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Chapter 13: Problem 119

Consider a large classroom with 90 students on a hot summer day. All the lights with \(2.0 \mathrm{~kW}\) of rated power are kept on. The room has no external walls, and thus heat gain through the walls and the roof is negligible. Chilled air is available at \(15^{\circ} \mathrm{C}\), and the temperature of the return air is not to exceed \(25^{\circ} \mathrm{C}\). The average rate of metabolic heat generation by a person sitting or doing light work is \(115 \mathrm{~W}\) ( \(70 \mathrm{~W}\) sensible and \(45 \mathrm{~W}\) latent). Determine the required flow rate of air that needs to be supplied to the room.

Short Answer

Expert verified
Answer: The required flow rate of air to maintain the maximum temperature of 25掳C in the classroom is approximately 1.03 m鲁/s.

Step by step solution

01

Calculate the total sensible heat

Calculate the total sensible heat generated in the room considering the heat from the students and the lights. Sensible heat generated by students = (90 students * 70 W) = 6300 W. Heat generated by the lights = 2000 W. So, the total sensible heat generated in the room = 6300 W (students) + 2000 W (lights) = 8300 W.
02

Calculate the total latent heat

Calculate the total latent heat generated in the room by considering the heat from the students: Latent heat generated by the students = 90 students * 45 W = 4050 W.
03

Calculate the heat gain ratio

Determine the heat gain ratio of sensible to latent heat by dividing the total sensible heat by the total latent heat. Heat gain ratio = Total Sensible Heat / Total Latent Heat = 8300 W (sensible) / 4050 W (latent) 鈮� 2.05.
04

Calculate the required cooling capacity

Determine the total heat that the cooling system must be able to remove to maintain the maximum return air temperature. Cooling capacity = Total Sensible Heat + Total Latent Heat = 8300 W (sensible) + 4050 W (latent) = 12350 W.
05

Calculate the temperature difference between the supply and return air

Assuming the air is cooled completely to 15掳C before being supplied and has to be maintained at 25掳C upon return, find the difference in temperatures. 鈭員 = 25掳C (return air temperature) 鈥� 15掳C (supply air temperature) = 10掳C.
06

Determine the specific heat of air

Assume the specific heat of air (Cp) is about 1005 J/kg路K.
07

Calculate the mass flow rate of air

Determine the mass flow rate of air (m) required to balance the heat gains in the room by using the formula: m = Cooling Capacity / (Cp * 鈭員). m = 12350 W / (1005 J/kg路K * 10掳C) 鈮� 1.23 kg/s.
08

Convert mass flow rate to volume flow rate

Using the formula: volume flow rate (Q) = mass flow rate (m) / air density (蟻). Assuming the air density (蟻) is 1.2 kg/m鲁, calculate the volume flow rate. Q = 1.23 kg/s / 1.2 kg/m鲁 鈮� 1.03 m鲁/s. Therefore, the required flow rate of air to be supplied to the room is approximately 1.03 m鲁/s.

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Key Concepts

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

Sensible Heat Calculation
Sensible heat refers to the energy needed to change the temperature of air without changing its phase. In our classroom scenario, the main sources of sensible heat come from the body's metabolic activities and electrical equipment like lights.

To determine the total sensible heat in the room, we add up the heat generated by each source:
  • Students: With 90 students producing 70 W of sensible heat each, the total heat contributed by students is 6300 W.
  • Lights: The room's lights add an additional 2000 W to the total heat load.
By combining these figures, we find the room has a total sensible heat load of 8300 W. This value is crucial for determining how much cooling is required.
Latent Heat Calculation
Latent heat represents the energy necessary to change the phase of air moisture content, contributing to humidity control without affecting temperature directly. In this case, the students inside the classroom are the primary contributors of latent heat.

Each student emits 45 W of latent heat, resulting in a total latent heat contribution from all students calculated as follows:
  • Latent heat from students: 90 students 脳 45 W = 4050 W
This value helps in assessing the overall cooling demand since both sensible and latent heats contribute to the total heat gain.
Air Flow Rate Calculation
Calculating the air flow rate is essential to determine how much air needs to be circulated to maintain desired indoor conditions. It's done to ensure that the cooling supplied by the air matches the total heat gains in the room.

Once the cooling capacity necessary to remove both the sensible (8300 W) and latent heat (4050 W) is known, you follow these steps:
  • Temperature Difference (螖T): The difference between the supply air (15掳C) and the return air (25掳C) is 10掳C.
  • Specific Heat of Air (Cp): Generally assumed to be 1005 J/kg路K.
Using the formula, the mass flow rate is calculated:
\[ m = \frac{12350 \, \text{W}}{1005 \, \text{J/kg路K} \times 10 \, ^\circ\text{C}} \approx 1.23 \, \text{kg/s} \]
Cooling Capacity
The cooling capacity signifies the total ability of a cooling system to remove heat from a designated area. It is the combination of both sensible and latent heat values required to maintain a comfortable environment.

In our classroom example, the cooling capacity necessary is the sum of all heat contributions:
  • Sensible Heat: 8300 W
  • Latent Heat: 4050 W
So, the total cooling capacity needed is 8300 W + 4050 W = 12350 W. Ensuring that cooling equipment can meet this capacity is vital for effective climate control.
Specific Heat of Air
The specific heat of air, denoted as \(Cp\), is the amount of energy required to raise the temperature of a unit mass of air by one degree Celsius (or one Kelvin). For typical calculations, the specific heat of air is assumed to be about 1005 J/kg路K.

This property is integral to thermal calculations because it links temperature changes to energy consumption. In our situation, \(Cp\) helps us find the mass flow rate, ensuring the cooling system can adequately adjust the air temperature.

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Most popular questions from this chapter

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.

Consider two concentric spheres with diameters \(12 \mathrm{~cm}\) and \(18 \mathrm{~cm}\) forming an enclosure. The view factor from the inner surface of the outer sphere to the inner sphere is (a) 0 (b) \(0.18\) (c) \(0.44\) (d) \(0.56\) (e) \(0.67\)

Consider a circular grill whose diameter is \(0.3 \mathrm{~m}\). The bottom of the grill is covered with hot coal bricks at \(950 \mathrm{~K}\), while the wire mesh on top of the grill is covered with steaks initially at \(5^{\circ} \mathrm{C}\). The distance between the coal bricks and the steaks is \(0.20 \mathrm{~m}\). Treating both the steaks and the coal bricks as blackbodies, determine the initial rate of radiation heat transfer from the coal bricks to the steaks. Also, determine the initial rate of radiation heat transfer to the steaks if the side opening of the grill is covered by aluminum foil, which can be approximated as a reradiating surface.

Two parallel disks of diameter \(D=0.6 \mathrm{~m}\) separated by \(L=0.4 \mathrm{~m}\) are located directly on top of each other. Both disks are black and are maintained at a temperature of \(450 \mathrm{~K}\). The back sides of the disks are insulated, and the environment that the disks are in can be considered to be a blackbody at \(300 \mathrm{~K}\). Determine the net rate of radiation heat transfer from the disks to the environment.

Two parallel black disks are positioned coaxially with a distance of \(0.25 \mathrm{~m}\) apart in a surrounding with a constant temperature of \(300 \mathrm{~K}\). The lower disk is \(0.2 \mathrm{~m}\) in diameter and the upper disk is \(0.4 \mathrm{~m}\) in diameter. If the lower disk is heated electrically at \(100 \mathrm{~W}\) to maintain a uniform temperature of \(500 \mathrm{~K}\), determine the temperature of the upper disk.
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