Volume Flow Rate: Open vs Closed Valves

AI Thread Summary
The discussion revolves around the impact of valve positions on volume flow rate calculations. When a valve transitions from open to closed, it affects the flow rate through the system, potentially leading to different distributions at the outlets. Participants clarify that if there is 1 m³/s flowing into the system, it can be evenly distributed to multiple outlets, such as 0.5 m³/s each for two outlets. This understanding is crucial for accurately solving related problems in fluid dynamics. Overall, the conversation emphasizes the importance of correctly interpreting flow rates in systems with varying valve states.
Bolter
Messages
262
Reaction score
31
Homework Statement
See below
Relevant Equations
volume flow rate = vA
Made an attempt at this Q but I'm unsure on how to do part c) or if I had even done part a) and b) right

Screenshot 2020-10-14 at 22.56.45.png

IMG_5364.JPG

IMG_5365.JPG


What difference does it make to the volume flow rate equation when the valve goes from open to closed?

Any help would be appreciated! Thanks
 
Physics news on Phys.org
How can there be 1m3/s flowing in and 1m3/s going out of both outlets?
 
haruspex said:
How can there be 1m3/s flowing in and 1m3/s going out of both outlets?

So do I assume that the inlet has 1 m^3/s flowing in
and both outlets at the end have 0.5 m^3/s flowing out
 
Bolter said:
So do I assume that the inlet has 1 m^3/s flowing in
and both outlets at the end have 0.5 m^3/s flowing out
For this system, yes.
 
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