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Chapter 18: Problem 18

Outside air at \(5^{\circ} \mathrm{C}\) and 20 percent relative humidity is introduced into a heating and air-conditioning plant where it is heated to 20 \({ }^{\circ} \mathrm{C}\) and its relative humidity is increased to a comfortable 50 percent. How many grams of water must be evaporated into a cubic meter of outside air to accomplish this? Saturated air at 5 \({ }^{\circ} \mathrm{C}\) contains \(6.8 \mathrm{~g} / \mathrm{m}^{3}\) of water, and at \(20^{\circ} \mathrm{C}\) it contains \(17.3 \mathrm{~g} / \mathrm{m}^{3}\). Mass/m \(^{3}\) of water vapor in air at \(5{ }^{\circ} \mathrm{C}=0.20 \times 6.8 \mathrm{~g} / \mathrm{m}^{3}=1.36 \mathrm{~g} / \mathrm{m}^{3}\) $$ \text { Comfortable mass/m }^{3} \text { at } 20^{\circ} \mathrm{C}=0.50 \times 17.3 \mathrm{~g} / \mathrm{m}^{3}=8.65 \mathrm{~g} / \mathrm{m}^{3} $$ \(1 \mathrm{~m}^{3}\) of air at \(5^{\circ} \mathrm{C}\) expands to \((293 / 278) \mathrm{m}^{3}=1.054 \mathrm{~m}^{3}\) at \(20^{\circ} \mathrm{C}\) Mass of water vapor in \(1.054 \mathrm{~m}^{3}\) at \(20^{\circ} \mathrm{C}=1.054 \mathrm{~m}^{3} \times 8.65 \mathrm{~g} / \mathrm{m}^{3}=9.12 \mathrm{~g}\) Mass of water to be added to each \(\mathrm{m}^{5}\) of air at \(5{ }^{\circ} \mathrm{C}=(9.12-1.36) \mathrm{g}=7.8 \mathrm{~g}\)

Short Answer

Expert verified
7.76 grams of water must be evaporated into each cubic meter of outside air.

Step by step solution

01

Determine Initial Water Content

Calculate the mass of water vapor in 1 m鲁 of outside air at 5掳C and 20% relative humidity. The formula is given by the product of relative humidity and saturated water content at 5掳C:\[\text{Mass of water vapor} = 0.20 \times 6.8 \, \text{g/m}^3 = 1.36 \, \text{g/m}^3\]
02

Determine Final Water Content

Calculate the mass of water vapor needed in 1 m鲁 of air at 20掳C to reach 50% relative humidity:\[\text{Mass of water vapor} = 0.50 \times 17.3 \, \text{g/m}^3 = 8.65 \, \text{g/m}^3\]
03

Adjust for Air Expansion

Calculate the final volume of 1 m鲁 of air initially at 5掳C when it expands to 20掳C\[\text{Final volume} = \frac{293}{278} \, \text{m}^3 = 1.054 \, \text{m}^3\]
04

Calculate Total Water Content Needed

Calculate the total mass of water vapor needed for the expanded volume of air at 20掳C:\[\text{Mass of water vapor in } 1.054 \, \text{m}^3 = 1.054 \, \text{m}^3 \times 8.65 \, \text{g/m}^3 = 9.12 \, \text{g}\]
05

Calculate Additional Water Required

Determine the mass of water to be added to every 1 m鲁 of air at 5掳C:\[\text{Additional water} = 9.12 \, \text{g} - 1.36 \, \text{g} = 7.76 \, \text{g}\]

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Key Concepts

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

Understanding Relative Humidity
Relative humidity is a measure of how much moisture the air contains compared to the maximum it could hold at a given temperature. This is expressed in percentage form. For instance, air with 20% relative humidity has 20% of the total water vapor it can possibly hold before reaching saturation at that temperature.

Imagine filling a sponge with water; the sponge can only hold so much water before it starts dripping. Similarly, air can only hold a certain amount of water vapor based on the temperature.
  • High relative humidity means the air is closer to being saturated.
  • Low relative humidity means there is less water vapor in the air compared to how much it can hold at that temperature.
Understanding this concept helps in solving problems related to how much moisture needs to be added to air to achieve desired comfort levels.
What is Saturated Air?
Saturated air occurs when the air holds the maximum amount of water vapor it can at a specific temperature. At this point, the relative humidity is 100% because the air cannot contain any more moisture, just like when a sponge cannot hold any more water without leaking.

Being aware of the saturation point is crucial in processes such as humidification and dehumidification, especially in heating, ventilation, and air conditioning (HVAC) systems.
  • At 5掳C, saturated air contains 6.8 g/m鲁 of water vapor.
  • At 20掳C, it holds up to 17.3 g/m鲁 of water vapor.
This understanding assists in determining how much water needs to be added or removed to achieve a particular humidity level.
Calculating the Mass of Water Vapor
To find out how much water vapor is in the air, we multiply the relative humidity by the saturated vapor density at the given temperature. For example, for air at 5掳C with 20% humidity, the mass of water vapor per cubic meter is \[\text{Mass of water vapor} = 0.20 \times 6.8 \, \text{g/m}^3 = 1.36 \, \text{g/m}^3\] This value indicates the amount of water vapor actually present in the air.

When solving thermodynamic problems related to conditioning air, it is necessary to know how much moisture each cubic meter of air starts with, and how much moisture must be added or removed to meet the desired humidity levels.
  • This calculation is pivotal in applications involving air heating, cooling, and humidification.
  • It helps to assess how effectively air-conditioning systems regulate indoor environments.
Factoring in Air Expansion
Air expansion comes into play when there's a temperature change. When air warms up, it expands. For a unit of air initially at 5掳C that heats up to 20掳C, it expands from 1 m鲁 to 1.054 m鲁.
  • The resulting expanded volume of air can hold more water vapor.
  • This is why the volume change must be taken into account when calculating humidity levels and is needed to determine the amount of additional water vapor required.
Understanding air expansion is crucial because:
  1. It affects how much water vapor the air can retain at different temperatures.
  2. It is essential when calibrating HVAC systems to ensure comfort through accurate temperature and humidity regulation.
Ensuring accuracy in these calculations leads to more efficient air conditioning solutions and better energy resource management.

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Most popular questions from this chapter

Suppose a 60-kg person consumes 2500 Cal of food in one day. If the entire heat equivalent of this food were retained by the person's body, how large a temperature change would it cause? (For the body, \(c=0.83 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\).) Remember that \(1 \mathrm{Cal}=1 \mathrm{kcal}=\) 1000 cal. The equivalent amount of heat added to the body in one day is $$ \Delta Q=(2500 \text { Cal })(1000 \mathrm{cal} / \mathrm{Cal})=2.5 \times 10^{6} \mathrm{cal} $$ Then, by use of \(\Delta Q=m c \Delta T\), $$ \Delta T=\frac{\Delta Q}{m c}=\frac{2.5 \times 10^{6} \mathrm{cal}}{\left(60 \times 10^{3} \mathrm{~g}\right)\left(0.83 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)}=50^{\circ} \mathrm{C} $$

How much heat is given up when \(20 \mathrm{~g}\) of steam at \(100{ }^{\circ} \mathrm{C}\) is condensed and cooled to \(20^{\circ} \mathrm{C}\) ? The steam loses an amount of heat to condense into water at 100 \({ }^{\circ} \mathrm{C}\) and more for the water at \(100{ }^{\circ} \mathrm{C}\) to drop in temperature to 20 \({ }^{\circ} \mathrm{C}\). From Table \(18-2, L_{v}=2259 \mathrm{~kJ} / \mathrm{kg}\). Thus the heat that must be removed is $$ Q=m L_{v}+c_{\text {wacer }} \Delta T_{\text {uateter }} $$ \(Q=-(0.020 \mathrm{~kg})(2259 \mathrm{~kJ} / \mathrm{kg})+(4.186 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(0.020 \mathrm{~kg})(-80 \mathrm{~K})\) $$ Q=-45.18 \mathrm{~kJ}-6.6976 \mathrm{~kJ} $$ and \(Q=-51.9 \mathrm{~kJ}\), or to two figures \(-52 \mathrm{~kJ}\). Alternative Method conter

How many calories are required to heat each of the following from \(15^{\circ} \mathrm{C}\) to \(65^{\circ} \mathrm{C} ?(\) a) \(3.0\) g of aluminum, \((b) 5.0 \mathrm{~g}\) of Pyrex glass, (c) 20 g of platinum. The specific heats, in \(\mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), for aluminum, Pyrex, and platinum are \(0.21,0.20\), and 0.032, respectively.

A \(3.00-\mathrm{g}\) bullet \(\left(c=0.0305 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}=128 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\) moving at 180 \(\mathrm{m} / \mathrm{s}\) enters a bag of sand and stops. By what amount does the temperature of the bullet change if all its KE becomes thermal energy that is added to the bullet? The bullet loses KE in the amount $$ \mathrm{KE}=\frac{1}{2} m v^{2}=\frac{1}{2}\left(3.00 \times 10^{-3} \mathrm{~kg}\right)(180 \mathrm{~m} / \mathrm{s})^{2}=48.6 \mathrm{~J} $$ This results in the addition of \(\Delta Q=48.6 \mathrm{~J}\) of thermal energy to the bullet. Then, since \(\Delta Q=m c \Delta T\), we can find \(\Delta T\) for the bullet: $$ \Delta T=\frac{\Delta Q}{m c}=\frac{48.6 \mathrm{~J}}{\left(3.00 \times 10^{-3} \mathrm{~kg}\right)\left(128 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)}=127^{\circ} \mathrm{C} $$ Notice that we had to use \(c\) in \(\mathrm{J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), and not in \(\mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\).

Suppose a person who eats 2500 Cal of food each day loses the heat equivalent of the food through evaporation of water from the body. How much water must evaporate each day? At body temperature, \(L_{v}\) for water is about \(600 \mathrm{cal} / \mathrm{g}\).
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