Needle and pressure (Bernoulli's equation?)

  • #1
~christina~
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[SOLVED] Needle and pressure (Bernoulli's equation?)

Homework Statement


A hypodermic syringe contains medicine having the density of water. Barrel has cross sectional area A= 2.50x10^-5 m^2 and the needle has a cross sectional area a= 1.00x10^-8 m^2. IN the absence of a force on the plunger the pressure everywhere is 1 atm.
A force of magnitude 2.00N acts on the plunger making medicine squirt from the needle. Determine the speed of the medicine as it leaves the needle's tip.

http://img123.imageshack.us/img123/1442/66319119wa1.th.jpg [Broken]


Homework Equations


A1v1= A2v2

[tex] P_1 + 1/2 \rho v^2 + \rho gy_1 = P_2 + 1/2\rho v^2 + \rho g y_2 [/tex]

P= F/A


The Attempt at a Solution



I'm not sure how to get started but I'm thinking that if the plunger is pushed with a force of 20N then

[tex]P_1= 20.0N/ 2.50x10^-5m^2 [/tex]

[tex] P_1= 8e5 Pa [/tex]

I was then thinking of plugging that into the Bernoulli's equation b/c it relates pressure however....I'm confused again...

using the bernoulli's equation... noting that no change in height

[tex]P_1 + 1/2 \rho v_1^2 = P_2 + 1/2 \rho v_2^2 [/tex]

[tex]v_1= (A_2/A_1) *v_2 [/tex] => no v1 or v2....:frown:

I'm basically not sure since I do not have v1 or v2 but I do have P1...
P2 is different as is v2. v1 I was initially thinking it was at rest but then I thought it could not be since the plunger moved with force of 20N
Now I'm starting to think it relates to :

[tex]\Sum F= F_{plunger} - Fatm= ma [/tex] => (where I can say P=F/A) but how can I do that if I don't have the m and not to mention I have no t or x so I wouldn't be able to use a as acceleration anyhow


I really would appreciate it if someone could help me out (especially with how to apply F to the equations to solve for v2)

Thanks alot
 
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Answers and Replies

  • #2
Doc Al
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I'm not sure how to get started but I'm thinking that if the plunger is pushed with a force of 20N then

[tex]P_1= 20.0N/ 2.50x10^-5m^2 [/tex]

[tex] P_1= 8e5 Pa [/tex]
The plunger force was only 2 N, not 20 N. Also, the pressure due to the plunger force is in addition to atmospheric pressure.

I was then thinking of plugging that into the Bernoulli's equation b/c it relates pressure however....I'm confused again...

using the bernoulli's equation... noting that no change in height

[tex]P_1 + 1/2 \rho v_1^2 = P_2 + 1/2 \rho v_2^2 [/tex]

[tex]v_1= (A_2/A_1) *v_2 [/tex] => no v1 or v2....:frown:
Since you know how v1 and v2 are related, you can write Bernoulli's equation with only one unknown variable. Everything else is known.

(Take advantage of the fact that A_2/A_1 << 1.)
 
  • #3
~christina~
Gold Member
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The plunger force was only 2 N, not 20 N. Also, the pressure due to the plunger force is in addition to atmospheric pressure.
Oops...

[tex]P_1= 2.0N/ 2.50x10^-5m^2 [/tex]

[tex] P_1= 8e4 Pa [/tex]

Hm...so then I guess I'd use this plus the atmospheric pressure and use that as P1
like so: P1= 8e4Pa+ 1.013x10^5Pa= 1.813e5


Since you know how v1 and v2 are related, you can write Bernoulli's equation with only one unknown variable. Everything else is known.

(Take advantage of the fact that A_2/A_1 << 1.)
okay so I would plug in for v1 since v1= (A2/A1)v2

What would I assume P2 to be though?

[tex]P_1 + 1/2 \rho v_1^2 + \rho gy_1= P_2 + 1/2 \rho v_2^2 = \rho gy_2 [/tex]

[tex] 1.813e5 Pa + 1/2 \rho [(A2/A1)^2 v_2^2]= P_2 + 1/2\rho v_2 [/tex]

[tex] 1.813e5 Pa + 1/2 \rho (1.6x10^{-7}) v_2^2= P_2 +1/2\rho v_2 [/tex]

But again...I'm stuck with what is P2?

Thank you :smile:
 
  • #4
Doc Al
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Since the needle end is still exposed to the atmosphere, P2 = atmospheric pressure.
 
  • #5
~christina~
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I almost forgot about this Q.. well here's what I did.

Since the needle end is still exposed to the atmosphere, P2 = atmospheric pressure.
[tex]P_1= 2.0N/ 2.50x10^-5m^2 [/tex]

[tex] P_1= 8e4 Pa [/tex]

Hm...so then I guess I'd use this plus the atmospheric pressure and use that as P1
like so: P1= 8e4Pa+ 1.013x10^5Pa= 1.813e5

okay so I would plug in for v1 since v1= (A2/A1)v2

[tex]P_1 + 1/2 \rho v_1^2 + \rho gy_1= P_2 + 1/2 \rho v_2^2 = \rho gy_2 [/tex]

[tex] 1.813e5 Pa + 1/2 \rho [(A2/A1)^2 v_2^2]= P_2 + 1/2\rho v_2 [/tex]

[tex] 1.813e5 Pa + 1/2 \rho (1.6x10^{-7}) v_2^2= P_2 +1/2\rho v_2 [/tex]

-----
now...using what you said: P2= atmospheric pres= 1.013x10^5 Pa

here's where I get stuck..

[tex] 1.813e5 Pa + 1/2 \rho (1.6x10^{-7}) v_2^2= P_2 +1/2\rho v_2 [/tex]

[tex] 1.813e5 Pa + 1/2 \rho (1.6x10^{-7}) v_2^2= 1.013x10^5Pa + 1/2\rho v_2[/tex]

[tex] 1.813e5 Pa -1.013x10^5Pa + 8e-5 v_2^2 - 500 v_2= 0 [/tex]

[tex] 8e-5 v_2^2- 500v_2 + 8e4 = 0 [/tex]

Do I have to use the quadradic eqzn??

Well I was thinking of using this eqzn which is probably the same but rearranged I did this...

[tex]P_1-P_2 = 1/2 \rho (v_2^2-v_1^2) + \rho g(y2-y1) [/tex]

[tex] 1.813e5Pa- 1.013x10^5Pa = 1/2(1000kg/m^3)(v_2^2- (A2/A1)v_2)^2 [/tex]

[tex] 8e4= 1/2(1000kg/m^3)v_2^2(1-(1.63e-7)^2)= 0 [/tex]

[tex] 8e-4 = 500kg/m^3v_2^2(1) [/tex]

[tex] 160 = v_2^2 [/tex]

[tex]v_2= 12.649 m/s[/tex]

hm..I'm not sure whether looks right but it looks funny somehow to me.

Is it incorrect?

Thanks very much Doc Al :smile:
 
  • #6
Doc Al
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I almost forgot about this Q.. well here's what I did.



[tex]P_1= 2.0N/ 2.50x10^-5m^2 [/tex]

[tex] P_1= 8e4 Pa [/tex]

Hm...so then I guess I'd use this plus the atmospheric pressure and use that as P1
like so: P1= 8e4Pa+ 1.013x10^5Pa= 1.813e5

okay so I would plug in for v1 since v1= (A2/A1)v2

[tex]P_1 + 1/2 \rho v_1^2 + \rho gy_1= P_2 + 1/2 \rho v_2^2 = \rho gy_2 [/tex]

[tex] 1.813e5 Pa + 1/2 \rho [(A2/A1)^2 v_2^2]= P_2 + 1/2\rho v_2 [/tex]

[tex] 1.813e5 Pa + 1/2 \rho (1.6x10^{-7}) v_2^2= P_2 +1/2\rho v_2 [/tex]
You forgot the ^2 in that last term (which is what caused your later problems); that last equation should read:

[tex] 1.813e5 Pa + 1/2 \rho (1.6x10^{-7}) v_2^2= P_2 +1/2\rho v_2^2 [/tex]

I would immediately compare the second terms on each side. I think we can safely ignore the second term on the left, since [itex]1 \pm 10^{-7} \approx 1[/itex]. :wink:

-----
now...using what you said: P2= atmospheric pres= 1.013x10^5 Pa

here's where I get stuck..

[tex] 1.813e5 Pa + 1/2 \rho (1.6x10^{-7}) v_2^2= P_2 +1/2\rho v_2 [/tex]

[tex] 1.813e5 Pa + 1/2 \rho (1.6x10^{-7}) v_2^2= 1.013x10^5Pa + 1/2\rho v_2[/tex]

[tex] 1.813e5 Pa -1.013x10^5Pa + 8e-5 v_2^2 - 500 v_2= 0 [/tex]

[tex] 8e-5 v_2^2- 500v_2 + 8e4 = 0 [/tex]

Do I have to use the quadradic eqzn??
You took a wrong turn due to your earlier error.

Well I was thinking of using this eqzn which is probably the same but rearranged I did this...

[tex]P_1-P_2 = 1/2 \rho (v_2^2-v_1^2) + \rho g(y2-y1) [/tex]

[tex] 1.813e5Pa- 1.013x10^5Pa = 1/2(1000kg/m^3)(v_2^2- (A2/A1)v_2)^2 [/tex]

[tex] 8e4= 1/2(1000kg/m^3)v_2^2(1-(1.63e-7)^2)= 0 [/tex]

[tex] 8e-4 = 500kg/m^3v_2^2(1) [/tex]

[tex] 160 = v_2^2 [/tex]

[tex]v_2= 12.649 m/s[/tex]

hm..I'm not sure whether looks right but it looks funny somehow to me.
Looks good to me. :smile:
 
  • #7
~christina~
Gold Member
715
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You forgot the ^2 in that last term (which is what caused your later problems); that last equation should read:

[tex] 1.813e5 Pa + 1/2 \rho (1.6x10^{-7}) v_2^2= P_2 +1/2\rho v_2^2 [/tex]

I would immediately compare the second terms on each side. I think we can safely ignore the second term on the left, since [itex]1 \pm 10^{-7} \approx 1[/itex]. :wink:
Hm so that's why it didn't work out right.

Looks good to me. :smile:

Thanks alot Doc Al :biggrin:
 

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