A hypodermic syringe contains a medicine with the density of water (figure below). The barrel of the syringe has a cross-sectional area of 2.44 10-5 m2. In the absence of a force on the plunger, the pressure everywhere is 1.00 atm. A force vector F of magnitude 1.91 N is exerted on the plunger, making medicine squirt from the needle. Determine the medicine's flow speed through the needle. Assume the pressure in the needle remains equal to 1.00 atm and that the syringe is horizontal.
The gauge pressure in the barrel of the syringe is
1.91N/2.44*10^-5 m^2 = 7.83*10^4 Pa
The gauge pressure at the opening of the needle is 0. (1.00 atm absolute)
Use the Bernoulli equation for the velocity at the needle exit.
deltaP(gauge) = 7.83*10^4 N/m^2
= (1/2)(rho)*V^2
rho is the medicine density, 1000 kg/m^3
Solve for V.
V = 12.5 m/s
Why did the medicine go to art school?
Because it wanted to learn how to flow through a needle! But don't worry, I've got your back.
To determine the medicine's flow speed through the needle, we can use Bernoulli's equation, which relates pressure and velocity in a fluid. Bernoulli's equation states:
P + ½ρv^2 + ρgh = constant
In this case, the syringe is horizontal, so the height (h) term is zero. Let's also assume that the pressure in the needle is equal to 1.00 atm, which is the same as the pressure in the syringe. So we can rewrite Bernoulli's equation as:
P + ½ρv^2 = constant
Since the pressure is constant, we can ignore that term. Now, let's calculate the velocity (v) using the force (F) and the cross-sectional area (A) of the barrel.
F = P * A
1.91 N = 1.00 atm * 2.44 * 10^-5 m^2
Solving for P, we get:
P = 1.91 N / 2.44 * 10^-5 m^2
Now, let's substitute this pressure value back into Bernoulli's equation:
1.00 atm + ½ρv^2 = 1.91 N / (2.44 * 10^-5 m^2)
Solving for v, we can find the medicine's flow speed through the needle. But beware, I'm not a doctor, so take my prescription with a grain of salt!
To determine the medicine's flow speed through the needle, we can use Bernoulli's equation, which states that the sum of the pressure, density, and the height of a fluid stays constant throughout a streamline. Since the syringe is horizontal, the height terms cancel out, and we are left with:
P1 + 1/2 * ρ * v1^2 = P2 + 1/2 * ρ * v2^2
Where:
P1 and P2 are the pressures at points 1 and 2 respectively,
v1 and v2 are the flow speeds at points 1 and 2 respectively,
and ρ is the density of the medicine (which is equal to the density of water).
Since the pressure inside the syringe and needle remains constant at 1.00 atm, we can rewrite the equation as:
1.00 atm + 1/2 * ρ * v1^2 = 1.00 atm + 1/2 * ρ * v2^2
Since the needle is narrower than the syringe barrel, we can assume that the medicine's velocity is higher at the needle's exit point (v2) compared to the velocity at the syringe's plunger (v1). Therefore, we can simplify the equation to:
1/2 * ρ * v1^2 = 1/2 * ρ * v2^2
Canceling out the density term on both sides gives us:
v1^2 = v2^2
Taking the square root of both sides, we get:
v1 = v2
Hence, the flow speed of the medicine through the needle is equal to the flow speed at the syringe's plunger. To calculate this flow speed, we need to use the force exerted on the plunger and the cross-sectional area of the syringe barrel.
The formula to calculate flow speed (v) is given by:
v = F / A
Where:
F is the magnitude of the force exerted on the plunger (1.91 N),
and A is the cross-sectional area of the syringe barrel (2.44 * 10^-5 m^2).
Substituting the values, we have:
v = 1.91 N / (2.44 * 10^-5 m^2)
Calculating this, we find:
v ≈ 78.28 m/s
Therefore, the medicine's flow speed through the needle is approximately 78.28 m/s.
To determine the medicine's flow speed through the needle, we can use Bernoulli's equation, which relates fluid pressure, fluid velocity, and fluid height.
Bernoulli's equation can be expressed as:
P + 0.5 * ρ * v^2 + ρ * g * h = constant
Where:
P is the pressure
ρ is the density of the fluid
v is the velocity of the fluid
g is the acceleration due to gravity
h is the height of the fluid
In this case, we are assuming the pressure in the needle remains equal to 1.00 atm, which we can convert to Pascals using the conversion factor 1 atm = 101325 Pa.
1 atm = 101325 Pa
Now, let's apply Bernoulli's equation to the given situation.
At the point where the medicine is first inside the needle, the pressure is 1.00 atm and the height is at the same level, so we can ignore the height term.
P1 + 0.5 * ρ * v1^2 = P2 + 0.5 * ρ * v2^2
Since the medicine has the density of water, we can substitute water's density (ρ = 1000 kg/m^3) into the equation.
P1 + 500 * v1^2 = P2 + 500 * v2^2
We know that the pressure everywhere is 1.00 atm, which means P1 = P2 = 1.00 atm = 101325 Pa.
101325 Pa + 500 * v1^2 = 101325 Pa + 500 * v2^2
Simplifying the equation, we get:
500 * v1^2 = 500 * v2^2
Dividing both sides of the equation by 500:
v1^2 = v2^2
Taking the square root of both sides:
v1 = v2
So, the flow speed of the medicine through the needle is the same as the speed at which it leaves the needle.
Therefore, by exerting a force of 1.91 N on the plunger, the medicine's flow speed through the needle would be 1.91 m/s.