How is work done in a nozzle zero

Fluid flow gets accelerated through a nozzle. As the flow is accelerating( and there is a change in momentum). According to Newton's second law we can say a net force is acting on the fluid through a distance. So why is the work done equal to zero? What am I missing?.

Applying Bernoulli's principle the mechanical energy along a streamline remains constant, which would imply no work is done. Is the energy in the system used to do work on the system here instead of external energy flowing into the system? This is confusing me. PLease help.

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.

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.

• 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.
• This is a similar situation to when people say that "magnetic fields do no work"; the more nuanced version is that magnetic fields (and nozzles) redirect work done by something else. – probably_someone Jun 25 '18 at 15:23

Applying Bernoulli's principle the mechanical energy along a streamline remains constant

That is not true. The sum

$$\frac{1}{2}\rho v^2 + \rho gh + p$$

is constant, but mechanical energy of a liquid element of volume $$\Delta V$$ is given by

$$\left(\frac{1}{2}\rho v^2 + \rho gh\right) \Delta V.$$

There is no contribution due to pressure. This is because the liquid assumed in Bernoulli's equation is incompressible. The pressure term in that equation comes in from work of surrounding liquid on the liquid element, not from energy of that liquid element.