6. A horizontal pipe 7 cm in diameter has a smooth reduction to a pipe 4 cm in diameter. If the pressure of the water in the larger pipe is 6 x 104 Pa and the pressure in the smaller pipe is 4 x 104 Pa, at what rate does water flow through the pipes? 7. A Venturi tube may be used as a fluid flow meter. If the difference in pressure is P1 - P2 = 21.0 kPa, find the fluid flow rate in cubic meters per second, given that the radius of the outlet tube is 1.00 cm, the radius of the inlet tube is 2.00 cm, and the fluid is gasoline (density = 7 g/cm3). 8. A Pitot tube can be used to determine the velocity of air flow by measuring the difference between the total pressure and the static pressure (Fig. 1). If the fluid in the tube is mercury, (density = 13 600 kg/m3), and ∆h = 5.00 cm, find the speed of air flow. (Assume that the air is stagnant at point A, and take the density of air to be 1.25 kg/m3)
To solve these problems related to fluid flow, we can use concepts like Bernoulli's equation and the Continuity equation.
For question 6, we can use the Continuity equation, which states that the mass flow rate of a fluid is conserved in a system. The equation is given as:
ρ1A1v1 = ρ2A2v2
Where:
ρ1 and ρ2 are the densities of the fluid in the large and small pipes, respectively.
A1 and A2 are the cross-sectional areas of the large and small pipes, respectively.
v1 and v2 are the velocities of the fluid in the large and small pipes, respectively.
In this case, the diameters of the pipes are known, and we can use the formula for the area of a circle (A = πr^2) to calculate the cross-sectional areas.
Once we have the density and velocities, we can calculate the rate of water flow (also known as the volume flow rate or flow rate) using the formula:
Q = Av
Where:
Q is the flow rate.
A is the cross-sectional area of the pipe.
v is the velocity of the fluid.
For question 7, we can use Bernoulli's equation to relate the pressure difference to the velocity of the fluid. Bernoulli's equation is given as:
P1 + (1/2)ρv1^2 = P2 + (1/2)ρv2^2
Where:
P1 and P2 are the pressures at two different points in the fluid flow.
ρ is the density of the fluid.
v1 and v2 are the velocities of the fluid at the two different points.
We can rearrange the equation to solve for the fluid flow rate (Q) using the formula:
Q = A1v1 = A2v2
Where:
A1 and A2 are the cross-sectional areas of the inlet and outlet tubes, respectively.
v1 and v2 are the velocities of the fluid at the inlet and outlet tubes, respectively.
For question 8, we can use Bernoulli's equation again, since the total pressure is given as the sum of the static pressure and the dynamic pressure. The equation is given as:
P1 + (1/2)ρv1^2 = P2 + (1/2)ρv2^2
Where:
P1 is the total pressure at point A in the Pitot tube (the sum of static pressure and dynamic pressure).
P2 is the static pressure at point B in the Pitot tube.
ρ is the density of the fluid.
v1 is the velocity of air flow at point A (what we need to find).
v2 is the velocity of air flow at point B, which is assumed to be zero (since the air is stagnant).
We can rearrange the equation to solve for the velocity of air flow (v1).