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

In summary: Making mistakes is part of learning, as long as they don't get repeated. Good luck with your studies.In summary, the conversation discusses determining the change in pressure required to reduce the volume of water by 0.3% at 70∘F given a bulk modulus of 319 kip/in^2. The equation E = dP/(dV/V) is used and re-arranged to solve for dP, resulting in units of pressure such as ksi or psi. The answer is determined to be 957 psi.
  • #1
leafjerky
43
7

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?
 
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  • #2
leafjerky said:

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.
 
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Likes leafjerky
  • #3
SteamKing said:
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.
 
  • #4
leafjerky said:
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.
 
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Likes leafjerky
  • #5
Just wanted to say thanks SteamKing for being so helpful, you've answered stuff for me before.
 
  • #6
You're welcome.
 

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

1. What is the formula for calculating change in pressure using bulk modulus?

The formula for calculating change in pressure (ΔP) using bulk modulus (K) is given by: ΔP = KΔV/V, where ΔV is the change in volume and V is the original volume.

2. How is bulk modulus related to change in pressure?

Bulk modulus is a measure of a substance's resistance to change in volume under pressure. It is directly proportional to change in pressure, meaning that as bulk modulus increases, the change in pressure also increases.

3. What units are used for bulk modulus?

Bulk modulus is typically measured in units of pressure, such as Pascals (Pa) or Newtons per square meter (N/m²). It can also be expressed in units of energy per unit volume, such as Joules per cubic meter (J/m³).

4. Can bulk modulus be negative?

No, bulk modulus cannot be negative. It is a measure of a substance's resistance to change in volume, so a negative value would indicate that the substance is expanding rather than compressing under pressure.

5. How is bulk modulus used in real-world applications?

Bulk modulus is an important factor in the design of structures and materials, particularly in engineering and geology. It is used to determine the compressibility of materials and can help predict how they will respond to external forces, such as pressure or vibrations. It is also used in the study of earthquakes and other natural phenomena that involve changes in pressure and volume.

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