Question

A 5-cm-external-diameter, 10-m-long hot-water pipe at \(80^{\circ} \mathrm{C}\) is losing heat to the surrounding air at \(5^{\circ} \mathrm{C}\) by natural convection with a heat transfer coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the rate of heat loss from the pipe by natural convection. Answer: \(2945 \mathrm{~W}\)

Short answer

Answer: The rate of heat loss from the pipe by natural convection is approximately 2945 W.

Step-by-step solution

Calculate the surface area of the pipe

The pipe is of cylindrical shape, so its surface area can be calculated using the formula \(A=2 \pi r L\), where \(r\) is the radius and \(L\) is the length of the pipe. Since the diameter is given, we can find the radius by dividing the diameter by two. So \(r = \frac{5}{2} \mathrm{~cm} = 0.025 \mathrm{~m}\). Now we can find the surface area of the pipe: \(A=2 \pi (0.025 \mathrm{~m})(10 \mathrm{~m}) \approx 1.571 \mathrm{~m}^2\).

Find the temperature difference

The temperature difference between the pipe and the surrounding air can be found by subtracting the surrounding air temperature (\(T_\mathrm{air}=5^{\circ} \mathrm{C}\)) from the pipe temperature (\(T_\mathrm{pipe}=80^{\circ} \mathrm{C}\)). So, the temperature difference is \(ΔT = T_\mathrm{pipe} - T_\mathrm{air} = 80^{\circ} \mathrm{C} - 5^{\circ} \mathrm{C} = 75 \mathrm{~K}\).

Calculate the rate of heat loss

Now we can use the formula for heat transfer by natural convection, which is \(q=h \cdot A \cdot ΔT\), where \(q\) represents the rate of heat loss, \(h\) is the heat transfer coefficient, \(A\) is the surface area of the pipe, and \(ΔT\) is the temperature difference. Plugging in the given and calculated values, we get: \(q = (25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}) \cdot (1.571 \mathrm{~m}^2) \cdot (75 \mathrm{~K}) \approx 2945 \mathrm{~W}\). The rate of heat loss from the pipe by natural convection is approximately \(2945 \mathrm{~W}\).

Key concepts

Natural Convection

Natural convection is a process where heat is transferred without any external aid, such as fans or pumps. This happens because of the movement caused by differences in temperature and density within the fluid surrounding the object. In the case of the hot-water pipe, the warm surface of the pipe heats the air nearby. As the air warms up, it becomes less dense and rises. Cooler, denser air then moves in to take its place. This cycle of rising warm air and falling cool air leads to natural convection currents.
The efficiency of heat transfer by natural convection depends on the surface area involved, the temperature difference between the surface and the surrounding fluid, and a property specific to the condition known as the heat transfer coefficient.

Heat Loss Calculation

Calculating heat loss involves determining how much heat escapes from an object into its surroundings. For objects like pipes, heat loss is primarily driven by the temperature difference between the pipe and its environment, as well as the medium through which the heat is lost. In natural convection, the formula used is:

• \[ q = h \times A \times \Delta T \]

where:

• \( q \) is the rate of heat loss in watts (W),
• \( h \) is the heat transfer coefficient in watts per square meter per kelvin (\( \mathrm{W} / \mathrm{m}^{2} \, \mathrm{K} \)),
• \( A \) is the surface area in square meters (\( \mathrm{m}^2 \)),
• \( \Delta T \) is the temperature difference in kelvin (K).

Understanding this formula helps in assessing the energy efficiency of piping systems and making necessary improvements to reduce energy costs.

Surface Area of Pipe

The surface area of a pipe significantly affects how much heat it can transfer to its surroundings. To find the surface area of a cylindrical pipe, the following formula is used:

• \[ A = 2 \pi r L \]

where \( r \) is the radius of the pipe and \( L \) is the length. Converting the diameter to radius is essential: given a diameter of 5 cm, the radius is half of that, or 2.5 cm (which is 0.025 meters for calculations).
The calculated surface area for a 10-meter-long pipe with this radius is approximately 1.571 square meters. A larger surface area allows for greater heat loss, making this a crucial value in heat transfer calculations.

Temperature Difference

A key driver in the rate of heat transfer by natural convection is the temperature difference between the object and its surroundings. The formula:

• \[ \Delta T = T_{\text{pipe}} - T_{\text{air}} \]

naturally arises from subtracting the temperature of the surrounding air from that of the pipe. In this scenario, the hot-water pipe is at \( 80^{\circ} \mathrm{C} \), and the external air is at \( 5^{\circ} \mathrm{C} \).
This difference of 75 Kelvin drives the convective heat transfer process. A greater temperature difference typically increases the rate of heat loss, as the energy moves more quickly from the warmer area to the cooler area to establish equilibrium.