Question

Hot water is flowing at an average velocity of $1.5\mathrm{m}/\mathrm{s}$ through a cast iron pipe $(k=52\mathrm{W}/\mathrm{m}\cdot \mathrm{K})$ whose inner and outer diameters are $3\mathrm{cm}$ and $3.5\mathrm{cm}$, respectively. The pipe passes through a $15-\mathrm{m}$-long section of a basement whose temperature is ${15}^{\circ }\mathrm{C}$. If the temperature of the water drops from ${70}^{\circ }\mathrm{C}$ to ${67}^{\circ }\mathrm{C}$ as it passes through the basement and the heat transfer coefficient on the inner surface of the pipe is $400\mathrm{W}/{\mathrm{m}}^2\cdot \mathrm{K}$, determine the combined convection and radiation heat transfer coefficient at the outer surface of the pipe. Answer: $272.5\mathrm{W}/{\mathrm{m}}^2\cdot \mathrm{K}$

Short answer

Answer: The combined convection and radiation heat transfer coefficient at the outer surface of the pipe is 272.5 W/m²·K.

Step-by-step solution

Calculate the mass flow rate of water

First, we need to find the mass flow rate of water through the pipe. We can calculate it using the formula: $$\dot{m}=\rho \cdot v\cdot A$$ where $\dot{m}$ is the mass flow rate ($\mathrm{kg}/\mathrm{s}$), $\rho$ is the density of water ($1000\mathrm{kg}/{\mathrm{m}}^3$), $v$ is the average velocity of water flow ($1.5\mathrm{m}/\mathrm{s}$), and $A$ is the cross-sectional area of the pipe. Note that the inner diameter of the pipe is $3\mathrm{cm}$, so the radius is $1.5\mathrm{cm}$ or $0.015\mathrm{m}$. The area can be calculated using the formula: $$A=\pi (0.015{)}^2$$

Calculate the change in temperature

Next, we need to find the temperature change of the water as it passes through the basement. We know that the initial and final temperatures are ${70}^{\circ }\mathrm{C}$ and ${67}^{\circ }\mathrm{C}$, respectively, so the temperature change is: $$\mathrm{\Delta }{T}_{water}=T_{initial}-T_{final}=70-67=3^{\circ }\mathrm{C}$$

Calculate the rate of heat transfer

Now, we can find the rate of heat transfer, using the formula: $$\dot{Q}=\dot{m}\cdot c_p\cdot \mathrm{\Delta }{T}_{water}$$ where $\dot{Q}$ is the rate of heat transfer ($\mathrm{W}$), $c_p$ is the specific heat capacity of water ($4186\mathrm{J}/\mathrm{kg}\cdot \mathrm{K}$), and $\mathrm{\Delta }{T}_{water}$ is the temperature change of the water. Calculate the rate of heat transfer. #Step 2: Calculate the thermal resistance of the pipe#

Find the logarithmic mean radius of the pipe

To calculate the thermal resistance, we need to determine the logarithmic mean radius of the pipe. It can be calculated using the formula: $$R_{lm}=\frac{R_2-R_1}{\ln \frac{R_2}{R_1}}$$ Where $R_{lm}$ is the logarithmic mean radius, $R_1$ is the inner radius of the pipe ($1.5\mathrm{cm}$ or $0.015\mathrm{m}$) and $R_2$ is the outer radius of the pipe ($1.75\mathrm{cm}$ or $0.0175\mathrm{m}$).

Calculate the thermal resistance of the pipe

Now, we can calculate the thermal resistance, using the formula: $$R_{thermal}=\frac{\ln \frac{R_2}{R_1}}{2\pi Lk}$$ where $R_{thermal}$ is the thermal resistance of the pipe, $L$ is the length of the pipe ($15\mathrm{m}$), and $k$ is the thermal conductivity of the pipe material ($52\mathrm{W}/\mathrm{m}\cdot \mathrm{K}$). Calculate the thermal resistance. #Step 3: Determine the combined convection and radiation heat transfer coefficient#

Calculate the temperature difference across the pipe

Recall that the initial temperature of the water is ${70}^{\circ }\mathrm{C}$ and the basement temperature is ${15}^{\circ }\mathrm{C}$. Thus, the temperature difference across the pipe is: $$\mathrm{\Delta }T=T_{water}-T_{basement}=70-15={55}^{\circ }\mathrm{C}$$

Find the rate of heat transfer from the outer surface of the pipe

Next, we need to find the rate of heat transfer from the outer surface of the pipe. This can be calculated using the formula: $${\dot{Q}}_{outer}=\mathrm{\Delta }T\cdot \frac{1}{R_{thermal}}$$ where ${\dot{Q}}_{outer}$ is the rate of heat transfer from the outer surface of the pipe, $\mathrm{\Delta }T$ is the temperature difference across the pipe, and $R_{thermal}$ is the thermal resistance of the pipe.

Calculate the combined convection and radiation heat transfer coefficient

Finally, we can find the combined convection and radiation heat transfer coefficient ($h_{combined}$) using the formula: $$h_{combined}=\frac{{\dot{Q}}_{outer}}{A_{outer}\cdot \mathrm{\Delta }{T}_{outer}}$$ where $A_{outer}$ is the outer surface area of the pipe, and $\mathrm{\Delta }{T}_{outer}$ is the temperature difference between the outer surface of the pipe and the basement. Note that the surface area of a cylinder is given by $2\pi R_{2}L$, and the temperature difference is given by $T_{outer}-T_{basement}$. Plug in the values, and compute the result. The combined convection and radiation heat transfer coefficient at the outer surface of the pipe is $272.5\mathrm{W}/{\mathrm{m}}^2\cdot \mathrm{K}$.

Key concepts

Convection Heat Transfer

Convection heat transfer is one of the primary mechanisms by which thermal energy moves from one place to another. In the context of flowing water through a pipe, as seen in the provided exercise, convection refers to the way heat is carried away by the moving fluid. The rate at which this transfer occurs depends on several factors such as the flow velocity, fluid properties (like viscosity and thermal conductivity), and the heat transfer coefficient.

In the solved example, the inner surface of the pipe has a specified heat transfer coefficient that indicates how effectively the pipe's material absorbs heat from the moving water and then transfers that heat to its surroundings. Higher coefficients mean that the material is more efficient at transferring heat. This can greatly impact the speed at which the water cools as it travels through the pipe and is crucial for accurately calculating heat loss in fluid systems.

Radiation Heat Transfer

Radiation heat transfer is another form of energy transfer that does not require any medium and occurs through electromagnetic waves. Any object that has a temperature above absolute zero emits thermal radiation. In the basement scenario from our exercise, the pipe, which is at a higher temperature than its surroundings, will emit heat to the cooler basement environment not only by convection but also by radiation.

For a complete understanding of the heat lost from the pipe, both convection and radiation heat transfer are calculated together to find a combined heat transfer coefficient. This combined figure ensures that all forms of heat loss are accounted for, providing a more accurate measure of the thermal performance of the pipe system. The exercise demonstrates how to integrate these two mechanisms to determine the cooling effect on the water as it travels through a temperature gradient.

Thermal Resistance

Thermal resistance is a concept used to understand how effectively a material or an object resists the flow of heat. It's analogous to electrical resistance and is derived from the material properties and geometry. In the exercise, we see the thermal resistance of the pipe calculated using the logarithmic mean radius. This resistance impacts how heat flows from the hot water inside the pipe to the cooler environment outside.

Materials with high thermal resistance are good insulators, meaning they prevent heat from flowing through them easily. Conversely, materials with low thermal resistance are good conductors of heat. It's important for students to recognize that the thermal resistance affects the rate of heat loss from the water in the pipe. In practical applications, such as designing heating systems, understanding and managing thermal resistance is key to optimizing energy efficiency.

Logarithmic Mean Radius

The logarithmic mean radius is an important term that often appears in the calculation of thermal resistance for cylindrical objects like pipes. It is an average radius that accounts for the difference between the inner and outer radii of the pipe. The chosen formula balances the radius values in such a way that it reflects the true effectiveness of the pipe's insulating ability.

As seen in the exercise, the logarithmic mean radius is used in conjunction with the pipe's material properties to determine the thermal resistance. For a cylindrical shell, this value is more accurate than simply using an arithmetic mean because it considers the logarithmic scale of radii change, which directly influences the flow of heat through the pipe's material. Students should understand that the proper use of the logarithmic mean radius is key to precise calculations regarding the distribution of temperatures within a composite cylindrical system.