2. Energy Management and Planning MSJ0210
Exercises 1
Eduard Latõšov

3. Contents
CHP benefits
Degree days/hours

4. CHP benefits

5. CHP plant average total efficiency is 85%
CHP benefits (1)
Initial data:
CHP plant average total efficiency is 85%
and average electrical efficiency is 25%.
1. In how many times fuel consumption for
separate production is higher than for CHP
in the case if CHP production will be replaced
by separate heat (average efficiency 90%)
and power production (average efficiency
32%)?
Tasks:

6. CHP benefits (1) 1,45 times CONSUMER 25 electricity (efficiency 32%)
FUEL: 78
60 electricity
(efficiency 90%)
60 (85 -25) heat
FUEL: 67
FUEL: 145
1,45 times
FUEL: 100

7. Initial data: Tasks: CHP benefits (2)
Annual heat consumption of the Office is 200 MWh.
Annual electricity consumption is 100 MWh.
Heat is produced in boiler house (efficiency 94%) and electricity in power plant (efficiency 34%).
After installation of local CHP it was able to produce 40 MWh of electricity and 150 MWh of heat. Missing energy is produced in boiler house and power plant as previously. Total efficiency of CHP plant is 92%.
Calculate fuel consumption before and after CHP installation.
Tasks:

8. CHP benefits (2) 60 electricity (efficiency 34%) CONSUMER FUEL: 177
100 MWh
Electricity
200 MWh
Heat
100 electricity
(efficiency 34%)
FUEL: 177
FUEL: 294
50 heat
(efficiency 94%)
FUEL: 53
200 heat
(efficiency 94%)
CHP
190 heat + electricity
(efficiency 92%)
FUEL: 213
FUEL: 207
FUEL: 507
FUEL: 437

9. Degree days/hours

10. Degree days/hours
Often, the seasonal energy requirements and fuel consumption of heating and air-conditioning systems (ventilation) must be estimated in order either to come close to an optimal design, in the sense of minimizing the life cycle cost of a building, or to assure compliance with codes or standards.

11. Degree days/hours
Several methods are available for the estimation of energy requirements. These methods usually vary in complexity, in number of separate ambient conditions taken into account, and in the time increment used for calculations.

12. Degree days/hours
Simpler methods use steady-state models, require less data and provide adequate results for simple systems and applications. They are defined as steady-state methods. The most widely known among them are the degree-day or degree-hour and the variable base degree-day or degree-hour approaches. They are appropriate for a quick estimation of the heating or cooling energy consumption of a building, especially in the case the utilization of the building and the efficiency of the heating equipment can be assumed constant.

13. Degree days/hours
Number of degrees by which the hourly (daily) average outdoor temperature is below or above a base temperature.

14. Degree days/hours

15. Degree days/hours
Temperature inside the building (ts) is achieved by heating/ventilation and free heat.
~21oC tS
Free heating, internal heat gain (people, electrical equipment)
~15oC tB
Heating and ventilation
base temperature of a building is the temperature below which that building needs heating.

16. Degree days/hours kus: QN – normalized heat consumption, MWh;
Qteg – real heat consumption, MWh;
SN – normalized year degree days/hours (degree days/hours, selected in accordance to building base temperature tB );
Steg – real degree days/hours (selected for the same base temperature  tB, as SN);
C - heat consumption which does not depend on degree days/hours, MWh.

17. Degree days/hours Example 1:
In the year 2009 old apartement house heat consumption for heating and ventilation was 500 MWh.
After renovation in the year 2010 heat consumption was 510 MWh.
Hot water is prepared by electrical boilers.
Where are energy savings?

18. Degree days/hours 2009. y. Before renovation
C = 0 (heat consumption is missing)
Qteg = 500 MWh
SN = 4220 degree days
Steg = 3999 degree days
2010. y. Peale renoveerimist
C = 0 (heat consumption is missing)
Qteg = 510 MWh
SN = 4220 degree days
Steg = 4606 degree days

19. 528 MWh 458 MWh Degree days/hours 2009. y. Before renovation
C = 0 (heat consumption is missing)
Qteg = 500 MWh
SN = 4220 degree hours
Steg = 3999 degree hours
QN = 500 x 4220/3999 =
528 MWh
SAVINGS
13%
2010. y. After renovation
C = 0 (heat consumption is missing)
Qteg = 510 MWh
SN = 4220 degree hours
Steg = 4606 degree hours
QN = 500 x 4220/4606 =
458 MWh

20. Degree days/hours Example 2:
The same type of cottage is constructed in:
1. Saaremaa island. Heat consumption 100 MWh y.
2. Tartu city. Heat consumption110 MWh aastal y.
Hot domestic water production by electrical boilers.

21. Recalculation to Tartu
Degree days/hours
Saaremaa
C = 0 (no hot domestic water production)
Qteg = 100 [MWh]
Steg = 2944 [degree days] (Base temp. 15oC)
Recalculation to Tartu
conditions
100 x 3280/2944 =
111 MWh
Tartu
C = 0 (no hot domestic water production)
Qteg = 110 [MWh]
Steg = 3280 [degree days] (Base temp. 15oC)

22. 0,98 [kW/oC] Degree days/hours
Example 3:
Normalized year heat consumption 100 MWh ( kWh).
SN = 4220 degree days [oC.days] (base temperatuure = 17 oC),
or 4220 x 24 = [oC.h]
Find between outside temperatuure and heating capacity.
[kWh] / [oC.h] = 0,98 [kWh/oC.h]
0,98 [kW/oC]

23. 0,98 [kW/oC] Degree days/hours Base temperature Capacity, P, [kW]
Outside temperture
tv, [oC]

24. Degree days/hours

25. Thank you