Heat transfer from air (at ambient temp.) to a liquid in an open container [closed]

Question

I’m designing a bath which should keep the temperature of the liquid inside (transformer oil) at a constant temperature (approx. 20 °C).

I have calculated the amount of energy needed to heat up the volume of liquid inside, however due to the bath having an open top, I also want to take into consideration the effect of ambient temperature. (the bath is small, so I'm worried that ambient temperature will affect its performance)

I know I should probably consider a simple model for this, as I don't want to over complicate by considering a lot of parameters.

By assuming that the air above the container is a reservoir whose temperature is constant and is in thermal contact with the surface of the liquid, would considering heat transfer through convection be enough?

Since the setup would be inside a laboratory, conditions should be stable enough (such as air velocity and humidity).

Answer 1

The low bath temperature of $20\text{°C}$ implies that radiation losses can be considered negligible and convection is the main mechanism for heat transfer.

In that case, you probably want to consider Newton's Law of Cooling (which incidentally works also for heating):

$$\dot{q}=-hA(T_{bath}-T_{amb})$$

where:

• $\dot{q}$ is the heat flux (flowing from the bath to the environment)
• $T_{bath}$ is the bath's temperature. It is assumed to be uniform throughout the bath
• $T_{amb}$ is the surroundings' temperature (considered constant)
• $A$ is the surface area of the bath exposed to the environment
• $h$ is the convection heat transfer coefficient (generally considered to be a constant)

If the bath has constant mass $m$ and the liquid has a heat capacity $c_p$ then the law can be rewritten slightly to:

$$mc_p\frac{\text{d}T_{bath}}{\text{d}t}=-hA(T_{bath}-T_{amb})$$

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One interesting application is finding the temperature evolution of the bath in time. Lets write $T_{bath}$ as $T(t)$ and $k=\frac{hA}{mc_p}$, so:

$$\frac{\text{d}T(t)}{\text{d}t}=-k(T(t)-T_{amb})$$

This is a simple differential equation that solves by separation of variables to:

$$\ln\Big({\frac{T(t)-T_{amb}}{T_0-T_{amb}}}\Big)=-kt$$ where $T_0$ is the bath temperature at $t=0$.

Rewritten we get:

$$\boxed{T(t)=T_{amb}+(T_0-T_{amb})\exp{(-kt)}}$$