Chap5 space heat load calculations

AIR CONDITIONING and VENTILATION SYSTEMS
NME 515
Heating, Ventilating, and Air
Conditioning
ANALYSIS and DESIGN
Engr. Charlton Inao, ME
TESDA TVET NC Level 3 RAC-CRE Certified
17/19/2020 Eng Charlton Inao

Space Heat Load
Engr. Charlton Inao, ME
TESDA TVET NC Level 3 RAC-
CRE Certified
2
Reference: Heating, Ventilating, and Air Conditioning
ANALYSIS and DESIGN by FC Mc Quiston and JD Parker

Space Heat Load
3
There are two kinds of heat losses: (1) the heat
transmitted through the walls, ceiling, floor, glass, or
other surfaces and (2) the heat required to warm
outdoor air entering the space. Transient analyses are
often used to study the actual energy requirements of a
structure in simulation studies. In such cases solar
effects and internal heat gains are taken into account.

7-1 OUTDOOR DESIGN CONDITIONS
4
The outdoor design temperature should generally be
the 97
1
2
percent value as specified by ASHRAE Energy
Standards. If the structure is of lightweight construction
(low heat capacity), is poorly insulated. or has
considerable glass or space temperature control is
critical, however, the 99 percent values should be
considered. The designer must remember that should
the outdoor temperature fall below the design value for
some extended period, the indoor temperature may do
likewise.

7-2 INDOOR DESIGN CONDITIONS
5
The indoor design temperature should be kept relatively
low so that the heating equipment will not be oversized.
A design temperature of 70 F or 22 C is commonly used
with relative humidity less than or equal to 30 percent.

6
Even properly sized equipment operates under partial
load. at reduced efficiency, most of the time; therefore,
any oversizing aggravates this condition and lowers the
overall system efficiency. The temperature in an
unheated space is needed to compute· the heat loss
and may be estimated by assuming steady-state heat
transfer and making an energy balance on the space.

7
EXAMPLE 7-1
Estimate the temperature in the crawl space of Fig. 7-1.
The conductance for the floor is 0.20 Btu/(hr-ft2-F)
including the air film on each side. The conductance for
the foundation wall including the insulation and inside
and outside air film resistances is 0.12 Btu/(hr-ft2-F).
Assume an indoor temperature of 70 F and an outdoor
temperature of 10 Fat a location where the degree days
are about 3000. The building dimensions are 50 × 75 ft.

9
SOLUTION
The first step is to make an energy balance on the crawl
space. Heat will be gained through the floor and heat
will be lost through the foundation wall and through the
ground around the perimeter much like a floor slab.

10
SOLUTION
Infiltration of outdoor air will also represent a heat loss
but that will be neglected in this example.

12
Now the area of the floor is 50 × 75 = 3750 ft2 and
assuming that the foundation wall averages a height of
2 ft, the area of the foundation wall is 2 (2 × 50) + (2
× 75) = 500 ft2. The perimeter of the building is (2 ×
50) + (2 × 75) = 250 ft. Referring to Table 5-12, the
heat-loss coefficient is estimated to be 0.8 Btu/(hr-ft-F).
Then

13
Now the area of the floor is 50 × 75 = 3750 ft2 and
assuming that the foundation wall averages a height of
2 ft, the area of the foundation wall is 2 (2 × 50) + (2
× 75) = 500 ft2. The perimeter of the building is (2 ×
50) + (2 × 75) = 250 ft. Referring to Table 5-12, the
heat-loss coefficient is estimated to be 0.8 Btu/(hr-ft-F).
Then

14
If the infiltration had been considered, the crawl-space
temperature would be lower. Many crawl spaces are
ventilated to prevent moisture problems. and infiltration
would be significant even when the vents are closed.

15
EXAMPLE 7-2
Estimate the temperature in the unheated room shown
in Fig. 7-2. The structure is built on a slab. The exterior
walls of the unheated room have an- overall heat-
transfer coefficient- of 0.2 Btu/(hr-ft2-F), the ceiling-
attic-roof combination has an overall coefficient of 0.07
Btu/(hr-ft2-F), and the interior walls have an overall
coefficient of 0.06 Btu/(hr-ft2-F). Inside and outdoor
design temperatures are 72 F and - 5 F, respectively.

16
SOLUTION
The heat transferred from the heated to the unheated
room has a single path, neglecting the door, and may
be represented as

17
SOLUTION
The heat transferred from the unheated room to the
outdoor air has parallel paths through the ceiling and
walls (neglecting the door and floor):

18
SOLUTION
The heat loss to the floor can be neglected because of
the anticipated low temperature in the room. Equation
7-2 may be written
where

20
By combining Eqs. 7-1 and 7-3. we obtain

21
Then from Eq. 7-6
The assumption of negligible heat transfer through the
slab is justified by the result.

7-3 TRANSMISSION HEAT LOSSES
22
The heat transferred through walls, ceiling, roof,
window glass, floors, and doors is all sensible heat
transfer, referred to as transmission heat loss and
computed from

7-4 INFILTRATION
23
All structures have some air leakage or infiltration. This
means a heat loss because the cold dry outdoor air
must be heated to the inside design temperature and
moisture must be added to increase the humidity to the
design value. The sensible heat required (to increase
the temperature) is given by

7-4 INFILTRATION
24
Infiltration is usually estimated on the basis of volume
flow rate at outdoor conditions. Equation 7-8 then
becomes

7-4 INFILTRATION
25
The latent heat required to humidify the air is given by

7-4 INFILTRATION
26
In terms of volume flow rate of air, Eq. 7-9 becomes

7-4 INFILTRATION
27
Various methods are used in estimating air infiltration in
building structures. In one method the estimate is
based on the characteristics of the windows, walls, and
doors and the pressure difference between inside and
outside. This is known as the crack method because of
the cracks around window sash and doors. The other
approach is the air-change method which is based on
an assumed number of air changes per hour based on
experience.

Air-change Method
28
Experienced engineers will often simply make an
assumption of the number of air-changes per hour
(ACH) that a building will experience based on their
appraisal of the building type, construction, and use.
The range will usually be from 0.5 ACH (very low) to
2.0 ACH (very high). The infiltration rate is related to
ACH and space volume as follows:

Air-change Method

Crack Method
30
Outdoor air infiltrates the indoor space through cracks
around doors, windows, lighting fixtures, and joints
between walls and floor and even through the building
material itself.

Crack Method
31
The pressure difference of Eq. 7-11 results from three
different effects:
All of the pressure differences are positive when each
causes flow of air to the inside of the building.

Crack Method

Crack Method
33
The pressure difference due to the wind results from an
increase or decrease in air velocity and is described by

Crack Method

Crack Method
35
Finally, Eq. 7-13a may be written

Crack Method

Crack Method
37
When larger openings predominate in the lower
portion of the building, the neutral pressure
level will be lowered., Similarly, the neutral
pressure level will be raised by larger openings
in the upper portion of the building. Normally
the larger openings will occur in the lower part
of the building because of doors.

Crack Method
38
The theoretical pressure difference with no internal
separations is given by

Crack Method
39
The floors in a conventional building offer resistance to
vertical air flow. Furthermore, this resistance varies
depending on how stairwells and elevator shafts are
sealed. When the resistance can be assumed equal for
each floor, a single correction, called the draft
coefficient, can be used to relate the actual pressure
difference ∆Ps, to the theoretical value ∆Pst:

Crack Method
40
The flow of air from floor to floor causes a decrease in
pressure at each floor; therefore, ∆Ps will be less than
∆Pst and Cd is less than one. Using the draft coefficient,
Eq. 7-15 becomes

Crack Method

Crack
Method

• Calculation Aids
44
The use of storm sash and storm doors is common. The
addition of a storm sash with crack length and a K value
equal to the prime window reduces infiltration by about
35 percent.

45
Single-bank doors open directly into the space;
however, there may be two or more doors at one
location.
Vestibule-type doors are best characterized as two
doors in series so as to form an air lock between them.
These doors often appear as two pairs of doors in
series, which amounts to two vestibule-type doors.

53
EXAMPLE 7-3
A 12-story office building is 120 ft tall with plan
dimensions of 120 × 80 ft. The structure is of
conventional curtain wan construction with all windows
fixed in place. There are double vestibule-type doors on
all four sides. Under winter design conditions, a wind of
15 mph blows normal to one of the long dimensions.
Estimate the pressure differences for all walls for the
first and twelfth floors. Consider only wind and stack
effects. The indoor-outdoor temperature difference is 60
F.

54
SOLUTION
The pressure difference for each effect must first be computed
and then combined to find the total. First consider the wind:
Equation 7-13b expresses the wind pressure difference where
the pressure coefficients may be obtained from Fig. 7-5 for a
normal wind. Then using standard sea level density:

55
SOLUTION
The wind effect will be assumed constant with respect
to height.

56
SOLUTION
Then for the first floor, h = 50 ft and

57
SOLUTION
For the twelfth floor, h = 70 ft and

58
SOLUTION
The negative sign indicates that the pressure is greater
inside the building than on the outside. The pressure
differences may now be summarized for each side
where ∆P = (∆Pw + ∆Ps) in. wg.

59
These results show that air will tend to infiltrate on
most floors on the windward wall. Infiltration will occur
on about the lower four floors on the leeward wall. All
other surfaces will have exfiltration.

60
EXAMPLE 7-4
Estimate the infiltration rate for the leeward doors of
Example 7-3. The doors have
1
8
-in. cracks and the traffic
rate is low except at 5:00 P.M., when the traffic rate is
350 people per hour per door for a short time.

61
SOLUTION
This problem is solved in two steps to account for crack
leakage and infiltration due to traffic. For the design
condition, the effect of traffic is negligible; however, it is
of interest to compute this component for 5:00 P.M.
Figure 7-9 pertains to crack leakage for commercial
swinging doors.

62
SOLUTION
For a pressure difference of 0.041 in. wg. and
1
8
-in.
cracks, the leakage rate is 8 cfm/ft. The crack length for
standard double swinging doors is

63
SOLUTION
Vestibule-type doors will tend to decrease the
infiltration rate somewhat like storm sash or a storm
door. Assume a 30 percent reduction; then

64
SOLUTION
Figures 7-10 and 7-11 are used to estimate the
infiltration due to traffic. The traffic coefficient C is read
from Fig. 7-11 for 350 people per hour and for
vestibule-type doors at 5000. Then, from Fig. 1-10 at a
pressure difference of 0.041 in. wg,

65
SOLUTION
A part of the crack leakage should be added to this;
however, this is somewhat academic. Care should be
exercised in including the traffic infiltration in the design
heat load. It will usually be a short-term effect.

66
EXAMPLE 7-5
Estimate the leakage rate for the twelfth floor of the
building in Example 1-3. Neglect the roof.

67
SOLUTION
Referring to the pressure differences computed in
Example 7-3, it is obvious that the leakage will be from
the inside out on the twelfth floor. Therefore, a great
deal of air must be entering the space from the
stairwells and elevator shafts. Because the twelfth floor
has no movable openings, except to the roof, all
leakage is assumed to be through the walls. Figure 7-8
gives data for this case where K = 0.66 for conventional
construction

68
SOLUTION

69
SOLUTION
where the negative sign indicates that the flow is from
the inside out. The net leakage flow of 630 cfm entered
the building at other locations where the heat loss
should be assigned.

70
EXAMPLE 7-6
A single-story building is oriented so that a 15-mph
wind approaches 45 degrees to the windward sides.
There are 120 ft of crack: for the windows and 20 ft of
crack for a door on the windward side. The sides have
130 ft of window cracks and 18 ft of door crack. All
windows and doors are average fitting. Estimate the
infiltration.

71
SOLUTION
The major portion of the infiltration for this kind of
building will be through the cracks. It is approximately
true that air will enter on the sides and flow out with
most of the heat loss imposed on the rooms where the
air enters. As suggested, we will use double the total
crack length and assume that most of the air leaves
through the ceiling area with a pressure difference
computed for a quartering wind on the windward side.
Using Eq. 7-13b, Fig. 7-4, and Table 7-1.

72
SOLUTION
where standard sea level air density bas been used.
From Tables 7-2 and 7-3. the K factor for the windows
and doors is read as 2.0. Then from Fig. 7-3, the
leakage per foot of crack is

73
SOLUTION
and the total infiltration for the space is

7-5 HEAT LOSSES FROM AIR DUCTS
74
The losses of a duct system can be considerable when
the ducts are not in the conditioned space. Proper
insulation will reduce these losses but cannot
completely eliminate them. The loss may be estimated
using die following relation:

75
EXAMPLE 7-7
Estimate the heat loss from 1000 cfm of air at 120 Ft
flowing in a l6-in. duct 25 ft in length. The duct has 1
in. of fibrous glass insulation and the overall coefficient
is 0.2 Btu/(hr-ft2-F). The environment temperature is 12
F.

76
SOLUTION
Equation 7-18 will be used to estimate the heat loss
assuming that the mean temperature difference is given
approximately by

77
SOLUTION
The surface area of the duct is

78
SOLUTION
The temperature of the air leaving the duct may be
computed from

79
SOLUTION
Although insulation and surface drastically reduces the
heat loss, the magnitude of the temperature difference
and surface area must be considered in each case.

7-6 AUXILLARY HEAT SOURCES
80
The heat energy supplied by people, lights, motors, and
machinery may be estimated. but any actual allowance
for these heat sources requires careful consideration.
People may not occupy certain spaces in the evenings,
or weekends, and during other periods, but these
spaces must generally be heated to a reasonably
comfortable temperature prior to occupancy.

7-6 AUXILLARY HEAT SOURCES

7-7 INTERMITTENTLY HEATED
STRUCTURES
82
When a structure is not heated on a continuous basis,
the beating equipment capacity may have to be
enlarged to assure that the temperature can be raised
to a comfortable level within a reasonable period of
time. The heat capacity of the building and occupant
comfort are important factors when considering the use
of intermittent beating.

7-8 SUPPLY AIR FOR SPACE HEATING
83
There are many cases when the air quantity is
conveniently computed from

7-8 SUPPLY AIR FOR SPACE HEATING
84
Obviously, the air quantity for each room should be
apportioned according to the heating load for that
space. Then

PROBLEMS
85
7-1. Select an indoor design relative humidity for structures located
in the cities given below. Assume an indoor design dry bulb
temperature from Table C-L. Windows in the building are double
glass. aluminum frame with thermal break. Other external surfaces
are well insulated.
(a) Sioux City, Iowa
(b) Atlanta, Georgia
(c) Syracuse, New York
(d) Denver, Colorado
(e) San Francisco. California
(f) Bismarck, North Dakota
(g) Rapid City, South Dakota7/19/2020 Eng Charlton Inao

PROBLEMS
86
7-2. Estimate the temperature in the crawl space of a
building with floor plan dimensions of 30 ×60 ft. The
concrete foundation has an average height of 2 ft and
the wall is 6-in. thick. Use winter design conditions for
Oklahoma City, Oklahoma.

PROBLEMS
87
7-3. Estimate the temperature in the knee space shown
in Fig. 7-12. The roof is equivalent to 0.5 in. (25 mm) of
wood on 2 × 6-in. rafters on 24-in. (0.6-01) centers.
The walls are 2 × 4-in. studs on 16-in., (0.4-m) centers
with 3.5 in. (90 mm) of insulation. The joist spaces all
have 6 in. (150 mm) of insulation. Inside and outside
temperatures are 70 F (21 C) and 10 F (-12 C).

PROBLEMS

PROBLEMS
89
7-4. Consider the knee space shown in Fig. 7-12. The
vertical dimension is 8 ft, the horizontal dimension is 3
ft. and the space is 20 ft long. The walls and roof
surrounding the space all have an overall heat-transfer
coefficient of about 0.09 Btu/(hr-ft2-F). Assuming an
outdoor temperature of 2 F and an indoor temperature
of 68 F, make a recommendation concerning the
placement of water pipes in the knee space.

PROBLEMS
90
7-5. Estimate the temperature in an unheated basement
that is completely below ground level with heated space
above. Assume no insulation and dimension of 20 × 20
× 7 ft. The basement is located in Lincoln, Nebraska.
Use standard design conditions.

PROBLEMS
91
7-6. Rework Example 7-1 assuming that the crawl space
experiences an infiltration rate of 0.25 air changes per
hour.

PROBLEMS
92
7-7. Rework Example 7-2 assuming an infiltration rate
of 0.2 air changes per hour.

PROBLEMS
93
7-8. A large single-story business office is fitted with
nine loose-fitting, double-hung woodsash windows 3 ft
wide by 6 ft high. If the outside wind is 15 mph at a
temperature of -10 F, what is the percent reduction in
sensible heat loss if the windows are weather stripped?
Assume an inside temperature of 70 F. Base your
solution on a quartering wind.

PROBLEMS
94
7-9. Using the crack method compute the infiltration for
a tight-fitting swinging door that is used occasionally.
The door bas dimensions of 0.9 × 2.0 m and is on the
windward side of a bouse exposed to a 9 m/s wind.
Neglect internal pressurization and stack effect.

PROBLEMS
95
7-10. A room in a single-story building has two 3 × 5-ft
wood, double-hung windows of average fit that are not
weather stripped. The wind is 20 mph and normal to
the wall with negligible pressurization of the room. Find
the infiltration rate assuming that the entire crack is
admitting air.

PROBLEMS
96
7-11. Compute the design infiltration rate and beat gain
for the bouse described in Problem 7-16 assuming an
orientation normal to a is 15 mph wind. The windows
and doors are tight fitting.

PROBLEMS
97
7-12. Refer to Example 7-3. (a) Estimate the total
pressure difference for each wall for the third and ninth
floors. (b) Using design conditions for Denver, Colorado,
estimate the heat load due to infiltration for the third
floors and ninth floors.

PROBLEMS
98
7-13. Refer to Examples 7-3 and 7-4. (a) Estimate the
infiltration rates for the windward and side doors for a
low traffic rate. (b) Estimate the curtain wall infiltration
for the first floor. (c) Compute the heat load due to
infiltration for the first floor if the building is located in
Cleveland. Ohio.

PROBLEMS
99
7-14. A 20-story office building has plan dimensions of 100 × 60 ft
and is oriented-at 45 degrees to a 20 mph wind. All windows are
fixed in place. There are double vestibule-type swinging doors on
the 6O-ft walls. The walls are average curtain wall construction and
the doors have about i-in. cracks. (a) Compute the pressure
differences for each wall due to wind and stack effect for the first,
fifth, fifteenth, and twentieth floors. (b) Plot pressure difference
versus height for each wall and estimate which surfaces have
infiltration and exfiltration. (c) Compute the total infiltration rate for
the first floor assuming a very low traffic rate. Compute the
infiltration rate for the fifteenth floor. (e) Compute the infiltration
rate for the twentieth floor. Neglect any leakage through the roof.

PROBLEMS
100
7-15. Refer to Problem 7-14. (a) Compute the heat gain due to
infiltration for the first floor with the building located in Dallas,
Texas. (b) Compute the heat gain due to infiltration for the
fifteenth floor. (c) what is the heat gain due to infiltration for the
twentieth floor?

PROBLEMS
101
7-17. Rework Problem 7-16 for Lincoln, Nebraska. Include
infiltration in the analysis.

PROBLEMS
102
7-18. Consider a building located in Syracuse, New York. Using
ASHRAE Standard recommendations, compute the transmission
loss for (a) a wall like Table 5-7a, case 2, which is 3 × 4 m with a
window of 1 m2 area, (b) a single pane window with aluminum
frame and no thermal break, (c) a roof-ceiling combination that is
like Table 5-7b, case 1, with an area of 16 m2.

PROBLEMS

PROBLEMS
105
7-20. A small commercial building has a computed heat
load of 200,000 Btu/hr sensible and 25,000 Btu/hr
latent. Assuming a 50 F temperature rise for the heating
unit, compute the quantity of air to be supplied by the
unit using the following methods: (a) Use a
psychrometric chart with room conditions of 70F and 30
percent relative humidity. (b) Calculate the air quantity
based on the sensible heat transfer.

Chap5 space heat load calculations

More Related Content

What's hot (20)

Similar to Chap5 space heat load calculations (20)

More from Charlton Inao (20)

Recently uploaded (20)

Chap5 space heat load calculations