Question

Liquid flows in a metal pipe with an inner diame\(D_{2}=32\) ter of $D_{1}=22 \mathrm{~mm}\( and an outer diameter of \)D_{2}=32 \mathrm{~mm}$. The thermal conductivity of the pipe wall is $12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. The inner surface of the pipe is coated with a thin polyvinylidene chloride (PVDC) lining. Along a length of \(1 \mathrm{~m}\), the pipe outer surface is exposed to convection heat transfer with hot gas, at \(T_{\infty}=100^{\circ} \mathrm{C}\) and $h=5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\(, and thermal radiation with a surrounding at \)T_{\text {surr }}=100^{\circ} \mathrm{C}\(. The emissivity at the pipe outer surface is \)0.3$. The liquid flowing inside the pipe has a convection heat transfer coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the outer surface of the pipe is at \(85^{\circ} \mathrm{C}\), determine the temperature at the PVDC lining and the temperature of the liquid. The ASME Code for Process Piping (ASME B31.3-2014, A.323) recommends a maximum temperature for PVDC lining to be \(79^{\circ} \mathrm{C}\). Does the PVDC lining comply with the recommendation of the code?

Short answer

Question: Determine the temperature of PVDC lining and the temperature of the liquid inside the pipe, and state if the lining temperature complies with ASME B31.3-2014 code recommendations. Answer: The temperature of PVDC lining is approximately \(79.23 ^\circ \mathrm{C}\) and the temperature of the liquid is approximately \(64.23 ^\circ \mathrm{C}\). The PVDC lining temperature slightly exceeds the ASME B31.3-2014 code recommendations of a maximum temperature of \(79 ^\circ \mathrm{C}\).

Step-by-step solution

Thermal Resistance of the pipe

Using the formula for the thermal resistance in a cylindrical wall: \(R_{th,pipe} = \frac{\ln(\frac{D_2}{D_1})}{2\pi kL} = \frac{\ln(\frac{0.032}{0.022})}{2 \pi \times 12 \times 1} = 0.0204~\mathrm{K/W}\)

Convection resistance at outer surface

For convection heat transfer on the outer surface of the pipe: \(R_{th,outer} = \frac{1}{h_{outer} A_{outer}} = \frac{1}{5 \times \pi (0.032) \times 1} = 0.0318~\mathrm{K/W}\)

Calculate the radiation heat transfer coefficient

We need to find the radiation heat transfer coefficient to find the combined resistance for radiation and convection at outer surface. Using the Stefan-Boltzmann law: \(h_r = \epsilon \sigma (T_{surface} + T_{surr})(T_{surface}^2 + T_{surr}^2)\), where \(\sigma = 5.67 \times 10^{-8}~\mathrm{W/m^2K^4}\) \(h_r =0.3 \times 5.67 \times 10^{-8}(85+100)((85)^2+(100)^2)= 3.802 \mathrm{~W/m^2K}\)

Combined Resistance at outer surface

Now we'll find the combined resistance for radiation and convection heat transfer at the outer surface: \(\frac{1}{R_{th,combined}} = \frac{1}{R_{th,outer}}+\frac{1}{h_r A_{outer}}\) \(R_{th,combined} = \frac{1}{(\frac{1}{0.0318}+\frac{1}{3.802\times \pi(0.032)\times 1})}= 0.0234~\mathrm{K/W}\)

Convection resistance at inner surface

For convection heat transfer on the inner surface of the pipe: \(R_{th,inner} = \frac{1}{h_{inner} A_{inner}}=\frac{1}{50\times \pi(0.022)\times 1} = 0.01445~\mathrm{K/W}\)

Temperature at PVDC Lining

With all of the thermal resistances values available, we can proceed to find the temperature at PVDC lining (\(T_{L}\)) using the temperature difference between outer surface and lining: \((T_{outer} - T_{L}) = (R_{th,combined} + R_{th,pipe}) \times q\) \(q = \frac{T_{outer} - T_{L}}{R_{th,combined} + R_{th,pipe}}\) Since liquid temperature \(T_{liq}\) is unknown, we cannot determine the heat transfer rate \(q\). So let's first find \(q\) in terms of \(T_{liq}\): \(q = (T_{liq} - T_{L}) / R_{th,inner}\)

Solve for \(T_{L}\)

Now, match the equations to find the value of \(T_L\): \(\frac{T_{outer} - T_{L}}{R_{th,combined} + R_{th,pipe}} = \frac{T_{liq} - T_{L}}{R_{th,inner}}\) \(T_L = \frac{(T_{outer} - T_{L})(R_{th,inner})+(T_{liq}- T_{L})(R_{th,combined} + R_{th,pipe}) }{R_{th,inner} + R_{th,combined} + R_{th,pipe}}\) \(T_L = \frac{(85 - T_{L})(0.01445)+(T_{liq} - T_L)(0.0234 + 0.0204)}{0.01445 + 0.0234 + 0.0204}\) Let this equation be Eq. (1) to find \(T_{liq}\).

Temperature of the liquid

We know temperature is continuous, so the temperature difference across PVDC lining and metal pipe wall is same: \((T_{outer} - T_{L}) = (T_{L} - T_{liq})\) After substituting the values of \(T_{outer}\), we get \(T_{liq} = T_{L} - 15^\circ \mathrm{C}\) Now, substitute equation \(T_{liq}\) in terms of \(T_L\) into Eq. (1), we get: \(-0.01445T_L^2 + 2.1345T_L - 102.495 = 0\) Solving this quadratic equation, we get \(T_L \approx 79.23 ^\circ \mathrm{C}\)

Temperature of the liquid Calculation

Now, we can determine the temperature of the liquid (\(T_{liq}\)) using the relation we found: \(T_{liq} = T_{L} - 15^\circ \mathrm{C} = 79.23 - 15 = 64.23^\circ \mathrm{C}\) Finally, the temperature of PVDC lining is found to be \(79.23^\circ \mathrm{C}\), and the temperature of the liquid is \(64.23^\circ \mathrm{C}\). In conclusion, PVDC lining temperature \(79.23^\circ \mathrm{C}\) slightly exceeds the ASME B31.3-2014 code recommendations of a maximum temperature of \(79 ^\circ \mathrm{C}\).