Question

A liquid of specific gravity 1.0 flows in a pipe at a rate of 900 lit/s, from point 1 to point 2 which is 1 m above point 1. The areas at section 1 and 2 are 0.6 m\(^2\) and 0.3 m\(^2\) respectively. If the pressure at section 1 is 20 kPa, determine the pressure at section 2. (Assume g = 10 m/s\(^2\))

• A. 6.625 kPa
• B. 3.3125 kPa
• C. 13.25 kPa
• D. 16.625 kPa

Hint:

Always double-check units during conversion.
Ensure that the flow rate is converted from liters/second to cubic meters/second by dividing by 1000, and keep all pressures in Pascals during the calculation to avoid errors.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The goal is to calculate the pressure at a downstream elevated location in a pipe flow using the continuity and Bernoulli equations.

Step 2: Key Formula or Approach:
First, convert the volumetric flow rate \(Q\) to standard SI units (\(\text{m}^3/\text{s}\)).
Find the velocities \(V_1\) and \(V_2\) using:
\[ Q = A_1 V_1 = A_2 V_2 \implies V = \frac{Q}{A} \] Apply Bernoulli's equation for steady, incompressible, frictionless flow:
\[ P_1 + \frac{1}{2}\rho V_1^2 + \rho g z_1 = P_2 + \frac{1}{2}\rho V_2^2 + \rho g z_2 \]

Step 3: Detailed Explanation:

•
Step 3.1: Convert and calculate velocities:
Flow rate, \(Q = 900\text{ lit/s} = 0.9\text{ m}^3/\text{s}\).
Density of the liquid (specific gravity = 1.0), \(\rho = 1000\text{ kg/m}^3\).
Velocity at section 1:
\[ V_1 = \frac{0.9}{0.6} = 1.5\text{ m/s} \] Velocity at section 2:
\[ V_2 = \frac{0.9}{0.3} = 3.0\text{ m/s} \]
•
Step 3.2: Define elevation and pressure values:
Let the reference datum be at section 1: \(z_1 = 0\text{ m}\).
Then elevation at section 2 is: \(z_2 = 1\text{ m}\).
Given pressure at section 1, \(P_1 = 20\text{ kPa} = 20000\text{ Pa}\).

•
Step 3.3: Substitute into Bernoulli's equation:
\[ 20000 + \frac{1}{2}(1000)(1.5^2) + 0 = P_2 + \frac{1}{2}(1000)(3^2) + (1000)(10)(1) \] \[ 20000 + 500(2.25) = P_2 + 500(9) + 10000 \] \[ 20000 + 1125 = P_2 + 4500 + 10000 \] \[ 21125 = P_2 + 14500 \] \[ P_2 = 21125 - 14500 = 6625\text{ Pa} = 6.625\text{ kPa} \]

Step 4: Final Answer:
The pressure at section 2 is \(6.625\text{ kPa}\).