# Streamline, stream-tube

1. Jan 1, 2008

### imy786

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

p + pgh + 1/2 pv^2

3. 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

2. Jan 1, 2008

### Staff: Mentor

Use the continuity equation.
Use Bernoulli's equation.

3. Jan 1, 2008

### imy786

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. Jan 1, 2008

### imy786

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. Jan 1, 2008

### Staff: Mentor

Careful with your notation. The nozzle area is A2 and the nozzle speed is v0. Redo this.
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. Jan 2, 2008

### imy786

v1*d*A1 = v2*d*A2

v1*d*A1 = vo*d*A2

v0 = ( v1 * A1 ) / A2

is this correct for the speed?

7. Jan 2, 2008

### Staff: Mentor

It would be, except that you should be solving for the speed in the hosepipe, which is v1, not v0.

8. Jan 2, 2008

### imy786

v1*d*A1 = v2*d*A2

v1*d*A1 = vo*d*A2

v1 = (vo * A2 ) / A1

9. Jan 2, 2008

### imy786

(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. Jan 2, 2008

### Staff: Mentor

Right.

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. Jan 5, 2008

### imy786

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. Jan 5, 2008

### Staff: Mentor

OK, now find p1 - p2.

Note: The speed in the nozzle is given as v0, so replace v2 with v0.

13. Jan 5, 2008

### imy786

(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. Jan 5, 2008

### Staff: Mentor

That last step is incorrect. Instead:

$$(\frac{v_0 A_2}{A_1})^2 = \frac{v_0^2 A_2^2}{A_1^2}$$

15. Jan 5, 2008

### imy786

Hi doc, what programme did you use to get your text with that type?

16. Jan 5, 2008

### Staff: Mentor

Last edited by a moderator: Apr 23, 2017