/*! 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 45 By supplying energy to a dwellin... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

By supplying energy to a dwelling at a rate of \(25,000 \mathrm{~kJ} / \mathrm{h}\), a heat pump maintains the temperature of the dwelling at \(20^{\circ} \mathrm{C}\) when the outside air is at \(-10^{\circ} \mathrm{C}\). If electricity costs 8 cents per \(\mathrm{kW} \cdot \mathrm{h}\), determine the minimum theoretical operating cost for each day of operation.

Short Answer

Expert verified
The minimum theoretical operating cost per day is 6 cents.

Step by step solution

01

Understand the problem and given data

The heat pump supplies energy at a rate of 25,000 kJ/h to maintain the temperature of a dwelling at 20°C while the outside temperature is -10°C. Electricity costs 8 cents per kWh. We need to find the minimum theoretical operating cost per day.
02

Convert energy rate to power

First, convert the given energy rate from kJ/h to kW. Power (P) in kW can be calculated as:
03

Determine the Coefficient of Performance (COP) for the heat pump

Calculate the Coefficient of Performance (COP) using the temperature values:
04

Calculate the work input needed

Using the COP, calculate the work input required to deliver the 25,000 kJ/h. Work input (W) can be found by rearranging the COP equation: 0
05

Convert work input to kWh

Convert the work input from kJ/h to kWh:
06

Calculate daily energy consumption

Calculate the energy consumption per day by multiplying the hourly work input by 24 hours: happen corn Lamborghini countyn
07

Determine the cost

Multiply the daily energy consumption by the cost per kWh to find the operating cost per day: car bear lamddit countyn kingnn0

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.

thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. It plays a crucial role in understanding how a heat pump operates.

In this exercise, we need to manage the indoor temperature at 20°C when the outdoor temperature is -10°C. This involves transferring heat from a colder environment (outside) to a warmer one (inside).

The principle governing this process is the second law of thermodynamics, which states that heat cannot spontaneously flow from a colder body to a warmer body. A heat pump uses work (energy) to transfer heat uphill against this natural direction.
  • For instance, think of it like a water pump that moves water uphill. Just as moving water requires energy, so does moving heat in thermodynamics.
coefficient of performance
The Coefficient of Performance (COP) is a measure of a heat pump's efficiency. It is defined as the ratio of the heat supplied to the dwelling to the work input. For heating, the COP can be calculated using the formula:
energy conversion
Energy conversion is the process of changing energy from one form to another. In the context of this exercise, we convert electrical energy into thermal energy.



We know that the heat pump supplies energy at a rate of 25,000 kJ/h. To understand how much electrical energy this corresponds to, we need to convert kJ/h to kW (since electricity usage is measured in kilowatt-hours, kWh).
  • To convert, note that 1 kW = 3,600 kJ/h. Hence, 25,000 kJ/h is approximately 6.94 kW.
operating cost calculation
Operating cost calculation helps determine the financial expense of running the heat pump.

After calculating the work input in kWh, we can find the total daily usage by multiplying the hourly energy usage by 24 hours.

Finally, to determine the cost, multiply the daily energy consumption by the cost per kWh. Given that electricity costs 8 cents per kWh:

  • If hourly work input = 6.94 kWh, then daily consumption = 6.94 kWh * 24
  • The daily cost is calculated as: daily consumption * cost per kWh

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

A heat pump operating at steady state is driven by a \(1-\mathrm{kW}\) electric motor and provides heating for a building whose interior is to be kept at \(20^{\circ} \mathrm{C}\). On a day when the outside temperature is \(0^{\circ} \mathrm{C}\) and energy is lost through the walls and roof at a rate of \(60,000 \mathrm{~kJ} / \mathrm{h}\), would the heat pump suffice?

Write a paper outlining the contributions of Carnot, Clausius, Kelvin, and Planck to the development of the second law of thermodynamics. In what ways did the now-discredited caloric theory influence the development of the second law as we know it today? What is the historical basis for the idea of a perpetual motion machine of the second kind that is sometimes used to state the second law?

To increase the thermal efficiency of a reversible power cycle operating between thermal reservoirs at temperatures \(T_{\mathrm{H}}\) and \(T_{C}\), would it be better to increase \(T_{\mathrm{H}}\) or decrease \(T_{\mathrm{C}}\) by equal amounts?

To increase the thermal efficiency of a reversible power cycle operating between reservoirs at \(T_{\mathrm{H}}\) and \(T_{C}\), would you increase \(T_{\mathrm{H}}\) while keeping \(T_{\mathrm{C}}\) constant, or decrease \(T_{\mathrm{C}}\) while keeping \(T_{\mathrm{H}}\) constant? Are there any natural limits on the increase in thermal efficiency that might be achieved by such means?

To maintain the passenger compartment of an automobile traveling at \(13.4 \mathrm{~m} / \mathrm{s}\) at \(21^{\circ} \mathrm{C}\) when the surrounding air temperature is \(32^{\circ} \mathrm{C}\), the vehicle's air conditioner removes \(5.275 \mathrm{~kW}\), by heat transfer. Estimate the amount of engine horsepower required to drive the air conditioner. Referring to typical manufacturer's data, compare your estimate with the actual horsepower requirement. Discuss the relationship between the initial unit cost of an automobile air-conditioning system and its operating cost.

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.