2018 A/L Physics Past Paper · Question 33 (English Medium)

Streamline flow in a curved tube — Bernoulli

Consider a non-viscous and incompressible fluid moving with steady streamline flow through a tube of non-uniform cross-sectional area in a vertical plane. Figure shows the vertical cross-section of the tube. $X, Y$ and $Z$ represent three positions of a streamline. Area of cross-section of the tube at $X$ is same as that at $Z$ . Consider the following inequalities for the relative magnitudes of the kinetic energies per unit volume $(KE_X, KE_Y, KE_Z)$ , potential energies per unit volume $(PE_X, PE_Y, PE_Z)$ and the fluid pressures $(P_X, P_Y, P_Z)$ at the positions $X, Y$ and $Z$ respectively.

(A) $KE_Z < KE_X < KE_Y$

(B) $PE_X < PE_Z < PE_Y$

Of the above inequalities,

(1) only A is true.
(2) only B is true.
(3) only A and B are true.
(4) only B and C are true.
(5) all A, B and C are true.

Answer and explanation

Answer: (4) only B and C are true.

What the question asks

We check three statements about the points $X$ , $Y$ and $Z$ in the tube:

the kinetic energy per unit volume, the potential energy per unit volume, and the pressure.

Relevant theory

On any point along the same streamline of a steady flow of an incompressible fluid of negligible viscosity, the sum of the pressure, the potential energy per unit volume and the kinetic energy per unit volume is a constant.

We know from the equation of continuity that when the cross – section gets narrow the velocity of flow increases.

Two rules help us.

Rule 1 (continuity): $Av$ stays the same. Where the tube is narrow, the fluid goes fast.

Rule 2 (Bernoulli): this total stays the same everywhere:

$p + \tfrac{1}{2}\rho v^2 + \rho gh$

$\tfrac{1}{2}\rho v^2$ is the kinetic energy per unit volume. $\rho gh$ is the potential energy per unit volume.

From the figure: $X$ is lowest, then $Z$ , and $Y$ is highest.

$X$ and $Z$ have the same area. The tube is narrow at $Y$ .

Reading the options

Each option is a mix of A, B and C.

Check each statement on its own. Then pick the option.

Reasoning and calculation

$KE_Z < KE_X < KE_Y$

$X$ and $Z$ have the same area, so the same speed. So $KE_X = KE_Z$ . " $KE_Z < KE_X$ " is false.

$PE_X < PE_Z < PE_Y$

Potential energy grows with height. $X$ is lowest, $Z$ is in the middle, $Y$ is highest. So $PE_X < PE_Z < PE_Y$ . True.

$P_Y < P_Z < P_X$

Pressure is what is left of the total after KE and PE. $X$ and $Z$ have the same KE, but $Z$ is higher. So $P_X > P_Z$ . At $Y$ both KE and PE are big, so $P_Y$ is the smallest. True.

A is false.

It says $X$ and $Z$ have different speeds.

But their areas are equal, so their speeds are equal.

Why the other options fail

This option says: only A is true.

But A is false.

And B and C are true, but it leaves them out.

So it is wrong.

This option says: only B is true.

B is true. But C is true too, and it leaves C out.

This option says: only A and B are true.

And C is true, but it leaves C out.

This option says: all A, B and C are true.

But A is false. So "all three" cannot be right.

Conclusion

A is false, because $KE_X = KE_Z$ .

B is true. C is true.

The answer is option (4).

Explore

The fluid flows steadily through the tube and runs faster where the tube is narrow.

The areas at $X$ and $Z$ are equal, so the speeds there are equal: (A) is false.

$X$ is lowest, $Y$ is highest and $Z$ is between them: (B) is true.

The total $P + \tfrac12\rho v^2 + \rho g h$ is the same everywhere, so $P_Y < P_Z < P_X$ : (C) is true.

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