Question

A long 8-cm-diameter steam pipe whose external surface temperature is ${90}^{\circ }\mathrm{C}$ passes through some open area that is not protected against the winds. Determine the rate of heat loss from the pipe per unit of its length when the air is at $1\mathrm{atm}$ pressure and $7^{\circ }\mathrm{C}$ and the wind is blowing across the pipe at a velocity of $50\mathrm{km}/\mathrm{h}$.

Short answer

Answer: The rate of heat loss from the steam pipe per unit length is approximately 2060 W/m.

Step-by-step solution

Convert given values to SI units

First, let's convert the given values to SI units: Pipe diameter: 8 cm = 0.08 m Wind velocity: 50 km/h = 13.89 m/s Air temperature: $7^{\circ }\mathrm{C}$ = 280.15 K Steam pipe temperature: ${90}^{\circ }\mathrm{C}$ = 363.15 K

Calculate the temperature difference

Next, let's calculate the temperature difference between the steam pipe surface and the air: $\mathrm{\Delta }T=T_{surface}-T_{air}=363.15-280.15=83K$

Calculate the Reynolds number

We need to calculate the Reynolds number to determine the flow regime and find the heat transfer coefficient. We can use the following formula for the Reynolds number: $Re=\frac{VD}{\nu }$ Here, we need to find the kinematic viscosity ($\nu$) of air at $7^{\circ }\mathrm{C}$. From air properties tables, we find that $\nu =1.49\times {10}^{-5}{m}^2/s$ at this temperature. Now, we can calculate the Reynolds number: $Re=\frac{(13.89m/s)(0.08m)}{1.49\times {10}^{-5}{m}^2/s}=74810$

Determine the Nusselt number

Since the Reynolds number is greater than 4000, the flow regime around the pipe is turbulent. We can use the Sieder-Tate correlation to find the Nusselt number for turbulent flows: $Nu=0.027R{e}^{4/5}P{r}^{1/3}$ We also need the Prandtl number (Pr), which represents the ratio of momentum diffusivity to thermal diffusivity. For air at $7^{\circ }\mathrm{C}$, we find that $Pr=0.707$ from the air properties tables. Now let's calculate the Nusselt number: $Nu=0.027(74810{)}^{4/5}(0.707{)}^{1/3}=328.8$

Calculate the convective heat transfer coefficient

Now, we can find the convective heat transfer coefficient (h) using the following formula: $h=\frac{Nuk}{D}$ Here, k is the thermal conductivity of air, which is equal to $0.024W/(mK)$ for air at $7^{\circ }\mathrm{C}$. Substituting the values, we get: $h=\frac{(328.8)(0.024W/(mK))}{0.08m}=98.64W/(m^{2}K)$

Calculate the rate of heat loss per unit length

Finally, we can calculate the rate of heat loss per unit length (Q) using the convective heat transfer formula: $Q=hA\mathrm{\Delta }T$ Here, A is the area of the pipe per unit length, which is equal to the pipe circumference: $A=\pi D=\pi (0.08m)$ Substituting the obtained values, we get: $Q=(98.64W/(m^{2}K))(\pi (0.08m))(83K)=2059.9W/m$ So, the rate of heat loss from the steam pipe per unit of its length is approximately $2060W/m$.

Key concepts

Convective Heat Transfer

In the study of thermodynamics, convective heat transfer is a fundamental concept that deals with the movement of heat between a solid surface and a fluid in motion, such as air or water. It plays a crucial role in various engineering scenarios, from HVAC systems in buildings to the cooling of electronic devices.

When a fluid, such as wind, moves across a surface at a different temperature—like our steam pipe in question—the fluid can absorb or lose heat. This process is known as convection, and it includes both 'free' or 'natural' convection, where the fluid motion is caused by density differences due to temperature gradients, and 'forced' convection, where the fluid movement is induced by external means, such as a fan or a pump. In our exercise, the wind blowing across the pipe induces forced convection.

The rate at which heat is transferred by convection can be expressed by the equation:
$Q=hA\text{texttviso}T$,
where $Q$ is the heat transferred per unit time, $h$ is the convective heat transfer coefficient (a value that characterizes how well the convection process transfers heat), $A$ is the area through which the heat is being transferred, and $\text{texttviso}T$ is the temperature difference between the surface and the fluid.

Determining the correct heat transfer coefficient is pivotal as it impacts the precision of our heat loss calculation. This coefficient is influenced by a variety of factors including the properties of the fluid, the velocity of the fluid movement, and the characteristics of the surface.

Reynolds Number

Understanding the behavior of fluid flow around objects is critical in calculating convective heat transfer, and that's where the Reynolds number comes into play. It's a dimensionless quantity used in fluid mechanics to predict flow patterns in different fluid flow situations. Named after Osborne Reynolds, who proposed it in 1883, it provides insight into whether the flow will be laminar (smooth and regular) or turbulent (chaotic and irregular).

The Reynolds number is given by the formula:
$$Re=\frac{VD}{u}$$,
where $V$ is the velocity of the fluid, $D$ is the characteristic length (like the diameter of a pipe), and $u$ is the kinematic viscosity of the fluid. By plugging in the appropriate measurements, engineers and scientists can predict the nature of the flow without complex calculations or experimentation.

In our exercise, the Reynolds number determined the flow regime around the steam pipe to be turbulent since it exceeded 4000. Turbulent flow usually results in a higher convective heat transfer coefficient compared to laminar flow, meaning more efficient heat exchange—critical for accurate heat loss calculations in our problem.

Nusselt Number

The Nusselt number is another crucial dimensionless parameter in the study of heat transfer, particularly convective heat exchange. It establishes a relationship between the convective and conductive heat transfer occurring in a fluid. Essentially, a higher Nusselt number indicates more effective convection relative to conduction.

The Nusselt number can be calculated using various correlations, depending on the nature of the flow and other conditions. One such correlation for turbulent flow, which was used in our exercise, is the Sieder-Tate correlation expressed as:
$$Nu=0.027R{e}^{4/5}P{r}^{1/3}$$.
It combines both the Reynolds number (Re) and the Prandtl number (Pr), the latter representing the ratio of momentum diffusivity (viscosity) to thermal diffusivity. Momentum diffusivity refers to how easily momentum exchanges within the fluid, while thermal diffusivity indicates how quickly heat can spread through the material.

By determining the Nusselt number in our exercise, we were able to calculate the convective heat transfer coefficient necessary for obtaining the rate of heat loss from the pipe. Since various correlations for Nusselt number exist, selecting the appropriate one is crucial for precise calculations, especially in complex real-world applications.