Building Heat Balance

The principles of heating, ventilation, and air-conditioning are all used to determine the amount of ventilation, heat, or air conditioning needed to maintain the temperature inside a building. The process of determining what changes are occurring in the environment inside a structure is call a heat balance. Expressed mathematically a heat balance is:

Three possible answers can result from using the heat balance equation. (1) The amount of heat gain is larger than the heat loss. This will result is a positive (+) heat gain. In this situation, the temperature of the building will increase. (2) The

±Heat ( —— I = Total heat gain (HG) -Total heat loss (HL)

heat gain and heat loss will be the same. When this is true, the answer from the heat balance equation is zero, and the temperature within the building will remain stable. (3) The heat loss is greater than heat gain. When this is true, the answer for the equation is negative (-) and the temperature inside the building will decrease.

The four factors needed to determine heat gain and loss are: mechanical equipment heat, animal heat, heat flow, and ventilation. Mechanical equipment, such as electric motors and heat lamps, and animals put heat into the building. Heat flows from areas of high temperature to areas of low temperature. The term heat flow is used to describe the movement of heat through the components of a building. Ventilation and its effect on the environment are discussed in a previous section.

These factors usually result in either a heat gain or a heat loss, depending on the season of the year and the temperatures. For a winter heat balance, the heat gain is the sum of the animal heat and the mechanical equipment heat, and the heat losses are due to heat flow and ventilation.

±HeatW = (Animal heat + Mechanical heat) - (Heat flow + Ventilation)

For a summer heat balance, the heat gain is the sum of the animal heat, mechanical equipment heat, and heat flow.

±HeatS = (Animal heat + Mechanical heat + Heat flow) - (Ventilation)

The following problem illustrates the procedure for calculating a winter heat balance for a building. Note: in this calculation the effects of radiation are ignored.

Problem: Determine the daytime heat balance (BTU/hr) for a 30.0 ft by 60.0 ft structure with 8.0 ft walls, which houses two hundred 150.0-lb feeder pigs. The building has two, 3.0-ft by 6.6-ft pine doors, 1.0 inch thick. In addition, it has ten 24.0-inch by 42.0-inch single-pane windows. The walls are 8.0-inch lightweight concrete block with the cores filled with lightweight perlite. The ceiling is 3/8-inch plywood with 6.0 inches of low-density batt insulation between the joists. The mechanical equipment includes nine, 100 W incandescent lights that operate 10 hr per day, two, 1/3 horsepower auger motors which operate 30 min per day, and ten, 1/4 horsepower ceiling fans that operate 5 hours per day. The attic space is well ventilated and is at the same temperature as the outdoors. The building is on an insulated concrete slab. The temperatures are 70°Fdb and 60°Fwb inside and 40° Fdb and 34° Fwb outside.

Solution: Two values are required to determine a heat balance—the total heat loss and the total heat gain. Begin by determining total heat loss. For winter conditions, total heat losses are heat flow and ventilation.

Heat losses: The heat flow equation used in the previous chapter is used to determine the total heat flow into or out of a building. The total heat flow is found by calculating and summing the heat flow for each component of the building having a unique ^-value. The different ^-values in this problem include those of the walls, windows, door, ceiling, and floor. Errors can be reduced by setting up a table of the required information. Because heat flow is a function of area, temperature difference, and thermal resistance, table columns are included for this

Table 23.2. Solution for building heat flow problem.

Component

Total R value

Area (ft2)

Walls (less openings)

Windows

Doors

Ceiling

Floor

1330 70 40 1800 180*

30 30 30 30 30

5215.68.. 2307.69. . 648.64. . 2657.48. . 2432.43. . 13,261.93. .

*The perimeter (ft) is used for slab floors, not the area (ft2).

information, Table 23.2. In addition, it may be helpful to sketch components, as in the example in Chapter 22. A sketch was not used in this example because each building component has a small number of parts. Using the conduction heat flow equation:

where Q = heat flow (BTU/hr); A = area of structure with similar R value (ft2); AT = difference in temperature between inside and outside (°F); R = thermal resistance of the building component.

With a temperature difference of 30°F, the total heat loss (flow) is 13,262 BTU/hr. These numbers were calculated as follows:

r-values: Wall:

Windows:

Door:

Ceiling:

Floor:

Outside air film Concrete blocks, filled cores Inside air film Total R-value for wall

Outside air film Single-pane glass Inside air film Total R-value for windows

Outside air film Wood

Inside air film Total R-value for door

Inside air film Insulation (6-in batt) Plywood Outside air film Total R-value for ceiling

Insulated concrete

2.22

Areas:

Walls:

ft2 = (area of endwalls + area of side walls)

= 1330 ft2

ft2 = Width x Height x Number of windows

Windows:

1 ft2

144 in2

Doors:

ft2 = Width x Height x Number of doors

40 ft2

Ceiling: ft2 = Length x Width = 60.0 ft x 30.0 ft = 1,800 ft2

Perimeter: Floor:

ft = (Length x 2) + (Width x 2) = (60.0 x 2) + (30.0 x 2) = 180 ft

Building heat flow (QT):

_ 1330 x 30.0 70.0 x 30.0 40.0 x 30.0 1800 x 30.0 180 x 30.0 = 7.65 + 0.91 + 1.85 + 20.32 + 2.22 = 5215.68 ... + 2307.69... + 648.64... + 2657.48 ... + 2432.43 ... BTU

This is a wintertime heat balance so ventilation is a source of heat loss. The ventilation heat loss is determined by the ventilation rate required to remove the excess moisture from the building and the difference in temperature between the inside and the outside air. A table should be used to organize the psychrometric data, Table 23.3.

Table 23.3. Psychrometric data for problem.

Properties Inside Outside Difference

Moisture (lb/lb) 0.0088 0.0026 0.0062

Specific volume (ft3/lb) 13.6 12.6

In this problem, the building is used to house livestock. Therefore, the ventilation rate will be determined by the amount of water vapor produced by the animals. Appendix X contains values that can be used to estimate the amount of heat and moisture produced by various animals, based on weight and species.

To determine the ventilation rate, begin by finding the amount of water released into the building. A 150-lb feeder pig produces 0.219 lb of moisture per hour, Appendix X. The total water produced is:

lb H2O

hr pig hr

Next, determine the amount of air needed to remove 43.8 lb of water per hour. This is accomplished by multiplying the amount of water that each pound of air can absorb as it passes through the building by the amount of water that needs to be removed. If we assume that the air is 100% saturated when it leaves the building, then the amount of water that the air will remove is determined by the difference between the moisture content of the air entering and that of the air exiting the building. From Table 23.3, this value is 0.0062 lb of water per pound of dry air. The amount of air needed is:

lb air 1 lb air 43.8 lb water lb air

hr 0.0062 lb water hr hr

Now that the ventilation rate is known, the amount of heat loss due to ventilation can be calculated. This is accomplished by multiplying the pounds of air per hour that will be moving through the building by the difference between the amount of heat in the inside air and that in the outside air. From Table 23.3, this value is 13.5 BTU/lb of air. The heat loss due to ventilation is:

BTU lb air BTU BTU

hr hr lb air hr

The total heat loss is obtained by adding the rates for heat flow and ventilation: BTU

= Heat flow + Ventilation losses BTU

BTU BTU

hr hr

Heat gain: Next determine the total heat gain. The sources of heat are the mechanical equipment and the heat given off by the animals, Appendix X. Mechanical equipment:

Heat gain — = Mechanical heat + animal heat hr

Mecahnical heat = Lights + Motors

light watt-hr hr

Motors = ( 2 motors x 0.33 —hp—) + (10 motors x 0.25

motor motor hp-hr hp-hr\ Btu

day day hp-hr

Btu Btu

day day

hr pig hr

Heat gain = Mechanical heat + Animal heat

Btu Btu

hr hr

Heat balance: All of the information needed to complete the heat balance is available.

± Heat balance-= Total heat gain - Total heat loss hr

BTU BTU

hr hr

A heat balance with a negative number indicates that the amount of heat gain in the building is less than the amount of heat being lost from the building. When this occurs, the temperature inside the building will decrease unless additional heat is added. For this problem, 22,100 BTU/hr of heat is needed to maintain the inside temperature of the building. Note: livestock building environments are dynamic systems. The heat gain and loss is never constant. In this problem the heat balance was calculated with the assumption that the auger motors and ceiling fans were operating. The auger motors only operate 30 min a day. What happens to the heat balance when they are turned off? What is the effect of turning off some of the ceiling fans, or one of the waterers developing a leak?

0 -1

Post a comment