Question:

Water flows in a horizontal pipe in a streamlined manner at an absolute pressure of \(4 \times 10^5 \, \text{Pa}\) and speed of \(6 \, \text{m/s}\). If it exits the pipe at a pressure of \(10^5 \, \text{Pa}\), the speed of water at the exit point is .......... m/s (Round off to 1 decimal place).
(The density of water is \(1000 \, \text{kg/m}^3\))

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For horizontal fluid flow, use Bernoulli's equation: \(P + \frac{1}{2}\rho v^2 = \text{constant}\). Always ensure consistent units.
Updated On: Dec 4, 2025
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Correct Answer: 25.1

Solution and Explanation

Step 1: Apply Bernoulli's theorem.
\[ P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 \] Substitute given values: \[ 4 \times 10^5 + \frac{1}{2}(1000)(6^2) = 10^5 + \frac{1}{2}(1000)v_2^2 \]

Step 2: Simplify.
\[ 400000 + 18000 = 100000 + 500v_2^2 \] \[ 500v_2^2 = 318000 $\Rightarrow$ v_2^2 = 636 $\Rightarrow$ v_2 = 24.2 \, \text{m/s} \]

Step 3: Conclusion.
Hence, the exit velocity of water = 24.2 m/s.

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