Calculating Pipe Length From Metal Mass Density And Dimensions A Step-by-Step Guide

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In this engineering problem, we delve into the practical application of density and volume calculations. An artisan possesses 63 kg of metal with a density of 7000 kg/m³. The artisan plans to utilize this metal to fabricate a rectangular pipe. The pipe's external dimensions are 12 cm by 15 cm, while the internal dimensions are 10 cm by 12 cm. Our primary objective is to determine the length of the pipe in meters. This problem combines concepts of density, volume, and unit conversions, making it a quintessential example of practical engineering calculations.

Before diving into the calculations, it's crucial to grasp the fundamental concepts involved. Density, a key property of matter, is defined as mass per unit volume. It essentially tells us how much "stuff" is packed into a given space. The formula for density (ρ{\rho}) is expressed as:

ρ=mV{\rho = \frac{m}{V}}

where:

  • ρ{\rho} represents density (typically in kg/m³ or g/cm³)
  • m{m} represents mass (typically in kg or g)
  • V{V} represents volume (typically in m³ or cm³)

In this scenario, we are given the mass of the metal (63 kg) and its density (7000 kg/m³). This information allows us to calculate the total volume of metal available. Volume, in general, is the amount of three-dimensional space a substance or object occupies. For a rectangular prism (which is the shape of our pipe), the volume (V{V}) is calculated as:

V=l×w×h{V = l \times w \times h}

where:

  • l{l} represents length
  • w{w} represents width
  • h{h} represents height

However, our pipe is not a solid rectangular prism; it's a hollow one. Therefore, we need to consider the difference between the external volume and the internal volume to find the volume of the metal used in the pipe. This difference will then be equated to the total volume of metal calculated from the density and mass.

Let's break down the problem into manageable steps:

1. Calculate the Total Volume of Metal

Using the density formula, we can rearrange it to solve for volume:

V=mρ{V = \frac{m}{\rho}}

Plugging in the given values:

V=63 kg7000 kg/m³=0.009 m³{V = \frac{63 \text{ kg}}{7000 \text{ kg/m³}} = 0.009 \text{ m³}}

So, the artisan has 0.009 cubic meters of metal to work with.

2. Convert Dimensions to Meters

Since our volume is in cubic meters, we need to ensure all dimensions are in meters. We are given the dimensions in centimeters, so we'll convert them:

  • External dimensions:
    • 12 cm = 0.12 m
    • 15 cm = 0.15 m
  • Internal dimensions:
    • 10 cm = 0.10 m
    • 12 cm = 0.12 m

3. Define Variables and Set Up the Equation

Let's denote the length of the pipe as L{L} (in meters), which is what we want to find. The external volume (V{V}) of the pipe can be calculated as:

Vexternal=L×0.12 m×0.15 m{V_{\text{external}} = L \times 0.12 \text{ m} \times 0.15 \text{ m}}

The internal volume (V{V}) of the pipe (the hollow space) can be calculated as:

Vinternal=L×0.10 m×0.12 m{V_{\text{internal}} = L \times 0.10 \text{ m} \times 0.12 \text{ m}}

The volume of the metal used in the pipe is the difference between the external and internal volumes:

Vmetal=Vexternal−Vinternal{V_{\text{metal}} = V_{\text{external}} - V_{\text{internal}}}

Substituting the expressions for external and internal volumes, we get:

Vmetal=(L×0.12 m×0.15 m)−(L×0.10 m×0.12 m){V_{\text{metal}} = (L \times 0.12 \text{ m} \times 0.15 \text{ m}) - (L \times 0.10 \text{ m} \times 0.12 \text{ m})}

We know that the volume of the metal must equal the total volume of metal we calculated in step 1 (0.009 m³). Therefore, we can set up the equation:

0.009 m³=(L×0.12 m×0.15 m)−(L×0.10 m×0.12 m){0.009 \text{ m³} = (L \times 0.12 \text{ m} \times 0.15 \text{ m}) - (L \times 0.10 \text{ m} \times 0.12 \text{ m})}

4. Solve for the Length (L)

Now, we solve the equation for L{L}:

0.009=L(0.12×0.15)−L(0.10×0.12){0. 009 = L(0.12 \times 0.15) - L(0.10 \times 0.12)}

0.009=L(0.018)−L(0.012){0. 009 = L(0.018) - L(0.012)}

0.009=L(0.018−0.012){0. 009 = L(0.018 - 0.012)}

0.009=L(0.006){0. 009 = L(0.006)}

L=0.0090.006{L = \frac{0.009}{0.006}}

L=1.5 m{L = 1.5 \text{ m}}

Therefore, the length of the rectangular pipe is 1.5 meters.

It's always a good practice to verify our result. We can plug the calculated length back into the volume equation to see if it matches the total volume of metal:

Vmetal=(1.5 m×0.12 m×0.15 m)−(1.5 m×0.10 m×0.12 m){V_{\text{metal}} = (1.5 \text{ m} \times 0.12 \text{ m} \times 0.15 \text{ m}) - (1.5 \text{ m} \times 0.10 \text{ m} \times 0.12 \text{ m})}

Vmetal=0.027 m³−0.018 m³{V_{\text{metal}} = 0.027 \text{ m³} - 0.018 \text{ m³}}

Vmetal=0.009 m³{V_{\text{metal}} = 0.009 \text{ m³}}

This matches our initial calculation of the total metal volume, confirming our result.

In a practical setting, there might be some material loss during the manufacturing process (e.g., cutting, welding). The artisan would need to account for this waste and potentially adjust the dimensions or length of the pipe to ensure they stay within the 63 kg of metal available. Furthermore, the strength and structural integrity of the pipe would also need to be considered, depending on its intended use. Thinner walls (smaller difference between external and internal dimensions) might lead to a longer pipe, but could also compromise its strength.

In conclusion, by applying the principles of density and volume calculation, we have successfully determined that the length of the rectangular pipe the artisan can create is 1.5 meters. This problem highlights the importance of understanding fundamental physics concepts in practical engineering applications. The ability to calculate volumes, work with densities, and perform unit conversions is essential for engineers and artisans alike. Moreover, we've touched upon the importance of considering real-world factors like material waste and structural integrity in practical scenarios. This comprehensive approach ensures not only the accuracy of the calculations but also the feasibility and durability of the final product. This exercise underscores how crucial a solid grasp of basic scientific principles is for tackling everyday engineering challenges. The interplay between theoretical calculations and practical considerations is what ultimately leads to successful engineering outcomes. Therefore, understanding density, volume, and their relationship is paramount in engineering design and problem-solving.

Keywords: Calculate the length of the pipe in metres given the mass and density of the metal, and the dimensions of the pipe?

Calculate Pipe Length from Metal Mass Density and Dimensions