A pump uses a piston of 15 cm diameter which moves 2 cm/s. What is the fluid velocity of a tube 3 cm in diameter?

Question

BlackTom AI · Accepted Answer

We start by identifying the principle: conservation of volumetric flow rate in an incompressible fluid means Q = A1 * v1 = A2 * v2.
Option 1 examines the piston side: the piston has a diameter of 15 cm, so its radius is 7.5 cm. The cross-sectional area is A1 = π(0.075 m)^2 ≈ 0.01767 m^2. The piston velocity is 2 cm/s = 0.02 m/s, so the flow rate is Q = A1 * v1 ≈ 0.01767 * 0.02 ≈ 0.000353 m^3/s.
Option 2 looks at the tube side: the tube diameter is 3 cm, so radius 1.5 cm. The cross-sectional area is A2 = π(0.015 m)^2 ≈ 0.0007069 m^2.
To find the fluid velocity in the tube, v2 = Q / A2 ≈ 0.000353 / 0.0007069 ≈ 0.5 m/s, which is 50 cm/s.
Option 3 proposes 6 cm/s. If we test this, the implied Q would be A2 * v2 = 0.0007069 m^2 * 0.06 m/s ≈ 0.0000424 m^3/s, which is far smaller than the piston’s Q ≈ 0.000353 m^3/s, indicating inconsistency with flow continuity.
Option 4 suggests 50 m/s. Converting, that would require Q = A2 * v2 = 0.0007069 m^2 * 50 m/s ≈ 0.0353 m^3/s, which is about 100 times larger than what the piston can supply, making this physically impossible in this system.
In summary, only the velocity that satisfies Q = A1*v1 on the piston side and Q = A2*v2 on the tube side is v2 ≈ 0.5 m/s (50 cm/s). The given calculations align with conservation of mass for an incompressible fluid, and the other options fail either due to dimensional mismatch or by violating the continuity of flow.

SPHY001 Graded Quiz 6 (Unit 11 and Unit 12)- Requires Respondus LockDown Browser

A pump uses a piston of 15 cm diameter which moves 2 cm/s. What is the fluid velocity of a tube 3 cm in diameter?

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