- #1

Fritz

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Please see the attached image to see what I'm talking about.

I understand how the forces (in terms of the pressures and areas) are equated, but then it says 'for small delta x' and from there on I'm confused.

It's something to do with differentiation, but I don't get it.

Can someone explain in simple terms what this all means?

Carrying out a force balance in the x and z direction.

x-direction

p(x). ∆z − p(x+∆x). ∆z = 0

For small ∆x,

p(x+∆x) = P(x) + (∂p/∂x). ∆x

Hence, ∂p/∂x = 0

Which confirms that the hydrostatic pressure does not change in the horizontal plane; i.e., p ≠ f(x,y).

z-direction

p(z). ∆x − p(z+∆z). ∆x − ∆W = 0

Where ∆W = ρ g ∆x∆z

assuming ρ is constant over the small element.

For small ∆z, and as p = f(z)

p(z+∆z) = p(z) + (dp/dz).∆z

Hence dp/dz = -ρg

Stuff in brackets after p is a subscript of p i.e. not a function p of...

I understand how the forces (in terms of the pressures and areas) are equated, but then it says 'for small delta x' and from there on I'm confused.

It's something to do with differentiation, but I don't get it.

Can someone explain in simple terms what this all means?

Carrying out a force balance in the x and z direction.

x-direction

p(x). ∆z − p(x+∆x). ∆z = 0

For small ∆x,

p(x+∆x) = P(x) + (∂p/∂x). ∆x

Hence, ∂p/∂x = 0

Which confirms that the hydrostatic pressure does not change in the horizontal plane; i.e., p ≠ f(x,y).

z-direction

p(z). ∆x − p(z+∆z). ∆x − ∆W = 0

Where ∆W = ρ g ∆x∆z

assuming ρ is constant over the small element.

For small ∆z, and as p = f(z)

p(z+∆z) = p(z) + (dp/dz).∆z

Hence dp/dz = -ρg

Stuff in brackets after p is a subscript of p i.e. not a function p of...

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