A venturi meter having an entrance diameter of 300mm and a throat diameter of 200mm is used to measure the volume of oil flowing through a pipe. The coefficient of discharge Cd is 0.95. If the density of oil is equal to 0.9kg / (m ^ 3) Making reasonable assumptions, calculate the volume flow per second when the pressure difference between the entrance and the throat as measured on a water U-tube is 63mm

To solve this problem, we will use the equation for the flow rate in a Venturi meter, which is given by:

Q = Cd * A1 * A2 * sqrt(2 * g * h / (ρ * (1 - (A1/A2)^2)))

Where:

• Q is the volume flow rate
• Cd is the coefficient of discharge
• A1 and A2 are the cross-sectional areas at the entrance and throat of the Venturi meter, respectively
• g is the acceleration due to gravity
• h is the pressure difference between the entrance and the throat
• ρ is the density of the fluid

Step 1: Calculate the cross-sectional areas A1 and A2

A1 = π * (D1/2)^2 = π * (0.3/2)^2 = 0.0707 m^2 A2 = π * (D2/2)^2 = π * (0.2/2)^2 = 0.0314 m^2

Step 2: Convert the pressure difference h from mm of water to m of oil

h = 63 mm * (1 m / 1000 mm) * (ρ_water / ρ_oil) = 0.063 m * (1000 kg/m^3 / 900 kg/m^3) = 0.07 m

Step 3: Substitute the known values into the equation for Q

Q = 0.95 * 0.0707 m^2 * 0.0314 m^2 * sqrt(2 * 9.81 m/s^2 * 0.07 m / (900 kg/m^3 * (1 - (0.0707 m^2 / 0.0314 m^2)^2)))

Solving this equation gives:

Q = 0.002 m^3/s or 2 litres/second

So, the volume flow rate of the oil is 2 litres per second.

To solve this problem, we will use the equation for the flow rate in a Venturi meter, which is given by:

Q = Cd * A1 * A2 * sqrt(2 * g * h / (ρ * (1 - (A1/A2)^2)))

Where:

• Q is the volume flow rate
• Cd is the coefficient of discharge
• A1 and A2 are the cross-sectional areas at the entrance and throat of the Venturi meter, respectively
• g is the acceleration due to gravity
• h is the pressure difference between the entrance and the throat
• ρ is the density of the fluid

First, we need to convert the diameters from mm to m:

D1 = 300mm = 0.3m D2 = 200mm = 0.2m

The cross-sectional areas can be calculated using the formula for the area of a circle, A = π * (D/2)^2:

A1 = π * (0.3/2)^2 = 0.0707 m^2 A2 = π * (0.2/2)^2 = 0.0314 m^2

The pressure difference h is given in mm of water, so we need to convert it to m of oil. The density of water is approximately 1000 kg/m^3, so:

h = 63mm * (0.9 kg/m^3 / 1000 kg/m^3) = 0.0567m

Now we can substitute all the values into the equation for Q:

Q = 0.95 * 0.0707 m^2 * 0.0314 m^2 * sqrt(2 * 9.81 m/s^2 * 0.0567m / (0.9 kg/m^3 * (1 - (0.0707 m^2 / 0.0314 m^2)^2)))

After calculating the above expression, we get:

Q = 0.0012 m^3/s

So, the volume flow per second is 0.0012 cubic meters.

To solve this problem, we will use the formula for the flow rate in a Venturi meter:

Q = Cd * A1 * A2 * sqrt(2 * g * h / ρ * (1 - (A2/A1)^2))

Where:

• Q is the volume flow rate
• Cd is the coefficient of discharge
• A1 and A2 are the cross-sectional areas at the entrance and throat respectively
• g is the acceleration due to gravity
• h is the pressure difference
• ρ is the density of the fluid

First, we need to convert the diameters to areas. The area of a circle is given by π * (d/2)^2. So, A1 = π * (0.3/2)^2 = 0.0707 m^2 and A2 = π * (0.2/2)^2 = 0.0314 m^2.

Next, we need to convert the pressure difference from mm of water to m of oil. 1 mm of water is equivalent to 9.81 * 10^-6 m of oil. So, h = 63 * 9.81 * 10^-6 = 0.000618 m.

Now we can substitute these values into the formula:

Q = 0.95 * 0.0707 * 0.0314 * sqrt(2 * 9.81 * 0.000618 / 0.9 * (1 - (0.0314/0.0707)^2))

Solving this gives Q = 0.0012 m^3/s or 1.2 L/s.