# What is dP in pressure?

I'm reading a textbook, chapter entitled "Fluid Mechanics".

This section is entirely devoted to fluid statics.

Assuming an element of fluid in equilibrium,

Consider a thin element of fluid with thickness dy. and density & gravity constant.

let p be pressure upward on the bottom of the surface, then total pressure on bottom of the surface is pA (pressure * Area).

Then, the pressure on the top surface is p+dp, and the total pressure on top of the surface is (p + dp) * A.

I have been looking into the web, but I am quite not sure what dp is meant when finding the top surface pressure.

My only guess is that dP is the difference of pressure due to change in altitude.

If my guess is right, is it notable to include such small change in altitude (considering the material has small dy).

• Note that "Is this right" type questions really are not a good fit for the Q&A format because the answer (yes or no) is too small to be an answer (minimum character count is 15). Could you modify your question to be less of a yes/no type question? – Kyle Kanos Jun 22 '15 at 1:53
• @KyleKanos I did! – hs2345 Jun 22 '15 at 1:58
• Such a "prefix", $d$, usually means a infinitesimal change, which in physics is often used for integrals. – fibonatic Jun 22 '15 at 2:13

Take for example, a submarine. The pressure against the walls of the submarine increments as this goes deeper.

For another example, take an airplane, as it goes higher the pressure against the walls decrease.

So yes, the pressure changes as the altitude (or depth) changes (if you are submerged on a fluid in a gravitational field). Even small amounts of change in altitude (or depth) will make a change in the pressure of the fluid against a real or imaginary wall.

dP is not a physicall quantity but a mathematical expression of something that changes an infinitesimal. In this particular case, dP is a small change (infinitesimal) in the pressure. And pressure is a dependant variable of 'w' (the weight of the fluid element), 'y' (the altitude or depth), and 'A' (the area of an horizontal plane).

Just another comment; take a look at the force diagram, you will see that all horizontal forces are counteracting forces, but the upper force is slightly smaller than the force pushing from below (The difference is F2-F1=dF, a very small amount of change in the force). As the system is in equilibrium, the vector sum of all forces must be zero. So the small force that you need to add is exactly A * dP.

• Can I ask one more question? Also my text book gave expression "pA - (p+dp)A - pgAdy = 0", now I get that pA - (p+dp)A = -dpA. so is dpA = pgA(dy)? I dont quite get what pgA(dy) is supposed to mean? I mean it's broad expression. My only guess here, too, is that pgA(dy) is gravitational pressure? – hs2345 Jun 22 '15 at 3:02
• @hs2345 You already did. Hahaha, C'mon ask whatever you need. – algolejos Jun 22 '15 at 3:04
• Edit: p in pgAdy is roh, density. – hs2345 Jun 22 '15 at 3:05
• pgAdy is really then (Weight/volume)(y2-y1)? – hs2345 Jun 22 '15 at 3:05
• you peeps seriously need to learn to use mathjax - it's really not very hard :) meta.math.stackexchange.com/questions/5020/… – danimal Jun 22 '15 at 10:55