Question

Water flows in a horizontal tube (see figure). The pressure of water changes by $700\, Nm^{-2}$ between $A$ and $B$ where the area of cross section are $40\, cm^2$ and $20\, cm^2$, respectively. Find the rate of flow of water through the tube. (density of water = $1000\, kgm^{-3}$)

• A. $3020\, cm^3/s$
• B. $2420\, cm^3/s$
• C. $2720\, cm^3/s$
• D. $1810\,cm^3/s$

Answer 1

The Correct Option is C

To solve the problem, we need to apply the principles of fluid dynamics, specifically the Bernoulli's equation and the continuity equation.

Step 1: Understanding Bernoulli's Equation

Bernoulli's equation for a flowing fluid is given by:

P + \frac{1}{2} \rho v^2 + \rho gh = \text{constant}

where P is the pressure, \rho is the density of the fluid, v is the velocity, and h is the height. Since the tube is horizontal, the height difference is zero and can be ignored.

Step 2: Pressure Difference and Continuity Equation

We are given:

• Pressure difference: \Delta P = 700\, Nm^{-2}
• Areas of cross-section:
  • A_1 = 40\, cm^2 = 40 \times 10^{-4}\, m^2
  • A_2 = 20\, cm^2 = 20 \times 10^{-4}\, m^2
• Density of water: \rho = 1000\, kgm^{-3}

Step 3: Applying the Continuity Equation

The continuity equation dictates that the flow rate must be constant across any two points:

A_1 v_1 = A_2 v_2

This can be rearranged to express one velocity in terms of the other:

v_1 = \frac{A_2}{A_1} v_2

Step 4: Applying Bernoulli's Equation for Horizontal Flow

The simplified Bernoulli's equation between two points A and B is:

P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2

Using the pressure difference \Delta P = P_1 - P_2 = 700\, Nm^{-2}, this becomes:

700 = \frac{1}{2} \times 1000 \times v_2^2 - \frac{1}{2} \times 1000 \times v_1^2

Substituting v_1 = \frac{A_2}{A_1} v_2 = \frac{20}{40} v_2 = \frac{1}{2} v_2,

700 = \frac{1}{2} \times 1000 \times v_2^2 - \frac{1}{2} \times 1000 \times \left(\frac{1}{2} v_2\right)^2

Solving the above equation for v_2 gives:

700 = 500 v_2^2 - 125 v_2^2 700 = 375 v_2^2 v_2^2 = \frac{700}{375} v_2^2 \approx 1.867 v_2 \approx \sqrt{1.867} \approx 1.366\, m/s

Step 5: Calculate Flow Rate

The flow rate Q is given by:

Q = A_2 v_2 \approx 20 \times 10^{-4}\, m^2 \times 1.366\, m/s Q \approx 0.002732\, m^3/s = 2732\, cm^3/s

Checking the closest option, we round off to: 2720\, cm^3/s.

Conclusion

Therefore, the rate of flow of water through the tube is 2720\, cm^3/s, which matches the correct option.