What Is a Stream-Tube in Fluid Dynamics?

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In summary, the hosepipe has an internal cross sectional area A1 and a nozzle with a hole whose cross-sectional area is A2. The speed of the water as it emerges from the nozzle is v0. If the speed of the water as it emerges from the nozzle is vo, determine its speed in the hosepipe. Also, determine the excess pressure inside the hosepipe, which is horizontal.
  • #1
imy786
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Homework Statement



q1.

a)Explain the terms stream-tube.

b)State the equation of continutity along a stream-tube for an incomperssible fluid.

c)A hosepipe with an internal cross sectional area A1 has at its end a nozzle with a hole whose cross-sectional area is A2 (A2<A1). If the speed of the water as it emerges from the nozzle is vo, determine its speed in the hosepipe. Also, determine the excess pressure inside the hosepipe, which is horizontal.



Homework Equations



p + pgh + 1/2 pv^2

The Attempt at a Solution



a) A thin bundle of adjacent streamlines forming a stream tube.

b) equation of continutity :

v1*d*A1 = v2*d*A2

for an incompressible fluid.

c)

p + Pgh + 1/2 Pv^2

v1*d*A1 = v2*d*A2

V1= vo

need help on this
 
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  • #2
imy786 said:
c)A hosepipe with an internal cross sectional area A1 has at its end a nozzle with a hole whose cross-sectional area is A2 (A2<A1). If the speed of the water as it emerges from the nozzle is vo, determine its speed in the hosepipe.
Use the continuity equation.
Also, determine the excess pressure inside the hosepipe, which is horizontal.
Use Bernoulli's equation.
 
  • #3
v1*d*A1 = v2*d*A2

v2= (vo *A1 ) / A2

----------------------------

p + Pgh + 1/2 Pv^2= constant

v2= (vo *A1 ) / A2

p + Pgh + 1/2 P ((vo *A1 ) / A2) ^2
 
  • #4
v1*d*A1 = v2*d*A2

v2= (vo *A1 ) / A2

----------------------------

p + Pgh + 1/2 Pv^2= constant

v2= (vo *A1 ) / A2

p + Pgh + 1/2 P ((vo *A1 ) / A2) ^2
 
  • #5
imy786 said:
v1*d*A1 = v2*d*A2

v2= (vo *A1 ) / A2
Careful with your notation. The nozzle area is A2 and the nozzle speed is v0. Redo this.
p + Pgh + 1/2 Pv^2= constant

v2= (vo *A1 ) / A2

p + Pgh + 1/2 P ((vo *A1 ) / A2) ^2
Redo this with the correct values for the speeds. Take advantage of the fact that the pipe is horizontal.

Set it up like this:
(p + Pgh + 1/2 Pv^2) for hosepipe = (p + Pgh + 1/2 Pv^2) for nozzle

Find the change in pressure.
 
  • #6
v1*d*A1 = v2*d*A2

v1*d*A1 = vo*d*A2

v0 = ( v1 * A1 ) / A2

is this correct for the speed?
 
  • #7
imy786 said:
v1*d*A1 = v2*d*A2

v1*d*A1 = vo*d*A2

v0 = ( v1 * A1 ) / A2

is this correct for the speed?
It would be, except that you should be solving for the speed in the hosepipe, which is v1, not v0. :wink:
 
  • #8
v1*d*A1 = v2*d*A2

v1*d*A1 = vo*d*A2

v1 = (vo * A2 ) / A1
 
  • #9
(p + Pgh + 1/2 Pv^2) for hosepipe = (p + Pgh + 1/2 Pv^2) for nozzle

if v1 = (vo * A2 ) / A1

now to find the change in the pressure-

(p + Pgh + 1/2 Pv^2) for hosepipe - (p + Pgh + 1/2 Pv^2) for nozzle = 0
 
  • #10
imy786 said:
v1*d*A1 = v2*d*A2

v1*d*A1 = vo*d*A2

v1 = (vo * A2 ) / A1
Right.

imy786 said:
(p + Pgh + 1/2 Pv^2) for hosepipe = (p + Pgh + 1/2 Pv^2) for nozzle

if v1 = (vo * A2 ) / A1
Plug in the values for hosepipe (p1, v1) and nozzle (p2, v0) and find the difference in pressure (p1 - p2); since the pipe is horizontal, the Pgh terms drop out.
 
  • #11
c) determine the excess pressure inside the hosepipe, which is horizontal.

(p1 + Pgh + 1/2 Pv1^2) for hosepipe - (p2 + Pgh + 1/2 Pv2^2) for nozzle = 0

since pipe is horizontal Pgh drops out.

(p1 + 1/2 Pv1^2) for hosepipe - (p2 + 1/2 Pv2^2) for nozzle = 0

if v1 = (vo * A2 ) / A1

then

(p1 + 1/2 P *v1^2) - (p2 + 1/2 Pv2^2) = 0

(p1 + 1/2 P*[(vo * A2 ) / A1]^2) - (p2 + 1/2 Pv2^2) = 0
 
  • #12
OK, now find p1 - p2.

Note: The speed in the nozzle is given as v0, so replace v2 with v0.
 
  • #13
(p1 + 1/2 P*[(vo * A2 ) / A1]^2) = (p2 + 1/2 Pv0^2)

[vo * A2 ) / A1]^2 = (vo * A2 ) / A1 * (vo * A2 ) / A1 = A1 ^-2 (vo^2 + 2A2 + A2^2)

p1 + 1/2 P*[(A1 ^-2 (vo^2 + 2A2 + A2^2)] = (p2 + 1/2 Pv0^2)

p1-p2= 1/2 Pv0^2 - 1/2 P*[(A1 ^-2 (vo^2 + 2A2 + A2^2)]
 
  • #14
imy786 said:
[vo * A2 ) / A1]^2 = (vo * A2 ) / A1 * (vo * A2 ) / A1 = A1 ^-2 (vo^2 + 2A2 + A2^2)
That last step is incorrect. Instead:

[tex](\frac{v_0 A_2}{A_1})^2 = \frac{v_0^2 A_2^2}{A_1^2}[/tex]
 
  • #15
Hi doc, what programme did you use to get your text with that type?
 
  • #16
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Related to What Is a Stream-Tube in Fluid Dynamics?

1. What is a streamline?

A streamline is a line that is tangent to the velocity vector at every point in a fluid flow. It represents the path of a fluid particle as it moves through the flow.

2. How are streamlines used in fluid mechanics?

Streamlines are used to visualize and analyze fluid flow patterns. They can help identify areas of high and low velocity, separation points, and potential areas of turbulence.

3. What is a stream-tube?

A stream-tube is a bundle of streamlines that represent the flow through a specific cross-sectional area. It is often used to calculate the mass flow rate of a fluid in a given system.

4. How do streamlines and stream-tubes differ?

Streamlines represent the path of individual fluid particles, while stream-tubes represent the flow through a specific cross-section. Stream-tubes can be thought of as a 3-dimensional representation of streamlines.

5. Can streamlines and stream-tubes intersect?

No, streamlines and stream-tubes cannot intersect. This is because a streamline represents the path of a single fluid particle, and a stream-tube represents the flow through a specific cross-section. Therefore, they cannot cross paths in a fluid flow.

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