3. Additional Pressure Calculation Problems

We have a 6 meter deep tank, which contains 4 meters of water and 2 meters of oil with a relative density of 0.88. We have to determine the pressure at the bottom of the tank.

This problem involves calculating the pressure at the bottom of a tank filled with two different fluids: water and oil. The total depth of the tank is 6 meters, with 4 meters of water on top and 2 meters of oil below it. We need to consider the pressure contributions from both fluids incrementally, starting from the top and moving downward to where we want to calculate the pressure, at the tank’s bottom.

Consider a glass of water with some honey at the bottom. The water on top exerts pressure on the honey due to its weight. In a similar manner, in our tank, the water above exerts pressure on the oil below, affecting the pressure at the tank's bottom.

To find P2, we use the formula P1 + pressure due to 2 meters of oil: P2 = P1 + (γ₀ × 2). Knowing P1 = 0, we calculate P2 as follows: P2 = 0 + (0.88 × 9790 × 2) = 17230.4 Newton per meter square.

First, we find the pressure where the water meets the oil (point P2). Since the pressure at the surface of the water is atmospheric pressure (which we consider 0 in this setup), the pressure due to the oil can be calculated using its density multiplied by the height of the oil column. Using the equation, we find that the pressure at point P2 is 17230.4 N/m² caused by the oil above.

Think of it as stacking books one on top of the other on a table. The pressure applied on the table increases with the number of books (weight) on top of it. Here, the oil acts like books, increasing the pressure at the water interface.

Next, we find P3 at the bottom of the tank, using P2 and calculating the pressure due to the 4 meters of water below: P3 = P2 + (γw × 4) = 17230.4 + (9790 × 4) = 56390.4 Newton per meter square.

Now, we calculate the pressure at the very bottom of the tank (P3). We take the pressure we just calculated at the water-oil interface (P2) and add the pressure contribution from the column of water above it (4 meters high). This approach gives us the total pressure at the bottom of the tank as 56390.4 N/m².

Imagine swimming in a deep pool. The deeper you go, the heavier the water above pushes down, making it harder to go deeper. Similarly, here, the depth of water above increases the pressure felt at the bottom of the tank.

We have to calculate the pressure difference between points M and N. The approach is to start at a point with known pressure and traverse to the other point, adjusting for the heights of the fluid columns.

In this part, we learn how to calculate the pressure difference between two points (M and N) using the relationship between them. We will account for the height of the fluid at both points and apply the principle that moving up through a fluid decreases pressure while moving down increases it. We set the relationship equation for pressure at M and N based on these principles.

It's like traveling up and down a staircase: the higher you go, the more you have to account for the climb (pressure decreases). When you go down, you feel the added weight from above (pressure increases). This analogy helps illustrate how pressures change in different fluid heights.

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