Needle and pressure (Bernoulli's equation?)

[~christina~]

[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

Homework Equations

A1v1= A2v2

$$P_1 + 1/2 \rho v^2 + \rho gy_1 = P_2 + 1/2\rho v^2 + \rho g y_2$$

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

$$P_1= 20.0N/ 2.50x10^-5m^2$$

$$P_1= 8e5 Pa$$

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

$$P_1 + 1/2 \rho v_1^2 = P_2 + 1/2 \rho v_2^2$$

$$v_1= (A_2/A_1) *v_2$$ => no v1 or v2...

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 :
$$\Sum F= F_{plunger} - Fatm= ma$$ => (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

 

[Doc Al]

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

$$P_1= 20.0N/ 2.50x10^-5m^2$$

$$P_1= 8e5 Pa$$

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

$$P_1 + 1/2 \rho v_1^2 = P_2 + 1/2 \rho v_2^2$$

$$v_1= (A_2/A_1) *v_2$$ => no v1 or v2...

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.)

 

[~christina~]

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...

$$P_1= 2.0N/ 2.50x10^-5m^2$$

$$P_1= 8e4 Pa$$

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?

$$P_1 + 1/2 \rho v_1^2 + \rho gy_1= P_2 + 1/2 \rho v_2^2 = \rho gy_2$$

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

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

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

Thank you

 

[Doc Al]

Since the needle end is still exposed to the atmosphere, P2 = atmospheric pressure.

 

[~christina~]

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.

$$P_1= 2.0N/ 2.50x10^-5m^2$$

$$P_1= 8e4 Pa$$

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

$$P_1 + 1/2 \rho v_1^2 + \rho gy_1= P_2 + 1/2 \rho v_2^2 = \rho gy_2$$

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

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

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

here's where I get stuck..

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

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

$$1.813e5 Pa -1.013x10^5Pa + 8e-5 v_2^2 - 500 v_2= 0$$

$$8e-5 v_2^2- 500v_2 + 8e4 = 0$$

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...

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

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

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

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

$$160 = v_2^2$$

$$v_2= 12.649 m/s$$

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

Is it incorrect?

Thanks very much Doc Al

 

[Doc Al]

I almost forgot about this Q.. well here's what I did.



$$P_1= 2.0N/ 2.50x10^-5m^2$$

$$P_1= 8e4 Pa$$

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

$$P_1 + 1/2 \rho v_1^2 + \rho gy_1= P_2 + 1/2 \rho v_2^2 = \rho gy_2$$

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

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

You forgot the ^2 in that last term (which is what caused your later problems); that last equation should read:

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

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

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

here's where I get stuck..

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

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

$$1.813e5 Pa -1.013x10^5Pa + 8e-5 v_2^2 - 500 v_2= 0$$

$$8e-5 v_2^2- 500v_2 + 8e4 = 0$$

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...

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

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

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

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

$$160 = v_2^2$$

$$v_2= 12.649 m/s$$

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

Looks good to me.

 

[~christina~]

You forgot the ^2 in that last term (which is what caused your later problems); that last equation should read:

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

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

Hm so that's why it didn't work out right.

Looks good to me.

Thanks a lot Doc Al