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Chapter 10: Problem 4

If it takes about \(335 \mathrm{~kJ}\) to freeze \(1 \mathrm{~kg}\) of water, how much ice could an ice maker having a 1-ton refrigeration capacity produce in 24 hours?

Short Answer

Expert verified
Approximately 909 kg of ice can be produced in 24 hours.

Step by step solution

01

Understand the ton of refrigeration

A ton of refrigeration is equivalent to the energy needed to freeze 1 ton (or 2000 pounds) of water at 0 degrees Celsius in 24 hours. This is standardized to be 12,000 BTU/hour.
02

Convert BTU to kJ

Since 1 BTU (British Thermal Unit) is equal to 1.055 kJ, calculate the total energy in kJ. \[ 12,000 \text{ BTU/hour} \times 24 \text{ hours} \times 1.055 \text{ kJ/BTU} = 304,560 \text{ kJ} \]
03

Determine the heat removal for freezing water

Given that it takes 335 kJ to freeze 1 kg of water, we can calculate the amount of ice produced by dividing the total energy in kJ by the energy required to freeze 1 kg of water. \[ \frac{304,560 \text{ kJ}}{335 \text{ kJ/kg}} \approx 909 \text{ kg} \]

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Key Concepts

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

ton of refrigeration
The 'ton of refrigeration' is a term used to measure the amount of heat removed by a refrigeration system. It is based on the amount of energy needed to freeze 1 ton (or 2,000 pounds) of water at 0 degrees Celsius in 24 hours. This energy amount is standardized at 12,000 BTU/hour. In the context of the exercise, this means that a refrigeration system with a 1-ton capacity should be able to remove enough heat from water to produce a ton of ice over a 24-hour period. This provides a way to quantify and compare the cooling power of refrigeration units.
energy conversion
Energy conversion is the process of changing one form of energy into another. In this problem, we need to convert British Thermal Units (BTU), which is the energy unit used in the refrigeration field, to kilojoules (kJ), which is often used in scientific contexts. For instance, we know that 1 BTU equals 1.055 kJ. Using this conversion factor, we can convert the energy measurement in BTU/hour to kJ for easier calculation. Specifically, the 1-ton refrigeration capacity of 12,000 BTU/hour translates to 304,560 kJ over a 24-hour period.
heat removal
Heat removal is a core concept in refrigeration. When a refrigeration unit cools down water, it removes heat from it. The amount of heat required to freeze water is known as its latent heat of fusion. In this problem, it takes 335 kJ to freeze 1 kg of water. The 1-ton refrigeration unit removes 304,560 kJ of heat in 24 hours. By dividing this total energy by the energy required to freeze 1 kg of water, we determine the mass of ice produced. This gives us a practical understanding of how much material (ice) results from the operation of the refrigeration unit.
mass of ice produced
The mass of ice produced is calculated by taking the total energy removed by the refrigeration system and dividing it by the energy needed to freeze 1 kg of water. From the earlier steps, we established that our 1-ton capacity unit removes 304,560 kJ in 24 hours. Knowing that it takes 335 kJ to freeze 1 kg of water, we can determine the mass of ice produced by dividing 304,560 kJ by 335 kJ/kg. This calculation yields approximately 909 kg of ice. This mass is a direct measure of the output capacity of the refrigeration system in a tangible form鈥攊ce.

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Most popular questions from this chapter

Why does the indoor unit of a central air-conditioning system have a drain hose?

A vapor-compression refrigeration system for a household refrigerator has a refrigerating capacity of \(0.3 \mathrm{~kW}\). Refrigerant enters the evaporator at \(-23^{\circ} \mathrm{C}\) and exits at \(-18^{\circ} \mathrm{C}\). The isentropic compressor efficiency is \(80 \%\). The refrigerant condenses at \(35^{\circ} \mathrm{C}\) and exits the condenser subcooled at \(32^{\circ} \mathrm{C}\). There are no significant pressure drops in the flows through the evaporator and condenser. Determine the evaporator and condenser pressures, each in \(\mathrm{kPa}\), the mass flow rate of refrigerant, in \(\mathrm{kg} / \mathrm{s}\), the compressor power input, in \(\mathrm{kW}\), and the coefficient of performance for (a) Refrigerant \(134 a\) and (b) propane as the working fluid.

In a vapor compression heat pump, Refrigerant 22 is compressed adiabatically from \(0.15 \mathrm{MPa},-20^{\circ} \mathrm{C}\) to \(1 \mathrm{MPa}\), \(80^{\circ} \mathrm{C}\). The Refrigerant enters the expansion valve at \(1 \mathrm{MPa}\), \(32^{\circ} \mathrm{C}\) in liquid state. At the exit of expansion valve pressure becomes \(0.15 \mathrm{MPa}\). Determine (a) the isentropic compressor efficiency. (b) the coefficient of performance.

Can the coefficient of performance of a heat pump have a value less than one?

An ideal vapor-compression refrigeration cycle, with ammonia as the working fluid, has an evaporator temperature of \(-22^{\circ} \mathrm{C}\) and a condenser pressure of 14 bar. Saturated vapor enters the compressor, and saturated liquid exits the condenser. The mass flow rate of the refrigerant is \(5 \mathrm{~kg} / \mathrm{min}\). Determine (a) the coefficient of performance. (b) the refrigerating capacity, in tons.
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