Another Simple Heat Loss Question

In summary, the goal is to size a heater in watts to heat a glass of water from ambient temperature (Ti) to a certain temperature (Tf) in a given time period of 1 hour. The heat-loss from the glass at Ti is 0 watts and at Tf is Y watts, with the assumption that heat loss increases linearly between Ti and Tf. The energy needed to heat the water from Ti to Tf in 1 hour is X watts assuming no heat-loss. The suggested heater size is (Y + X) watts, but this may result in the water being heated faster than necessary due to initial lack of heat-loss from the glass. The question remains if there is a more accurate way to size the heater.
  • #1
TheH
12
0
I'm trying to size (in watts) a heater to heat a glass of water from ambient (Ti) to a certain temperature (Tf) in a given time period (say, 1 hour).

The heat-loss from the glass at Ti is 0 watts.
The heat-loss from the glass at Tf is Y watts.
I'm assuming that the heat loss increases linearly between Ti and Tf (I hope this is reasonable).
The energy needed (assuming no heat-loss) to heat the water from Ti->Tf in 1 hour is X watts.

Naturally, I can size the heater to be (Y + X) watts. However, this will heat the water up faster than necessary because initially there will be no heat-loss from the glass.

Is there a clever way to more accurately size this heater?

Thanks for your time.
 
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  • #2
Note that here it is possible for X to be much larger than Y (when I want to heat the water up rapidly) and Y to be much larger than X (when I want to heat the water up slowly).
 

Related to Another Simple Heat Loss Question

1. How do you calculate heat loss?

Heat loss can be calculated by using the formula Q = U x A x (Ti - To), where Q is the heat loss, U is the overall heat transfer coefficient, A is the surface area, Ti is the inside temperature, and To is the outside temperature.

2. What factors affect heat loss?

Some factors that affect heat loss include the temperature difference between the inside and outside, the surface area, the type and thickness of insulation, and the overall heat transfer coefficient of the material.

3. How can I reduce heat loss in my home?

To reduce heat loss in your home, you can improve insulation, seal any air leaks, use energy-efficient windows and doors, and maintain a consistent indoor temperature.

4. What is the difference between conductive and convective heat loss?

Conductive heat loss occurs when heat is transferred through a solid material, while convective heat loss occurs when heat is transferred through a fluid (such as air or water). Both can contribute to overall heat loss.

5. How can I measure heat loss in my home?

One way to measure heat loss in your home is by using a thermal imaging camera, which can detect temperature differences and show areas of heat loss. You can also consult with a professional energy auditor for a more comprehensive assessment.

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