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Chapter 13: Problem 121

An average person produces \(0.50 \mathrm{lbm}\) of moisture while taking a shower and \(0.12 \mathrm{lbm}\) while bathing in a tub. Consider a family of four who shower once a day in a bathroom that is not ventilated. Taking the heat of vaporization of water to be \(1050 \mathrm{Btu} / \mathrm{lbm}\), determine the contribution of showers to the latent heat load of the air conditioner in summer per day.

Short Answer

Expert verified
Answer: The contribution of showers to the latent heat load of the air conditioner in summer per day for a family of four is 2100 Btu/day.

Step by step solution

01

Calculate the total moisture produced by the family

In order to calculate the total moisture produced by the family, we will multiply the amount of moisture produced during one shower by the number of family members taking showers. Total moisture produced = (moisture produced per shower) x (number of family members) Total moisture produced = (0.50 lbm/shower) x (4 family members) = 2 lbm/day
02

Calculate the daily latent heat load due to showers

Now that we have the total amount of moisture produced per day, we can calculate the daily latent heat load due to showers by multiplying the total moisture produced by the heat of vaporization of water. Daily latent heat load = (total moisture produced) x (heat of vaporization of water) Daily latent heat load = (2 lbm/day) x (1050 Btu/lbm) = 2100 Btu/day So, the contribution of showers to the latent heat load of the air conditioner in summer per day is 2100 Btu/day.

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Key Concepts

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

Heat of Vaporization
Understanding the heat of vaporization is crucial for many engineering applications, including air conditioning. It is the amount of heat energy required to change a substance from a liquid phase to a vapor phase without changing its temperature. The heat of vaporization for water is typically around 1050 Btu per pound-mass (lbm) at normal room temperature.

When individuals shower, the water used eventually evaporates into the air, and to do so, it absorbs this specific amount of heat from its surroundings. This absorption is not something that can be casually observed, but it significantly affects the indoor environment and, consequently, the operation of an air conditioning system. In the context of our exercise, calculating the latent heat load involves using the heat of vaporization to determine how much heat energy is transferred to the air as the water evaporates.
Moisture Production
Moisture production in an indoor environment can come from various sources such as people, plants, cooking, and as highlighted in our example, showers. The amount of moisture produced by an average person during a shower is 0.50 lbm, and this seems minor, but when considering multiple individuals and repeated activities, this moisture adds up.

For a family of four taking daily showers, the combined moisture production is 2 lbm per day. It's essential to consider this moisture because it's not just water; it's a medium that carries energy in the form of the heat of vaporization. In air conditioning, controlling moisture levels is a part of managing thermal loads. Too much moisture leads to higher humidity, which can make the air feel warmer and require the air conditioning system to work harder to maintain comfort levels.
Air Conditioning Thermal Load
The air conditioning thermal load is the amount of energy an air conditioning system needs to remove from a space to maintain a comfortable environment. There are two components to this load: the sensible load, which relates to the dry-bulb temperature, and the latent load, associated with moisture and humidity control.

In the provided exercise, the calculation focused on the latent heat load due to moisture added by showers. Showers contribute to the latent load because the air conditioner must remove the heat absorbed by the vaporized water to maintain a comfortable humidity level. The 2100 Btu/day calculated in the exercise represents the additional energy the air conditioning system needs to remove daily due to the showers alone, highlighting how everyday activities influence the thermal loads on our comfort systems.

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Most popular questions from this chapter

Two square plates, with the sides \(a\) and \(b\) (and \(b>a\) ), are coaxial and parallel to each other, as shown in Fig. P13-132, and they are separated by a center-to-center distance of \(L\). The radiation view factor from the smaller to the larger plate, \(F_{a b}\), is given by $$ F_{a b}=\frac{1}{2 A}\left\\{\left[(B+A)^{2}+4\right]^{0.5}-\left[(B-A)^{2}+4\right]^{0.5}\right\\} $$ where, \(A=a / L\) and \(B=b / L\). (a) Calculate the view factors \(F_{a b}\) and \(F_{b a}\) for \(a=20 \mathrm{~cm}\), \(b=60 \mathrm{~cm}\), and \(L=40 \mathrm{~cm}\). (b) Calculate the net rate of radiation heat exchange between the two plates described above if \(T_{a}=800^{\circ} \mathrm{C}\), \(T_{b}=200^{\circ} \mathrm{C}, \varepsilon_{a}=0.8\), and \(\varepsilon_{b}=0.4\). (c) A large square plate (with the side \(c=2.0 \mathrm{~m}, \varepsilon_{c}=0.1\), and negligible thickness) is inserted symmetrically between the two plates such that it is parallel to and equidistant from them. For the data given above, calculate the temperature of this third plate when steady operating conditions are established.

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