Calculating Fluid Specific Weight: A Comprehensive Guide
Hey guys! Let's dive into a cool fluid mechanics problem. This one's all about figuring out the specific weight of a fluid, given its relative specific weight and the specific weight of water. It's super important for understanding how fluids behave, especially in things like engineering and physics. So, let's break it down step by step and make sure you get a handle on it.
Understanding Specific Weight and Relative Specific Weight
Okay, before we jump into the problem, let's make sure we're on the same page with the key terms. We're talking about specific weight and relative specific weight, right? So, what do these terms even mean?
Firstly, the specific weight of a substance is essentially how heavy it is per unit volume. Think of it like this: if you have a box, the specific weight tells you how much that box full of fluid would weigh. It's usually measured in units like Newtons per cubic meter (N/m³). Now, water, being a common reference point, has a specific weight of about 9800 N/m³ (at standard conditions). This is crucial because it helps us compare different fluids. It shows how much gravitational force acts on a fluid's mass within a defined space.
Now, let's talk about relative specific weight. This is a dimensionless number that compares the specific weight of a fluid to the specific weight of a reference fluid, typically water. So, if a fluid has a relative specific weight of 0.9, it means the fluid is 0.9 times as heavy as water, volume for volume. It's a handy way to quickly assess how dense a fluid is compared to something we already know well, like water. The relative specific weight tells us the ratio of the fluid's weight to the weight of an equal volume of water. For example, if the relative specific weight is greater than 1, the fluid is heavier than water. If it is less than 1, the fluid is lighter than water. This concept is fundamental in comparing different fluids and understanding their behavior in various applications, like buoyancy and fluid dynamics. To fully understand these terms is to understand how the weight of a fluid is distributed in a specific volume relative to that of water.
The relative specific weight is also known as the specific gravity. The use of this concept simplifies calculations by providing a reference point.
So, with these definitions in mind, let's tackle the problem.
Problem Setup and Given Information
Alright, let's get down to the actual problem. We're told that a fluid has a relative specific weight of 0.9. We also know that the specific weight of water is 9800 N/m³. Our mission? To determine the specific weight of the fluid. The problem also provides the value of gravitational acceleration (g = 9.8 m/s²), though, in this case, we won't need it directly because we're given the specific weight of water, not just the density.
- Relative specific weight of the fluid: 0.9
- Specific weight of water: 9800 N/m³
- Gravitational acceleration (g): 9.8 m/s² (Though not directly used in this problem)
This setup allows us to understand the relationship between different fluid weights easily.
The Calculation: Finding the Specific Weight
Here’s where we get to the core of the problem: finding the specific weight of the fluid. Since we know the relative specific weight (0.9) and the specific weight of water (9800 N/m³), we can use a simple formula:
Specific Weight of Fluid = Relative Specific Weight × Specific Weight of Water
Plugging in the numbers:
Specific Weight of Fluid = 0.9 × 9800 N/m³ = 8820 N/m³
So there you have it! The specific weight of the fluid is 8820 N/m³. This tells us that, for every cubic meter of this fluid, you'd have a weight of 8820 Newtons. The ease of this calculation highlights the usefulness of relative specific weight.
Choosing the Correct Answer and Why
Now, let's look at the multiple-choice options you provided:
- A) 882 N/m³
- B) 1089 N/m³
- C) 8820 N/m³
- D) 980 N/m³
- E) 9800 N/m³
Our calculated answer is C) 8820 N/m³. This is because we multiplied the relative specific weight (0.9) by the specific weight of water (9800 N/m³). This matches our calculations and confirms our understanding of specific weight concepts.
Additional Considerations and Real-World Applications
This type of problem might seem straightforward, but it has some really important real-world applications. Understanding specific weight is crucial in many areas, including:
- Engineering: Designing structures that interact with fluids (like dams, ships, or pipelines) relies heavily on knowing the specific weight of the fluids involved.
- Buoyancy: Figuring out whether an object will sink or float depends on the specific weight of both the object and the fluid it's in. This is why ships, which are made of heavy materials, can float.
- Environmental Science: Specific weight is used to analyze pollutants in water bodies, understand the movement of fluids, and manage water resources. For example, knowing the specific weight of different substances helps scientists predict how they will spread and interact in aquatic environments.
- Industrial Processes: Specific weight is a key factor in industrial applications, such as the separation of different substances based on density, the design of storage tanks, and the control of fluid flow.
Basically, if you're working with fluids, knowing their specific weight is pretty fundamental. It helps you predict how they'll behave under different conditions and design systems that work effectively.
Conclusion: Wrapping It Up
Awesome, you’ve now successfully calculated the specific weight of a fluid! You've seen how to use the relative specific weight and the specific weight of a reference fluid (water) to find the specific weight of any other fluid. This basic concept unlocks understanding in many areas. Remember, understanding the fundamentals, like specific weight, is the cornerstone to success in fluid mechanics and related fields. Keep practicing these problems, and you'll be a pro in no time! Keep up the great work, and happy calculating!