Question

Cryogenic fluid flows inside a 10-mm-diameter metal tube. The metal tube is enclosed by a concentric polypropylene tube with a diameter of \(15 \mathrm{~mm}\). The minimum temperature limit for polypropylene tube is \(-18^{\circ} \mathrm{C}\), specified by the ASME Code for Process Piping (ASME B31.3-2014, Table B-1). The gap between the concentric tubes is a vacuum. The inner metal tube and the outer polypropylene tube have emissivity values of \(0.5\) and \(0.97\), respectively. The concentric tubes are placed in a vacuum environment, where the temperature of the surroundings is \(0^{\circ} \mathrm{C}\). Determine the lowest temperature that the inner metal tube can go without cooling the polypropylene tube below its minimum temperature limit of $-18^{\circ} \mathrm{C}$. Assume both tubes have thin walls.

Short answer

Answer: The lowest temperature that the inner metal tube can reach without cooling the polypropylene tube below its minimum temperature limit is approximately -38°C.

Step-by-step solution

Identify the variables needed for the radiative heat transfer equation

To calculate the radiative heat transfer between two concentric tubes, we need to know the area of the inner tube, the emissivity values of both tubes, their temperatures, and the Stefan-Boltzmann constant. Based on the information given, we have the following variables: - Diameter of inner metal tube (D1) = \(10 \mathrm{~mm}\) - Diameter of outer polypropylene tube (D2) = \(15 \mathrm{~mm}\) - Emissivity of metal tube (ε1) = \(0.5\) - Emissivity of polypropylene tube (ε2) = \(0.97\) - Minimum temperature of polypropylene tube (T2) = \(-18^{\circ} \mathrm{C}\) - Surrounding temperature (Ts) = \(0^{\circ} \mathrm{C}\) - Stefan-Boltzmann constant (σ) = \(5.67 \times 10^{-8} \mathrm{W / m^2K^4}\) Now we will convert the temperature values to Kelvin (K).

Convert temperatures to Kelvin

To convert temperatures from Celsius to Kelvin, we add 273.15 Minimum temperature of polypropylene tube in Kelvin (T2): \(T2 = -18 + 273.15 = 255.15 \mathrm{K}\) Surrounding temperature in Kelvin (Ts): \(Ts = 0 + 273.15 = 273.15 \mathrm{K}\)

Find the radiative heat transfer coefficient

The effective radiative heat transfer coefficient (hr) can be calculated using the following formula: \(hr = \frac{1 - \epsilon_1}{A_1(1 - \frac{D_1}{D_2})} + \frac{1}{\epsilon_1A_1} \bigg( \frac{\epsilon_2}{\epsilon_1} - 1 \bigg) \) Where \(A_1\) is the area of the first surface (inner metal tube) and can be calculated as: \(A_1 = \pi D_1 L\) Since we are not given the length (L) of the tubes, we can use the ratio of area to length (A1/L) in our formula: \(\frac{A_1}{L} = \pi D_1\) Now substituting the emissivity values, and diameters, our formula becomes: \(hr = \frac{1 - 0.5}{\pi (10 \times 10^{-3})(1 - \frac{10 \times 10^{-3}}{15 \times 10^{-3}})}+ \frac{1}{0.5\pi(10 \times 10^{-3})}\bigg(\frac{0.97}{0.5} - 1 \bigg)\) Calculate hr: \(hr \approx 28.076 \mathrm{W/m^2K}\)

Determine the lowest temperature of the inner metal tube without cooling the polypropylene tube below its minimum temperature limit

Now, we can use the radiative heat transfer equation: \(q = hr \times A_1 (T1^4 - T2^4)\) We want to find T1 when the heat transfer (q) between the concentric tubes is equal to the heat transfer from the polypropylene tube to its surroundings (T2). So we can write: \(q = hr \times A_1 ((T1^4 - Ts^4) - (T2^4 - Ts^4))\) Rearrange the formula for T1: \(T1^4 = \frac{q + hr \times A_1 (T2^4 - Ts^4)}{hr \times A_1} + Ts^4\) For our case, since the minimum temperature limit of the polypropylene tube is reached, the radiative heat transfer between the tubes is zero (q = 0): \(T1^4 = \frac{0 + hr \times A_1 (T2^4 - Ts^4)}{hr \times A_1} + Ts^4\) Substitute the known values and calculate T1: \(T1^4 = \frac{28.076 \times \pi (10 \times 10^{-3}) (255.15^4 - 273.15^4)}{28.076 \times \pi (10 \times 10^{-3})} + 273.15^4\) Solve for T1: \(T1 \approx 235.17 \mathrm{K}\) Now, convert T1 back to Celsius: \(T1 = 235.17 - 273.15\) \(T1 \approx -38^{\circ}\mathrm{C}\) Therefore, the lowest temperature that the inner metal tube can go without cooling the polypropylene tube below its minimum temperature limit is approximately \(-38^{\circ}\mathrm{C}\)