/*! 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 14 A heat pump is an electrical dev... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A heat pump is an electrical device that heats a building by pumping heat in from the cold outside. In other words, it's the same as a refrigerator, but its purpose is to warm the hot reservoir rather than to cool the cold reservoir (even though it does both). Let us define the following standard symbols, all taken to be positive by convention: \(T_{h}=\) temperature inside building \(T_{c}=\) temperature outside \(Q_{h}=\) heat pumped into building in 1 day \(Q_{c}=\) heat taken from outdoors in 1 day \(W=\) electrical energy used by heat pump in 1 day (a) Explain why the "coefficient of performance" (COP) for a heat pump should be defined as \(Q_{h} / W\) (b) What relation among \(Q_{h}, Q_{c},\) and \(W\) is implied by energy conservation alone? Will energy conservation permit the COP to be greater than \(1 ?\) (c) Use the second law of thermodynamics to derive an upper limit on the \(\mathrm{COP},\) in terms of the temperatures \(T_{h}\) and \(T_{\mathrm{c}}\) alone. (d) Explain why a heat pump is better than an electric furnace, which simply converts electrical work directly into heat. (Include some numerical estimates.)

Short Answer

Expert verified
(a) COP is defined as \( Q_{h} / W \) because it shows efficiency. (b) By energy conservation, \( Q_{h} = Q_{c} + W \); COP can be >1. (c) \( \text{COP}_{\text{max}} = \frac{T_{h}}{T_{h} - T_{c}} \). (d) Heat pump's COP can exceed 1, unlike an electric furnace.

Step by step solution

01

Understanding COP for Heat Pumps

The Coefficient of Performance (COP) for a heat pump is defined as the ratio of useful heat output to the work input. Therefore, the COP is given by \( \text{COP} = \frac{Q_{h}}{W} \). This definition is chosen because the purpose of a heat pump is to provide as much heat as possible to the building (\( Q_{h} \)) for a given amount of work (\( W \)) from the electrical energy used.
02

Applying Energy Conservation

By the First Law of Thermodynamics (energy conservation), the total energy input into the system must equal the total energy output. Thus, we have \( Q_{h} = Q_{c} + W \). This implies that the heat pumped into the building (\( Q_{h} \)) is the sum of the heat extracted from the outside (\( Q_{c} \)) and the work done by the heat pump (\( W \)). Because \( Q_{h} > W \), the COP can be greater than 1.
03

Finding Upper Limit on COP (Using Second Law)

The Second Law of Thermodynamics provides an upper limit to efficiency which is described by the Carnot cycle for a reversible process. The maximum COP is given by the reversible condition: \( \text{COP}_{ ext{max}} = \frac{T_{h}}{T_{h} - T_{c}} \). Here, \( T_{h} \) and \( T_{c} \) must be in absolute temperatures (Kelvin). This shows that the closer \( T_{h} \) and \( T_{c} \) are, the higher the possible COP.
04

Comparing Heat Pump with Electric Furnace

An electric furnace directly converts electrical energy into thermal energy with an efficiency close to 1 (every unit of work results in one unit of heat: \( \text{COP} \approx 1 \)). However, a heat pump can have a COP greater than 1 because it transfers additional energy from the external environment. For example, if a heat pump has a COP of 3, it can deliver 3 units of heat energy per unit of electrical energy, making it more efficient than an electric furnace.

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.

Coefficient of Performance
In the world of heat pumps, the Coefficient of Performance (COP) is a crucial measurement. It tells us how efficiently the system is working by comparing the heat output to the energy input. For a heat pump, we calculate this as
  • \( \text{COP} = \frac{Q_{h}}{W} \),
where \(Q_{h}\) is the heat delivered to the building and \(W\) is the electrical energy used. This ratio is significant because it shows the system's ability to transfer heat relative to the energy it consumes. A higher COP means the heat pump is more efficient at producing more heat with less energy, directly influencing its cost-effectiveness and energy-saving capabilities.

There’s another important aspect to keep in mind: the practical range of COP values. For a heat pump, the COP can exceed 1, meaning it delivers more heat energy than the electrical energy it uses. This is contrary to many other systems, where energy efficiency typically maxes out at a ratio of 1.
First Law of Thermodynamics
The First Law of Thermodynamics, also known as the principle of energy conservation, plays a pivotal role in understanding heat pumps. It tells us that the total energy entering a system should equal the total energy leaving it. For heat pumps, this principle translates to:
  • \( Q_{h} = Q_{c} + W \).
This equation explains that the heat delivered inside the building \(Q_{h}\) comes from two sources:
  • the heat extracted from the outside \(Q_{c}\),
  • and the work done by the heat pump \(W\).
By this law, as long as \(Q_{h}\) is greater than \(W\), the COP can be larger than 1. This shows the power of heat pumps: they can warm spaces efficiently, using external heat rather than relying solely on electrical energy, which sets them apart from traditional heating methods.
Second Law of Thermodynamics
The Second Law of Thermodynamics helps us understand the efficiency limits of heat pumps. It provides insights into why no machine can be perfectly effective and sets a theoretical maximum for the COP based on temperature differences. This maximum is described using the Carnot cycle. The equation:
\[\text{COP}_{\text{max}} = \frac{T_{h}}{T_{h} - T_{c}}\]
shows us the highest efficiency a heat pump can achieve when running in a reversible process. This formula takes into account:
  • \(T_{h}\) - the temperature inside the building,
  • \(T_{c}\) - the temperature outside,
expressed in Kelvin. As the outside and inside temperatures become closer, the COP increases. Although achieving the Carnot efficiency practically is impossible due to irreversible losses, this formula gives valuable insights into optimizing heat pump performance.
Carnot Cycle
The Carnot cycle represents an idealized thermodynamic cycle that sets an upper limit for the efficiency of heat engines. Understanding the Carnot cycle is instrumental for designing efficient heat pumps. It utilizes four key processes: two isothermal (constant temperature) and two adiabatic (no heat exchange) processes. In simplest terms, it outlines the most efficient way possible to convert heat into work and vice versa.

When we consider heat pumps, the Carnot efficiency informs us about the theoretical best performance we can aim for. It tells us that if we approach this efficiency, we minimize energy waste. For practical purposes, while real-world devices cannot reach Carnot efficiency due to factors like friction and heat loss, this cycle is the benchmark.

Thus, using the principles derived from the Carnot cycle, engineers strive to make devices as efficient as possible by reducing energy losses and maximizing the heat transfer between reservoirs effectively.
Energy Efficiency
Energy efficiency in the context of heat pumps refers to their ability to use less energy to achieve the desired heating effect. The COP provides an indication of efficiency, positioning heat pumps as more efficient alternatives to conventional heating systems, such as electric furnaces. This high efficiency stems from their capability to move heat, instead of generating it directly from electrical work.

To illustrate this: if a heat pump has a COP of 3, it means it delivers three times more heat than the electrical energy it consumes. Contrast this with an electric furnace, whose efficiency usually peaks at a COP of 1, delivering heat equivalent to the input of electrical energy.

So, in terms of raw numbers, the higher COP indicates more heat output per unit of energy, representing not only a cost-effective but also a more environmentally sustainable heating solution. This efficiency can help lower energy bills and decrease the carbon footprint of heating systems in buildings, aligning well with modern environmental goals.

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

Derive a formula for the efficiency of the Diesel cycle, in terms of the compression ratio \(V_{1} / V_{2}\) and the cutoff ratio \(V_{3} / V_{2}\). Show that for a given compression ratio, the Diesel cycle is less efficient than the Otto cycle. Evaluate the theoretical efficiency of a Diesel engine with a compression ratio of 18 and a cutoff ratio of 2

Prove that \(if\) you had a refrigerator whose COP was better than the ideal value \((4.9),\) you could hook it up to an ordinary Carnot engine to make an engine that produces no waste heat.

Liquid HFC-134a at its boiling point at 12 bars pressure is throttled to 1 bar pressure. What is the final temperature? What fraction of the liquid vaporizes? $$\begin{array}{rrrrrr}\begin{array}{r}P \\\\(\text { bar })\end{array} & \begin{array}{r}T \\\\\left(^{\circ} \mathrm{C}\right)\end{array} & \begin{array}{r}H_{\text {liquid }} \\\\(\mathrm{kJ})\end{array} & \begin{array}{r} H_{\text {gas }} \\\\(\mathrm{kJ})\end{array} & \begin{array}{r}S_{\text {liquid }} \\\\(\mathrm{kJ} / \mathrm{K}) \end{array} & \begin{array}{r}S_{\text {gas }} \\\\(\mathrm{kJ} / \mathrm{K})\end{array} \\\\\hline 1.0 & -26.4 & 16 & 231 & 0.068 & 0.940 \\\1.4 & -18.8 & 26 & 236 & 0.106 & 0.932 \\\2.0 & -10.1 & 37 & 241 & 0.148 & 0.925 \\ 4.0 & 8.9 & 62 & 252 & 0.240 & 0.915 \\\6.0 & 21.6 & 79 & 259 & 0.300 & 0.910 \\\8.0 & 31.3 & 93 & 264 & 0.346 & 0.907 \\\10.0 & 39.4 & 105 & 268 & 0.384 & 0.904 \\\12.0 & 46.3 & 116 & 271 & 0.416 & 0.902 \end{array}$$ Properties of the refrigerant HFC-134a under saturated conditions (at its boiling point for each pressure). All values are for 1 kg of fluid, and are measured relative to an arbitrarily chosen reference state, the saturated liquid at \(-40^{\circ} \mathrm{C} .\) Excerpted from Moran and Shapiro (1995). Properties of superheated (gaseous) refrigerant HFC-134a. All values are for \(1 \mathrm{kg}\) of fluid, and are measured relative to the same reference state as in Table 4.3. Excerpted from Moran and Shapiro (1995).

To get more than an infinitesimal amount of work out of a Carnot engine, we would have to keep the temperature of its working substance below that of the hot reservoir and above that of the cold reservoir by non-infinitesimal amounts. Consider, then, a Carnot cycle in which the working substance is at temperature \(T_{h w}\) as it absorbs heat from the hot reservoir, and at temperature \(T_{\mathrm{cw}}\) as it expels heat to the cold reservoir. Under most circumstances the rates of heat transfer will be directly proportional to the temperature differences: $$\begin{aligned} &\frac{Q_{h}}{\Delta t}=K\left(T_{h}-T_{h w}\right) \quad \text { and } &\frac{Q_{c}}{\Delta t}=K\left(T_{c w}-T_{c}\right) \end{aligned}$$ I've assumed here for simplicity that the constants of proportionality ( \(K\) ) are the same for both of these processes. Let us also assume that both processes take the same amount of time, so the \(\Delta t^{\prime}\) s are the same in both of these equations. (a) Assuming that no new entropy is created during the cycle except during the two heat transfer processes, derive an equation that relates the four temperatures \(T_{h}, T_{c}, T_{h w},\) and \(T_{c w}\) (b) Assuming that the time required for the two adiabatic steps is negligible, write down an expression for the power (work per unit time) output of this engine. Use the first and second laws to write the power entirely in terms of the four temperatures (and the constant \(K\) ), then eliminate \(T_{c w}\) using the result of part (a). (c) When the cost of building an engine is much greater than the cost of fuel (as is often the case), it is desirable to optimize the engine for maximum power output, not maximum efficiency. Show that, for fixed \(T_{h}\) and \(T_{c}\) the expression you found in part (b) has a maximum value at \(T_{h w}=\) \(\frac{1}{2}\left(T_{h}+\sqrt{T_{h} T_{c}}\right) .\) (Hint: You'll have to solve a quadratic equation.) Find the corresponding expression for \(T_{c w}\) (d) Show that the efficiency of this engine is \(1-\sqrt{T_{c} / T_{h}}\). Evaluate this efficiency numerically for a typical coal-fired steam turbine with \(T_{h}=600^{\circ} \mathrm{C}\) and \(T_{c}=25^{\circ} \mathrm{C},\) and compare to the ideal Carnot efficiency for this temperature range. Which value is closer to the actual efficiency, about \(40 \%\) of a real coal-burning power plant?

At a power plant that produces \(1 \mathrm{GW}\left(10^{9} \text { watts }\right)\) watts) of electricity, the steam turbines take in steam at a temperature of \(500^{\circ} \mathrm{C}\), and the waste heat is expelled into the environment at \(20^{\circ} \mathrm{C}\) (a) What is the maximum possible efficiency of this plant? (b) Suppose you develop a new material for making pipes and turbines, which allows the maximum steam temperature to be raised to \(600^{\circ} \mathrm{C}\). Roughly how much money can you make in a year by installing your improved hardware, if you sell the additional electricity for 5 cents per kilowatt- hour? (Assume that the amount of fuel consumed at the plant is unchanged.)

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.