/*! 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 46 A heating system must maintain t... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 46

A heating system must maintain the interior of a building at \(20^{\circ} \mathrm{C}\) during a period when the outside air temperature is \(5^{\circ} \mathrm{C}\) and the heat transfer from the building through its roof and walls is \(3 \times 10^{6} \mathrm{~kJ}\). For this duty heat pumps are under consideration that would operate between the dwelling and (a) the ground at \(15^{\circ} \mathrm{C}\). (b) a pond at \(10^{\circ} \mathrm{C}\). (c) the outside air at \(5^{\circ} \mathrm{C}\). For each case, evaluate the minimum theoretical net work input required by any such heat pump, in kJ.

Short Answer

Expert verified
(a) 51,148.74 kJ, (b) 102,383.73 kJ, (c) 153,544.64 kJ

Step by step solution

01

- Identify the temperatures in Kelvin

Convert the given temperatures from Celsius to Kelvin using the formula: \[ T(K) = T(^{\text{掳}C}) + 273.15 \]1. Interior temperature: \[ T_{\text{int}} = 20 + 273.15 = 293.15 \text{ K} \]2. Ground temperature: \[ T_{\text{ground}} = 15 + 273.15 = 288.15 \text{ K} \]3. Pond temperature: \[ T_{\text{pond}} = 10 + 273.15 = 283.15 \text{ K} \]4. Outside air temperature: \[ T_{\text{air}} = 5 + 273.15 = 278.15 \text{ K} \]
02

- Determine the coefficient of performance (COP) for a heat pump

The COP for a heat pump operating between two reservoirs is given by:\[ \text{COP} = \frac{T_{\text{hot}}}{T_{\text{hot}} - T_{\text{cold}}} \]Here, \( T_{\text{hot}} \) is the interior temperature, and \( T_{\text{cold}} \) is the temperature of the heat source (ground, pond, or air).
03

- Calculate COP for each case

Use the formula from Step 2 to calculate the COP for each heat source:(a) Ground at 15掳C: \[ \text{COP}_{\text{ground}} = \frac{293.15}{293.15 - 288.15} = \frac{293.15}{5} \ \text{COP}_{\text{ground}} = 58.63 \](b) Pond at 10掳C: \[ \text{COP}_{\text{pond}} = \frac{293.15}{293.15 - 283.15} = \frac{293.15}{10} \ \text{COP}_{\text{pond}} = 29.32 \](c) Outside air at 5掳C: \[ \text{COP}_{\text{air}} = \frac{293.15}{293.15 - 278.15} = \frac{293.15}{15} \ \text{COP}_{\text{air}} = 19.54 \]
04

- Calculate the minimum theoretical net work input

The net work input \( W \) required can be found using the relationship:\[ Q = \text{COP} \times W \]Given that the heat transfer \( Q \) is \( 3 \times 10^6 \text{ kJ} \), rearrange to solve for \( W \):\[ W = \frac{Q}{\text{COP}} \](a) For the ground at 15掳C:\[ W_{\text{ground}} = \frac{3 \times 10^6 \text{ kJ}}{58.63} = 51,148.74 \text{ kJ} \](b) For the pond at 10掳C:\[ W_{\text{pond}} = \frac{3 \times 10^6 \text{ kJ}}{29.32} = 102,383.73 \text{ kJ} \](c) For the outside air at 5掳C:\[ W_{\text{air}} = \frac{3 \times 10^6 \text{ kJ}}{19.54} = 153,544.64 \text{ kJ} \]

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Key Concepts

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

Coefficient of Performance (COP)
The coefficient of performance (COP) is a key measure of a heat pump's efficiency. It indicates how well a heat pump can transfer heat in relation to the energy input.
To determine COP, use the formula: \[ \text{COP} = \frac{T_{\text{hot}}}{T_{\text{hot}} - T_{\text{cold}}} \] where:
  • \( T_{\text{hot}} \) is the interior or target temperature.
  • \( T_{\text{cold}} \) is the source temperature (ground, pond, or air).

For our case:
  • The interior temperature is \( 293.15 \text{ K} \).
  • The source temperatures are \( 288.15 \text{ K} \) (ground), \( 283.15 \text{ K} \) (pond), and \( 278.15 \text{ K} \) (air).
The higher the COP, the more efficient the heat pump is at transferring heat with less work.
Heat Transfer
Heat transfer is the movement of thermal energy from one place to another. In heating systems, it generally moves from a heat source (like the ground, pond, or air) to the space being heated (like a building).
The amount of heat needed is given in kilojoules (kJ). For this problem, the heat transfer requirement is \( 3 \times 10^6 \text{ kJ} \).
Understanding heat transfer is crucial because it directly ties to the heat pump's workload. The efficiency of this transfer affects how much energy is required to maintain a comfortable interior temperature.
Theoretical Net Work Input
The theoretical net work input is the minimum energy required to operate a heat pump. Calculation of this value involves the heat requirement and the COP.
The formula is: \[ W = \frac{Q}{\text{COP}} \]where:
  • \( W \) is the net work input in kJ
  • \( Q \) is the heat transfer requirement (\( 3 \times 10^6 \text{ kJ} \))
  • \( \text{COP} \) is the coefficient of performance
For example, with ground temperature:
\[ W_{\text{ground}} = \frac{3 \times 10^6}{58.63} = 51,148.74 \text{ kJ} \]
This indicates how much work the heat pump must do to maintain the desired interior temperature with different sources.
Temperature Conversion
Temperature conversion is calculating temperature in different scales, such as Celsius to Kelvin. For these calculations, it's essential to use Kelvin.
The conversion formula is: \[ T(K) = T(^{\text{掳}C}) + 273.15 \] For example:
  • Comfortable indoor temperature: \( 20^{\text{掳}C} + 273.15 = 293.15 \text{ K} \)
  • Ground temperature: \( 15^{\text{掳}C} + 273.15 = 288.15 \text{ K} \)
  • Pond temperature: \( 10^{\text{掳}C} + 273.15 = 283.15 \text{ K} \)
  • Outside air temperature: \( 5^{\text{掳}C} + 273.15 = 278.15 \text{ K} \)
Using Kelvin ensures accuracy and consistency in thermodynamic calculations.
Thermal Energy Sources
Thermal energy sources for heat pumps include the ground, water sources (ponds), and the air. Each has different temperatures which affect heat pump efficiency:
  • Ground temperature: \( 15^{\text{掳}C} \). Heat pumps using the ground have high COP (58.63).
  • Pond temperature: \( 10^{\text{掳}C} \). Heat pumps using a pond have a moderate COP (29.32).
  • Outside air temperature: \( 5^{\text{掳}C} \). Heat pumps using air have the lowest COP (19.54).

The choice of thermal energy source impacts the heat pump's work requirement and efficiency. The warmer the source, the less work is needed by the pump (indicated by higher COP).

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Most popular questions from this chapter

At steady state, a refrigeration cycle operating between hot and cold reservoirs at \(300 \mathrm{~K}\) and \(275 \mathrm{~K}\), respectively, removes energy by heat transfer from the cold reservoir at a rate of \(600 \mathrm{~kW}\). (a) If the cycle's coefficient of performance is 4 , determine the power input required, in \(\mathrm{kW}\). (b) Determine the minimum theoretical power required, in \(\mathrm{kW}\), for any such cycle.

The data listed below are claimed for a power cycle operating between hot and cold reservoirs at \(1500 \mathrm{~K}\) and \(450 \mathrm{~K}\), respectively. For each case, determine whether the cycle operates reversibly, operates irreversibly, or is impossible. (a) \(Q_{\mathrm{H}}=600 \mathrm{~kJ}, W_{\text {cycle }}=300 \mathrm{~kJ}, Q_{\mathrm{C}}=300 \mathrm{~kJ}\) (b) \(Q_{\mathrm{H}}=400 \mathrm{~kJ}, W_{\text {cycle }}=280 \mathrm{~kJ}, Q_{\mathrm{C}}=120 \mathrm{~kJ}\) (c) \(Q_{\mathrm{H}}=700 \mathrm{~kJ}, W_{\text {cyck }}=300 \mathrm{~kJ}, Q_{\mathrm{C}}=500 \mathrm{~kJ}\) (d) \(Q_{\mathrm{H}}=800 \mathrm{~kJ}, W_{\text {cycle }}=600 \mathrm{~kJ}, Q_{\mathrm{C}}=200 \mathrm{~kJ}\)

A power cycle operates between hot and cold reservoirs at \(600 \mathrm{~K}\) and \(300 \mathrm{~K}\), respectively. At steady state the cycle develops a power output of \(0.45 \mathrm{MW}\) while receiving energy by heat transfer from the hot reservoir at the rate of \(1 \mathrm{MW}\). (a) Determine the thermal efficiency and the rate at which energy is rejected by heat transfer to the cold reservoir, in MW. (b) Compare the results of part (a) with those of a reversible power cycle operating between these reservoirs and receiving the same rate of heat transfer from the hot reservoir.

A heating system must maintain the interior of a building at \(T_{\mathrm{H}}=20^{\circ} \mathrm{C}\) when the outside temperature is \(T_{\mathrm{C}}=2^{\circ} \mathrm{C}\). If the rate of heat transfer from the building through its walls and roof is \(16.4 \mathrm{~kW}\), determine the electrical power required, in \(\mathrm{kW}\), to heat the building using (a) electrical-resistance heating, (b) a heat pump whose coefficient of performance is \(3.0\), (c) a reversible heat pump operating between hot and cold reservoirs at \(20^{\circ} \mathrm{C}\) and \(2^{\circ} \mathrm{C}\), respectively.

A power cycle receives 1000 Btu by heat transfer from a reservoir at \(1000^{\circ} \mathrm{F}\) and discharges energy by heat transfer to a reservoir at \(300^{\circ} \mathrm{F}\). The thermal efficiency of the cycle is \(75 \%\) of that for a reversible power cycle operating between the same reservoirs, (a) For the actual cycle, determine the thermal efficiency and the energy discharged to the cold reservoir, in Btu. (b) Repeat for the reversible power cycle.
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