How does Boyles law apply to air in motion in a uniform tube?

AI Thread Summary
The discussion centers around applying Boyle's law to air in motion within a uniform tube, focusing on the mathematical derivations related to fluid dynamics. Participants explore the velocity components of a three-dimensional flow and how to demonstrate that the streamlines are helices. The conversation also addresses the irrotational velocity field associated with a given potential and the incompressible continuity equation. Additionally, the relationship between velocity potential and stream function in two-dimensional incompressible flow is examined. The thread concludes with participants seeking guidance on applying Bernoulli's equation to derive specific fluid dynamics equations.
Mechdude
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Homework Statement



1.a) the velocity components of a 3-d flow are
u= \frac{ax}{x^2+y^2}
v= \frac{ay}{x^2+y^2}
\omega = c
where a and c are arbitrary constants . show that the streamlines of this
flow are helics
x=acos(t) ;
y=asin(t) ;
x=a*c*t

a. what is the irrotational velocity filed associated with the potential
\phi = 3x^2 -3x +3y^2 + 16t^2 +12zt ? does the flow satisfy the
incompresible continuity equation \nabla \vec{q} = 0 where q is the velocity.

2. b) the velocity potential of a 2D incompressible flow is
\phi = \frac{1}{2} log \left( \frac{(x+a)^2 +y^2}{(x-a)^2+y^2 } \right)
show that the stream function \psi is given by:
\psi = \arctan \frac{y} {x-a} - \arctan \frac{y}{x-a}

3 a)
Air obeying Boyles law p=k \rho is in motin in a uniform tube of small
cross-sectional area. show that if \rho is the density and u is the velocity
at a distance x from a fixed point a; and t is time , this is true:
\frac{ \partial^2 \rho}{\partial t^2} = \frac {\partial^2 (u^2 + k) \rho}{\partial x^2}

3 b. A steam is rushing from a boiler throught a conical pipe, the diameters of the ends a
being D and d . if v and u are the corresponding velocities of the steam an if the motion is
supposed to be that of divergence from the vortex of the cone prove that
\begin{displaymath} \frac{u}{v} = \frac {D^2} {d^2} e^{\frac{u^2-v^2}{2k} } \end{displaymath}
where k is the pressure divide by the density and its a constant ie k= \frac{p}{\rho}
note its getting in at one end with a velocity v and density \rho_1
and out the other side with u and \rho_2






Homework Equations



\nabla \vec{q} = 0
Boyles law: p=k \rho

The Attempt at a Solution


i do not know how to start this stuff with all honesty.
 
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Mechdude said:

Homework Statement



1.a) the velocity components of a 3-d flow are
u= \frac{ax}{x^2+y^2}
v= \frac{ay}{x^2+y^2}
\omega = c
where a and c are arbitrary constants . show that the streamlines of this
flow are helics
x=acos(t) ;
y=asin(t) ;
x=a*c*t

You've got u, v, w as well as x, y, z. To prove the equations for u,v,w are equivalent to those of x,y,z, you have to either integrate the velocity equations or derive the position equations. Integrating is hard; deriving is much easier.

a. what is the irrotational velocity filed associated with the potential
\phi = 3x^2 -3x +3y^2 + 16t^2 +12zt ? does the flow satisfy the
incompresible continuity equation \nabla \vec{q} = 0 where q is the velocity.

The velocity field is the gradient of the potential, so just find the gradient of that equation. As for the continuity equation, q is the gradient of the potential and the continuity equation takes the curl of q. What's the curl of a gradient always equal to?
2. b) the velocity potential of a 2D incompressible flow is
\phi = \frac{1}{2} log \left( \frac{(x+a)^2 +y^2}{(x-a)^2+y^2 } \right)
show that the stream function \psi is given by:
\psi = \arctan \frac{y} {x-a} - \arctan \frac{y}{x-a}

The curl of the stream function is the velocity field. Do you know how to get from a velocity field to a stream function?
 
Thanks 4 the reply , i managed to do problems, 1. b) & 2.b) , any clues for the rest, part of my problem is i don't know where to start, take 3a) &b) I am thinking of using bernoullis equation, but i don't know how to arrive at the second derivative of density nor the exponent respectively
 
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