Hydrolic Pressure and Bulk Modulus

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Homework Help Overview

The discussion revolves around a problem involving the calculation of pressure required to compress a solid copper cube, given its initial and final edge lengths and the bulk modulus of copper. The subject area includes concepts of fluid mechanics and material properties.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculations for volume change and pressure, questioning the interpretation of the bulk modulus value and its application in the formula. There is also a focus on the significance of using accurate values and significant figures in calculations.

Discussion Status

Some participants have provided guidance on the calculations, and there is acknowledgment of a successful resolution by one participant who clarified their approach to finding the volume change. Multiple interpretations of the bulk modulus value have been explored, but no explicit consensus has been reached.

Contextual Notes

There is a noted confusion regarding the correct interpretation of the bulk modulus value, whether it is 1.4 x 10^10 Pa or 1.4 x 10^11 Pa, which has implications for the calculations. Additionally, the importance of significant figures in the calculations has been highlighted.

vMaster0fPuppet
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Homework Statement



A solid copper cube has an edge length of 87.0 cm. How much pressure must be applied to the cube to reduce the edge length to 85.7 cm? The bulk modulus of copper is 1.4 multiplied by 1011 N/m2.

Homework Equations



Vcube= s^3

F/A=(B x deltaV)/V
p=F/A
p=(B x deltaV)/V

The Attempt at a Solution

V= (.870m)^3= .659m^3
Vn= (.857m)^3= .629m^3
deltaV= V-Vn= .03m^3

p=(1.4e^10N/m^2)x(.03m^3)/(.659m^3)= 6.37e^9N/m^2This is apparently not the answer?
 
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vMaster0fPuppet said:

Homework Statement



A solid copper cube has an edge length of 87.0 cm. How much pressure must be applied to the cube to reduce the edge length to 85.7 cm? The bulk modulus of copper is 1.4 multiplied by 1011 N/m2.

Homework Equations



Vcube= s^3

F/A=(B x deltaV)/V
p=F/A
p=(B x deltaV)/V

The Attempt at a Solution

V= (.870m)^3= .659m^3
Vn= (.857m)^3= .629m^3
deltaV= V-Vn= .03m^3

p=(1.4e^10N/m^2)x(.03m^3)/(.659m^3)= 6.37e^9N/m^2This is apparently not the answer?

Is that 1.4 x 1010 Pa or 1.4 x 1011 Pa.
I think think it is 1.4 x 1011 Pa for copper as you originally gave in the problem.
 
Its 10^11, and I did the calculations with 10^11.

Thanks anyway, I actually figured the problem out. When I used .870^3-.857^3 for deltaV the calculations worked just fine. I didn't realize that making the calculation easier by breaking it up would cause such a problem if I used the correct number of significant figures.
 
vMaster0fPuppet said:
Its 10^11, and I did the calculations with 10^11.

Thanks anyway, I actually figured the problem out. When I used .870^3-.857^3 for deltaV the calculations worked just fine. I didn't realize that making the calculation easier by breaking it up would cause such a problem if I used the correct number of significant figures.

OK then rightio. I didn't see a problem with your method.