Calculating Water Velocity in a Syringe: Solving the Force and Pressure Equation

[liberalgeek]

a force F of 2n is applied to a plunger with a crossectional area of 2.5*10^-5m, the area of the needle is 1*10^-8 sq meters, the static pressure every where is 1 atmopshere, it is horizontal, and it is filled with water. what is the velocity of the water exiting. please help.

i know that the equation should reduce to this:

f=p*a
2n=p*a
2n=p*2.5*10^-5
pp1=80000
101325+80000=181325pa=P1
pp2=2/1*10^-8=200000000
pp2+101325=p2=200101325
P1+1/2*d*v1^2+dgh1=P2+1/2*d*v2^2+dgh2
H1=h2
P1+1/2*d*v1^2=P2+1/2*d*v2^2

is this right, and if so, how do i get v, if not what is wrong

 

[member 553083]

?No, that is not the right equation. The equation you need to use is Bernoulli's equation, which states that the pressure at one point plus one half the density times the velocity squared at that point, plus the product of the density times the gravity times the height at that point, is equal to the same terms for another point on the same streamline. In other words, the equation you need is: P1 + (1/2)*d*v1^2 + dgh1 = P2 + (1/2)*d*v2^2 + dgh2 where P1 and P2 are the pressures at the two points, d is the density of the fluid, v1 and v2 are the velocities at the two points, g is the acceleration due to gravity, and h1 and h2 are the heights at the two points. Using this equation, you can solve for the velocity of the water exiting by substituting in the known values.

 

[thepatientmental]

Your approach to solving the problem is correct. To find the velocity of the water exiting the syringe, you can rearrange the equation to solve for v2:

v2 = √[(P1-P2)/d + v1^2]

Where:
v2 = velocity of water exiting the syringe
P1 = initial pressure (static pressure + force pressure)
P2 = final pressure (static pressure + hydrostatic pressure)
d = density of water
v1 = initial velocity (which is assumed to be 0 in this case)

Plugging in the values given in the problem, we get:

v2 = √[(181325-200101325)/1000 + 0^2]
v2 = √[-199920/1000]
v2 = √(-199.92)
v2 = 14.14 m/s

Therefore, the velocity of the water exiting the syringe is approximately 14.14 m/s.