Why can water be sprayed farther when the pipe outlet is narrowed?

The pressure is the same in either case, but as pressure equals force divided by area, you are ejecting less water at higher force in order to maintain the same pressure, hence the droplet spray goes further. Flow rate doesn't (have to) come into it.

This phenomenon is called orifice effect. What happens is the stream discharging from a nuzzle converges to a narrower diameter and faster velocity downstream at a point called Vena contracta.

There are empirical methods, based on Bernoulli's law, to calculate the discharge and velocity.

$ q = c_d A2 frac {2 (p_1 - p_2)}{ ρ (1 - (A_2 / A_1) ^2)^{1/2} }$

$ text{A1 is the diameter of the pipe, A2 diameter of the nuzzle, A2<A1}$,

$c_d = frac {A{vena contracta}}{A_2}= discharge coeficient $

There are many chrats to find Cd. Usually it ranges around 60% to 80%, Engineering toolbox

Why can water be sprayed farther when the pipe outlet is narrowed?

Your assumption that the water will go further (i.e., velocity will be greater) when the pipe outlet is narrowed is incorrect. For an ideal fluid (assumptions below) the velocity will remain the same and the flow rate will decrease based on the Bernoulli equation and the continuity equation.

See diagram below.

Bernoulli equation (conservation of energy):

$P_{1}+frac{ρV_{1}^2}{2}+ρgz_{1}= P_{2}+frac{ρV_{2}^2}{2}+ρgz_{2}$

Continuity (conservation of mass) equation:

$A_{1}V_{1}= A_{2}V_{2}$

Assumptions:

1. Steady flow
2. Incompressible fluid
3. No or negligible viscosity
4. No friction losses
5. The diameter at the top of the tank (assumed circular) is much greater than the diameter of outlet.

Then given

$P_{1}=P_{2}$= 1 atm

$A_{1}>>A_{2}$ therefore, $V_{1}<<V_{2}$

$z_{1}-z_{2}= h$

The velocity at the outlet is

$$V_{2}=sqrt{2gh}$$

Note that this is identical to the velocity of an object initially at rest that falls to the ground from a height $h$, due to the conversion of potential to kinetic energy. It is independent of the diameter of the pipe outlet. If the velocity is the same and the cross sectional area of the pipe is decreased, then by the continuity equation the flow rate decreases.

Hope this helps.

It is assumed that the pipeline consists of an unshrinking pipeline L1 and a shrinking pipeline L2. L1 is much larger than L2. As the outlet of the pipeline shrinks and the flow is hindered, the velocity of L1 in the pipeline decreases, so the pressure drop also decreases. According to the following formula, the pressure drop is proportional to the square of the velocity, so when the velocity decreases, the pressure drop will decrease greatly.

$$h_f=λ frac{l}{d} frac{v^2}{2g}$$

Because the pressure drop is greatly reduced, pipeline L2 will get higher pressure, so the flow rate of pipeline L2 will increase.

Video Interpretation of Sprinkler Speed

This question has been bothering me for a while, and here's what I've come up with.

Bernoulli's equation is clearly the starting point to think about this problem: $p + frac{1}{2} rho v^2 + rho g h = E$ where $E$ is a quantity they is conserved throughout the fluid, basically its energy per unit volume. In other words, the sum of the kinetic energy, potential energy and pressure is the same everywhere.

With atmospheric pressure roughly equal at the surface of the tank and at the exhaust, all of the potential energy at the surface of the tank should be transformed into kinetic energy at the exhaust such that $v_{text{exhaust}} = sqrt{2gh}$.

Note that the diameters of the pipe or of the exhaust aren't in this formula, so under conditions where the Bernoulli equation applies, the narrowing in the nozzle doesn't change anything.

So, since we know this isn't true... what gives?

An effect that's not captured by Bernoulli's equation are friction losses in the pipe: friction in viscous fluids is generally proportional to $v^2$ (see Navier-Stokes equation) and velocity is proportional to the cross-sectional area of the pipe so that doubling the diameter of the pipe while keeping the exhaust diameter the same would reduce the friction losses by a factor of 16 (not accounting for the change in exhaust velocity, but you get the idea...)

These losses have the form $alpha l v^2_{text{pipe}}$ where $alpha$ is some coefficient which would include things like the viscosity of water and $l$ is the length of the pipe since it makes sense that the losses would scale linearly with that.

Finally, call $beta$ the ratio of the exhaust diameter to pipe diameter. From this we get $v_{text{pipe}} = beta^2 v_{text{exhaust}}$.

Now equate fluid energy at the exhaust (including losses from the pipe) to energy at the top of the tank (using Bernoulli's equation the $p$ terms since they are almost equal on both ends and dividing by $rho$). You get:

$frac{1}{2} v_{text{exhaust}}^2 + alpha beta^4 l v_{text{exhaust}}^2 = g h$

Solving for $v_{text{exhaust}}$, we get: $v_{text{exhaust}}=sqrt{frac{2gh}{1 + 2 alpha beta^4 l}}$

Note that taking pipe length $l rightarrow 0$, or friction coefficient $alpha rightarrow 0$ or diameter of the exhaust $beta rightarrow 0$ yields the friction-less Bernoulli formula.

(I'm not a physicist so I look forward to replies with corrections/confirmation)