SHeat m,d,hþ1 ¼ ε H Á SHeat m,d,h þ
X
i
IHeat i,m,d,h
À
X
u
OHeat m,d,h,u 8m, d, h if h
6 ¼
0 24
0
ð6:26Þ
SHeat m,d,1þ1 ¼ ε H Á SHeat m,d,24 þ
X
i
IHeat i,m,d,24
À
X
u
OHeat m,d,24,u 8m, d, if d
6 ¼
0 MDays
0
ð6:27Þ
SHeat mþ1,1,1 ¼ ε H Á SHeat m,MDays m ,24 þ
X
i
IHeat i,m,MDays m ,24
À
X
u
OHeat m,MDays m ,24,u 8m if m
6 ¼
0 12
0
ð6:28Þ
OHeat m,d,h,u ¼ 0 8m, d, h if u 2 electricity only
f
g
ð6:29Þ
SHeat m,d,h HSmax 8m, d, h
ð6:30Þ
SHeat 1,1,1 ¼ 0
ð6:31Þ
SHeat 12,d,24 ¼ 0
ð6:32Þ
OHeat 1,1,1,u ¼ 0
ð6:33Þ
δ u Á OHeat m,d,h,u þ
X
i
γ i,u Á RHeat i,m,d,h,u
X
i
α i Á γ i,u Á OPerate i,m,d,h Á F max p i 8m, d, h, u
ð6:34Þ
X
u
OHeat m,d,h,u SHeat m,d,h 8m, d, h
ð6:35Þ
Constraints are also needed to ensure appropriate operation of the battery.
Equation (6.36) shows the electricity inventory balance constraint. It states that the
total amount of electricity stored at the beginning of an hour is equal to the
non-dissipated electricity stored at the beginning of the previous hour, plus the
electricity charged during the hour, minus stored electricity discharged to meet
end-use loads during the hour. Equation (6.37) is the specific case of Eq. (6.36)
for the first hour of the day. Here, the total electricity stored at the beginning of the
first hour of the day is equal to the non-dissipated electricity stored at the beginning
of the last hour of the previous day, plus the inflow and minus the outflow of
electricity during the hour. Equation (6.38) is similar to Eq. (6.37). Equation
130
H. Ren and W. Zhou