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I was playing around, and I found something unexpected. If we are analyzing a simple fluid jet:
We can apply Bernoulli's (which is Conservation of Energy) and arrive at:
$$ P_{1_{B}} = \frac{1}{2} \rho \left( v_2^2 - v_1^2 \right) = \frac{1}{2} \rho ( v_2 - v_1 ) ( v_2+ v_1) $$
It would seem you could also use the Momentum Equation:
$$ \sum \boldsymbol F = \frac{d}{dt} \int_{cv} \boldsymbol v \rho~dV\llap{-}+ \int_{cs} \boldsymbol v \rho \boldsymbol V \cdot d\boldsymbol A $$
Which says in words:
$$ \left[ \begin{array} ~sum~ of~ forces \\ ~acting ~on ~matter \\ ~in~ control ~volume \end{array} \right] = \left[ \begin{array}~time ~rate~ of \\ ~change ~of~ momentum \\ ~in ~control ~volume \end{array} \right] + \left[ \begin{array} ~ net ~outflow~ rate \\ ~of ~momentum \\ ~through~ control ~surface \end{array} \right]$$
Assuming constant properties across the control surface and steady flow (changing only in position, not time) for that nozzle its reduced to:
$$ P_{1_{M}} = \rho v_1 ( v_2 - v_1) $$
Comparing the two taken independently you get differing results for ##P_1##
$$ \frac{P_{1_{M}} }{P_{1_{B}}} = \frac{\rho v_1 ( v_2 - v_1)}{\frac{1}{2} \rho ( v_2 - v_1 ) ( v_2+ v_1)} = \frac{v_1}{\frac{1}{2} ( v_2 + v_1 )}$$
This implies that:
$$ P_{1_{M}} < P_{1_{B}}$$
I guess the difference comes from the fact that I've subtlety left out something in the momentum analysis. In order for the control volume to not be accelerating there must be a force acting on the nozzle in the momentum analysis that was not taken into account, perhaps this is where the discrepancy lies? I notice that in my fluids text they use momentum to solve for the force acting on the nozzle to keep it in place, by first invoking Bernoulli's to solve for the velocities ##v_1,v_2## given pressure ##P_1##.
We can apply Bernoulli's (which is Conservation of Energy) and arrive at:
$$ P_{1_{B}} = \frac{1}{2} \rho \left( v_2^2 - v_1^2 \right) = \frac{1}{2} \rho ( v_2 - v_1 ) ( v_2+ v_1) $$
It would seem you could also use the Momentum Equation:
$$ \sum \boldsymbol F = \frac{d}{dt} \int_{cv} \boldsymbol v \rho~dV\llap{-}+ \int_{cs} \boldsymbol v \rho \boldsymbol V \cdot d\boldsymbol A $$
Which says in words:
$$ \left[ \begin{array} ~sum~ of~ forces \\ ~acting ~on ~matter \\ ~in~ control ~volume \end{array} \right] = \left[ \begin{array}~time ~rate~ of \\ ~change ~of~ momentum \\ ~in ~control ~volume \end{array} \right] + \left[ \begin{array} ~ net ~outflow~ rate \\ ~of ~momentum \\ ~through~ control ~surface \end{array} \right]$$
Assuming constant properties across the control surface and steady flow (changing only in position, not time) for that nozzle its reduced to:
$$ P_{1_{M}} = \rho v_1 ( v_2 - v_1) $$
Comparing the two taken independently you get differing results for ##P_1##
$$ \frac{P_{1_{M}} }{P_{1_{B}}} = \frac{\rho v_1 ( v_2 - v_1)}{\frac{1}{2} \rho ( v_2 - v_1 ) ( v_2+ v_1)} = \frac{v_1}{\frac{1}{2} ( v_2 + v_1 )}$$
This implies that:
$$ P_{1_{M}} < P_{1_{B}}$$
I guess the difference comes from the fact that I've subtlety left out something in the momentum analysis. In order for the control volume to not be accelerating there must be a force acting on the nozzle in the momentum analysis that was not taken into account, perhaps this is where the discrepancy lies? I notice that in my fluids text they use momentum to solve for the force acting on the nozzle to keep it in place, by first invoking Bernoulli's to solve for the velocities ##v_1,v_2## given pressure ##P_1##.
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