# Work calculation from applied pressure

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Miraboreasu 2022 年 9 月 24 日
コメント済み: William Rose 2022 年 9 月 30 日
A spherical is meshed by many little triangles.A time-dependent pressure (p=10*t) is equally applied to the inner surface of a spherical. After t1=0.1s, the spherical broken and each little triangle is disconnected.
The data I have is the nodes locations (m) of each little triangle like
point1 2.48309 2.51276 2.45388
point2 2.4875 2.50415 2.45103
point3 2.47773 2.50283 2.45452
and velocity (m/s) of each nodes
v1 -11.352 4.68846 -58.9501
v2 -10.2788 -1.54017 -60.6666
v3 12.043 6.94501 -34.1632
How can I calculate the work has been done by pressure p
I think need to ignore the work done by the interation of each triangle

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### 採用された回答

William Rose 2022 年 9 月 24 日
You ask "How can I calculate the work has been done by pressure p".
Do you mean the work done from time t=0 to time t=0.1 seconds, when the sphere breaks apart?
The pressure-volume work from time 0 to 0.1 is
W=integral from t=0 to 0.1 of {P(t)dV(t)}
where P(t) is the pressure at time t, (and you told us that P=10*t), and V(t) is the volume at time t.
If the triangle has mass, then the pressure may also do mechanical work, accelerating the triangle. But you did not tell us the triangle's mass or its acceleration. If the triangle is made of elastic material, then the pressure may do work stretching the triangle. But you did not tell us the triangle's spring constant.
The computation of pressure-volume work requires knowing the total volume of the sphere at each instant. We cannot determine this from knowing just how the corners of one triangle are moving. The overall movement may be a combination of inflation/deflation plus translation plus rotation. We do not even know on which side of the triangle the pressure is applied. I assume the pressure on the "other" side is zero.
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William Rose 2022 年 9 月 30 日
Thank you for accepting this answer.

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