Differences in Water Pressure in a Horizontal Pipe with Changing Radius

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The discussion focuses on calculating the pressure difference in a horizontal pipe with varying radius, specifically from 6 cm to 4 cm, while maintaining a volume flow rate of 0.008 m³/s. The continuity equation indicates that the flow velocity must increase as the pipe narrows, leading to a change in pressure according to Bernoulli's principle. The participants suggest using the areas of the pipe sections to determine the velocities and subsequently apply these values to the pressure equation. The key challenge is to find the necessary force to produce the acceleration of water as it flows through the narrower section. Ultimately, the solution involves calculating the velocities and substituting them into the pressure equation to find the difference in pressure between the two regions.
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Homework Statement


Water is flowing through a horizontal pipe that has a radius of 6 cm gradually narrowing to a
radius of 4 cm. The volume flow rate is 0.008 m3/s. What is the difference in the water pressurebetween the 6 cm radius region and the 4 cm radius region of the pipe?


Homework Equations



P + .5pv^2 + pgy = constant


The Attempt at a Solution



y is 0 so that part cancels

p(density) is 1000 kg/m^3 for water

P1 + .5pv1^2 = P2 + .5pv2^2

Not sure where to go from there
 
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Since water can't "build up" any where in the pipe, the same volume of water must flow through both 6 cm radius and 4 cm radius portions in the pipe at the same time. Since the area of of the 4 cm radius portion is smaller than the 6 cm radius portion, the water must flow faster through the 4 cm portion to do that. That is, it must accelerate. How much faster must it go? What force is necessary to product that acceleration? That force, divided by the area is the pressure difference.
 
HallsofIvy said:
What force is necessary to product that acceleration?

Any hints?

choice 1: pressure?

If so, how?
 
Last edited:
The volume flow rate Q = A1*v1 = A2*v2.
from the radii of the pipes you can find A1 and A2. From that you can find v1 and v2. Substitute these values in the relevant equation to find P1 - P2.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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