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Chapter 1: Problem 23

Water is heated in an insulated, constant diameter tube by a \(5-\mathrm{kW}\) electric resistance heater. If the water enters the heater steadily at \(15^{\circ} \mathrm{C}\) and leaves at \(60^{\circ} \mathrm{C}\), determine the mass flow rate of water.

Short Answer

Expert verified
Answer: The mass flow rate of the water in the tube is approximately 0.7077 kg/s.

Step by step solution

01

Write down the given information

The given information can be written as: - Electric heater power: \(P = 5\,\mathrm{kW}\) - Water temperature at inlet: \(T_1 = 15^{\circ}\mathrm{C} \) - Water temperature at outlet: \(T_2 = 60^{\circ}\mathrm{C} \)
02

Convert temperatures to Kelvin

To work with temperatures in calculations, it is necessary to convert them to Kelvin because Kelvin is an absolute temperature scale: $$T_1(K) = T_1(^{\circ}\mathrm{C}) + 273.15$$ $$T_2(K) = T_2(^{\circ}\mathrm{C}) + 273.15$$
03

Write down the energy balance equation

The energy balance equation is given as follows: $$P = \dot{m}c\Delta T$$ Where: - \(P\) is the power (in Watts) - \(\dot{m}\) is the mass flow rate (in \(\mathrm{kg/s}\)) - \(c\) is the specific heat capacity of water at constant pressure (\(4.18\,\mathrm{kJ/kg\cdot K}\)) - \(\Delta T = T_2 - T_1\) is the temperature difference between the outlet and inlet
04

Convert the power to watts

The power of the electric heater is given in kilowatts, so we need to convert it to watts: $$P (\mathrm{W}) = P (\mathrm{kW}) \times 1000$$ Where \(P(\mathrm{W})\) is the power in watts, and \(P(\mathrm{kW})\) is the power in kilowatts.
05

Solve for the mass flow rate

We can rearrange the energy balance equation to solve for the mass flow rate: $$\dot{m} = \frac{P}{c\Delta T}$$ Plug in the values: $$\dot{m} = \frac{P(\mathrm{W})}{c\Delta T}$$ $$\dot{m} = \frac{P(\mathrm{W})}{c(T_2(K) - T_1(K))}$$
06

Calculate the mass flow rate

Use the given and calculated values to find the mass flow rate \(\dot{m}\): $$\dot{m} = \frac{5000}{4.18\times (333.15-288.15)}$$ $$\dot{m} = \frac{5000}{4.18\times45}$$ $$\dot{m} = 0.7077\,\mathrm{kg/s}$$ Thus, the mass flow rate of the water in the tube is approximately \(0.7077\,\mathrm{kg/s}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Energy Balance Equation
In order to calculate the mass flow rate of water heated by an electric resistance heater, the energy balance equation is invaluable. This equation expresses the conservation of energy principle. In simple terms, it states that the energy input into a system must equal the energy output. This is crucial for understanding how much energy is transferred into heating the water. The formula used is:\[P = \dot{m}c\Delta T\]- **P** represents the power provided by the heater in watts.- **\(\dot{m}\)** is the mass flow rate of the fluid, which is what you are solving for.- **c** is the specific heat capacity, a property of water that shows how much energy is needed to raise the temperature of a unit mass by one degree Kelvin.- **\(\Delta T\)** signifies the change in temperature from the inlet to the outlet.Using this simple yet powerful equation, we can determine how much water can be heated per second by the electrical power provided.
Electric Resistance Heater
An electric resistance heater converts electrical energy into thermal energy. This appliance often uses metal wires that heat up when an electric current passes through them. This property makes electric resistance heaters efficient for heating fluids like water in this exercise. When current flows through the heater, it generates a specific power output measured in kilowatts (kW). In our case, the heater provides a power of 5 kW. For ease of calculation, it is important to remember that 1 kW equals 1,000 watts. The entire produced energy is utilized to heat the water as the process is insulated. Thus, the heat loss is negligible, which means the system efficiency is approximately 100%. This makes it ideal for calculating the energy input for the water heating, needed in conjunction with the energy balance equation to determine the mass flow rate.
Specific Heat Capacity
The specific heat capacity (\(c\)) is an intrinsic property of any substance. For water, this is a functional parameter when calculating how much energy is needed to increase its temperature. In this exercise, the specific heat capacity of water is approximately 4.18 kJ/kg路K, which signifies that it requires 4.18 kilojoules to raise 1 kilogram of water by 1 Kelvin.This value is crucial in the energy balance equation, influencing how much water can be heated in a specific time given the energy available. This property allows us to convert energy input into a temperature change, helping to find the mass flow rate.
Temperature Conversion
Temperature conversion is necessary when working with equations that require absolute temperature. In thermodynamics, Kelvin is the preferred scale because it starts at absolute zero, making calculations more straightforward. For instance, converting Celsius to Kelvin is easy and done by adding 273.15:- **Celsius to Kelvin conversion.**\[T(K) = T(^{\circ}\mathrm{C}) + 273.15\]In this exercise, you'll convert the water temperature from 15掳C and 60掳C to Kelvin before using them in the energy balance equation. This is critical for accuracy in thermodynamic equations, ensuring all energy unit calculations are consistent.

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Most popular questions from this chapter

The outer surface of a spacecraft in space has an emissivity of \(0.8\) and a solar absorptivity of \(0.3\). If solar radiation is incident on the spacecraft at a rate of \(950 \mathrm{~W} / \mathrm{m}^{2}\), determine the surface temperature of the spacecraft when the radiation emitted equals the solar energy absorbed.

An ice skating rink is located in a building where the air is at \(T_{\text {air }}=20^{\circ} \mathrm{C}\) and the walls are at \(T_{w}=25^{\circ} \mathrm{C}\). The convection heat transfer coefficient between the ice and the surrounding air is \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The emissivity of ice is \(\varepsilon=0.95\). The latent heat of fusion of ice is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\) and its density is \(920 \mathrm{~kg} / \mathrm{m}^{3}\). (a) Calculate the refrigeration load of the system necessary to maintain the ice at \(T_{s}=0^{\circ} \mathrm{C}\) for an ice rink of \(12 \mathrm{~m}\) by \(40 \mathrm{~m}\). (b) How long would it take to melt \(\delta=3 \mathrm{~mm}\) of ice from the surface of the rink if no cooling is supplied and the surface is considered insulated on the back side?

Hot air at \(80^{\circ} \mathrm{C}\) is blown over a 2-m \(\times 4\) - \(\mathrm{m}\) flat surface at \(30^{\circ} \mathrm{C}\). If the average convection heat transfer coefficient is \(55 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat transfer from the air to the plate, in \(\mathrm{kW}\).

A 2-kW electric resistance heater submerged in 30-kg water is turned on and kept on for \(10 \mathrm{~min}\). During the process, \(500 \mathrm{~kJ}\) of heat is lost from the water. The temperature rise of water is (a) \(5.6^{\circ} \mathrm{C}\) (b) \(9.6^{\circ} \mathrm{C}\) (c) \(13.6^{\circ} \mathrm{C}\) (d) \(23.3^{\circ} \mathrm{C}\) (e) \(42.5^{\circ} \mathrm{C}\)

A series of experiments were conducted by passing \(40^{\circ} \mathrm{C}\) air over a long \(25 \mathrm{~mm}\) diameter cylinder with an embedded electrical heater. The objective of these experiments was to determine the power per unit length required \((\dot{W} / L)\) to maintain the surface temperature of the cylinder at \(300^{\circ} \mathrm{C}\) for different air velocities \((V)\). The results of these experiments are given in the following table: $$ \begin{array}{lccccc} \hline V(\mathrm{~m} / \mathrm{s}) & 1 & 2 & 4 & 8 & 12 \\ \dot{W} / L(\mathrm{~W} / \mathrm{m}) & 450 & 658 & 983 & 1507 & 1963 \\ \hline \end{array} $$ (a) Assuming a uniform temperature over the cylinder, negligible radiation between the cylinder surface and surroundings, and steady state conditions, determine the convection heat transfer coefficient \((h)\) for each velocity \((V)\). Plot the results in terms of \(h\left(\mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\) vs. \(V(\mathrm{~m} / \mathrm{s})\). Provide a computer generated graph for the display of your results and tabulate the data used for the graph. (b) Assume that the heat transfer coefficient and velocity can be expressed in the form of \(h=C V^{m}\). Determine the values of the constants \(C\) and \(n\) from the results of part (a) by plotting \(h\) vs. \(V\) on log-log coordinates and choosing a \(C\) value that assures a match at \(V=1 \mathrm{~m} / \mathrm{s}\) and then varying \(n\) to get the best fit.
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