Where does this equation come from?

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The discussion centers on the derivation of the volumetric flow rate equation for air through a tube, specifically Q = -pi*d^4/(128*viscosity)*dp/dx. The equation relates flow rate (Q) to the pressure gradient (dp/dx) and viscosity, with a focus on how velocity (v) can be expressed through the simplified equation of motion. It is noted that the velocity profile varies across the tube's cross-section, being zero at the walls and maximum at the center. The relationship between mass density and flow is also highlighted, emphasizing the integration over the tube's area. The conversation aims to clarify the origins and implications of the equation in fluid dynamics.
hermano
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Hi,

In a textbook I found an equation for the volumetric flow rate of air through a channel (tube). I can't find from where the equation is coming, someone an idea?

The equation which I mean is: Q = - pi*d^4/(128*viscosity)* dp/dx

I know that Q = A*v = pi*d^2/4 *v

Also from the simplified equation of motion 'v' can be calculated as:

dp/dx = viscosity * ddv/ddy (dd = second derivative)

from this equation v = ?? if I fill v in, in Q = pi*d^2/4 *v will this give Q = - pi*d^4/(128*viscosity)* dp/dx ??
 
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Work is being done by the Pressure gradient, to friction with the walls, due to the viscosity. ... your Q is mass density times v dot dA, integrated over the tube's cross-section Area. the velocity goes to zero at the tube surface, and is maximum in its center.
 
the density comes from massXvolume

and the mass comes from 1/2mv^2=Ek
i hope this helped a little bit
 
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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