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I have the following equation for the torque exerted on a turbine blade.

The paper that I am reading says, "the corresponding torque is given by..."

$$dT = ρV_2ωr^22πrdr$$

My problem is that while I do know that $d$ is the derivative operator in this case, how do I apply it? Why is the corresponding torque given by $dT$ as opposed to $T$? I am trying to find an expression for finding $T$ based on the variables needed.

I realize this may be simple, however, I ask because every google search I did yielded $\frac{dM}{dT}$ or something of the sort, not a simple $dM$.

My background: I am in high school and I have taught myself parts of differential and integral calculus. I am going to learn differential equations next, I would just like to know how to apply this operator.

Good day!

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    $\begingroup$ I'd recommend becoming well-versed in the single-variable calculus before moving on to diff eq-- pretty advanced for high school, but doable. In addition, multivariable calculus will give you a good understanding of the total derivative and total differential. $\endgroup$ – terrygarcia Mar 16 at 16:27
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This expression $dT$ isn't an operator, it's a differential form, specifically a 1-form. Loosely speaking, it's a quantity that can be integrated over one variable and represents a differential (read: "infinitesimally small piece of...") torque at a point. Divide by $dr$ on both sides and you get $dT/dr$, the total derivative of torque with respect to the radial coordinate. Integrate with respect to $r$ over the volume/surface/whatever domain $\Omega$ (in your case the turbine blade, I believe) to get the total torque, i.e. $$T = \int_{\Omega} dT = \int_{\Omega} \dfrac{dT}{dr} dr$$ Make sure you fully grasp the (definite) integral as an infinite sum of infinitesimally small "pieces," and this should start to make a whole lot of sense.

Technically speaking, the operator $d$ is the exterior derivative, which corresponds with the total differential in real vector spaces.

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  • $\begingroup$ I'd put "Divide" in quotation marks, too. $\endgroup$ – Klaus Mar 16 at 16:42
  • $\begingroup$ @Klaus I thought about it, but that use of language is completely valid in the context of differential forms and is justified in calculus with a few assumptions on the behavior of $T$ $\endgroup$ – terrygarcia Mar 16 at 16:44
  • $\begingroup$ Thank you very much, this is actually very helpful! To clarify, since I get: $$\int_{\Omega}{\rho v_2\omega r^3 2\pi dr}$$, I now integrate with respect to $r$? Also, how would I integrate with only a lower bound? And what domain is $\Omega$ in this case? $\endgroup$ – limitsandlogs224 Mar 16 at 16:44
  • $\begingroup$ @limitsandlogs224 it's not a lower bound; it represents the whole domain of integration. For example, if $r$ goes from $a$ to $b$ then you'd just write $\Omega=[a,b]$, then the symbol $\int_{\Omega}$ becomes $\int_{a}^{b}$ $\endgroup$ – terrygarcia Mar 16 at 16:45
  • $\begingroup$ @terrygarcia I understand. I just have one more question, although it is a little different than my original question, would it be correct to pull out the constants and multiply them by $\int_{\Omega}{r^3 dr}$. Thank you for your time. $\endgroup$ – limitsandlogs224 Mar 16 at 16:50

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