What Is a Stream-Tube in Fluid Dynamics?

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Homework Help Overview

The discussion revolves around the concept of a stream-tube in fluid dynamics, specifically focusing on the continuity equation and Bernoulli's principle as they apply to a hosepipe and nozzle system. Participants are exploring the relationship between fluid speed and pressure in a horizontal pipe with varying cross-sectional areas.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants attempt to define a stream-tube and apply the continuity equation to relate the speeds and areas of the hosepipe and nozzle. Questions arise regarding the correct application of Bernoulli's equation and the interpretation of pressure differences in the system.

Discussion Status

Multiple interpretations of the problem are being explored, with some participants providing guidance on notation and setup. There is an ongoing examination of the relationships between variables, particularly concerning the speeds and pressures in the hosepipe and nozzle.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available for discussion. The problem involves assumptions about incompressible fluid flow and horizontal pipe conditions, which are being questioned and clarified throughout the thread.

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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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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.
 
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
 
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
 
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.
 
v1*d*A1 = v2*d*A2

v1*d*A1 = vo*d*A2

v0 = ( v1 * A1 ) / A2

is this correct for the speed?
 
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:
 
v1*d*A1 = v2*d*A2

v1*d*A1 = vo*d*A2

v1 = (vo * A2 ) / A1
 
(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?
 

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