/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 Water is heated in an insulated,... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 \(0.7077\,\mathrm{kg/s}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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·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°C and 60°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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What is the caloric theory? When and why was it abandoned?

A flat-plate solar collector is used to heat water by having water flow through tubes attached at the back of the thin solar absorber plate. The absorber plate has a surface area of \(2 \mathrm{~m}^{2}\) with emissivity and absorptivity of \(0.9\). The surface temperature of the absorber is \(35^{\circ} \mathrm{C}\), and solar radiation is incident on the absorber at \(500 \mathrm{~W} / \mathrm{m}^{2}\) with a surrounding temperature of \(0^{\circ} \mathrm{C}\). Convection heat transfer coefficient at the absorber surface is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the ambient temperature is \(25^{\circ} \mathrm{C}\). Net heat rate absorbed by the solar collector heats the water from an inlet temperature \(\left(T_{\text {in }}\right)\) to an outlet temperature \(\left(T_{\text {out }}\right)\). If the water flow rate is \(5 \mathrm{~g} / \mathrm{s}\) with a specific heat of \(4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), determine the temperature rise of the water.

Consider a flat-plate solar collector placed horizontally on the flat roof of a house. The collector is \(5 \mathrm{ft}\) wide and \(15 \mathrm{ft}\) long, and the average temperature of the exposed surface of the collector is \(100^{\circ} \mathrm{F}\). The emissivity of the exposed surface of the collector is \(0.9\). Determine the rate of heat loss from the collector by convection and radiation during a calm day when the ambient air temperature is \(70^{\circ} \mathrm{F}\) and the effective sky temperature for radiation exchange is \(50^{\circ} \mathrm{F}\). Take the convection heat transfer coefficient on the exposed surface to be \(2.5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\).

How are heat, internal energy, and thermal energy related to each other?

Air enters the duct of an air-conditioning system at \(15 \mathrm{psia}\) and \(50^{\circ} \mathrm{F}\) at a volume flow rate of \(450 \mathrm{ft}^{3} / \mathrm{min}\). The diameter of the duct is 10 inches and heat is transferred to the air in the duct from the surroundings at a rate of \(2 \mathrm{Btu} / \mathrm{s}\). Determine \((a)\) the velocity of the air at the duct inlet and (b) the temperature of the air at the exit.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.