https://file.aiquickdraw.com/mathbot/file/1726554667991iw4z8573.jpg

The problem you're dealing with involves a manometer connected to a pipe containing oil, and it asks for the pressure at the center of the pipe. Let's break it down step by step using the data provided in the diagram.

### Known Data:
- The height of the mercury column above the oil: \(1 \, \text{m}\)
- The height of the oil column: \(1.5 \, \text{m}\)
- Specific gravity of oil, \(s_{\text{oil}} = 0.8\)

### Steps for Solving:

#### 1. **Pressure Relationship in a Manometer**:
The pressure difference between two points in a fluid system can be calculated using the manometer equation:
\[
\Delta P = \rho g h
\]
Where:
- \(\Delta P\) is the pressure difference.
- \(\rho\) is the density of the fluid.
- \(g\) is the acceleration due to gravity.
- \(h\) is the height of the fluid column.

#### 2. **Apply for the Oil Column**:
- Specific gravity of oil: \(s_{\text{oil}} = 0.8\)
- Hence, density of oil, \(\rho_{\text{oil}} = 0.8 \times 1000 \, \text{kg/m}^3 = 800 \, \text{kg/m}^3\).

Pressure due to oil column of height 1.5 m:
\[
P_{\text{oil}} = \rho_{\text{oil}} \cdot g \cdot h_{\text{oil}} = 800 \times 9.81 \times 1.5 = 11772 \, \text{Pa}
\]

#### 3. **Apply for the Mercury Column**:
The pressure difference due to the mercury column needs to be added. The density of mercury (\(\rho_{\text{Hg}}\)) is much higher than that of oil, approximately \(13600 \, \text{kg/m}^3\).

Pressure due to mercury column of height 1 m:
\[
P_{\text{Hg}} = \rho_{\text{Hg}} \cdot g \cdot h_{\text{Hg}} = 13600 \times 9.81 \times 1 = 133416 \, \text{Pa}
\]

#### 4. **Total Pressure Difference**:
The total pressure at the center of the pipe can be determined by adding the pressure contributions from both the oil and mercury columns:
\[
P_{\text{total}} = P_{\text{oil}} + P_{\text{Hg}} = 11772 + 133416 = 145188 \, \text{Pa}
\]

Thus, the pressure at the center of the pipe is approximately **145,188 Pa (or 145.19 kPa)**.

---

Would you like a more detailed explanation of any part of this solution, or do you have further questions?

### Follow-up Questions:
1. How does specific gravity affect pressure calculations in fluids?
2. What assumptions are made when calculating pressure in manometers?
3. Why is mercury often used in manometers instead of other liquids?
4. How does the height of the liquid column affect the pressure difference?
5. What is the relationship between absolute and gauge pressure in this context?

**Tip:** Remember, specific gravity is simply the ratio of the density of a fluid to the density of water, and it helps to simplify calculations for pressure.

For the manometer shown, determine the pressure at the center of the pipe.

Determine the pressure at the center of a pipe using a manometer with oil (specific gravity = 0.8) and mercury. This problem involves calculating pressure differences based on fluid heights and specific gravities, suitable for Fluid Mechanics studies.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation