# reynold's transport theorom contradicting bernoulli equation

I am considering a funnel with two areas $A_1$ and $A_2$ with velocity at the interface as $V_1$ and $V_2$ respectively. let pressures at those areas be $P_1$ and $P_2$ . For simplification we assume no gravity. therefore our bernoulli equation becomes $$\frac{P_1}{\rho}+\frac{1}{2}{V_1}^2=\frac{P_2}{\rho}+\frac{1}{2}{V_2}^2$$ therefore $$P_1=P_2+{\frac{\rho}{2}({V_2}^2-{V_1}^2)}$$ now we solve the same question using reynold's transport theorem. i have taken the funnel as my control volume. $$F=d/dt{\int_{CV}{\rho}V\forall}+\int_{CS}{V}{\rho}{\vec{V}.d{\vec{A}}}$$ now as it's a steady state first integral turns $0$ and on solving the second integral we can see $$F=\int_{CS}{V}{\rho}{\vec{V}.d{\vec{A}}}={{\rho}({V_2}^2A_2-{V_1}^2A_1)}$$ now we can see that $$F=F_s+F_b$$ where $F_s$ is surface forces and $F_b$ is body force. now $F_b=0$ as we are neglecting body forces and assuming $0$ viscosity $$F_s={{\rho}({V_2}^2A_2-{V_1}^2A_1)}=P_1A_1 -P_2A_2$$ therefore from the above equation $$P_1=P_2\frac{A_2}{A_1}+{{\rho}({V_2}^2\frac{A_2}{A_1}-{V_1}^2)}$$ now equating both equations we get , we find , $$P_2+{\frac{\rho}{2}({V_2}^2-{V_1}^2)}=P_2\frac{A_2}{A_1}+{{\rho}({V_2}^2\frac{A_2}{A_1}-{V_1}^2)}$$ now we find $$P_2(1-\frac{A_2}{A_1})={{\rho}({V_2}^2\frac{A_2}{A_1}-{V_1}^2)}-{\frac{\rho}{2}({V_2}^2-{V_1}^2)}$$thus $$P_2=\frac{{{\rho}({V_2}^2\frac{A_2}{A_1}-{V_1}^2)}-{\frac{\rho}{2}({V_2}^2-{V_1}^2)}}{(1-\frac{A_2}{A_1})}$$ now this seems very absurd to me as for values of velocities and areas we are getting a finite value of $P$ which should not be the case as system should depend on pressure difference rather than $P$ value at a particular area. This seems counter-intuitive to me. Have i made a conceptual error or the result that i have obtained is right?

• Why do you say $\rho\left(V_2^2A_2-V_1^2A_1\right)=P_1A_1-P_2A_2$ in the expression for $F_s$? Specifically how do you relate $P$ to $\rho V^2$? – nluigi Feb 19 '16 at 17:55

Your mistake is in writing the equation: $$F_s={{\rho}({V_2}^2A_2-{V_1}^2A_1)}=P_1A_1 -P_2A_2$$
It should read: $$P_1A_1 -P_2A_2-F_s={{\rho}({V_2}^2A_2-{V_1}^2A_1)}$$
• Ah, i forgot about the fact that we would need a force to keep the system intact. i thought $F_s$ accounts for surface forces thus simply wrote the equation. Btw is it possible to derive bernoulli from reynold's transport theorem ? . Thanks for clearing my doubt , it was an insight i did not have before :) – avz2611 Feb 19 '16 at 18:15
• $F_s$ is mainly the result of the pressure forces imposed by the slanted surfaces of the funnel. These pressure forces have a component in the axial direction. Usually, there is also a drag contribution to Fs. It is not possible to derive the bernoulli equation from the reynolds transport theorem. Bernoulli expresses conservation of mechanical energy, and reynolds transport expresses conservation of momentum. – Chet Miller Feb 19 '16 at 19:45