If in a fluid of density 1000 kg/m^3, confined to run in a series of full pipes, the velocity of the fluid changes from 9 m/s to 7.7 m/s with no change in altitude, what will be the change in pressure? Assume that no dissipative forces act on the fluid.
`rho g y will not change, since there is no change in altitude y. Thus .5 `rho v^2 + P will remain constant, so that the change in pressure will be the negative of the change in .5 `rho v^2.
The change in .5 `rho v^2 will be .5 `rho v2^2 - .5 `rho v1^2 = .5 `rho (v2^2-v1^2) = .5 * 1000 kg / m^3 * [ ( 7.7 m/s)^2 - ( 9 m/s)^2 ] = -10855 N/m^2.
The pressure change will be the negative of this, or 10855 N / m^2.
When the velocity of a confined fluid changes at constant altitude the change in pressure `rho g y remains unchanged to .5 `rho v^2 + P remains constant, and therefore
It follows that `dP = -`d(.5 `rho v^2) = - [ .5 `rho v2^2 - .5 `rho v1^2 ] = .5 `rho (v1^2 - v2^2).
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