Steel Section Of The Alaskan Pipeline Had A Length Of 65 M

A Steel Section Of The Alaskan Pipeline Had A Length Of 65 M And A Tem

A Steel Section Of The Alaskan Pipeline Had A Length Of 65 M And A Tem

These questions encompass various physics concepts such as thermal expansion, temperature conversions, wave mechanics, sound propagation, and heat transfer. Below, each problem is addressed with detailed explanations and calculations based on fundamental principles and typical values.

Paper For Above instruction

Problem 1: Thermal Expansion of the Pipeline

The problem states that a steel pipeline section is initially 65 meters long at 21°C, and asks for the change in length when the temperature drops to -46°C. The coefficient of linear expansion for steel is approximately 11 x 10-6 /°C.

The change in length, ΔL, can be calculated using the formula:

ΔL = α × L₀ × ΔT

Where:

  • α = coefficient of linear expansion = 11 x 10-6 /°C
  • L₀ = original length = 65 m
  • ΔT = temperature change = T_final - T_initial = -46°C - 21°C = -67°C

Calculating:

ΔL = 11 x 10-6 /°C × 65 m × (-67°C) = -0.0478 m

Therefore, the pipeline will contract by approximately 0.048 meters when cooled from 21°C to -46°C.

Problem 2: Liquid Nitrogen Temperature Conversion

The temperature of liquid nitrogen is given as 77 K.

(a) To convert kelvin to Celsius:

°C = K - 273.15

°C = 77 - 273.15 = -196.15°C

(b) To convert Celsius to Fahrenheit:

°F = (°C × 9/5) + 32

°F = (-196.15 × 9/5) + 32 ≈ -321.07°F

Thus, liquid nitrogen's temperature is approximately -196.15°C or -321.07°F.

Problem 3: Coefficient of Linear Expansion of a Coin

A coin's diameter increases by 1.0 x 10-3 m for a temperature increase of 76°C. The initial diameter is 1.8 x 10-2 m.

The linear expansion formula:

ΔD = α × D₀ × ΔT

Rearranged to find α:

α = ΔD / (D₀ × ΔT)

Calculating:

α = 1.0 x 10-3 m / (1.8 x 10-2 m × 76°C) ≈ 7.3 x 10-4 /°C

This coefficient is unusually large for metals, indicating a high sensitivity or an approximation for this specific material. Typically, metals have α values around 10-5 to 10-6 /°C. Nonetheless, based on the data provided, α ≈ 7.3 x 10-4 /°C.

Problem 4: Wave Speed on a Longer Wire

Two wires are stretched with the same tension, but the second wire is twice as long as the first. The wave speed v on a string depends on the tension T and mass per unit length μ:

v = √(T / μ)

Since tension is the same and both wires have the same mass, the wave speed depends on μ, which is proportional to the length and density. For a wire with uniform density and cross-sectional area, the mass per unit length remains constant.

The speed of wave propagation in the shorter wire is 300 m/s. The relation between length and wave speed simplifies to considering linear density. As length doubles, the mass per unit length remains constant (assuming same material). Therefore, the wave speed stays the same if tensions are equal and density is uniform.

However, if the tension or mass distribution changes, the wave speed can be affected. Given the problem's specifics, since tension and mass are equal, the wave speed on the longer wire remains approximately 300 m/s.

Problem 5: Echoes and Distance Between Cliffs

The hunter hears three echoes: the second echo arrives 2.5 seconds after the first, and the third 0.5 seconds after the second. Both echoes are reflections from the cliffs, with sound speed c = 343 m/s.

The total time for the wave to travel to the cliff and back for the first echo:

t₁ = total time to reach the cliff and return.

Similarly, the total travel times for subsequent echoes can be summed to find the distances.

Let D be the distance to the closer cliff, and D' to the farther (or the same, depending on setup). Because:

Time for the first echo:

t₁ = 2D / c

Time for the second echo:

t₂ = 2D' / c

The differences in arrival times indicate the extra distances traveled. Since the second echo arrives 2.5 seconds after the first, the path difference gives:

Δt = (2D' - 2D) / c = 2.5 s

Similarly, the third echo's extra delay of 0.5 s indicates the difference between the echoes, but for simplicity, using the first two echoes, the total distance D + D' is derived from the time differences:

2D / c = time for first echo

2D' / c = time for second echo

Using the difference:

2D' / c - 2D / c = 2.5 s → 2(D' - D) / c = 2.5 s → D' - D = (2.5 s × c) / 2

D' - D = (2.5 × 343) / 2 ≈ 429.   meters

The total distance between the cliffs is D + D', which combines with the additional delay to yield the total separation. Assuming the closer cliff is D and the farther D + 429 meters, and solving based on the echoes, the approximate distance between the cliffs is around 858 meters.

Problem 6: Heat Transfer Through Refrigerator Insulation

The refrigerator's surface area A = 4.9 m2. The insulation thickness d = 0.082 m, thermal conductivity k = 0.030 J/(s·m·°C). The temperature difference ΔT = 25°C - 5°C = 20°C.

The heat transfer rate Q̇ is given by Fourier’s law:

Q̇ = (k × A × ΔT) / d

Calculating:

Q̇ = (0.030 × 4.9 × 20) / 0.082 ≈ (0.030 × 98) / 0.082 ≈ 2.94 / 0.082 ≈ 35.85 J/s

Thus, approximately 36 Joules of heat are being removed per second to maintain the interior temperature.

Conclusion

These physics problems illustrate the application of fundamental principles in real-world contexts, including thermal expansion, temperature conversions, wave mechanics, echo analysis, and heat transfer. Accurate calculations rely on understanding the underlying formulas and assumptions, and are essential for engineering and scientific problem-solving.

References

  • Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers with Modern Physics. Cengage Learning.
  • Halliday, D., Resnick, R., & Walker, J. (2014). Fundamentals of Physics. Wiley.
  • Tipler, P. A., & Mosca, G. (2008). Physics for Scientists and Engineers. W. H. Freeman.
  • Young, H. D., & Freedman, R. A. (2019). University Physics. Pearson.
  • NASA Glenn Research Center. (n.d.). Thermal Expansion of Materials. https://www.grc.nasa.gov
  • Hibbeler, R. C. (2017). Engineering Mechanics: Dynamics. Pearson.
  • Incropera, F. P., DeWitt, D. P., Bergman, T. L., & Lavine, A. S. (2007). Fundamentals of Heat and Mass Transfer. Wiley.
  • Feynman, R. P., Leighton, R. B., & Sands, M. (2011). The Feynman Lectures on Physics. Addison-Wesley.
  • Brillouin, L. (1970). Wave Propagation in Periodic Structures. Dover Publications.
  • Mechanical Properties of Steel. (2020). ASM International. https://www.asminternational.org/

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