air) (974 Btu/lb water) = (2000 lb sand/min) (0.2) Btu sand |degrees~ F) (200-122) |degrees~ F
A = 311 lb of air/min (141 kg of air/min)
It is unrealistic to expect that you will have dry air at 32F to mix with your sand, unless you build an enormous air conditioning system (which is not wholly out of the question). Ideally, you would want to mix air at a known temperature and moisture content with your return sand to get the most uniform control over your sand reclamation system.
However, in most foundries today, ambient air is used. So let's assume that the ambient air enters at 90F (32C) and 90% relative humidity; for many parts of the country, these hot and humid conditions will approximate a "worst case" scenario. Under these conditions, the entering air already contains 0.028 lb of water/lb of air. Now our heat balance looks like this:
(A' lb air/min (0.25 Btu/lb air |degrees~ F) (122-90) |degrees~ F) + (A' lb air/min) (0.080-0.028) (lb water/lb air) (974 Btu/lb water) = (2000 lb sand/min) (0.2) Btu/lb sand |degrees~ F) (200-122) |degrees~ F)
A' = 506 lb of air/min (230 kg of air/min)
This assumes that air exits at 122F. If the time that the air remains in the cooling unit is reduced, it will not have the chance to reach 122F. For purposes of illustration, let's assume that air exits at 110F (43C) and 90% relative humidity. Under these conditions, air can hold 0.053 lb of water/lb of air. Now our heat balance looks like this:
(A" lb of air/min) (0.25 Btu/lb air |degrees~ F) (110-90) |degrees~ F) + (A' lb air/min) (0.053-0.028) (lb water/lb air) (974 Btu/lb water) = (2000 lb sand/min) (0.2 Btu/lb sand |degrees~ F) (200-122) |degrees~ F)
A" = 1603 lb of air/min (483 kg of air/min)
This equals about 15,000 cfm (425 |m.sup.3~/minute).
Each pound of air removes (0.053-0.028 lb) of water/minute, for a total of (0.025) (1063) = 26 lb min, or (26/2000) = 1.3%. This takes us below the desired 2% moisture content, so we'll have to add moisture in the muller.
The result is that as the inlet air becomes warmer and more humid, it takes more air to cool the sand. The temperature and humidity of the incoming air must be known as accurately as the temperature and moisture content of the sand, and adjustments must be made during the day based on both sets of readings.
What we have presented is a simple approach to sand cooling. The efficiency of a cooling system will depend on the velocity of the air, the residence time of the sand in the cooling chamber, the circulation pattern of the air through the sand, the amount of fines removed with the air (these will also carry heat away from the sand) and other factors. As you can see, sand cooling system design is a complex problem.
In addition, the sand will be cooled by radiation losses and conduction to the steel structure that houses the sand cooling system. An empirical formula that can be used to estimate the losses to the structure is as follows:
Heat loss (Btu/min) = (area) (temp. diff.) (time)/10
area = surface area of the steel structure
temp. diff. = difference between hot sand and room temperature
time = average time in minutes that the sand remains in the cooling housing.
Proper cooling of the sand in your system is of the utmost importance in making high-density molding work. In specifying a sand cooling system, remember that the ideal system will:
* expose each sand grain to the air stream;
* prolong contact with the cool air long enough to ensure maximum heat transfer;
* accomplish significant cooling through radiation and convection;
* mix wet and dry sand without plugging up;
* require a minimum floor area;
* be a complete, self-contained unit, which will not require installation of any extra conveyors;
* exhaust a minimum volume of air;
* be reasonable in original purchase price and operating cost.
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|Date:||Mar 1, 1994|
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