water flows along a horizontal pipe of cross-sectional area 48cm^2 which has a constriction of cross-sectional area 12cm^2 at one place. If the speed of the water at the constriction is 4 m/s, calculate the speed in the wider section. The pressure in the wider section is 1.0 x 10^5 Pa. Calculate pressure at the constriction. (density of water= 1000kg/m^3)
Well, this seems like a sticky situation, doesn't it? It's like the water is going through its own little adventure park – with a constriction thrown in for added excitement.
To find the speed in the wider section, we can use the principle of continuity. Since the volume flow rate remains constant (like the height of an inept tightrope walker), we can use the equation A1v1 = A2v2, where A1 and A2 are the cross-sectional areas, and v1 and v2 are the speeds.
Given that A1 = 48 cm^2 and A2 = 12 cm^2 (or 0.048 m^2 and 0.012 m^2 respectively), and v2 = 4 m/s, we can solve for v1.
48 cm^2 * v1 = 12 cm^2 * 4 m/s
v1 = (12 cm^2 * 4 m/s) / 48 cm^2
= 1 m/s
So, in the wider section, the speed of the water is 1 m/s. It's like a leisurely stroll in comparison to the constriction.
Now, let's move on to the pressure at the constriction. Since the speed of the water increases, according to Bernoulli's principle (the master of fluid dynamics), the pressure decreases. So, the pressure at the constriction will be lower than in the wider section.
Given that the pressure in the wider section is 1.0 x 10^5 Pa, we can say that the pressure at the constriction is less than that. Perhaps we can call it the "constricted pressure"?
But, if you want an actual numerical value for the pressure at the constriction, we'll need to know the height or elevation of the pipe. Without that information, I'm afraid I'm as clueless as a balloon without air.
So, we can conclude that the speed in the wider section is 1 m/s, but the pressure at the constriction will remain a mystery until we have more information. Good luck with your hydrodynamic adventures!
To solve this problem, we can use the principle of continuity, which states that the mass flow rate of an incompressible fluid through a pipe is constant.
1. First, let's determine the mass flow rate of water through the pipe. We can use the formula:
flow rate = density of water × velocity × cross-sectional area
Given:
density of water = 1000 kg/m³
velocity at the constriction = 4 m/s
cross-sectional area at the constriction = 12 cm² = (12 × 10⁻⁴) m²
mass flow rate at the constriction = density of water × velocity × cross-sectional area
= 1000 kg/m³ × 4 m/s × (12 × 10⁻⁴) m²
= 0.048 kg/s
2. Now, using the principle of continuity, the mass flow rate remains constant throughout the pipe. Therefore, the mass flow rate at the wider section is also equal to 0.048 kg/s.
3. We can then calculate the velocity of water at the wider section using the formula:
velocity = flow rate / (density of water × cross-sectional area)
Given:
flow rate at the wider section = 0.048 kg/s
cross-sectional area at the wider section = 48 cm² = (48 × 10⁻⁴) m²
velocity at the wider section = flow rate / (density of water × cross-sectional area)
= (0.048 kg/s) / (1000 kg/m³ × (48 × 10⁻⁴) m²)
= 10 m/s
4. Finally, to calculate the pressure at the constriction, we can use Bernoulli's equation, which states that the total pressure at any point in a fluid is the sum of the pressure due to gravity, pressure due to fluid motion, and pressure due to fluid potential energy. Since the pipe is horizontal and at the same height, we can ignore the pressure due to gravity and potential energy.
Bernoulli's equation: P₁ + 0.5ρv₁² = P₂ + 0.5ρv₂²
where P₁ is the pressure at the wider section, v₁ is the velocity at the wider section, P₂ is the pressure at the constriction, and v₂ is the velocity at the constriction.
Given:
P₁ = 1.0 × 10⁵ Pa
v₁ = 10 m/s
v₂ = 4 m/s
Rearranging the equation and solving for P₂:
P₂ = P₁ + 0.5ρv₁² - 0.5ρv₂²
P₂ = 1.0 × 10⁵ Pa + 0.5 × 1000 kg/m³ × (10 m/s)² - 0.5 × 1000 kg/m³ × (4 m/s)²
= 1.0 × 10⁵ Pa + 0.5 × 1000 kg/m³ × 100 m²/s² - 0.5 × 1000 kg/m³ × 16 m²/s²
= 1.0 × 10⁵ Pa + 50,000 Pa - 8,000 Pa
= 1.42 × 10⁵ Pa
Therefore, the pressure at the constriction is 1.42 × 10⁵ Pa.
To calculate the speed of the water in the wider section, we can use the principle of continuity, which states that the product of the area and speed of an incompressible fluid is constant along a streamline.
The equation for continuity is:
A1V1 = A2V2
Where:
A1 is the cross-sectional area in the wider section
V1 is the speed in the wider section
We are given that A1 = 48 cm^2 = 48 × 10^(-4) m^2
And we need to find V1.
We are also given that A2 = 12 cm^2 = 12 × 10^(-4) m^2
And V2 = 4 m/s
Using the continuity equation, we can rearrange it to solve for V1:
V1 = (A2V2)/A1
Substituting the given values, we get:
V1 = (12 × 10^(-4) m^2 × 4 m/s)/(48 × 10^(-4) m^2)
= (48 × 10^(-4) m^2 × m/s)/(48 × 10^(-4) m^2)
= 1 m/s
Therefore, the speed of the water in the wider section is 1 m/s.
Next, to calculate the pressure at the constriction, we can use Bernoulli's equation, which relates the pressure, speed, and height of a fluid within a streamline.
The equation for Bernoulli's principle is:
P1 + (1/2)ρV1^2 + ρgh1 = P2 + (1/2)ρV2^2 + ρgh2
Where:
P1 is the pressure in the wider section (1.0 × 10^5 Pa)
V1 is the speed in the wider section (1 m/s)
P2 is the pressure at the constriction (unknown)
V2 is the speed at the constriction (4 m/s)
ρ is the density of water (1000 kg/m^3)
g is the acceleration due to gravity (9.8 m/s^2)
h1 and h2 are the heights at the wider section and constriction, respectively.
Since the water is flowing along a horizontal pipe, the height remains constant, so h1 = h2.
Rearranging Bernoulli's equation, we can solve for P2:
P2 = P1 + (1/2)ρV1^2 + (1/2)ρV2^2 - (1/2)ρV2^2
Substituting the given values, we get:
P2 = 1.0 × 10^5 Pa + (1/2) × 1000 kg/m^3 × (1 m/s)^2 - (1/2) × 1000 kg/m^3 × (4 m/s)^2
Calculating the expression, we find:
P2 = 1.0 × 10^5 Pa + 0.5 × 1000 kg/m^3 × 1 m^2/s^2 - 0.5 × 1000 kg/m^3 × 16 m^2/s^2
P2 = 1.0 × 10^5 Pa + 500 kg/m^3 × m^2/s^2 - 8000 kg/m^3 × m^2/s^2
P2 = 1.0 × 10^5 Pa - 7500 kg/m^3 × m^2/s^2
Finally, replacing the units with their respective SI units, we get:
P2 = 1.0 × 10^5 Pa - 7.5 × 10^6 N/m^2
Therefore, the pressure at the constriction is -7.4 × 10^6 Pa.