Steady state heat transfer - calculating delta T
August 15, 2022 7:45 AM Subscribe
I need to know how to calculate the delta t between "inside" and "outside" given a u factor of either .0555 or .0333. Inside of the cooler is to be at 1 degree Celsius, ambient outside the cooler is 23.5 degrees Celsius.
I used to know how to do this, but I cannot remember
Response by poster: Sorry, I wan't quite clear snough.
I solved the above equation for 1 sq. meter, I get 1.2 somethings. Watts I assume.
The issue is that the cooler is sweating on the exterior. Since inside and out temp are constant, if I know what the temp of the exterior surface would be, I could then look up what the max humidity "outside" would be before the skin starts to sweat
I need the delta T, maybe I don't understand how to get there from the above equation.
posted by rudd135 at 9:52 AM on August 15, 2022
I solved the above equation for 1 sq. meter, I get 1.2 somethings. Watts I assume.
The issue is that the cooler is sweating on the exterior. Since inside and out temp are constant, if I know what the temp of the exterior surface would be, I could then look up what the max humidity "outside" would be before the skin starts to sweat
I need the delta T, maybe I don't understand how to get there from the above equation.
posted by rudd135 at 9:52 AM on August 15, 2022
Take a look at this, particularly Eq. 17..22. You're trying to find Tw2.
To do this, you need to know something about heat transfer through the boundary region of length δ'2. Here, they lump this phenomenon into the constant h2. Unfortunately, there's no a priori way to find the value of this constant. Nor is there a simple way to predict the value of δ'2. You have to measure, use some kind of empirical correlation, or run a simulation.
posted by mr_roboto at 11:49 AM on August 15, 2022
To do this, you need to know something about heat transfer through the boundary region of length δ'2. Here, they lump this phenomenon into the constant h2. Unfortunately, there's no a priori way to find the value of this constant. Nor is there a simple way to predict the value of δ'2. You have to measure, use some kind of empirical correlation, or run a simulation.
posted by mr_roboto at 11:49 AM on August 15, 2022
Can you explain what you mean by cooler? Is there a source of cold, or do you mean a cool-box? Because the latter won't be steady-state.
Generally you have 3 different transfer of heat - between the fluid on the inside of the cooler and the inner wall, across the wall, and between the outer wall and the bulk air.
What's not clear is what your U-value relates to - just the heat transfer across the wall, or the combination of everything? Either way, you are missing information.
posted by scorbet at 11:56 AM on August 15, 2022
Generally you have 3 different transfer of heat - between the fluid on the inside of the cooler and the inner wall, across the wall, and between the outer wall and the bulk air.
What's not clear is what your U-value relates to - just the heat transfer across the wall, or the combination of everything? Either way, you are missing information.
posted by scorbet at 11:56 AM on August 15, 2022
Can you explain what you mean by cooler? Is there a source of cold, or do you mean a cool-box? Because the latter won't be steady-state.
If it's filled with an equilibrium water-ice mixture it will stay clamped at 0 °C for a long time (until the ice is completely melted).
posted by mr_roboto at 11:59 AM on August 15, 2022
If it's filled with an equilibrium water-ice mixture it will stay clamped at 0 °C for a long time (until the ice is completely melted).
posted by mr_roboto at 11:59 AM on August 15, 2022
Response by poster: Cooler will be full of dead people. It is for a coroner's office, it is a big walk in cooler like one sees at restaurants.
The fluid is air.
The U value I gave is the U value of the wall where condensation is happening. I am neglecting the air layer on the inside and outside as it is typically negligible
Yes convective transfer.
Measuring the actual temp of the outside of the wall may be the easiest solution.
posted by rudd135 at 6:07 AM on August 16, 2022
The fluid is air.
The U value I gave is the U value of the wall where condensation is happening. I am neglecting the air layer on the inside and outside as it is typically negligible
Yes convective transfer.
Measuring the actual temp of the outside of the wall may be the easiest solution.
posted by rudd135 at 6:07 AM on August 16, 2022
Best answer: I am neglecting the air layer on the inside and outside as it is typically negligible
If you neglect that layer, the temperature of the outside wall of the cooler is 23.5 °C.
posted by mr_roboto at 12:30 PM on August 16, 2022
If you neglect that layer, the temperature of the outside wall of the cooler is 23.5 °C.
posted by mr_roboto at 12:30 PM on August 16, 2022
This thread is closed to new comments.
So if the u-value is 0.0555 W/m2 . K then the heat loss rate is going to be (23.5 - 1) * 0.0555 * area.
To keep the temperature difference steady this is the amount of heat you would need to pump out of the cooler; otherwise if you know the thermal mass in the cooler you can solve a PDE to get the time constant. This will tell you how long it takes to get halfway to the temperature equalizing.
posted by larkery at 7:58 AM on August 15, 2022