# Needle and pressure (Bernoulli's equation?)

1. Feb 6, 2008

### ~christina~

[SOLVED] Needle and pressure (Bernoulli's equation?)

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

2. Relevant 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

3. 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

2. Feb 6, 2008

### Staff: Mentor

The plunger force was only 2 N, not 20 N. Also, the pressure due to the plunger force is in addition to atmospheric pressure.

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. Feb 6, 2008

### ~christina~

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

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

4. Feb 6, 2008

### Staff: Mentor

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

5. Feb 15, 2008

### ~christina~

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

$$P_1= 8e4 Pa$$

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..

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

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

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

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

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

6. Feb 15, 2008

### Staff: Mentor

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 $1 \pm 10^{-7} \approx 1$.

You took a wrong turn due to your earlier error.

Looks good to me.

7. Feb 15, 2008

### ~christina~

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

Thanks alot Doc Al