https://file.aiquickdraw.com/mathbot/file/1725503883272ymda1mn5.jpg

The inverted U-tube manometer shown below is used for measuring pressure differences in liquids. The manometer contains two fluids, with one fluid having a specific gravity of 13 (SG1 = 13). Given that PA - PD = 32 kPa, determine the specific gravity of the second fluid. Assume the atmospheric pressure is 100 kPa, acceleration due to gravity g=9.8 m/s2, and the density of water is 1000 kg/m3.  (35 marks: 10 marks for the correct answer and 25 marks for showing the correct working)

To solve the problem, we need to determine the specific gravity (SG2) of the second fluid using the information provided in the problem and the diagram.

### Given:
- Specific gravity of the first fluid, \( SG_1 = 13 \).
- Pressure difference, \( P_A - P_D = 32 \) kPa.
- Atmospheric pressure, \( P_{\text{atm}} = 100 \) kPa.
- Acceleration due to gravity, \( g = 9.8 \) m/s².
- Density of water, \( \rho_{\text{water}} = 1000 \) kg/m³.

### Assumptions:
- The fluids are incompressible.
- Hydrostatic pressure equation is applicable.

### Step 1: Understand the relationship between pressure and specific gravity.
Specific gravity (SG) is defined as the ratio of the density of the fluid to the density of water. Mathematically:
\[
SG = \frac{\rho}{\rho_{\text{water}}}
\]
where \( \rho \) is the density of the fluid.

For the manometer, the pressure difference is determined by the height difference of the fluids and their densities.

### Step 2: Write the pressure equation for the U-tube.
Considering that pressure difference between points A and D is due to the height of fluid columns in the manometer:
\[
P_A = P_{\text{atm}} + \rho_1 g h_1
\]
\[
P_D = P_{\text{atm}} + \rho_2 g h_2
\]
Where:
- \( \rho_1 \) and \( \rho_2 \) are the densities of the two fluids.
- \( h_1 \) and \( h_2 \) are the heights of the two columns.

The pressure difference between points A and D is:
\[
P_A - P_D = \rho_1 g h_1 - \rho_2 g h_2
\]
Given that \( P_A - P_D = 32 \) kPa, we have:
\[
32 \times 10^3 = \rho_1 g h_1 - \rho_2 g h_2
\]

### Step 3: Relate densities to specific gravities.
\[
\rho_1 = SG_1 \times \rho_{\text{water}} = 13 \times 1000 = 13000 \, \text{kg/m}^3
\]
Let \( SG_2 \) be the specific gravity of the second fluid:
\[
\rho_2 = SG_2 \times \rho_{\text{water}} = SG_2 \times 1000
\]

### Step 4: Express the equation in terms of SG2.
Now substitute the densities into the equation:
\[
32 \times 10^3 = 13000 \times 9.8 \times h_1 - SG_2 \times 1000 \times 9.8 \times h_2
\]
Simplifying the equation:
\[
32 \times 10^3 = 127400 \times h_1 - 9800 \times SG_2 \times h_2
\]
### Step 5: Final Solution.
To solve for \( SG_2 \), we need the ratio \( \frac{h_1}{h_2} \). If this information is known or provided in the diagram, substitute it in to find \( SG_2 \).

If you can provide the ratio or confirm the values of \( h_1 \) and \( h_2 \), we can continue with the calculation.

Let me know if you need further clarification or details on any part of the solution. Below are some questions to help deepen your understanding:

1. What would happen to the specific gravity of the second fluid if the pressure difference \( P_A - P_D \) was higher?
2. How does the atmospheric pressure influence the readings in a U-tube manometer?
3. What assumptions are necessary for the hydrostatic pressure equation to be valid in this setup?
4. How would the result change if the specific gravity of the first fluid were different?
5. What physical properties of the fluids could cause errors in the pressure measurements?

**Tip:** Always cross-check the dimensions when working with pressure-related problems to ensure consistency across all units.

The inverted U-tube manometer shown below is used for measuring pressure differences in liquids. The manometer contains two fluids, with one fluid having a specific gravity of 13 (SG1 = 13). Given that PA - PD = 32 kPa, determine the specific gravity of the second fluid. Assume the atmospheric pressure is 100 kPa, acceleration due to gravity g=9.8 m/s2, and the density of water is 1000 kg/m3.  (35 marks: 10 marks for the correct answer and 25 marks for showing the correct working).

Learn how to determine the specific gravity of a second fluid in an inverted U-tube manometer. This problem involves key concepts of fluid mechanics and hydrostatics, such as the hydrostatic pressure equation, specific gravity, and density relations. Suitable for undergraduate engineering or physics students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation