How much is the cross sectional area for the tube to fill up the compartment?

AI Thread Summary
To determine the cross-sectional area (A) for a tube to fill a compartment with a volume of 0.0001 m³ in 60 seconds, the volume flow rate equation, which relates flow rate to cross-sectional area and water velocity, is essential. The volume flow rate can be calculated as the change in volume over time. However, additional factors such as tube length and pressure difference also influence flow rate and should be considered, especially regarding whether the flow is turbulent or laminar. Resources like Bernoulli's equation and various online calculators can provide further insights into these relationships. Understanding these principles will help accurately calculate the required cross-sectional area for the tube.
aiklone1314
Messages
25
Reaction score
0

Homework Statement


abstract question, but we can assume the
volume,V to be filled is 0.0001m3 and
time for it to filled up water is 60seconds,
water density is 1000kg/m3,
so how much should be the cross sectional area,A for the tube?

Homework Equations


Volume flow rate=delta Volume/delta time
Volume flow rate= cross sectional area*velocity of water

Relavant Picture:
When 0.0001m3 of water is filled up in the compartment, the compartment of the verge will move downward as shown in the rightside of the picture.
http://img339.imageshack.us/img339/8789/21885237.jpg
 
Last edited by a moderator:
Physics news on Phys.org
You need an equation for how flow rate depends on length of tube, cross-sectional area of tube, pressure difference along the tube and properties of the fluid. See what you can find online.
 
haruspex said:
You need an equation for how flow rate depends on length of tube, cross-sectional area of tube, pressure difference along the tube and properties of the fluid. See what you can find online.

So is this equation is needed to be used?

Volume flow rate= cross sectional area*velocity of water
 
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