Example Problem: Determining Pressure at Bottom of Tank
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Understanding Pressure Variation
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Good morning class! Today, we will explore how pressure varies in fluids at different depths. Can anyone tell me how pressure changes when we go deeper in a fluid?
I think pressure increases as we go deeper!
Exactly! The pressure increase can be calculated using the formula: P = h * γ, where 'h' is the depth and 'γ' is the specific weight of the fluid. Can anyone remind me what specific weight means?
Isn't specific weight the weight per unit volume of the fluid?
That's correct! As we dive deeper into this topic, we will apply these principles to calculate pressure at different points in our example tank.
Setting Up the Problem
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Let’s consider a tank that is 6 meters deep, containing 4 meters of water and 2 meters of oil. Who can provide me the initial steps to find the pressure at the tank's bottom?
First, we need to determine the pressure at the water-oil interface, right?
Absolutely! We can denote this pressure as P2. If we start with atmospheric pressure at the top, how should we calculate P2?
It should be P1 plus the pressure due to the oil column!
Correct! We will calculate P2 using the specific weight of oil. Remember the formula: P2 = P1 + h_oil * γ_oil.
Calculating the Pressures
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Now, let’s compute P2 with our known values. What is the specific weight of the oil if its relative density is 0.88?
It would be 0.88 times the specific weight of water, which is approximately 9790 N/m³.
Excellent! So, what’s the specific weight of the oil?
It’s around 8615.2 N/m³.
Now, using this value, let’s calculate P2 as we discussed earlier.
Summing Up the Calculations
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Great work so far! After computing P2 as 17230.4 N/m², how do we find P3?
We add the pressure from the water column to P2. So, P3 = P2 + (4m * γ_water).
Exactly! And what is the result?
The final pressure at the bottom, P3, is 56390.4 N/m² or 56.39 kPa!
Well done! This demonstrates how we can calculate pressures in a hydraulic system.
Introduction & Overview
Read summaries of the section's main ideas at different levels of detail.
Quick Overview
Standard
The section covers the methodology for calculating pressure at different points in a fluid system through a practical example involving a tank with water and oil. Key formulas and relationships between pressure, depth, and specific weight are emphasized.
Detailed
Detailed Summary
This section of the chapter on Hydraulic Engineering revolves around determining the pressure at the bottom of a tank that contains a mixture of water and oil. By utilizing the principles of fluid mechanics, specifically the concept of hydrostatic pressure variation, the section guides through the calculation step-by-step. The process begins with the understanding of how pressure varies with depth in a fluid and applies this principle to an example problem where a tank has different fluid layers: 4 meters of water topped by 2 meters of oil with a specific density of 0.88. The section provides the necessary equations for calculating pressure at key points (P2 at the water-oil interface and P3 at the bottom of the tank), leading to an answer for the overall pressure at the tank bottom being 56.39 kPa. Thus, this section effectively illustrates the application of theoretical fluid mechanics in practical scenarios.
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Introduction to the Problem
Chapter 1 of 5
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Chapter Content
So, we have a 6 meter deep tank, so this is 6 meter, right? And contains 4 meters of water, so this is as it is written here very clear 4 meter and 2 meter of oil of relative density 0.88.
Detailed Explanation
In this section, we are introduced to a specific problem related to fluid mechanics. We have a tank that is 6 meters deep. Within this tank, there are layers of different fluids: there are 4 meters of water and 2 meters of oil having a relative density (or specific gravity) of 0.88. The relative density helps us understand how the density of the oil compares to the density of water, which is an essential property in fluid mechanics.
Examples & Analogies
Think of a glass filled with different colored liquids. The heavier liquid settles at the bottom, while the lighter one floats on top. Just like the oil floating on the top of the water in this tank, understanding how these different liquids interact helps us comprehend concepts such as pressure and buoyancy.
Calculating Pressure at the Water Interface
Chapter 2 of 5
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Chapter Content
First, determine the pressure while water interface, that is, p2, so p2 is written as p1 plus pressure due to 2 meters of oil, very nice. So or p1 plus what is the pressure due to 2 meters of oil, gamma 0 into 2.
Detailed Explanation
To find the pressure at the interface between the water and the oil (denoted as p2), we need to consider the pressure exerted by the oil. The pressure p2 can be calculated by starting from the known pressure at the top of the oil layer (p1, which is 0 at the open tank surface) and adding the pressure created by the 2 meters of oil above that point. This is calculated using the formula: pressure = density × gravitational acceleration × height (p = ρgh).
Examples & Analogies
Imagine pressing down on a full balloon filled with water; the pressure you feel at the bottom of the balloon increases as you apply more force. Similarly, in our tank, the weight of the oil above pushes down on water below, increasing the pressure at the interface.
Finding the Pressure Due to Oil
Chapter 3 of 5
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Chapter Content
Here p1 is equal to 0, right? Whereas gamma not which is the pressure of the oil it is 0.88 specific gravity into 9790 that is 8615.2 Newton per meter square.
Detailed Explanation
In calculating p2, we substitute our known value of p1 (which is 0) and calculate the pressure due to the 2 meters of oil (using its specific gravity relative to water). The unit weight (gamma) of the oil is calculated to understand how much pressure it exerts on the underlying water layer. The calculation provides us with a specific value in Newtons per square meter, which reflects the amount of force being applied over any given area.
Examples & Analogies
Consider the way we use scales to measure weight. Just as an object of known weight can exert a certain force on the scale, the heavy oil presses down on the water below, creating pressure—this is part of the foundation of how we measure force and pressure in fluids.
Calculating Total Pressure at the Bottom of the Tank
Chapter 4 of 5
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Chapter Content
So p3 will be p2 + pressure due to 4 meters of water, right.
Detailed Explanation
After calculating p2 (the pressure at the bottom of the oil layer), the next step involves calculating p3, which is the pressure at the bottom of the tank. To do this, we add the pressure contribution from the 4 meters of water above the interface. This cumulative approach helps us determine the total pressure acting at the bottom of the tank, incorporating contributions from all fluid layers above.
Examples & Analogies
Think of layers in a cake, where each layer adds to the overall height and weight of the cake. Similarly, the water layer contributes to the total pressure felt at the bottom of the tank. Each layer influences the pressure just like each cake layer affects the cake's weight.
Final Result of Pressure Calculation
Chapter 5 of 5
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Chapter Content
The answer was P 3 is equal to 56.39 kilo Pascal.
Detailed Explanation
After performing all necessary calculations, we arrive at the final result for p3, which is the pressure at the bottom of the tank. Expressed as 56.39 kilopascals, this pressure is a crucial parameter for engineers when designing tanks and systems to ensure that structures can handle the weight of the fluids.
Examples & Analogies
When filling a car tire, the pressure inside is measured in kilopascals (kPa). If the pressure is too low, the tire may not perform well; too high, and it could burst. Similarly, understanding the pressure at the tank's bottom is vital for ensuring safety and functionality in engineering applications.
Key Concepts
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Pressure Variation: Pressure increases with the depth of the fluid column.
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Hydraulic Equation: P = h * γ, where h is the depth and γ is specific weight.
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Differential Manometer: A device that measures pressure differences between two points in a fluid.
Examples & Applications
Example of pressure variation: A tank with water at different depths shows increased pressure with depth.
Example of differential manometers: Using a manometer to measure the pressure difference between two liquid points.
Memory Aids
Interactive tools to help you remember key concepts
Rhymes
Pressure down low, in fluids, will grow; the deeper you go, the more you'll know!
Stories
Imagine diving under water; the deeper you go, the heavier it feels. Just like pressure, it builds as you go down!
Memory Tools
Remember: "Penny’s Heavy Weight" (P = h * γ) to recall how to calculate pressure.
Acronyms
Pressure Increase – P.I. (Pressure increases with depth).
Flash Cards
Glossary
- Hydrostatic Pressure
The pressure exerted by a fluid at equilibrium due to the force of gravity.
- Specific Weight
The weight per unit volume of a substance, typically expressed in N/m³.
- Relative Density
The ratio of the density of a substance to the density of a reference material, usually water.
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