change 30kPa h2 50cm h1 8cm L1 and L2 6cm gage fluid SG 2.4

To solve the problem with the updated parameters, let's rework the calculations based on the new values.

### Updated Given Data:
1. **Gauge pressure of air:** \( P_{\text{air}} = 30 \, \text{kPa} \)
2. **Height of water column \( h_2 \):** \( h_2 = 50 \, \text{cm} = 0.50 \, \text{m} \)
3. **Height of the pipe above the water surface \( h_1 \):** \( h_1 = 8 \, \text{cm} = 0.08 \, \text{m} \)
4. **Length \( L_1 \) and \( L_2 \) of the fluid columns in the manometer:** \( L_1 = L_2 = 6 \, \text{cm} = 0.06 \, \text{m} \)
5. **Specific Gravity (SG) of the manometer fluid:** SG = 2.4
6. **Density of water \( \rho_{\text{water}} \):** Approximately \( 1000 \, \text{kg/m}^3 \)
7. **Acceleration due to gravity \( g \):** \( 9.81 \, \text{m/s}^2 \)

### Key Equations:
1. **Manometer fluid density \( \rho_{\text{gage}} \):**
   \[
   \rho_{\text{gage}} = \text{SG} \times \rho_{\text{water}} = 2.4 \times 1000 \, \text{kg/m}^3 = 2400 \, \text{kg/m}^3
   \]
   
2. **Pressure at the water surface inside the container:**
   \[
   P_{\text{surface}} = P_{\text{air}} = 30 \, \text{kPa} = 30 \times 10^3 \, \text{Pa}
   \]

3. **Pressure difference due to the manometer fluid (using the manometer readings \( L_1 \) and \( L_2 \)):**
   The difference in pressure indicated by the manometer is:
   \[
   \Delta P_{\text{gage}} = \rho_{\text{gage}} \times g \times (L_2 - L_1) 
   \]
   However, since \( L_1 = L_2 \), there is no differential pressure contribution from the manometer fluid itself. The difference will be zero due to the symmetric setup:
   \[
   \Delta P_{\text{gage}} = 0
   \]

4. **Pressure at the base of the water column \( h_2 \):**
   \[
   P_{\text{base}} = P_{\text{surface}} + \rho_{\text{water}} \times g \times h_2
   \]
   Substituting values:
   \[
   P_{\text{base}} = 30 \times 10^3 \, \text{Pa} + 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 0.50 \, \text{m}
   \]
   \[
   P_{\text{base}} = 30000 \, \text{Pa} + 4905 \, \text{Pa} 
   \]
   \[
   P_{\text{base}} \approx 34905 \, \text{Pa}
   \]

5. **Pressure in the pipe (at height \( h_1 \) above the water surface):**
   \[
   P_{\text{pipe}} = P_{\text{base}} - \rho_{\text{water}} \times g \times h_1
   \]
   Substituting the values:
   \[
   P_{\text{pipe}} = 34905 \, \text{Pa} - 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 0.08 \, \text{m}
   \]
   \[
   P_{\text{pipe}} = 34905 \, \text{Pa} - 784.8 \, \text{Pa}
   \]
   \[
   P_{\text{pipe}} \approx 34120.2 \, \text{Pa}
   \]

### Final Answer:
The pressure in the pipe is approximately **34.12 kPa**.

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Would you like any further explanations or have any additional questions? Here are some related questions to consider:

1. What would happen if the lengths \( L_1 \) and \( L_2 \) were different, introducing a differential pressure?
2. How does the specific gravity of the manometer fluid affect the pressure reading in this scenario?
3. If the pipe's height \( h_1 \) were further increased, how would that influence the pressure?
4. How would the results change if the fluid in the container were oil with a different density?
5. Can this manometer setup be used for highly compressible fluids like gases? Why or why not?

**Tip:** Always ensure that you account for the specific gravity when dealing with manometric fluids, as it directly influences the pressure readings by altering the density factor in the calculations.

Learn how to solve a fluid mechanics problem involving a manometer. Calculate pressure using given parameters: 30 kPa air pressure, 50 cm water column height, 8 cm pipe height, and 6 cm manometer fluid columns (SG 2.4).

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