Question

It is proposed to have a water heater that consists of an insulated pipe of 7.5 -cm diameter and an electric resistor inside. Cold water at \(20^{\circ} \mathrm{C}\) enters the heating section steadily at a rate of \(24 \mathrm{L} / \mathrm{min}\). If water is to be heated to \(48^{\circ} \mathrm{C}\), determine \((a)\) the power rating of the resistance heater and \((b)\) the average velocity of the water in the pipe.

Short answer

Answer: (a) The power rating of the resistance heater is approximately 46,823.2 W, and (b) the average velocity of the water in the pipe is approximately 0.09046 m/s.

Step-by-step solution

Calculating the mass flow rate

We're given the volumetric flow rate, which is 24 L/min. We need to calculate the mass flow rate to find the power rating. First, we need to convert L/min to m³/s: \(\text{Volumetric flow rate} = 24 \frac{\text{L}}{\text{min}} \times \frac{1 \text{ m}^3}{1000 \text{ L}} \times \frac{1 \text{min}}{60 \text{s}} = 0.0004 \frac{\text{m}^3}{\text{s}}\) Now we can calculate the mass flow rate by multiplying the volumetric flow rate by the density of water, using the density of water at \(20^{\circ} \mathrm{C}\), which is approximately \(998 \mathrm{kg/m}^3\): \(\text{Mass flow rate} = \text{Volumetric flow rate} \times \text{Density} = 0.0004 \frac{\text{m}^3}{\text{s}} \times 998 \mathrm{kg/m}^3 = 0.3992 \mathrm{kg/s}\)

Finding the required power

To find the power required to heat the water to the desired temperature, we can use the equation \(\text{Power} = \text{Mass flow rate} \times \text{Specific heat} \times \Delta \text{Temperature}\) The specific heat of water is approximately \(4,184 \mathrm{J/(kg \cdot K)}\). The temperature change is \(\Delta \text{Temperature} = 48^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C} = 28^{\circ} \mathrm{C}\). Now we can calculate the required power: \(\text{Power} = 0.3992 \mathrm{kg/s} \times 4,184 \mathrm{J/(kg \cdot K)} \times 28 \mathrm K = 46,823.2\,\mathrm {J/s}\) To convert to watts, we have \(\text{Power} = 46,823.2 \,\mathrm{J/s} \times \frac{1 \,\mathrm{W}}{1 \,\mathrm{J/s}} = 46,823.2 \,\mathrm{W}\) So the power rating of the resistance heater is approximately \(46,823.2 \,\mathrm W\).

Calculating the average velocity

To find the average velocity of water in the pipe, we can use the equation \(\text{Average velocity} = \frac{\text{Volumetric flow rate}}{\text{Area}}\) First, we find the area of the pipe using the given diameter: \(\text{Area} = \pi (\frac{\text{Diameter}}{2})^2 = \pi (\frac{0.075 \,\mathrm{m}}{2})^2 = 0.004419 \,\mathrm{m}^2\) Next, we can calculate the average velocity: \(\text{Average velocity} = \frac{0.0004 \,\mathrm{m^3/s}}{0.004419\,\mathrm{m}^2} = 0.09046 \,\mathrm{m/s}\) The average velocity of the water in the pipe is approximately \(0.09046 \,\mathrm{m/s}\). In conclusion, \((a)\) the power rating of the resistance heater is approximately \(46,823.2 \,\mathrm W\), and \((b)\) the average velocity of the water in the pipe is approximately \(0.09046 \,\mathrm{m/s}\).

Key concepts

Mass Flow Rate

The mass flow rate is a critical concept in thermodynamics, particularly when analyzing the efficiency of heating systems like water heaters. It represents the amount of mass passing through a certain point in a system per unit time. In the context of our water heater exercise, we calculate the mass flow rate by converting the volumetric flow rate of water supplied to the heater into a mass-based metric. This is done using the equation:

\[\begin{equation}\text{Mass flow rate} = \text{Volumetric flow rate} \times \text{Density} \end{equation}\]

Since we know the density of water at room temperature is approximately 998 kg/m³, we can easily find the mass of the water moving through the heater every second. Understanding the mass flow rate is pivotal because it dictates how much water the heater needs to warm up within a certain time frame. Therefore, it directly influences the design and power requirements of the heating element to achieve the desired temperature rise.

Specific Heat

Specific heat, often denoted as 'c,' is a property that describes the amount of heat required to raise the temperature of a unit mass of a substance by a unit of temperature (usually one degree Celsius or Kelvin). The specific heat of water is quite high, around 4,184 Joules per kilogram Kelvin, which means it takes a significant amount of energy to change its temperature.

This inherent characteristic of water affects calculations for heating systems. For instance, when determining the power rating for our resistance heater, we used the specific heat of water in the equation:

\[\begin{equation}\text{Power} = \text{Mass flow rate} \times \text{Specific heat} \times \Delta \text{Temperature} \end{equation}\]

The specific heat value is essential here because it quantifies the energy needed per kilogram of water to achieve the 28°C temperature rise stated in the exercise. It is a fundamental parameter that allows us to translate a temperature change into an energy requirement.

Heat Transfer

Heat transfer is the process of thermal energy moving from a warmer area to a cooler area. It plays an integral role in systems like water heaters, where we want to raise the temperature of water by transferring heat from an electric resistor to the water. In our exercise, we're looking at how much heat the water absorbs as it flows through the heater.

There are three modes of heat transfer: conduction, convection, and radiation. However, for the insulated pipe scenario, we focus primarily on conduction and convection, as the pipe's insulation limits the impact of radiation. When calculating the power requirement for the heater, we essentially determine how much energy the water must receive (via conduction from the resistor and subsequent convection within the water) to achieve the target temperature. The power, then, can be thought of as the rate of heat transfer required to maintain the desired water heating rate, as defined by the mass flow rate and specific heat of water.