What is the pressure change in water going from a 3.0- cm-diameter pipe to a 1.8- cm-diameter pipe if the velocity in the smaller pipe is 3.0 m/s?
Im using the equation pv2^2(A2^2-A1^2)/2A1^2 and get (1000)(3)^2(1.8^2-3^20/(2(3)^2) and get -2880 Pa but I'm getting wrong? Thanks
I have no idea what you are doing.
pstatic1+rho*V1^2/2=pstatic2+rhoV2^2/2
you want pstatic1-pstatic1
So you need V2: Using the law of continuity,
Area1*V1=area2*V2 or V2=V1(1.8/3.0)^2
or V2=1.08m/s
pressure difference=rho(V2^2-V1^2)/2
=1000*(1.08^2-3^3)/2
check my thinking
To solve this problem, we can use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid at different points in a flow.
The equation you stated is a simplified version of Bernoulli's equation, but there seems to be a mistake in your calculation. Let's go through the steps together:
Given:
Diameter of the larger pipe (D1): 3.0 cm
Diameter of the smaller pipe (D2): 1.8 cm
Velocity in the smaller pipe (v2): 3.0 m/s
First, let's convert the diameters to meters:
D1 = 3.0 cm = 0.03 m
D2 = 1.8 cm = 0.018 m
Now, we can calculate the areas of the pipes:
A1 = π * (D1 / 2)^2 = π * (0.03 / 2)^2 ≈ 0.00070685 m^2
A2 = π * (D2 / 2)^2 = π * (0.018 / 2)^2 ≈ 0.00025447 m^2
Next, we'll use your equation, which is derived from Bernoulli's equation:
ΔP = (ρ * v2^2 * (A2^2 - A1^2)) / (2 * A1^2)
Substituting the values:
ρ = density of water ≈ 1000 kg/m^3
ΔP = (1000 * 3^2 * (0.00025447^2 - 0.00070685^2)) / (2 * 0.00070685^2)
Calculating this equation, we get:
ΔP ≈ -536.54 Pa
So, the pressure change in water going from a 3.0 cm diameter pipe to a 1.8 cm diameter pipe, with a velocity of 3.0 m/s in the smaller pipe, is approximately -536.54 Pa.
Please double-check your calculation steps and ensure you have used the correct values in your equation to determine the pressure change.
To find the pressure change in water going from a 3.0 cm-diameter pipe to a 1.8 cm-diameter pipe, you can use Bernoulli’s principle, which states that the total mechanical energy of a fluid system remains constant along a streamline.
The equation you mentioned, pv^2(A2^2 - A1^2) / 2A1^2, is a valid form of Bernoulli's principle equation, applied to this situation. However, there seem to be errors in your calculation.
Let's go through the correct calculation step by step:
1. First, convert the diameters of the pipes to their respective radii:
Diameter of the 3.0 cm pipe: 3.0 cm = 0.03 m (converted to meters)
Radius of the 3.0 cm pipe: 0.03 m / 2 = 0.015 m
Diameter of the 1.8 cm pipe: 1.8 cm = 0.018 m (converted to meters)
Radius of the 1.8 cm pipe: 0.018 m / 2 = 0.009 m
2. Plug the given values into the equation:
Density of water, ρ = 1000 kg/m^3 (given)
Velocity of the smaller pipe, v = 3.0 m/s (given)
Area of the larger pipe, A1 = π(0.015 m)^2 = 0.00070685 m^2
Area of the smaller pipe, A2 = π(0.009 m)^2 = 0.00025447 m^2
Now we can substitute these values into the equation:
ΔP = ρv^2(A2^2 - A1^2) / (2A1^2)
= (1000 kg/m^3)(3.0 m/s)^2((0.00025447 m^2)^2 - (0.00070685 m^2)^2) / (2(0.00070685 m^2)^2)
3. Calculate the pressure change:
ΔP = (1000)(9.0)(0.0000000169 - 0.0002958) / (2 x 0.000000498) ≈ -22104 Pa
So, the correct pressure change in water going from the 3.0 cm-diameter pipe to the 1.8 cm-diameter pipe is approximately -22104 Pa (or -22.1 kPa). Note that the negative sign indicates a decrease in pressure.