Question

Saturated refrigerant-134a vapor undergoes conASTM A268 TP443 stainless steel tube at \(133 \mathrm{kPa}\). The tube is \(50 \mathrm{~cm}\) long and has a diameter of \(15 \mathrm{~mm}\). According to the ASME Code for Process Piping, the minimum temperature suitable for ASTM A268 TP443 stainless steel tube is \(-30^{\circ} \mathrm{C}\) (ASME B31.3-2014, Table A-1M). Determine whether the condensation of refrigerant- \(134 \mathrm{a}\) on the tube surface at a rate of \(3 \mathrm{~g} / \mathrm{s}\) would cool the surface below the minimum suitable temperature set by the ASME Code for Process Piping.

Short answer

Answer: Yes, the condensation at a rate of 3 g/s on the stainless steel tube would cool the surface below the minimum suitable temperature set by the ASME Code for Process Piping, since the temperature of the refrigerant inside the tube will naturally be lower than the minimum temperature suitable for the ASTM A268 TP443 stainless steel tube.

Step-by-step solution

Find heat transfer rate during condensation

First, we need to find the heat transfer rate during condensation. To do this, we'll need the rate of condensation (\(3 \mathrm{~g/s}\)) and the heat of vaporization of refrigerant-134a. The heat of vaporization of refrigerant-134a can be obtained from a thermodynamic table or similar resources. It is approximately \(215 \mathrm{kJ/kg}\). Thus, the heat transfer rate can be calculated as follows: \(Q = \dot{m} \times L_v\) Where \(Q\) is the heat transfer rate, \(\dot{m}\) is the mass flow rate of the condensation, and \(L_v\) is the heat of vaporization. First, convert the mass flow rate to kg/s: \(\dot{m} = 3 \times 10^{-3} \mathrm{kg/s}\) Now, we can calculate the heat transfer rate: \(Q = (3 \times 10^{-3}) \times 215000\) \(Q = 645 \mathrm{W}\)

Calculate the convective heat transfer coefficient

The convective heat transfer coefficient (h) can be calculated using Nusselt's condensation theory as follows: \(h = \frac{Q_{total}}{(T_{sat} - T_{refrigerant})A}\) Here, the surface area of the tube (A) can be calculated using its diameter and length: \(A = \pi \times d \times L\) Where, d is the diameter of the tube, and L is its length. Plugging in the given values: \(A = \pi \times 0.015 \times 0.5\) \(A \approx 0.0236 \mathrm{m^2}\) From the problem statement, we have: \(T_{sat} = -30^{\circ} \mathrm{C}\) and \(Q_{total} = 645 \mathrm{W}\) Now rewrite the formula and solve for \(T_{refrigerant}\): \((T_{sat} - T_{refrigerant}) = \frac{Q_{total}}{h \times A}\) Plugging the values into the formula: \((-30 - T_{refrigerant}) = \frac{645}{h \times 0.0236}\) Add T_refrigerant to both sides and multiply the right side: \(-30 = \frac{15.6h}{1000}T_{refrigerant}\)

Determine whether the condensation cools the tube surface below the minimum suitable temperature

In order to assess whether the condensation cools the tube surface below the minimum suitable temperature, we need to figure out the values of temperature that satisfy our equation: If \((-30 - T_{refrigerant}) < \frac{15.6h}{1000}T_{refrigerant}\), then condensation would cool the tube surface below the minimum suitable temperature. This inequality is true if and only if \(T_{refrigerant} > -30^{\circ} \mathrm{C}\). As the temperature of the refrigerant inside the tube will naturally be lower than the minimum temperature suitable for the ASTM A268 TP443 stainless steel tube, the condensation of the refrigerant on the tube surface at a rate of \(3 \mathrm{~g/s}\) would cool the surface below the minimum suitable temperature set by the ASME Code for Process Piping.