Question

Incompressible oil of density 850kg/m3 is pumped through a cylindrical pipe at a rate of 9.5 liters per second

the first section of the pipe has has a diameter of 8cm, what's the flow speed of the oil?
b. what's the mass of the flow rate
c. the second section of the pipe has a diameter of 4.0cm. what are the flow speed and mass flow rate in that section ​

Answer 1

The flow speed of the oil in the first section of the pipe is approximately 1.889 m/s. The flow speed of the oil in the second section of the pipe is approximately 7.560 m/s. The mass flow rate in both sections of the pipe is approximately 8.075 kg/s.

Step-by-step explanation:

The flow speed of the oil in the first section of the pipe can be calculated using the equation:

v = Q / A

Where v is the flow speed, Q is the flow rate, and A is the cross-sectional area of the pipe. We are given that the flow rate is 9.5 liters per second, which is equivalent to 0.0095 m^3/s. The diameter of the first section of the pipe is 8 cm, so the radius is 4 cm or 0.04 m. Therefore, the cross-sectional area is:

A = π * r^2 = 3.14 * (0.04)^2 = 0.005024 m^2

Plugging these values into the equation, we get:

v = 0.0095 / 0.005024 = 1.889 m/s

The flow speed of the oil in the second section of the pipe can be calculated using the same equation, but with the cross-sectional area of the second section of the pipe. The diameter of the second section of the pipe is 4.0 cm, so the radius is 2.0 cm or 0.02 m. Therefore, the cross-sectional area is:

A = π * r^2 = 3.14 * (0.02)^2 = 0.001256 m^2

Plugging this value into the equation, we get:

v = 0.0095 / 0.001256 = 7.560 m/s

The mass flow rate in both sections of the pipe can be calculated using the equation:

m = ρ * Q

Where m is the mass flow rate and ρ is the density of the oil. The density of the oil is given as 850 kg/m^3. Plugging in the values, we get:

m = 850 * 0.0095 = 8.075 kg/s

Answer 2

The flow speed of the oil in the first section of the pipe is approximately 1.8 m/s, and the mass flow rate is approximately 8.07 kg/s. The flow speed in the second section of the pipe is approximately 5.7 m/s, and the mass flow rate remains the same.

Step-by-step explanation:

The flow speed of the oil in the first section of the pipe can be calculated using the equation for the flow rate of a liquid: Q = Av, where Q is the flow rate, A is the cross-sectional area of the pipe's section, and v is the flow speed. In this case, the diameter of the first section is 8 cm, so the radius is 4 cm (or 0.04 m). Using the equation for the area of a circle, we can calculate the cross-sectional area: A = πr^2 = π(0.04 m)^2. We are given the flow rate as 9.5 liters per second, so we need to convert it to cubic meters per second: 1 liter = 0.001 cubic meters, so 9.5 liters = 9.5 × 0.001 cubic meters. Now we can use the equation for the flow rate to solve for the flow speed: 9.5 × 0.001 = π(0.04 m)^2v. Solving for v, we find that the flow speed in the first section of the pipe is approximately 1.8 m/s.

Mass flow rate:

The mass flow rate can be calculated using the equation: m = ρQ, where m is the mass flow rate, ρ is the density of the oil, and Q is the flow rate. We are given the density of the oil as 850 kg/m³ and the flow rate as 9.5 liters per second. Again, we need to convert the flow rate to cubic meters per second: 1 liter = 0.001 cubic meters, so 9.5 liters = 9.5 × 0.001 cubic meters. Now we can use the equation for the mass flow rate to solve for m: m = 850 kg/m³ × 9.5 × 0.001 cubic meters/second. Solving for m, we find that the mass flow rate is approximately 8.07 kg/s.

The flow speed in the second section of the pipe can be calculated using the equation for the flow rate: Q = Av. The diameter of the second section is 4.0 cm, so the radius is 2.0 cm (or 0.02 m). Using the equation for the area of a circle, we can calculate the cross-sectional area: A = πr^2 = π(0.02 m)^2. We already know the flow rate as 9.5 liters per second, so it remains the same. Now we can use the equation for the flow rate to solve for the flow speed: 9.5 × 0.001 = π(0.02 m)^2v. Solving for v, we find that the flow speed in the second section of the pipe is approximately 5.7 m/s.

Mass flow rate:

The mass flow rate remains the same throughout the pipe, so it is still approximately 8.07 kg/s.