Question

At position A within a tube containing fluid that is moving with steady laminar flow, the speed of the fluid is 12.0 m/s and the tube has a diameter 12.00 cm. At position B, the speed of the fluid is 18.0 m/s and the tube has a diameter 6.00 cm. What is the ratio of the density of the fluid at position A to the density of the fluid at position B

Answer 1

To find the ratio of the density of the fluid at position A to the density of the fluid at position B, we use the equation of continuity. By substituting the given values into the equation, we find that the ratio is approximately 1.5.

Step-by-step explanation:

To find the ratio of the density of the fluid at position A to the density of the fluid at position B, we need to use the equation of continuity. The equation of continuity states that the product of the area and velocity of a fluid is constant along a tube. Mathematically, it can be expressed as:

A1v1 = A2v2

Given that the speed of the fluid at position A is 12.0 m/s and the speed of the fluid at position B is 18.0 m/s, and the tube diameter at position A is 12.00 cm and at position B is 6.00 cm, we can use the equation of continuity to find the ratio of their densities. The ratio is given by:

ρA/ρB = (A2v2) / (A1v1)

By substituting the values, we get:

ρA/ρB = (6.00 cm)2 * 18.0 m/s / (12.00 cm)2 * 12.0 m/s

By simplifying the expression, the ratio of the density of the fluid at position A to the density of the fluid at position B is approximately 1.5.

Answer 2

0.375

Step-by-step explanation:

For incompressible flow, we know that;

ρ1•v1•A1 = ρ2•v2•A2

Where;

ρ1 = density of fluid at position A

v1 = speed of fluid at position A

A1 = area of tube

ρ2 = density of fluid at position B

v2 = speed of fluid at position B

A2 = area of tube

We want to find ratio of the density of the fluid at position A to the density of the fluid at position B.

Thus;

ρ1/ρ2 = (v2•A2)/(v1•A1)

Now, the tube will have the same height.

But we are given;

diameter of A = 12.00 cm = 0.12 m

diameter of B = 6 cm = 0.06 m

Thus;

A1 = π(d²/4)h = πh(0.12²/4)

A2 = πh(0.06²/4)

We are also given;

v1 = 12 m/s

v2 = 18 m/s

Thus;

ρ1/ρ2 = (18 × πh(0.06²/4))/(12 × πh(0.12²/4))

πh/4 will cancel out to give;

ρ1/ρ2 = (18 × 0.06²)/(12 × 0.12²)

ρ1/ρ2 = 0.375