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Water flows steadily through a horizonta...

Water flows steadily through a horizontal pipe of a variable cross-section. If the pressure of the water is p at a point , where the speed of the flow is v. What is the pressure at another point , where the speed of the flow is 2 v ? Let the density of water be `1 g cm ^(-3)`.

A

`p+(3/2)rhov^2`

B

`p-2rhov^2`

C

`p - 3rhov^2`

D

`p- (3/2) rhov^2`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we will use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid in a streamline flow. Since the pipe is horizontal, we can ignore the height terms. ### Step-by-Step Solution: 1. **Understand Bernoulli's Equation**: The equation for a horizontal flow is given by: \[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \] where: - \( P_1 \) is the pressure at point 1, - \( v_1 \) is the velocity at point 1, - \( P_2 \) is the pressure at point 2, - \( v_2 \) is the velocity at point 2, - \( \rho \) is the density of the fluid. 2. **Assign Known Values**: From the problem: - Let \( P_1 = P \) (pressure at point 1), - Let \( v_1 = v \) (velocity at point 1), - Let \( v_2 = 2v \) (velocity at point 2). 3. **Substitute Known Values into Bernoulli's Equation**: Substitute \( P_1 \), \( v_1 \), and \( v_2 \) into the equation: \[ P + \frac{1}{2} \rho v^2 = P_2 + \frac{1}{2} \rho (2v)^2 \] 4. **Simplify the Equation**: The term \( (2v)^2 \) simplifies to \( 4v^2 \): \[ P + \frac{1}{2} \rho v^2 = P_2 + \frac{1}{2} \rho (4v^2) \] This can be rewritten as: \[ P + \frac{1}{2} \rho v^2 = P_2 + 2 \rho v^2 \] 5. **Rearranging to Solve for \( P_2 \)**: Rearranging gives: \[ P_2 = P + \frac{1}{2} \rho v^2 - 2 \rho v^2 \] \[ P_2 = P - \frac{3}{2} \rho v^2 \] 6. **Substituting the Density of Water**: Given that the density of water \( \rho = 1 \, \text{g/cm}^3 = 1000 \, \text{kg/m}^3 \) (for SI units), we can leave it in terms of \( \rho \) for now: \[ P_2 = P - \frac{3}{2} \rho v^2 \] ### Final Answer: Thus, the pressure at the second point where the speed of flow is \( 2v \) is: \[ P_2 = P - \frac{3}{2} \rho v^2 \]
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