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I am using Lienhard & Lienhard's A Heat Transfer Textbook (freely available here) to study Heat Transfer.

On page 19, the heat flux is written as $q=\bar{h}\cdot{(T_{body}(t)-T_\infty)}$, where $\bar{h}$, the average heat transfer coefficient along a body's surface, is assumed to be positive. But then on page 21, an equation for heat convection is derived from the First Law of thermodynamics:

$Q=\frac{dU}{dt}$

Where:

$Q=q\cdot{A}$

And:

$\frac{dU}{dt}=mc\cdot{\frac{dT}{dt}}=\rho\cdot{V}\cdot{c}\cdot{\frac{dT}{dt}}$

Where $T=T_{body}(t)$.

Now, naively plugging in the first equation for $q$ into $q\cdot{A}=\rho\cdot{V}\cdot{c}\cdot{\frac{dT}{dt}}$ leads to this first-order differential equation:

$\bar{h}\cdot{A}\cdot{(T-T_\infty)}=\rho\cdot{V}\cdot{c}\cdot{\frac{dT}{dt}}$

And when you solve for $T(t)$ given $T(0)=T_i$, you get that:

$T=T(t)=(T_i-T_\infty)\cdot{e^{\frac{\bar{h}\cdot{A}}{\rho \cdot{V} \cdot{c}}\cdot{t}}}+T_\infty$

Since the heat transfer coefficient, surface area, density, volume and the heat capacity of a material is assumed to be positive, this solution implies that the temperature $T(t)$ will increase or decrease indefinitely as time progresses, depending on whether the medium you're using to change $T_i$ is hotter or cooler than $T_i$, which makes no sense at all.

So in the book they derive $T$ using $Q=-\bar{h}\cdot{(T-T_\infty)}$, which results in a more intuitive solution:

$T=T(t)=(T_i-T_\infty)\cdot{e^{-\frac{\bar{h}\cdot{A}}{\rho \cdot{V} \cdot{c}}\cdot{t}}}+T_\infty$

Because this expression implies that the temperature difference between the medium and the body will get smaller as time progresses.

So it turns out that taking negative $q$ leads to a reasonable expression for $T(t)$, but the expression for flux was originally stated without flipping any signs. Is there an explanation or intuition behind taking the negative flux for this derivation, or is it just a matter of interpreting the maths and convincing yourself that the sign change was necessary to model reality more correctly?

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2 Answers 2

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In the case where the + sign is used, you are determining the heat flux from the body to the surroundings, and, in the case where the - sign is used, you are determining the heat flux from the surroundings to the body. The heat flux from the body to the surroundings is, of course, minus the heat flux from the surroundings to the body. So, in summary, the two versions are totally consistent.

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  • $\begingroup$ But does the first derivation not imply that a hot object in a cool room will increase the room's temperature indefinitely? Should this increase in time not drop off at some point? $\endgroup$ Mar 23, 2020 at 12:13
  • $\begingroup$ The room is assumed to have a large capacity to absorb heat. But, even if it didn't, the temperature of the room would not increase indefinitely. The body's temperature would be decreasing, and the room's temperature would be increasing. So the temperuatre difference and the heat flux would be decreasing with time. That would cause the temperature changes of the room and body to slow down and eventually become zero (when the temperatures were finally equal). $\endgroup$ Mar 23, 2020 at 13:35
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$$\dot{q}=\bar{h}\cdot{(T_{body}(t)-T_\infty)}$$

where:

$$\dot{q}=\frac{\text{d}Q}{\text{d}t}=\frac{\text{d}(mcT)}{\text{d}t}$$

$$\dot{q}=\frac{\text{d}(mcT)}{\text{d}t}=mc\frac{\text{d}T}{\text{d}t}$$

But the body is cooling, so $\text{d}T < 0$, thus the correct expression is:

$$\boxed{\dot{q}=-\bar{h}\cdot{(T_{body}(t)-T_\infty)}}$$

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  • $\begingroup$ But this equation should also hold when the body is being heated up, as a matter of fact, the first example in the book concerning convection talks about a small metal bead being exposed to hot air. So that's where I'm kind of stuck :/ $\endgroup$ Mar 23, 2020 at 13:41
  • $\begingroup$ If the body is heating up, then $T_{body}(t)-T_{\infty} <0$, so the relation still holds. $\endgroup$
    – Gert
    Mar 23, 2020 at 13:52

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