Water flows through a horizontal tube of diameter 2.9cm that is joined to a second horizontal tube of diameter 1.5 cm. The pressure difference between the tubes is 7.2 kPa.
A)Find the speed of flow in the first tube.
To find the speed of flow in the first tube, we can use Bernoulli's equation, which relates the pressure difference and the velocity of a fluid flowing through a horizontal tube.
Bernoulli's equation:
P1 + 1/2ρv1^2 + ρgh1 = P2 + 1/2ρv2^2 + ρgh2
Where:
P1 and P2 are the pressures in the two tubes,
v1 and v2 are the velocities of the fluid in the two tubes,
ρ is the density of the fluid, and
h1 and h2 are the heights of the fluid in the two tubes (since they are horizontal, we can set h1 and h2 to be the same).
In this case, we are given:
P1 = ?
P2 = P1 + 7.2 kPa = P1 + 7.2 × 10^3 Pa (since 1 kPa = 10^3 Pa)
ρ = the density of water = 1000 kg/m^3
h1 = h2 = 0 (since the tubes are horizontal)
We need to find v1. So, let's rearrange the equation:
P1 + 1/2ρv1^2 = P1 + 7.2 × 10^3 + 1/2ρv2^2
Since h1 = h2 = 0, the equation becomes:
P1 + 1/2ρv1^2 = P1 + 7.2 × 10^3
Now, we can cancel out P1 on both sides:
1/2ρv1^2 = 7.2 × 10^3
Next, substitute the given values:
1/2 × 1000 kg/m^3 × v1^2 = 7.2 × 10^3 Pa
Simplify the equation:
v1^2 = (2 × 7.2 × 10^3) / 1000
v1^2 = 14.4
Finally, take the square root of both sides to find the speed of flow in the first tube:
v1 = √14.4
v1 ≈ 3.8 m/s
Therefore, the speed of flow in the first tube is approximately 3.8 m/s.
To find the speed of flow in the first tube, you can use Bernoulli's equation. Bernoulli's equation states that the total pressure of a fluid is constant along a streamline. In this case, we can ignore any changes in elevation and focus on the pressure difference between the two tubes. The equation takes the form:
P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂
where P₁ and P₂ are the pressures in the first and second tubes, ρ is the density of the fluid (assuming it is constant), v₁ and v₂ are the velocities of the fluid in the first and second tubes, and h₁ and h₂ are the heights of the fluid in the first and second tubes.
In this case, the tubes are horizontal, so the height terms (ρgh₁ and ρgh₂) cancel out. We are given that the pressure difference between the tubes is 7.2 kPa, so we can rewrite the equation as:
P₁ + ½ρv₁² = P₂ + ½ρv₂² + 7.2 kPa
Since the tubes are connected, the pressure at any given point in both tubes is the same. Therefore, P₁ = P₂, and the equation simplifies to:
½ρv₁² = ½ρv₂² + 7.2 kPa
Now, we can rearrange the equation to solve for v₁:
v₁² = (v₂² + 7.2 kPa) / ρ
To find the speed of flow in the first tube, we need to know the value of ρ, the density of the fluid. Depending on the fluid, this value can vary. For water, the density is approximately 1000 kg/m³. If we assume the fluid in the tubes is water, we can substitute this value and calculate the speed of flow in the first tube.
Please note that the unit of pressure used in the equation is Pascal (Pa), and 1 kPa = 1000 Pa. So, you'll need to convert the given pressure difference of 7.2 kPa to Pascal before substituting it into the equation.
Data
d1=2.9cm
d2=1.5cm
dP=7.2kPa
required, V1=?
from,
A1V1=A2V2
Hence
A1/A2=V2/V1
A1=V2
A2=V1
hence
A2=((3.14*1.5*1.5)/(4))
A2=1.77cm2=V1
V1=1.77cm/s