An open cylindrical tank contains 1.20 m of saltwater (s=1.03), 0.80 m of gasoline (s= 0.70) and 1 m of oil (s=0.80.)compute the pressures at the interfaces and at 2.5 m below the liquid surface

To solve this problem, we need to calculate the pressures at the interfaces between the different liquids and at a point 2.5 m below the liquid surface in an open cylindrical tank.

### Given:
- \( h_1 = 1.20 \) m (height of saltwater)
- \( h_2 = 0.80 \) m (height of gasoline)
- \( h_3 = 1.00 \) m (height of oil)
- \( s_1 = 1.03 \) (specific gravity of saltwater)
- \( s_2 = 0.70 \) (specific gravity of gasoline)
- \( s_3 = 0.80 \) (specific gravity of oil)
- \( g = 9.81 \) m/s² (acceleration due to gravity)
- \( \rho_{\text{water}} = 1000 \) kg/m³ (density of water)

### Step 1: Determine the densities of the liquids
\[ \rho_1 = s_1 \times \rho_{\text{water}} = 1.03 \times 1000 \, \text{kg/m}^3 = 1030 \, \text{kg/m}^3 \]
\[ \rho_2 = s_2 \times \rho_{\text{water}} = 0.70 \times 1000 \, \text{kg/m}^3 = 700 \, \text{kg/m}^3 \]
\[ \rho_3 = s_3 \times \rho_{\text{water}} = 0.80 \times 1000 \, \text{kg/m}^3 = 800 \, \text{kg/m}^3 \]

### Step 2: Calculate the pressure at the interfaces

#### 1. Pressure at the saltwater-gasoline interface
\[ P_{\text{interface 1}} = \rho_1 \times g \times h_1 \]
\[ P_{\text{interface 1}} = 1030 \times 9.81 \times 1.20 \]
\[ P_{\text{interface 1}} = 12116.76 \, \text{Pa} \]

#### 2. Pressure at the gasoline-oil interface
First, calculate the pressure due to gasoline and saltwater:
\[ P_{\text{interface 2}} = P_{\text{interface 1}} + \rho_2 \times g \times h_2 \]
\[ P_{\text{interface 2}} = 12116.76 + (700 \times 9.81 \times 0.80) \]
\[ P_{\text{interface 2}} = 12116.76 + 5493.6 \]
\[ P_{\text{interface 2}} = 17610.36 \, \text{Pa} \]

### Step 3: Pressure at 2.5 m below the liquid surface
To calculate the pressure at a point 2.5 m below the liquid surface:

Since the total height of the liquids is \( h_1 + h_2 + h_3 = 3 \) m, and we need the pressure 2.5 m below the surface, the point is 0.5 m above the bottom of the tank. 

This point is in the oil layer, so the pressure will be the pressure at the gasoline-oil interface plus the pressure due to the 0.5 m of oil:

\[ P_{2.5m} = P_{\text{interface 2}} + \rho_3 \times g \times h \text{(remaining oil height)} \]
\[ P_{2.5m} = 17610.36 + (800 \times 9.81 \times 0.5) \]
\[ P_{2.5m} = 17610.36 + 3924 \]
\[ P_{2.5m} = 21534.36 \, \text{Pa} \]

### Final Answers:
- **Pressure at saltwater-gasoline interface:** \( 12116.76 \, \text{Pa} \)
- **Pressure at gasoline-oil interface:** \( 17610.36 \, \text{Pa} \)
- **Pressure at 2.5 m below the liquid surface:** \( 21534.36 \, \text{Pa} \)

Would you like further details or explanations? Here are some related questions:

1. How would the pressure change if the tank were closed instead of open?
2. What if the densities of the liquids were different, how would it affect the pressure?
3. How is specific gravity related to density?
4. How can you calculate the total force on the bottom of the tank?
5. What is the significance of atmospheric pressure in this context?

**Tip:** When calculating pressures in fluids, always consider the height and density of each fluid layer to ensure accuracy.

Learn how to calculate pressures at liquid interfaces and at specific depths in a multi-layered tank with saltwater, gasoline, and oil. Detailed step-by-step solution covering fluid mechanics, pressure calculations, and density considerations.

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