Question

Consider a chilled-water pipe of length \(L\), inner radius \(r_{1}\), outer radius \(r_{2}\), and thermal conductivity \(k\). Water flows in the pipe at a temperature \(T_{f}\), and the heat transfer coefficient at the inner surface is \(h\). If the pipe is well insulated on the outer surface, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the pipe, and (b) obtain a relation for the variation of temperature in the pipe by solving the differential equation.

Short answer

Question: Derive a relation for the temperature variation in a well-insulated water pipe. Answer: The temperature variation in the well-insulated water pipe in the radial direction can be given by the relation: \(T(r) = \frac{h}{k}(r_1 - r) + T_f\)

Step-by-step solution

Part (a): Derive the differential equation and boundary conditions for heat conduction in the pipe

To derive the differential equation for steady, one-dimensional heat conduction through the pipe, we will use the general heat conduction equation in cylindrical coordinates. For one-dimensional heat conduction in the radial direction, the equation will be: \(\frac{1}{r}\frac{d}{dr}(r\frac{dT}{dr})=0\) where T is the temperature and r is the radial distance. Now, we will derive the boundary conditions at the inner surface (r=r1) and outer surface (r=r2), where the pipe is well-insulated. At the inner surface, the heat transfer from the water to the pipe will be related to the heat transfer coefficient, h: \(q_{in} = h(T_f - T(r_1))\) At the outer surface (r=r2), the pipe is well-insulated, which means no heat is transferred to the surroundings, thus: \(\frac{dT}{dr}\Big|_{r=r_2} = 0\) Now we have our differential equation and boundary conditions.

Part (b): Obtain a relation for the variation of temperature in the pipe by solving the differential equation.

To obtain a relation for the temperature variation in the pipe, we will solve the following differential equation: \(\frac{1}{r}\frac{d}{dr}(r\frac{dT}{dr})=0\) Integrating with respect to r, we get: \(\int{\frac{1}{r}}d(r\frac{dT}{dr})=C_1\) \(r\frac{dT}{dr}=C_1\) Now, we will apply the boundary condition at the outer surface (r=r2), where \(\frac{dT}{dr}\Big|_{r=r_2} = 0\): \(C_1 = 0\) So our current equation is: \(r\frac{dT}{dr}=0\) The only way for this equation to hold true is if \(\frac{dT}{dr}\) is a constant value, let's call it C: \(\frac{dT}{dr}=C\) Integrating this equation with respect to r, we get: \(T(r)=Cr+D\) Now, we will apply the boundary condition at the inner surface (r=r1), where \(q_{in} = h(T_f - T(r_1))\): \(h(T_f - T(r_1)) = -k\frac{dT}{dr}\Big|_{r=r_1}=-kC\) After rearranging and substituting temperature expression at r=r1: \(h(T_f - (Cr_1+D)) = -kr_1C\) \(h(T_f - T_f + (Cr_1 + D)) = -kr_1C\) \(Cr_1 + D = \frac{h}{k}r_1\) Finally, we have obtained a relation for the temperature variation in the pipe: \(T(r) = \frac{h}{k}(r_1 - r) + T_f\) This equation represents the temperature distribution throughout the pipe in the radial direction while considering the well-insulated outer surface and the heat transfer at the inner surface.