SIMPLIFIED ANNUAL ENERGY

ESTIMATION METHODS

Hourly analysis for peak

design heating and cooling Sizing HVAC equipment
load calculations

Annual energy calculations Energy Consumption

Energy Audits

Repeating Quicker
hourly methods
analysis

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The yearly energy consumption Qyr is needed to evaluate the
operating cost.

Consumption Qyr is the time integral of the instantaneous

consumption over the heating (or cooling season); the
instantaneous consumption is the instantaneous heating load
Qh divided by the instantaneous efficiency ηh of the heating (or
cooling) equipment.

• degree-day methods

• bin methods

Degree-day methods are the simplest and these are appropriate if the
building is operated in the same manner (for example, the indoor
thermostat is set at a constant value throughout the year) and if the
efficiency of the HVAC equipment can be considered constant for all the
hours of operation over the year or the season.

For situations where efficiency or conditions of utilization vary with

outdoor temperature, one can calculate the consumption for different
values or bins of the outdoor temperature and multiply it by the
corresponding number of hours over the year or season. This is the bin
methods.

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One needs the value of outdoor dry-bulb air temperature To
below which heating becomes necessary (or above which
cooling is needed); that is called the “balance-point
temperature”

The severity of a climate can be characterized concisely in

terms of degree-days.

Degree-Day Method

Balance-Point Temperature

Consider a small building in a cold location maintained at an

indoor air temperature Ti with “free” heat gains Qgain
(from the sun, occupants, lights, etc.).

The balance-point temperature Tbal of this building is defined

as that value of the outdoor dry-bulb temperature To where,
for the specified value of Ti, the total heat loss is equal to the
free heat gain.

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 If the peak winter design load of the building is known, it can be
conveniently determined from design temperatures:

The balance-point temperature is

Tbal is the outdoor dry-bulb temperature above which cooling is

required and below which heating is required

Heating Degree-Day Method

Heating is needed only when To drops below Tbal.

The rate of energy consumption of the heating system is

Plus sign
where ηh is the efficiency of the heating system indicates that
only positive
numbers are
The annual heating consumption is considered.

Note that ηh, Tbal, and Ktot are constant for the simplification.

The integral of the temperature difference is approximated by a sum of

averages over short-time intervals (daily or hourly), and the result is called
“degree-days” or “degree-hours.”

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 If daily average values of outdoor temperature To are used for evaluating
the integral, one obtains the degree-days for heating HDD(Tbal) as

with dimensions of K · days

For shorter time intervals

In terms of degree-days, the annual

heating consumption is

Degree-days or degree-hours for a balance-point temperature of 18°C

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Isıtma derece-gün bölgelerinin

tespitinde 18 C denge noktası
sıcaklığındaki derece-gün
değerleri esas alınmıştır.
Başlangıçta derece-gün değerleri
artan şekilde sıralanmıştır. Daha
sonra aşağıdaki 1000 C.gün
aralıklarına göre bölgeler
oluşturulmuştur.

I. Bölge ≤ 1250
1251≤ II. Bölge ≤2250
2251≤ III. Bölge ≤3250
3251≤ IV. Bölge ≤4250
V. Bölge ≥4251

ULIBTK’07 16. Ulusal Isı Bilimi ve Tekniği Kongresi, 30

Mayıs-2 Haziran 2007, KAYSERİ TÜRKİYE İÇİN ISITMA
VE SOĞUTMA DERECE-GÜN BÖLGELERİ Hüsamettin
BULUT , Orhan BÜYÜKALACA, Tuncay YILMAZ

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Example: Annual Heating Energy

Find the annual heating bill for a house in Ankara under the
following conditions.

Given: Heat transmission coefficient Ktot = 205 W/K

Heat gain = 569 W
Indoor temperature Ti = 21.1°C
Efficiency of heating system ηh = 0.75
Fuel price $8/GJ

Assumptions: Ktot , , Ti , and ηh are constant.

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The balance-point temperature:

= 205 W/ K x 2677 K × days X 24 h/day x 3600 s/h / 0.75

= 47.415 GJ / 0.75 = 63.22 GJ

The annual cost is 63.22 GJ × $8/GJ = $ 505.8 / year.

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Cooling Degree-Day Method

The numerical values of degree-days for cooling (CDD) are generally

different from HDD because Ti, Qgain, and Ktot can be different.

Annual heating and cooling degree-days, if based on the same Tbal, are
related by

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 Outdoor average daily
temperatures To for the months
of March (a mild month) and
July (a very hot month) are
plotted along with the 65°F
line as shown. The hatched
areas are indicative of the
CDD (65°F) values for the 2
months.

For the month of March, there

are days when To < 65°F, and
such days do not contribute to
the CDD value, i.e., no cool-
ing is necessary and the air
conditioner does not operate.
During the month of July, we
note that To > 65°F for all the
days, and so the air
conditioner has to run
continuously.

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If an air conditioner operates continously the CDD equation can

be simplified to

where Nm is the number of days in the month.

The annual cooling energy use is

Not precise due to

• Opening windows (occupant behaviour)
• İncreasing ventilation rate in industrial
buildings

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 Therefore, the air conditioner is needed only when the outdoor
temperature exceeds the threshold Tmax.

Ktot (closed window) is

replaced by Kmax

The solid line is the load

with open windows or
increased ventilation; the
dashed line is the load if
Ktot is kept constant.

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The annual cooling load for this mode is the sum of the cooling load
during the days when To > Tmax with windows closed plus that needed to
cool down the building from Tmax to Tbal during these same days:

where CDD(Tmax) is the cooling degree-days for base Tmax and Nmax is the
number of days during the season when To rises above Tmax.

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 Example: Annual Cooling Energy

Estimate the annual cooling degree-days in New York for a base of

Tbal = 23.1°C (73.6°F) and for a base of Tbal = 18.1°C (64.7°F).

Find: CDD(Tbal)

Lookup values: To,yr = 12.5°C, from the weather data

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HDD(23.1°C) = 4080 K × days

HDD(18.1°C) » 2778 K × days

CDD(23.1°C) = HDD(23.1°C) - 365 days x [(23.1 - 12.5) K] = 208 K × days

Likewise for 18.1°C,

CDD(18.1°C) = HDD(18.1°C) - 365 days x [(18.1 - 12.5) K] = 734 K × days

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Example: Effect of Window Opening on Cooling Energy

Estimate the annual cooling load, with and without a window opening, for
a house in New York under the following conditions:

Given: Heat transmission coefficient Ktot= 205 W/K with closed windows at
0.5 air change/h

Kcond = 145 W/K due to conduction alone

Heat gains Qgain = 1200 W
Ti = 24.0°C
Average air change rate with open windows = 10 air changes/h

Find: Qc,yr

Assumptions: Ktot, Kmax, Qgain, and Ti are constant.

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The balance-point temperature for cooling is

From previous example, CDD(Tbal ) = 734 K × days

If windows were always closed, the annual cooling load would be

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If windows are opened for ventilative cooling, we first have to find Kmax.
Noting that at 0.5 air change/h, the contribution of the air change is

Hence, at 10 air changes/h this term must be multiplied by 10/0.5,

giving a value

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The required number of cooling degree-days was calculated in the

previous Example as

If Nmax = 56.5 days

It is found from the graph given in the previous example. Outside air
temperature greater than 24.2 C gives 1004 hours. That is approximately
equal to the 56.5 days.

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By opening windows the energy consumption for cooling decreases
from 13 GJ to 8.7 GJ.

That is the reason why actual residental cooling loads are highly
sensetive to occupant behaviour.

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Bin Method

There are problems in the application of degree day method.

• Effect of solar loads

• The efficiency of a heat pump and directly related to the To
• The efficiency of the HVAC equipment indirectly related to the To
especially for the chillers and boilers
• Occupancy
• Indoor temperature
• Ventilation rate

If different temperature intervals and time periods are evaluated

separately a steady state calculation (bin method) can yield good
results.

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The consumption is calculated for several values of the outdoor
temperature To and multiplied by the number of hours Nbin in the
temperature interval (or bin) centered on that temperature:

The plus subscript on the parentheses indicates that only positive

values are to be counted; no heating is needed when To is above Tbal.
This equation is evaluated for each bin, and the total consumption is
the sum of the Qbin values over all bins.

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In the United States, the necessary data, called “bin data,” have been
generated in temperature intervals of 5°F (2.8°C) A bin data for
Gaziantep with 3°C temperature difference was given below.

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MIDTERM

1. Bin Method to Determine Annual Heating Needs of an Office Building

Find the heating energy consumption of a simple commercial building in Gaziantep

in December, January, and February for the following conditions:

Given: Occupied 7 days per week, 8:00 to 18:00; unoccupied otherwise

Ktot = 2.0 kW/K occupied
= 1.0 kW/K unoccupied (reduced because ventilation system shut off)

Efficiency of heating system η = 0.94

2. Find the heating energy consumption for the same building with the
Degree-day method

3. Compare the results.

Last submission date and time:

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