Finding pressure at narrow end of segment

  • Thread starter Thread starter sona1177
  • Start date Start date
  • Tags Tags
    Pressure
AI Thread Summary
The discussion centers on calculating the pressure at the narrow end of a tapered pipe segment, given the pressure and speed at the larger end. The user initially attempts to use the equation P = ρg(d) but struggles with the definition of "d" and its relevance in this context. Clarification reveals that "d" was mistakenly interpreted as the length of the narrow segment instead of its diameter. Participants emphasize the need to apply Bernoulli's equation correctly, focusing on the fluid's density, velocity, and pressure without considering potential energy since the pipe is horizontal. The conversation highlights the importance of understanding the relationship between cross-sectional area and fluid velocity in pressure calculations.
sona1177
Messages
171
Reaction score
1

Homework Statement


A horizontal segment of pipe tapers from a cross-sectional area of 50.0 cm^2. The pressure at the larger end of the pipe is 1.20 x 10^5 Pa and the speed is .040 m/s. What is the pressure at the narrow end of the segment?


Homework Equations


I know how to do the problem using Bernoulli's equation but why can't I manage to get the correct answer using P=rho(g)(d). where rho=1000, g=9.8, and d=.00798 m


The Attempt at a Solution


 
Physics news on Phys.org
Could you explain what d is in your equation.
As the pipe is horizontal there is no potential energy consideration.
I also note that there is no information given about the change in the cross section area.
Maybe you could clarify the question?
 
Stonebridge said:
Could you explain what d is in your equation.
As the pipe is horizontal there is no potential energy consideration.
I also note that there is no information given about the change in the cross section area.
Maybe you could clarify the question?

d is the length of the narrow segment which I obtained by dividing area/pi and taking the sq root to get the radius. I then multiplied that by 2.
 
Stonebridge said:
Could you explain what d is in your equation.
As the pipe is horizontal there is no potential energy consideration.
I also note that there is no information given about the change in the cross section area.
Maybe you could clarify the question?

Ooops! Sorry about leaving that info out. It tapers from 50-> 5 cm^2
 
I don't quite understand what you are doing here.
The expression ρhg in the Bernoulli formula includes the terms
ρ the density of the fluid
g acceleration due to gravity
h the height of the liquid above the reference level
You have written ρdg but stated that d is "the length of the narrow segment". However, your calculation looks like it is for the diameter of the narrow segment.
Neither of these values, length or diameter, are the ones in the expression.
As the tube is horizontal, you don't need this term anyway, you just need the terms
½ρv² and p to find the pressure.
The value of v is found from the volume flowing per second being equal in both parts of the tube.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
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}...
Back
Top