1.A hypodermic needle of lenght 0.020m and inner radius 3.0 x10^-4 is used to force water at 20 degree celcious into the air at a flow rate of 1.0 x10^-7m^2 s-1 (a) what is the average velocity in the needle?(assume laminar flow), (b) what is the pressure drop necessary to achieve the flow rate?
2. An artery has an inner radius of 2.0 x 10^-3m. if the temperature is 37 degree celcius, the average velocity of the blood is 0.030ms^-1 and the flow is laminar flow. find (a)the maximum velocity, (b) the flow rate and (c) the pressure drop in 0.050m, if the artery is horizontal.
To solve these problems, we will use the Hagen-Poiseuille equation for laminar flow in a tube:
ΔP = 8ηLQ / πr^4
where ΔP is the pressure difference between the ends of the tube, η is the dynamic viscosity of the fluid, L is the length of the tube, Q is the volumetric flow rate, and r is the inner radius of the tube.
For water, the dynamic viscosity at 20°C is approximately η = 1.0 x 10^-3 Pa⋅s. For blood, the dynamic viscosity at 37°C is approximately η = 4.0 x 10^-3 Pa⋅s.
1(a). To find the average velocity in the needle, we can use the flow rate Q and the cross-sectional area A of the needle:
Q = Av
where A = πr^2, and v is the average velocity.
1.0 x 10^-7 m³/s = π(3.0 x 10^-4 m)²v
Solving for v, we get:
v ≈ 3.54 m/s
1(b). Now we can use the Hagen-Poiseuille equation to find the pressure drop:
ΔP = 8ηLQ / πr^4
ΔP = 8(1.0 x 10^-3 Pa⋅s)(0.020 m)(1.0 x 10^-7 m³/s) / π(3.0 x 10^-4 m)⁴
ΔP ≈ 5924 Pa
2(a). In laminar flow, the maximum velocity occurs at the center of the tube and is twice the average velocity:
v_max = 2 * 0.030 m/s = 0.060 m/s
2(b). We can find the flow rate using the average velocity and the cross-sectional area:
Q = Av
Q = π(2.0 x 10^-3 m)² * 0.030 m/s
Q ≈ 3.77 x 10^-7 m³/s
2(c). Finally, we can find the pressure drop using the Hagen-Poiseuille equation:
ΔP = 8ηLQ / πr^4
ΔP = 8(4.0 x 10^-3 Pa⋅s)(0.050 m)(3.77 x 10^-7 m³/s) / π(2.0 x 10^-3 m)⁴
ΔP ≈ 47 Pa
To solve both of these problems, we can use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid in a system. Bernoulli's equation in simplified form for incompressible fluids is:
P + 1/2 ρv^2 + ρgh = constant,
where P is pressure, ρ is the density of the fluid, v is the velocity of the fluid, g is the acceleration due to gravity, and h is the height above a reference point.
Let's solve both problems step-by-step:
1. Hypodermic needle:
(a) To find the average velocity in the needle, we'll start by rearranging Bernoulli's equation in terms of the velocity:
P + 1/2 ρv^2 + ρgh = constant.
Since the needle is a thin, open-ended tube, the pressure at both ends can be considered atmospheric pressure (P0), and the height difference (h) is negligible. So the equation simplifies to:
P0 + 1/2 ρv^2 = constant.
Since the flow is assumed to be laminar, we can use Poiseuille's equation to relate the flow rate (Q) to the pressure difference ΔP and the radius (r) of the needle:
Q = (πr^4ΔP) / (8ηL),
where η is the dynamic viscosity of water and L is the length of the needle.
(b) To find the pressure drop necessary to achieve the flow rate, we can rearrange the equation above to solve for ΔP:
ΔP = (8ηLQ) / (πr^4).
Now, let's substitute the given values and solve the problems.
2. Artery:
(a) To find the maximum velocity, we can use the same equation as in problem 1 to relate the pressure difference to the velocity:
P0 + 1/2 ρvmax^2 = constant.
(b) The flow rate can be calculated by multiplying the maximum velocity (vmax) by the cross-sectional area of the artery:
Q = A * vmax,
where A is the cross-sectional area of the artery.
(c) To find the pressure drop in 0.050m of the artery, we'll use the same equation as problem 1:
ΔP = (8ηLQ) / (πr^4).
Again, let's substitute the given values and solve the problem.
Remember to use consistent units throughout the calculations.
To solve these problems, we will use the principles of fluid dynamics and the equations related to laminar flow.
1. (a) To find the average velocity in the needle, we can use the equation for volume flow rate (Q):
Q = A * v
where Q is the flow rate, A is the cross-sectional area, and v is the velocity.
Given:
- Inner radius (r) of the needle = 3.0 x 10^-4 m
- Flow rate (Q) = 1.0 x 10^-7 m^3/s
The cross-sectional area of the needle can be calculated using the formula for the area of a circle:
A = π * r^2
Substituting the values into the equation, we have:
Q = π * r^2 * v
Rearranging the equation to solve for v:
v = Q / (π * r^2)
Plug in the values:
v = (1.0 x 10^-7 m^3/s) / (π * (3.0 x 10^-4 m)^2)
Calculate the value of v using this formula.
(b) To find the pressure drop necessary to achieve the flow rate, we can use Poiseuille's Law. For laminar flow in a cylindrical tube, it states that the pressure drop (∆P) across a length (L) is given by the equation:
∆P = 8μQ / (πr^4),
where μ is the dynamic viscosity of the fluid, Q is the flow rate, and r is the radius.
Given:
- Inner radius (r) of the needle = 3.0 x 10^-4 m
- Flow rate (Q) = 1.0 x 10^-7 m^3/s
The dynamic viscosity (μ) of water at 20 degrees Celsius can be obtained from reference material (approximately 0.001 Pa·s).
Plug in the values into the equation:
∆P = (8 * 0.001 Pa·s * (1.0 x 10^-7 m^3/s)) / (π * (3.0 x 10^-4 m)^4)
Calculate the pressure drop (∆P) using this formula.
2. (a) To find the maximum velocity, we need to consider the continuity equation, which states that the product of cross-sectional area and velocity is constant along a streamline:
A1 * v1 = A2 * v2,
where A1 and A2 are the cross-sectional areas, and v1 and v2 are the velocities at two points in a streamline.
At the narrowest point of the artery (assuming circular cross-section), the velocity is maximum, and hence the cross-sectional area is minimum. Thus, this cross-section can be used to find the maximum velocity.
Given:
- Inner radius (r) of the artery = 2.0 x 10^-3 m
- Average velocity (v1) of the blood = 0.030 m/s
Using the formula for the area of a circle:
A1 = π * r^2
Rearranging the equation and substituting the values, we have:
v2 = (A1 * v1) / A2
v2 = (π * (2.0 x 10^-3 m)^2 * 0.030 m/s) / A2
Calculate the value of v2 using this formula.
(b) To find the flow rate, we can use the same equation as in problem 1(a):
Q = A * v
where Q is the flow rate, A is the cross-sectional area, and v is the velocity.
Given:
- Inner radius (r) of the artery = 2.0 x 10^-3 m
- Average velocity (v) of the blood = 0.030 m/s
Using the formula for the area of a circle:
A = π * r^2
Substituting the values, we have:
Q = π * (2.0 x 10^-3 m)^2 * 0.030 m/s
Calculate the value of Q using this formula.
(c) To find the pressure drop across the 0.050 m length of the artery, we can still use Poiseuille's Law from problem 1(b).
Given:
- Inner radius (r) of the artery = 2.0 x 10^-3 m
- Flow rate (Q) = value calculated in part (b)
Using the previously mentioned formula:
∆P = (8 * 0.001 Pa·s * Q) / (π * (2.0 x 10^-3 m)^4)
Calculate the pressure drop (∆P) using this formula.