# How do I rearrange this equation?

[a]1. How do I make dp the subject?

Q=[(dp/dx)/2k]^(1/n).[pi/(1/n)+3].R^(1/n)+3

^= power of

I also have attached the equation.

Thanks
M.Stephens

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I am trying to rearrange to work out dp. To make dp the subject.

Ray Vickson
Homework Helper
Dearly Missed
If dp/dx stands for the derivative of p with respect to x, you can't just isolate dp [although you could consider dp to be dp/dx * dx. Anyway, you have Q = [(dp/dx)/2k]^(1/n) * S + 3, where S = some stuff. If you set y = dp/dx, can't you see how to solve for y, in terms of Q, S, k and n?

RGV

No the dp = drop in pressure. Does that make alot of difference.

Thanks
MS

Q=[(dp/dx)/2k]^(1/n).[pi/(1/n)+3].R^(1/n)+3
Why do you write this? Is it:

$$\frac{\pi}{\frac{1}{n}} = n \, \pi$$

or did you mean something else? Also, there is no attachment.

The equation is for flow rate down a circular tube, in my question i am asked to rearrange the equation to find the pressue drop(dp).

I have attached a picture of the equation.

Thanks

#### Attachments

You can leave $\left(\frac{dP/dx}{2 k}\right)^{1/n}$ on the r.h.s. and move everything else on the l.h.s.:

$$\frac{\left(3 + \frac{1}{n}\right)}{\pi} \, Q \, R^{-3 - \frac{1}{n}} = \left[\frac{dP/dx}{2 k}\right]^{\frac{1}{n}}$$

What would be the next step?

sorry Dickfore but that doesnt make n e sense to me?

What does tex n frac mean?

Does the equation in the attachment look correct?

#### Attachments

Mark44
Mentor
sorry Dickfore but that doesnt make n e sense to me?
In post #8, Dickfore multiplied both sides of the equation by the reciprocal of what was multiplying [(dP/dx)/(2k)]^(1/n).

What does tex n frac mean?
He is using LaTeX commands to format the equation.
No the dp = drop in pressure. Does that make alot of difference.
I'm almost certain you are wrong here. dP/dx is a derivative, which in this case means the rate of change in pressure per unit of pipe length. dP would be the differential of the pressure.

Also, it's given in this problem that n = 0.5, so 1/n = 2. Making that substitution would simplify some of the exponents and fractions in your equation.

Last edited:
sorry Dickfore but that doesnt make n e sense to me?
You should enable JavaScript in your browser to look at the rendered equations. Click Refresh/Reload.

I think it should be done simpler in this case. Replace the infinitesimal change with a finite change ($\alpha\equiv n^{-1}$):

$$Q=\left(\frac{\Delta P}{2k\Delta x}\right)^\alpha\frac{\pi R^{\alpha+3}}{\alpha+3}$$
$$\Rightarrow\Delta P=2k\Delta x\left(\frac{(\alpha+3)Q}{\pi R^{\alpha+3}}\right)^{1/\alpha}$$

Filling in all numerical values should give you the pressure drop.

Ok yes think I've got the correct answer, i do apologise it was the derivative of pressure. thank you all for your help.
MS

I think it should be done simpler in this case. Replace the infinitesimal change with a finite change
Actually, there is no approxmation here, since the pressure is a linear function of the pipe length and the slope is constant.