# Needle and pressure (Bernoulli's equation?)

• ~christina~
In summary: Then I would rewrite this as: 1.813e5 Pa + 1/2 \rho (1.6x10^{-7}) v_2^2 -1.013x10^5Pa= 1/2\rho v_2^2 8.00e4 Pa + 1/2 \rho (1.6x10^{-7}) v_2^2 = 1/2\rho v_2^2 Keep going... 8.00e4 Pa = 1/2\rho v_2^2 - 1/2 \rho (1.6x10^{-7}) v_2^2 8.00e4 Pa
~christina~
Gold Member
[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

Last edited by a moderator:
~christina~ said:
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.)

Doc Al said:
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

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

Doc Al said:
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..

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

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

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

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

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

~christina~ said:

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

$$P_1= 8e4 Pa$$

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..

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

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

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

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

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.

Doc Al said:
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

## 1. What is Bernoulli's equation and how does it relate to needle and pressure?

Bernoulli's equation is a fundamental principle in fluid dynamics that describes the relationship between fluid pressure, velocity, and elevation in a flow. It states that as the velocity of a fluid increases, its pressure decreases. This is relevant to needle and pressure because as a needle moves through a fluid, the fluid's velocity changes and therefore its pressure also changes according to Bernoulli's equation.

## 2. Why is Bernoulli's equation important in understanding the behavior of fluids around needles?

Bernoulli's equation is important because it helps us understand how the pressure and velocity of a fluid are related. In the case of a needle, it helps us understand how the fluid flows around the needle and how changes in the needle's velocity can affect the pressure of the fluid.

## 3. How does the shape and size of a needle affect the pressure around it?

The shape and size of a needle can affect the pressure around it because it can change the velocity of the fluid as it flows around the needle. For example, a thinner needle may create a faster flow of fluid, resulting in lower pressure according to Bernoulli's equation.

## 4. What other factors can influence the pressure around a needle?

In addition to the shape and size of the needle, other factors that can influence the pressure around it include the fluid's viscosity, density, and the angle at which the needle is inserted into the fluid. These factors can all impact the fluid's velocity and therefore its pressure.

## 5. How can Bernoulli's equation be applied in real-world situations involving needles and pressure?

Bernoulli's equation can be applied in a variety of real-world situations involving needles and pressure, such as in medical applications where needles are used to inject fluids into the body. Understanding how changes in needle velocity can affect fluid pressure can help ensure accurate and safe injections. It can also be applied in engineering and aerodynamics to design more efficient and streamlined structures, such as airplane wings or wind turbines.

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