Steel Section Of Alaskan Pipeline Was 65 M Long

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 meters and a temperature of 21°C when it was installed. What is its change in length when the temperature drops to a frigid -46°C? Dermatologists often remove small precancerous skin lesions by freezing them quickly with liquid nitrogen, which has a temperature of 77 K. What is this temperature on the (a) Celsius scale? (b) Fahrenheit scale? When the temperature of a coin is raised by 76°C, the coin's diameter increases by 1 millimeter. If the original diameter is 1.8 × 10-2 meters, find the coefficient of linear expansion (°C)-1. A wire is stretched between two posts. Another wire is stretched between two posts that are twice as far apart. The tension in the wires is the same, and they have the same mass. A transverse wave travels on the shorter wire with a speed of 300 m/s. What would be the speed of the wave on the longer wire? meters per second. A hunter is standing on flat ground between two vertical cliffs that are directly opposite one another. He is closer to one cliff than to the other. He fires a gun and, after a while, hears three echoes. The second echo arrives 2.5 seconds after the first, and the third echo arrives 0.5 seconds after the second. Assuming that the speed of sound is 343 m/s and that there are no reflections of sound from the ground, find the distance between the cliffs in meters. A refrigerator has a surface area of 4.9 m2. It is lined with 0.082 meters thick insulation whose thermal conductivity is 0.030 J/(s·m·°C). The interior temperature is kept at 5°C, while the temperature at the outside surface is 25°C. How much heat per second is being removed from the unit?

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The problem involves multiple physical concepts, including thermal expansion, temperature conversions, wave speed on varying mediums, echoes and distance calculations, and heat transfer. Each part requires understanding of the respective physics principles and applying formulas accurately.

Firstly, the change in length of the steel pipeline segment due to temperature variation can be calculated using the coefficient of linear expansion formula: ΔL = L₀·α·ΔT, where ΔL is the change in length, L₀ is the initial length, α is the linear expansion coefficient, and ΔT is the temperature change.

Given: L₀ = 65 m, T_initial = 21°C, T_final = -46°C, so ΔT = -46°C - 21°C = -67°C. To find ΔL, we need α, which is unknown; subsequent calculations would involve typical steel expansion coefficients (~11 × 10-6 /°C).

Next, for temperature conversions: Liquid nitrogen has a temperature of 77 K. To convert this to Celsius, use T(°C) = T(K) - 273.15, resulting in 77 - 273.15 ≈ -196.15°C. To convert to Fahrenheit, use T(°F) = (T(°C) × 1.8) + 32, which gives approximately -321.07°F.

The coefficient of linear expansion, α, of the coin is derived from the change in diameter (Δd) related to the temperature change (ΔT): Δd = d₀·α·ΔT. Using the given data: Δd = 1 mm = 1 × 10-3 m, d₀ = 1.8 × 10-2 m, ΔT = 76°C, then α = Δd / (d₀·ΔT) ≈ (1 × 10-3) / (1.8 × 10-2 × 76), which yields approximately 7.3 × 10-4 /°C.

For wave speed on the longer wire, since tension T and mass per unit length μ are the same, wave speed v is proportional to the inverse square root of μ. The relation v ∝ 1/√μ and μ ∝ mass / length. If the length doubles, the mass doubles assuming density remains constant, so μ doubles, and v₁ / v₂ = √(μ₂ / μ₁) = √(2), thus v₂ = v₁ / √2 ≈ 300 / 1.414 ≈ 212.13 m/s.

The distance between cliffs based on echoes: The second echo arrives 2.5 s after the first, and the third 0.5 s after the second, totaling 3 seconds after the first. The difference in time corresponds to the extra distance sound travels. For the second echo, total time for round trip from closer cliff: t₂ = 2.5 s, distance to closer cliff d₁ = (v_sound × t) / 2 = (343 m/s × 2.5 s) / 2 ≈ 429.75 m. For the third echo, total time from the farther cliff: t₃ = (2 × total time) - t₂ = 3 s total, implying the distance to the farther cliff d₂ = (343 m/s × 3 s) / 2 ≈ 514.5 m. The distance between the cliffs is d₂ - d₁ ≈ 84.75 meters.

For the heat transfer through insulation, the heat transfer rate Q is given by Fourier's law: Q = (k × A × ΔT) / d, where k is thermal conductivity, A is surface area, ΔT is temperature difference, d is thickness. Substituting values: k = 0.030 J/(s·m·°C), A=4.9 m2, ΔT=20°C (25°C - 5°C), d=0.082 m, Q = (0.030 × 4.9 × 20) / 0.082 ≈ 35.85 J/s.

This comprehensive breakdown demonstrates the physics principles and calculations necessary to address each aspect of the multi-part problem.

References

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