In the derivation of Bernoulli's equation through conservation of energy, as shown in http://www.4physics.com/phy_demo/bernoulli-effect-equation.html, the net work done on a section of fluid is calculated and equated to the change in its mechanical energy. According to the site, two forces do work on the fluid; namely, an external force $F_1$ driving the fluid forward, and a force $F_2$ exerted at the right end of the fluid. How is the force $F_2$ exerted on the fluid, and why is it directed opposite to the motion?
[Physics] Derivation of Bernoulli’s Equation
bernoulli-equationenergyenergy-conservationfluid dynamics
Related Solutions
Your assumption that the work done is zero is wrong.
Work is being done everywhere on the fluid by an external force: the pressure in the pipe. To see this, let's derive Bernoulli's Equation: $P_1 + 1/2 \rho v_1^2 + \rho g y_1 = P_2 + 1/2 \rho v_2^2 + \rho g y_2 $
Lets assume your pipe narrows like a nozzle and begin with conservation of energy and see what happens to a small section of incompressible fluid of density $\rho$ and volume $V$: $$ W_{external} = \Delta K + \Delta U $$ The pressure in the pipe is the external agent here causing work to be done. $W = F \cdot \Delta x = (P A) \Delta x$. If the fluid is flowing from high pressure to low pressure, in one direction the force will be with $\Delta x$: $$ W_1 = P_1 A_1 \Delta x_1$$
and opposite for the low pressure side: $$ W_2 = - P_2 A_2 \Delta x_2$$
Since the fluid is incompressible, $A\Delta x = V$ for any part of the pipe. Putting this into the work-energy relationship: $$ P_1 V - P_2 V = 1/2 (\rho V)v_2^2 - 1/2 (\rho V)v_1^2 + (\rho V)g y_2 - (\rho V) g y_1 $$ Where the mass of the fluid is $m= \rho V$. Clearly the Volume cancels out and leaves you with Bernoulli's equation.
answer
Work is being done by whatever is supplying the pressure. This work is accounted for by the change in pressure in the fluid. When someone writes $P + 1/2 \rho v^2 + \rho g y = Constant$ what they are really writing is $W/V + K/V + U/V = E/V$, where $E/V$ is the energy per unit volume of fluid (J/m^3). So you see, the work done is hidden in the change in pressure.
additional edits:
- It is true to say that nozzles do no work in the same way its true normal forces do no work: their forces are always perpendicular to the direction of motion. It is wrong however to conclude that no work is done.
- It is true to say that the energy / volume doesn't change on a streamline. As long as that definition of energy includes work done on/by the fluid. So either write $\Delta K + \Delta U_{gravity} + \Delta U_{pressure} = 0$ or $\Delta K + \Delta U_{gravity} = \Sigma W_{pressure}$. But to say $\Delta K + \Delta U_{gravity} = 0$ is in general wrong.
Best Answer
The forces are provided by the pressure of fluid further on. The force $F_1$ is caused by the pressure of the fluid 'behind' the body of fluid considered, the force $F_2$ is caused by the fluid 'ahead' pushing back. Pressure is an internal force for a fluid, any surface within a fluid volume will have the force of pressure acting across it.