Oil of specific gravity 0.8 flows in a horizontal pipe at a height of 3 m above a datum plane. At a section of the pipe, diameter is 120 mm and the pressure intensity is 125 KN/m2 . If the total energy at the section is 25 m, find the rate of flow of oil.

To find the rate of flow of oil in the given problem, we can use the principles of fluid mechanics, specifically the Bernoulli equation and the continuity equation.

Given:
- Specific gravity of oil, \( SG = 0.8 \)
- Height above datum plane, \( z = 3 \) m
- Diameter of pipe, \( D = 120 \) mm = 0.12 m
- Pressure intensity, \( P = 125 \) kN/m² = 125,000 N/m²
- Total energy head, \( H = 25 \) m

First, let's convert the specific gravity to density:
\[ \text{Density of oil,} \ \rho = SG \times \text{Density of water} \]
Since the density of water is \( 1000 \ \text{kg/m}^3 \):
\[ \rho = 0.8 \times 1000 = 800 \ \text{kg/m}^3 \]

The Bernoulli equation at the section of the pipe can be written as:
\[ \frac{P}{\rho g} + \frac{v^2}{2g} + z = H \]

Where:
- \( P \) is the pressure intensity
- \( \rho \) is the density of oil
- \( g \) is the acceleration due to gravity (9.81 m/s²)
- \( v \) is the velocity of the oil
- \( z \) is the height above the datum plane
- \( H \) is the total energy head

We need to find the velocity \( v \) of the oil. Rearrange the Bernoulli equation to solve for \( v \):
\[ \frac{v^2}{2g} = H - \frac{P}{\rho g} - z \]

Plug in the given values:
\[ \frac{v^2}{2 \times 9.81} = 25 - \frac{125000}{800 \times 9.81} - 3 \]

Calculate each term:
\[ \frac{125000}{800 \times 9.81} = \frac{125000}{7848} \approx 15.93 \]
\[ \frac{v^2}{19.62} = 25 - 15.93 - 3 \]
\[ \frac{v^2}{19.62} = 6.07 \]
\[ v^2 = 6.07 \times 19.62 \]
\[ v^2 = 119.02 \]
\[ v = \sqrt{119.02} \]
\[ v \approx 10.91 \ \text{m/s} \]

Next, we use the continuity equation to find the rate of flow \( Q \):
\[ Q = A \times v \]
where \( A \) is the cross-sectional area of the pipe:
\[ A = \pi \left(\frac{D}{2}\right)^2 = \pi \left(\frac{0.12}{2}\right)^2 \approx 0.0113 \ \text{m}^2 \]

Then, the rate of flow \( Q \) is:
\[ Q = 0.0113 \times 10.91 \approx 0.123 \ \text{m}^3/\text{s} \]

So, the rate of flow of oil is approximately \( 0.123 \ \text{m}^3/\text{s} \).

Do you want more details or have any questions? Here are some related questions you might find helpful:

1. What is the Bernoulli equation and how is it derived?
2. How does specific gravity relate to the density of a fluid?
3. What are the assumptions made in the Bernoulli equation?
4. Can you explain the concept of total energy head in fluid mechanics?
5. How do you calculate the flow rate in different types of pipes?
6. What are the practical applications of the Bernoulli equation?
7. How does the diameter of a pipe affect the velocity of fluid flow?
8. What are the limitations of using the Bernoulli equation for real-life fluid flow problems?

**Tip:** When solving fluid mechanics problems, always ensure that units are consistent. Converting units at the

Learn how to calculate the rate of flow of oil in a horizontal pipe using the principles of fluid mechanics. Solve a problem involving specific gravity, pressure intensity, and total energy head. Understand the application of the Bernoulli equation and continuity equation in fluid dynamics.

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