Heat loss whilst in steady state

In summary, the system reaches a steady state by a balance in the flows through the opening and the heat loss through the fabric of the upper layer.
  • #1
smithy360
8
0
Hi all,

I have a system which involves a ventilated box with a heat source located inside, if the heat source inside is say 1KW and I know it eventually reaches a steady state, in an ideal world it would reach this steady state by effectively ventilating 1KW of heat through the opening and assuming the boundaries of the box were adiabatic. However, obviously in the real world this isn't the case as there will be some heat loss from the container itself, dependent on material, if the heat loss through the fabric is relatively small, say 10W, how would I account for this heat loss. I should be more clear, if I am heating a box of fluid immersed in a large reservoir of ambient fluid held at a constant 20 degrees C and I know that if the box of fluid where perfectly insulated it would heat up the fluid in the container to 25 degrees C before reaching steady state as described above, i.e. as much heat gets vented as I am putting in. How would I determine the temperature reached in the container when accounting for the heat loss from the fabric? Naively I would say that as the heat loss is 1% of the heat flux going into the container I could just say that the temperature reached would be 99% of the 25 degrees C it is expected to reach. I say naively as I am not sure this is a fair assumption at all, is it better to say that now the flux going in, the 1KW, is now being balanced by the flux going out through the vent plus the heat loss, i.e. the vent flux is now equal to 990W?

Any help of guidance would be much appreciated, and I apologise in advance if this makes no sense!
 
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  • #2
I am having a hard time understanding your system, but I think you are on the right track...

If you have some kind of electric element in the middle of some kind of box and air is going in on one and and going out on the other end, then, yes, some heat is going to be lost through the box to the surrounding air and hence not all the heat generated by the electric element will be part of the air leaving the box.

The heat lost through the box to the surrounding air, though, will come from the air inside the box, right? This will happen more toward the exit where the air is hot, and not at the entrance, though...you know what I am saying?

This "steady state" that you talk about, though, is not your typical one where you could treat the air as being stationary and being heated from the center towards the surround box...instead, the air is always flowing and heated as it flows...
 
  • #3
Thanks for taking the time to reply. Sorry I know it was a little bit hard to understand as written. The system ends up in a steady state with a layer of hot air sitting on a layer of cool air, so the box has a simple two layer stratification. The flow in the box gets to the steady state by a balance in the flows through the openings, ambient air being drawn in the lower opening and hot air being exhausted through the top vent, and the flow from the plume emanating from the heat source at the base of the tank. In this way the temperature also reaches a steady state, all of the plumes energy goes into keeping the upper warm layer well mixed so it is of uniform temperature (still sat above the cooler ambient layer), and as said earlier the heat put into the upper layer is matched by the heat being extracted from the upper vent, hence keeping the temperature steady.

My question still remains on how to work out what the temperature of this upper layer will be in steady state but with additional heat loss through the container itself, if I calculate it as being 25 degrees C is it right to just say it will be 99% of that, or is a slightly more involved idea of now saying the plume heat flux entering the bottom of the upper layer is matched by the sum of the heat lost through the opening and the heat loss through the fabric of the upper layer (obviously only over the surface area in contact with the upper layer).

Thanks again.
 
  • #4
Just from thinking about it all logically I assume that I am correct in my thinking in terms of equating the relevant fluxes, it also seems to match up to some data I have so hopefully it is correct. Thanks again for the help and feel free to respond if my thinking is definitely not correct!

All the best.
 
  • #5


I would first like to commend you for considering the real-world factors and not just relying on idealized conditions. It is important to account for all variables when studying a system. In this case, the heat loss from the container itself is a crucial factor in determining the temperature reached in the container.

To accurately account for this heat loss, you would need to know the specific material and its thermal conductivity. This will allow you to calculate the rate of heat transfer through the material. You can then use this value to adjust your calculations for the temperature reached in the container. Simply assuming a percentage reduction may not accurately reflect the actual heat loss and could lead to incorrect conclusions.

Furthermore, you may also need to consider other factors such as the surface area of the container and the surrounding environment, as well as any insulation or other barriers that may affect the heat transfer. These variables can also impact the temperature reached in the container and should be taken into account in your calculations.

In summary, to accurately determine the temperature reached in the container while accounting for heat loss, you will need to consider the specific material and its thermal conductivity, as well as other variables that may affect the heat transfer. It is important to carefully analyze and calculate these factors in order to accurately understand and predict the behavior of your system.
 

Related to Heat loss whilst in steady state

1. What is "steady state" in terms of heat loss?

Steady state refers to a condition where the rate of heat loss from an object or system is equal to the rate at which it is being heated. This means that the temperature of the object or system remains constant over time.

2. How does heat loss occur in steady state?

Heat loss in steady state occurs through conduction, convection, and radiation. Conduction is the transfer of heat through direct contact, while convection is the transfer of heat through the movement of fluids. Radiation is the transfer of heat through electromagnetic waves.

3. What factors affect heat loss in steady state?

The rate of heat loss in steady state is affected by the temperature difference between the object or system and its surroundings, the surface area of the object, and the thermal conductivity of the material it is made of.

4. How can we reduce heat loss in steady state?

To reduce heat loss in steady state, we can use insulation materials such as foam or fiberglass to decrease the rate of heat conduction. We can also minimize air movement around the object or system to reduce heat loss through convection. Additionally, using reflective surfaces can reduce heat loss through radiation.

5. Why is understanding heat loss in steady state important?

Understanding heat loss in steady state is important in various industries such as building construction, energy production, and climate control. It allows us to design more efficient systems and reduce energy consumption, ultimately leading to cost savings and environmental benefits.

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