Bulk modulus given, find change in P [Fluid Mechanics]

[leafjerky]

Homework Statement

If the bulk modulus for water at
70∘F is 319 kip/in^2, determine the change in pressure required to reduce its volume by 0.3%.

Homework Equations

E = dP/(dV/V)
E - Bulk Modulus
dP - change in pressure
dV - change in volume
V - volume

The Attempt at a Solution

Well I just said 319 kip/in^2 = dP/(.003/1 in^2) so then dP = .957 kip or 957 lb. But it's looking for an answer using US customary dimensions for pressure. Any ideas? What did I overlook?

 

[SteamKing]

Homework Statement

If the bulk modulus for water at
70∘F is 319 kip/in^2, determine the change in pressure required to reduce its volume by 0.3%.

Homework Equations

E = dP/(dV/V)
E - Bulk Modulus
dP - change in pressure
dV - change in volume
V - volume

The Attempt at a Solution

Well I just said 319 kip/in^2 = dP/(.003/1 in^2) so then dP = .957 kip or 957 lb. But it's looking for an answer using US customary dimensions for pressure. Any ideas? What did I overlook?

First of all, units of pounds indicate force, rather than pressure in USCS, which are given in pounds per square inch, or psi, usually.

Since the volume of the sample is reduced by 0.3%, a pure number without units, then re-arranging the original equation thus:

$E = \frac{dP}{dV/V}$

$E ⋅ (dV/V) = dP$

should result in units of pressure, whether they be ksi (= kip / in²) or psi, by suitable conversion.

 

[leafjerky]

First of all, units of pounds indicate force, rather than pressure in USCS, which are given in pounds per square inch, or psi, usually.

Since the volume of the sample is reduced by 0.3%, a pure number without units, then re-arranging the original equation thus:

$E = \frac{dP}{dV/V}$

$E ⋅ (dV/V) = dP$

should result in units of pressure, whether they be ksi (= kip / in²) or psi, by suitable conversion.

I figured it should be, but for some reason I kept getting it as just lb or kip. So would the answer be 957 psi? I only have one attempt left.

 

[SteamKing]

I figured it should be, but for some reason I kept getting it as just lb or kip. So would the answer be 957 psi? I only have one attempt left.

Yes.

 

[leafjerky]

Just wanted to say thanks SteamKing for being so helpful, you've answered stuff for me before.

 

[SteamKing]

You're welcome.