# Homework Help: Need help with finding flow speed

1. Apr 10, 2010

### yang09

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

The hypodermic syringe contains a medicine with the same density of water. The barrel of the syringe has a cross-sectional area of 3.69595 × 10^−5 m^2(A1). The cross-sectional area of the needle is 1.14537 × 10^−8 m^2(A2). In the absence of a force on the plunger, the pressure everywhere is 1.0 atm. A force of magnitude 2.29077 N is exerted on the plunger, making medicine squirt from the needle. Find the medicine’s flow speed through the needle. Assume that the pressure in the needle remains at atmospheric pressure, that the syringe is horizontal, and that the speed of the emerging fluid is the same as the speed of the fluid in the needle.

2. Relevant equations

Forcein/Areain = Forceout/Areaout
Mass Flow Rate = (ρ)(Area)(Velocity)

3. The attempt at a solution

I do not know where to start so any help at all is appreciated.

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

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 10, 2010

### jack action

With the force applied on the plunger and its area, you can get the pressure inside the barrel.

Then, regarding the barrel vs the needle, you have 2 equations (Bernoulli and mass flow rate) with 2 unknowns (velocity in barrel and velocity in needle).

3. Apr 10, 2010

### yang09

I got my Pressure, dividing the force applied on the plunger by the Area of the barrel. But you're not given the Mass Flow Rate so how would you find the velocity using the MFR equation

4. Apr 10, 2010

### jack action

Isolate the velocity of the barrel in one equation (say, the mass flow rate equation), this will give you equation #1. Replace the velocity of the barrel in the other equation (the Bernoulli equation) with equation #1. This will give you equation #2. Isolate the needle velocity in equation #2: it is your only unknown.

N equations with N unknowns means you can find all unknowns.

5. Apr 10, 2010

### yang09

I think I get what you're saying. I tried doing it your way but instead of using the Mass Flow Rate equation, I used the continuity equation, A1V1 = A2V2. I solved for V1 and got V1 = (A2V2) / A1. I plugged this equation into V1 in the Bernoulli's equation but squared the (A2V2) / A1 because the V1 in Bernoulli's is squared. I manipulated the new equation and got V2 by itself. Here's my final equation: V2^2 = [(2)(A1^2)(Pressure2 - Pressure1)] / [(ρ)((A2^2) - (A1^2))].
But after plugging in my numbers, I had a square root of a negative number, which is wrong. What am I doing wrong?

6. Apr 10, 2010

### jack action

The continuity equation IS related to mass flow rate ($$\rho_1 = \rho_2$$)

The equation is good, make sure you put the right numbers (2 --> needle, 1 --> barrel)

7. Apr 10, 2010

### yang09

So what you're saying is that I'm supposed to relate the continuity equation with the Mass Flow equation. To plug in the v1 into the Mass Flow equation?

I did use the right numbers though. The negative part is coming from the (A2^2) - (A1^2). A2 is the area for the needle, 1.14537 × 10^−8 m^2, and A1 is the area for the barrel, 3.69595 × 10^−5 m^2. A2 is smaller than A1 and that's why I'm getting a negative number. If the equation seems good, then what am I doing wrong?

8. Apr 10, 2010

### jack action

The continuity states that whatever mass comes in must come out, so the mass flow rate is the same everywhere:

$$\dot{m_1}=\dot{m_2}$$

$$\rho_1 V_1 A_1=\rho_2 V_2 A_2$$

But if $$\rho_1=\rho_2$$, then:

$$V_1 A_1=V_2 A_2$$

But P2 - P1 is also negative.

9. Apr 10, 2010

### yang09

Ok, so the way I found my P2 and P1 is probably wrong then. Here's how I solved for them and tell me where I went wrong.
P1:
I did P1 = Forcein / Area1.
2.29077 was the Forcein and 3.69595 × 10^−5 m^2 was the Area1.

P2:
I did P2 = Forcein / Area2.
I assumed the force was constant throughout the syringe so I plugged in 2.29077 into the force and 1.14537 × 10^−8 m^2 was the Area2.
I guess now that the force is not constant because P2 should be smaller P1
So how would I solve for P2 in this case? Dont just tell me how to solve for it, but can you give me hints on solving for it so it'll give me a chance to think about it

Last edited: Apr 10, 2010
10. Apr 10, 2010

### jack action

11. Apr 10, 2010

### yang09

Ok so since the pressure in the needle remains at atmospheric pressure, would the P2 be 1 atm?
If it is, then I would still get a positive answer because my P1 is .61185 atm and the P2 would be 1 atm

12. Apr 10, 2010

### jack action

This means that, at rest, there is a pressure of 1 atm on the barrel end as well as on the needle end. So the atmospheric pressure is still acting when you push on the syringe, hence you must add the pressure due to the force to the atmospheric pressure.

13. Apr 10, 2010

### yang09

O I see. So I'm supposed to add 1 atm to my P1. So my thinking process originally was half right. I just forgot to add 1 atm to my new pressure after the force was acted on it.

14. Apr 10, 2010

### jack action

There you go.

15. Apr 10, 2010

### yang09

Ok so with P2 being 1 atm and P1 being 1.61185 atm, I got a V2 of 1.3537 X 10^-3 (m^2 atm) / kg. I then converted atm into Pascal because 1 Pa = 1 kg / (m s^2). After doing the calculations, I ended up with 11.7120764, which is not correct. Am I just calculating wrong?

Last edited: Apr 10, 2010
16. Apr 10, 2010

### jack action

Nope. Check your units again in the equation you gave me:

pressure/density*area/area = (velocity)²

17. Apr 10, 2010

### yang09

I checked again and my units are still right. After multiplying the numerator out, my units were m^4 atm and my denominator had units of kg m. After multiplying by the inverse of my denominator, to get rid of the bottom units, my new units were (m^3 atm) / kg. I converted my answer from atm to Pa so I then got (m^3 Pa) / kg. I know 1 Pa = 1 (kg m) / s^2 so I then got m^2 / s^2. I square rooted the answer and got a final units in m/s

18. Apr 10, 2010

### jack action

Numerator = (Area)² X Pressure = (Area)² X Force / Area = Area X Force = Area X Mass X Acceleration = kg.m³/s²

Denominator = Density X (Area)² = Mass / Volume X (Area)² = kg.m

Then:

(kg.m³/s²) / (kg.m) = (m/s)²

19. Apr 10, 2010

### yang09

Ya and then you square root the (m/s)² to get m/s which is the correct units.

So if you do it that way, then you didn't have to convert the answer from atm to Pascal.
But can you convert the answer to Pascal though?

20. Apr 10, 2010

### jack action

When you use a SI equation, the best way is to convert all your units to basic SI before doing the calculations and you get your answer in basic SI units. In this case:

Pressure: Pa
density: kg/m³
area: m²
velocity: m/s

21. Apr 10, 2010

### yang09

So the answer I got was .0349814652 m/s. Before I enter that answer in, does it seem reasonable?

22. Apr 10, 2010

### jack action

23. Apr 10, 2010

### yang09

Here's how I got .0349814652 m/s:
My numerator was -1.672 X 10^-9
My denominator was -1.366 X 10^-6
My answer after division was 1.224 X 10^-3
My answer after square rooting it was 3.498 X10^-2

24. Apr 10, 2010

### jack action

$$V_2=\sqrt{\frac{2\left(3.69595X10^{-5}\right)^2\left(101325-\left(101325+\frac{2.29077}{3.69595X10^{-5}}\right)\right)}{1000\left(\left(1.14537X10^{-8}\right)^2-\left(3.69595X10^{-5}\right)^2\right)}} = 11.13 m/s$$