Approximation using first order term

Question

I'm trying to understand some equations concerning isentropic nozzle flow, found here (under "flow analysis"):

https://en.wikipedia.org/wiki/Isentropic_nozzle_flow

We have this continuity equation:

$$\rho AV = (\rho+d\rho)(A+dA)(V+dV)$$

Then the next step is:

If only the first-order terms in a differential quantity are retained, continuity takes the form

$$ \frac{dV}{V}+\frac{dA}{A}+\frac{d\rho}{\rho} = 0 $$

I don't understand how this step was made. I understand the concept of approximating a function with Taylor series and neglecting higher order terms when working in small intervals, but I'm still having hard time following how the step was made. Could somebody help me and explain this in more detail? Thank you!

Answer 1

The way one usually goes about showing this is by imagining $dV$, $dA$, and $d\rho$ are themselves first-order small quantities. Then expand: $$\begin{align} \rho AV &= (\rho + d\rho)(A + dA)(V+dV) \\&= \rho A V + \rho A dV + \rho V dA + AVd\rho + \rho dA dV + A d\rho dV + V d\rho dA + d\rho dA dV \end{align}$$ Now the terms $dV dA$ or any other term with at least two first-order small quantities multiplied together is a second-order small quantity (or higher-order) and can be assumed to be zero if the first-order small quantity is small. So we have: $$ \rho AV = \rho A V + \rho A dV + \rho V dA + AVd\rho $$ Divide both sides by $\rho AV$: $$ 1 = 1 + \frac{dV}{V} + \frac{dA}{A} + \frac{d\rho}{\rho}$$ And finally subtract $1$ from both sides: $$ 0 = \frac{dV}{V} + \frac{dA}{A} + \frac{d\rho}{\rho}$$

This presentation is a bit informal, but it gets the point across.

Answer 2

$$ (a + da)(b + db)(c + dc) = (ab + a\cdot db + b\cdot da + da\cdot db)(c + dc) $$ we drop terms higher than first order, the ones where the delta terms are combined i.e. $da\cdot db$ etc. $$\require{enclose} (ab + a\cdot db + b\cdot da + \enclose{horizontalstrike}{da\cdot db})(c + dc) = (ab + a\cdot db + b\cdot da)(c + dc) = (abc + ac\cdot db + bc\cdot da + ab\cdot dc + \enclose{horizontalstrike}{a\cdot db\cdot dc} + \enclose{horizontalstrike}{b\cdot da\cdot dc}) = abc + ac\cdot db + bc\cdot da + ab\cdot dc $$ now we have $$ abc = abc + ac\cdot db + bc\cdot da + ab\cdot dc $$ we can re-arrange to get what you want (with the change in variable labels ;) )

Answer 3

$$(\rho+d\rho)(A+dA)(V+dV) = \rho AV+\rho AdV+\rho V dA+AVd\rho+\rho dAdV+Vd\rho dA+AdVd\rho+dAdVd\rho$$ Now if you keep only terms of the first order (or less) you get : $$\rho AV=\rho AV+\rho AdV+\rho V dA+AVd\rho$$ Simplyfying and dividing by $AV\rho$ you get the expected result : $$\frac{dV}{V}+\frac{dA}{A}+\frac{d\rho}{\rho}=0$$