Physics Asked by Gaurav L on January 5, 2021

Does it mean adding really small pressures on one side of the particle and subtracting from the other side to cancel out?

Also, what’s with the delta sign and it’s reverse? What’s it called? (Sign on the lhs and rhs)

There are some inconsistencies in the setup of the problem however the author does reach the correct conclusion. Let me clarify...

Identify the average pressure $p$ being at the center of the particle of size $delta s$ and assume that there is a pressure drop $delta p_s$ over the particle, then at the faces of the particle we have: $$pleft(spmfrac{delta s}{2}right) = pleft(sright)pmfrac{delta p_s}{2}$$ In other words, the pressure at the faces compared to the center are decreased and increased respectively according to the pressure drop over the particle.

The Taylor expansion is after substituting the previous result: $$pleft(spmfrac{delta s}{2}right) = pleft(sright)pm left.frac{partial p}{partial s}right|_sleft(frac{delta s}{2}right) + O(s^2)$$ $$delta p_s = frac{partial p}{partial s}delta s$$ Notice the factor 2 compared to the result in the text.

The force on the particle: $$delta F_{sn} =left(p-frac{delta p_{s}}{2}right)delta ndelta y-left(p+frac{delta p_{s}}{2}right)delta ndelta y =-delta p_{s}delta ndelta y =-frac{partial p_{s}}{partial s}delta sdelta ndelta y =-frac{partial p_{s}}{partial s}delta V$$ which is the same result as in the text

Answered by nluigi on January 5, 2021

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