DC power system sizing

We can size the dc power system using three method

 Ericson method: - Using the discharge power per cell for x hours to 1.8V
stop voltage. This method need to have the constant power discharge( watts per
cell) to 1.8V data table of the battery type to be configured

The method follow 7 procedure to size the battery and rectifier quantity

1. Sum up all the DC load wattages (PLoad) including future expansion

if any
2. Choose load battery backup time (user defined), Tbackup
3. Select the battery capacity from the product information sheet
provided and corresponding discharge power per cell (watt/cell)
4. Then, determine required number of battery strings/groups
required (NBB)
𝑃
𝑁𝐵𝐵 = 𝐿𝑜𝑎𝑑⁄[𝑁
𝑐𝑒𝑙𝑙𝑠/𝑠𝑡𝑟𝑖𝑛𝑔 ∗ 𝑃𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒𝑑/𝑐𝑒𝑙𝑙 ∗ 𝐷𝑂𝐷]

𝑵𝑩𝑩 = Number of battery bank required to handle the load for the user-defined time
𝑷𝑳𝒐𝒂𝒅 =The total site load
𝑵𝒄𝒆𝒍𝒍𝒔/𝒔𝒕𝒓𝒊𝒏𝒈 =the number of battery cell per bank
𝑷𝒅𝒊𝒔𝒄𝒉𝒂𝒓𝒈𝒆𝒅/𝒄𝒆𝒍𝒍 =the amount of constant power that single battery cell can supply for the defined time
DOD= the percent of the battery capacity that can be used without affecting the nominal battery cycle recommended by
company
5. Determine watts required power for charging the batteries (PCharging)
𝑃𝐶ℎ𝑎𝑟𝑔𝑖𝑛𝑔 = 𝐾𝐶 ∗ 𝐶𝑇𝐵𝑎𝑡𝑡 −𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒 ∗ 𝑉𝑛
Where, Kc- Charging Coefficient & Vn- Nominal system voltage

6. Determine total power output required (PTOTAL)

𝑃𝑇𝑂𝑇𝐴𝐿 = 𝑃𝐿𝑜𝑎𝑑 + 𝑃𝐶ℎ𝑎𝑟𝑔𝑖𝑛𝑔
7. Determine number of rectifier modules required (NR modules) by using N+1
scenario
𝑃
𝑁𝑅_𝑀𝑜𝑑𝑢𝑙𝑒𝑠 = 𝑇𝑜𝑡𝑎𝑙⁄𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 (𝑤𝑎𝑡𝑡)
/𝑚𝑜𝑑𝑢𝑙𝑒

Example: Power system sizing using watts per cell method

Size a DC power system for a BTS site with network elements of total
power consumption 3000 watt by considering backup time of 3 hours.
Select 190AH/2V battery from the above watts per cell constant power
discharge. Note: assume module capacity is 3000 watt & use N+1
scenario for rectifier module quantity selection and 80 % DOD & Kc=0.1.
Solution

1 Sum up all the DC load wattages (PLoad) including future expansion if any
=3000 watts (given)
2 Choose load battery backup time (user defined), Tbackup
= 3 hours (given)
3 Select the battery capacity from the product information sheet provided and
corresponding discharge power per cell (watt/cell)
= 110.5 Watts (referenced from the above table)

4 Then, determine required number of battery strings/groups required (N BB)

𝑃
𝑁𝐵𝐵 = 𝐿𝑜𝑎𝑑⁄[𝑁
𝑐𝑒𝑙𝑙𝑠/𝑠𝑡𝑟𝑖𝑛𝑔 ∗ 𝑃𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒𝑑/𝑐𝑒𝑙𝑙 ∗ 𝐷𝑂𝐷]

Nbb=3000 / (24*110.5*0.8)
=3000/2121.6
=1.41
Since we have 190Ah/2v battery on hand, we need two bank of 190Ah/2v battery to handle the load for 3 hours
 5 Determine watts required power for charging the batteries (PCharging)
𝑃𝐶ℎ𝑎𝑟𝑔𝑖𝑛𝑔 = 𝐾𝐶 ∗ 𝐶𝑇𝐵𝑎𝑡𝑡 −𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒 ∗ 𝑉𝑛

Pcharging = 0.1(2*190Ah*0.8)*48
=1459.2 Watt
Note:- the unit of Kc is per hour and the 0.8 is inserted into the equation to match the discharged power of the
battery which is the same amount of the DOD

6. Determine total power output required (PTOTAL)

𝑃𝑇𝑂𝑇𝐴𝐿 = 𝑃𝐿𝑜𝑎𝑑 + 𝑃𝐶ℎ𝑎𝑟𝑔𝑖𝑛𝑔

Ptotal = 3000 Watt + 1459.2 Watt

= 4459.2 Watts
7. Determine number of rectifier modules required (NR modules) by using N+1 scenario
𝑃𝑇𝑜𝑡𝑎𝑙
𝑁𝑅_𝑀𝑜𝑑𝑢𝑙𝑒𝑠 = ⁄𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 (𝑤𝑎𝑡𝑡)
/𝑚𝑜𝑑𝑢𝑙𝑒

NR_modules =4459.2watt/3000watt
= 1.49
We need two rectifier modules to charge and handle the load and the number of rectifier module
to be configured for the site is 3 *3000watt using the N+1 scenario
Huawei Method: - The method uses the Discharge Time and Discharge
Adjustment Coefficient.
Note the amount of usable power from the battery capacity depends on the discharge rate.

Table: Relation of Discharge Time and Discharge Adjustment Coefficient

Discharge
0.5 1 1.5 2 3 4 5 6 7 8 9 10 12
Time (h)

η 0.4 0.55 0.63 0.7 0.75 0.8 0.85 0.9 0.92 0.95 0.98 1 1.01

 Using The method we can size the battery capacity and rectifier modules quantity
in 6 procedure
Method 2-1: Discharge Time and Discharge Adjustment Coefficient Method
1. Determine max. power of the DC-powered load (Pmax) in the unit of watt
2. Choose load battery backup time (user defined), Tbackup
3. Determine battery capacity (C)
[𝑃𝑚𝑎𝑥 ∗ 𝑇𝑏𝑎𝑐𝑘𝑢𝑝 ]
C= ⁄[𝐾
𝑎𝑔𝑖𝑛𝑔 ∗ 𝐾𝑣 ∗ 𝑉𝑛 ∗ 𝑛 ∗ 𝜂]

Where,

𝐊 𝐚𝐠𝐢𝐧𝐠 - Aging coefficient=0.8

𝐊 𝐯 - Voltage coefficient=0.96

𝐕𝐧 - Nominal system voltage

n- Number of battery units (optional, generally n = 1, 2, or 4)

𝛈 – Battery capacity coefficient, which depends on Tbackup

4. Calculate the maximum battery charge power (Pcmax)

𝑃𝑐𝑚𝑎𝑥 = 𝑉𝑒𝑞𝑢𝑎𝑙𝑖𝑧𝑒 ∗ 𝐾𝑐 ∗ C
Where,

𝐾𝑐 - charging coefficient

C- Total battery capacity

5. Determine the maximum output power (POut) required
𝑃𝑂𝑢𝑡 = 𝑃𝑚𝑎𝑥 + 𝑃𝑐𝑚𝑎𝑥
6. Determine quantity of rectifier modules required (N+1,..)
𝑃𝑂𝑢𝑡
𝑁𝑚𝑜𝑑𝑢𝑙𝑒𝑠 = ⁄𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 (𝑤𝑎𝑡𝑡)
/𝑚𝑜𝑑𝑢𝑙𝑒

Example 2-1: Sizing using “Discharge Time-Discharge Adjustment Coefficient

method”
 Size a DC power system for a BTS site with network elements of total power
consumption 3000 watt by considering backup time of 3 hours. Assume on hand
battery resource is 190AH/2V & use table above showing Discharge Time-
Discharge Adjustment Coefficient relation. Note: assume module capacity is
3000 watt & use N+1 scenario for rectifier module quantity selection and
Vequalize=56.4 & Kc=0.1.

Solution
1. Determine max. power of the DC-powered load (Pmax) in the unit of watt
=3000Watt (given)
2. Choose load battery backup time (user defined), Tbackup
=3 hours (given)
3. Determine battery capacity (C)
[𝑃𝑚𝑎𝑥 ∗ 𝑇𝑏𝑎𝑐𝑘𝑢𝑝 ]
C= ⁄[𝐾
𝑎𝑔𝑖𝑛𝑔 ∗ 𝐾𝑣 ∗ 𝑉𝑛 ∗ 𝑛 ∗ 𝜂]

Where,

𝐊 𝐚𝐠𝐢𝐧𝐠 - Aging coefficient=0.8

𝐊 𝐯 - Voltage coefficient=0.96

𝐕𝐧 - Nominal system voltage

n- Number of battery units (optional, generally n = 1, 2, or 4)

𝛈 – Battery capacity coefficient, which depends on Tbackup

 C = (3000*3)/ (0.8*0.96*48*0.75)
= 9000/27.648
=325.52
Since we have 190Ah/2v on hand, we use two bank to handle the load for 3 hours

4. Calculate the maximum battery charge power (Pcmax)

𝑃𝑐𝑚𝑎𝑥 = 𝑉𝑒𝑞𝑢𝑎𝑙𝑖𝑧𝑒 ∗ 𝐾𝑐 ∗ C
Where,

𝐾𝑐 - charging coefficient

C- Total battery capacity

Pcmax = 56.4*0.1*2*190Ah
=2143.2 watt
5. Determine the maximum output power (POut) required
𝑃𝑂𝑢𝑡 = 𝑃𝑚𝑎𝑥 + 𝑃𝑐𝑚𝑎𝑥
Pout =3000W +2143.2 W
=5143,2 watt
6. Determine quantity of rectifier modules required (N+1)
𝑃𝑂𝑢𝑡
𝑁𝑚𝑜𝑑𝑢𝑙𝑒𝑠 = ⁄𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 (𝑤𝑎𝑡𝑡)
/𝑚𝑜𝑑𝑢𝑙𝑒

NR_modules =5143,2/ 3000

=1.71
We need two rectifier module to charge the battery and handle the load, using the N+1 scenario to
get 3 *3000W rectifier modules
ZTE method: - the method uses Discharge Time and Discharge Time Factor
to size the load and battery charging current.

Table: Relation of Discharge Time and Discharge Time Factor

Note: - the discharge time factor is higher than one for time less than 10 hours to compensate the power loss due to fast
discharging effect

Procedures:

Method 2-2: Discharge Time and Discharge Time Factor Method

1 Determine max. load current of the DC-powered load (Imax) in the unit of A
𝑃𝑚𝑎𝑥
𝐼𝑚𝑎𝑥 = ⁄𝑉
𝑛

2 Choose load battery backup time (user defined), Tbackup

3 Determine battery capacity (C)
C= 𝐴 ∗ 𝐾𝑇 ∗ 𝐼𝑚𝑎𝑥 ∗ 𝑇𝑏𝑎𝑐𝑘𝑢𝑝
Where,

A- Life time compensating time =1.25

K T - Discharge time factors which depends on Tbackup

Vn - Nominal system voltage, (48V)

4 Calculate total battery charging current (Ic)

𝐼𝑐 = 𝐾𝑐 ∗ C
Where,

𝑲𝒄 - battery charging coefficient

C- Total battery capacity

5. Determine the maximum output current (IOut) required

𝐼𝑂𝑢𝑡 = 𝐼𝑚𝑎𝑥 + 𝐼𝑐
6. Determine quantity of rectifier modules required (N+1)
𝐼𝑂𝑢𝑡
𝑁𝑚𝑜𝑑𝑢𝑙𝑒𝑠 = ⁄𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 (𝐴)
/𝑚𝑜𝑑𝑢𝑙𝑒
Example 2-2: Sizing using “Discharge Time-Discharge Adjustment Coefficient
method”
 Size a DC power system for a BTS site with network elements of total power
consumption 3000 watt by considering backup time of 3 hours. Assume on hand
battery resource is 190AH/2V & use table below showing Discharge Time-
Discharge Time Factor relation. Note: assume module capacity is 3000 watt &
 use N+1 scenario for rectifier module quantity selection and Kc=0.1.

1 Determine max. load current of the DC-powered load (Imax) in the unit of A
𝑃𝑚𝑎𝑥
𝐼𝑚𝑎𝑥 = ⁄𝑉
𝑛

Imax=3000w/48v
=62.5A
2 Choose load battery backup time (user defined), Tbackup
3 Determine battery capacity (C)
C= 𝐴 ∗ 𝐾𝑇 ∗ 𝐼𝑚𝑎𝑥 ∗ 𝑇𝑏𝑎𝑐𝑘𝑢𝑝
Where,

A- Life time compensating time =1.25

K T - Discharge time factors which depends on Tbackup

Vn - Nominal system voltage, (48V)

C=1.25*1.34*62.5A*3h
=314.0625 Ah
Since we have 190Ah/2V on hand, we need 2*190Ah/2v battery to handle the load for three hours

4 Calculate total battery charging current (Ic)

𝐼𝑐 = 𝐾𝑐 ∗ C
Where,

𝑲𝒄 - battery charging coefficient

C- Total battery capacity

Ic=0.1/h*(2*190Ah)
=38A
5. Determine the maximum output current (IOut) required
𝐼𝑂𝑢𝑡 = 𝐼𝑚𝑎𝑥 + 𝐼𝑐
= 62.5A +38A
=100.5A
6. Determine quantity of rectifier modules required (N+1)
𝐼𝑂𝑢𝑡
𝑁𝑚𝑜𝑑𝑢𝑙𝑒𝑠 = ⁄𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 (𝐴)
/𝑚𝑜𝑑𝑢𝑙𝑒

=100.5/ 50
=2.01
We need two rectifier module to charge battery and handle the site load, Using the N+1 scenario we need to configure the
site with 3*3000W rectifier modules

Cable Sizing
1. Cable sizing can be done by referring the current caring capacity table such us
below
 2. Using cable sizing formula below:- the method follow three procedure
I. Calculate the theoretical cross section area(St)
𝐼∗𝐿
𝑆𝑡 =
𝜌 ∗ 𝛥𝑉
𝜌 = 𝑡ℎ𝑒 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑣𝑖𝑡𝑦 𝑜𝑓 𝑐𝑜𝑝𝑝𝑒𝑟 = 57
𝛥𝑉 = 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑑𝑟𝑜𝑝 𝑜𝑓 𝑏𝑎𝑡𝑡𝑒𝑟𝑦 = 1 𝑓𝑜𝑟 20𝑚
II. Calculate the cross section area(S) using engineering practice
𝑆 = 𝐾 ∗ 𝑆𝑡
𝐾 = 1.2 𝑢𝑠𝑒𝑑 𝑡𝑜 𝑝𝑟𝑜𝑡𝑒𝑐𝑡 𝑐𝑎𝑏𝑙𝑒 𝑓𝑟𝑜𝑚 𝑐𝑎𝑟𝑟𝑖𝑛𝑔 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑐𝑢𝑟𝑟𝑒𝑛𝑡(𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟)
III. Select the Cross sectional area from the standard cable from the
ware house
Example: Sizing and Selection of Cables & Breakers

Assume a rectifier system is to be configured with one battery group of 300Ah/2V and the battery loop is
20m in length; the max. rated input current of load is 35A and the load loop is 30m in length.

1. Determine CSA of the battery cable (Use Kc=0.15)

2. Determine CSA of the working ground & earthing cable for battery cabinet

3. Determine CSA of the load cable

4. Determine CSA of the working ground of load cable and earthing cable

5. Determine the capacity of breakers for a battery and load

Solution
1 Determine CSA of the battery cable (Use Kc=0.15)

ICharging = 0.15/h(300Ah)
=45A
𝐼∗𝐿
𝑆𝑡 =
𝜌 ∗ 𝛥𝑉
St= (45A*20)/ (57*1)
=900/57
= 15.79 mm2
S= 1.2*15.79mm2
=18.95mm2
Since we have no 18.95mm2 cable standard in the warehouse, choose 25mm2
2 Determine CSA of the working ground & earthing cable for battery cabinet

The CSA of the working ground of battery =25mm2 and the CSA of earthing cable =16mm2

3 Determine CSA of the load cable

𝐼∗𝐿
𝑆𝑡 =
𝜌 ∗ 𝛥𝑉
St= (35A*30)/ (57*1.7)
=1050/96.9
= 10.8 mm2
S= 1.2*10.8mm2
=13.0mm2
Since we have no 13mm2 cable standard in the warehouse, choose 16mm2
4 Determine CSA of working ground cable of load and earthing cable
The CSA of the working ground of load =16mm2 and the CSA of earthing cable =16mm2
 5 Determine the capacity of breakers for a battery and load
𝐶𝐵𝑟𝑒𝑎𝑘𝑒𝑟 = 1.2 ∗ 𝐼𝑀𝑎𝑥._𝐿𝑜𝑎𝑑
CB-breaker=1.2*45A
=54A
Since we have no standard 54A breaker in the warehouse, choose 63A breaker for
battery
𝐶𝐿−𝑏𝑟𝑒𝑎𝑘𝑒𝑟 = 1.2 ∗ 𝐼𝑀𝑎𝑥._𝐿𝑜𝑎𝑑
CB-breaker=1.2*35A
=42A
Since we have no standard 42A breaker in the warehouse, choose 50A breaker for load

Solar Power System Sizing

To size the solar power system, we need to have the annual solar radiation power of
particular site. We can get the data from the following NASA web site using the
coordinate of the site.
• https://power.larc.nasa.gov/data-access-viewer/
Note: - we use the power of the month that we collect the minimum solar radiation power data

We can size the solar power system using the 14 procedure as follow
Solar Power System Sizing Procedures:
1. Collect data from national metrology & determine average solar insolation
(kWh/m2/day). At STC), solar radiation intensity of PV modules is 1kWh/m2 or 1000
Wh/m2.
Note: NASA surface metrological data of specific site location (lat., long.) can be used as option

2. Determine average sunshine time of the year (T) (Note: Minimum sunshine time of
the year is to be chosen to make the design more accurate). Example: let average solar
insolation of the site is 4.6 kWh/m2/day, and then average sunshine time of the day (T)
will be
4.6𝑘𝑊ℎ/𝑚2⁄
𝑑𝑎𝑦
𝑇= = 4.6ℎ/𝑑𝑎𝑦
1𝑘𝑊ℎ/𝑚2
 3. Sum up all the DC load wattages (PLoad) including future expansion if any
𝑃 ∗𝑇 𝑃 ∗𝑇 𝑃 ∗𝑇
𝑃𝐿𝑜𝑎𝑑 = 124 1 + 224 2 + ⋯ + 𝑛24 𝑛

Where,

PLoad- Total load DC power

Pn- DC power consumption of the equipment

Tn- Running time of the DC power consuming equipment

4. Choose load battery backup time (user defined), T backup

5. Determine load battery capacity required, C (Ah)
𝑃𝐿𝑜𝑎𝑑 ∗𝑇𝑏𝑎𝑐𝑘𝑢𝑝
C= 𝑉𝑛 ∗𝐷𝑜𝐷

where,

Vn- Nominal System Voltage

DoD- Depth of Discharge

6. Then, determine required number of battery strings/groups (NBB)

𝐶
𝑁𝐵𝐵 =
𝐶ℎ𝑜𝑠𝑒𝑛 𝐵𝑎𝑡𝑡𝑒𝑟𝑦 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 (𝐴ℎ)
6. Determine required number of PV modules (NPV)
(𝑃𝐿𝑜𝑎𝑑 +𝑃𝐶 ) ∗ 𝑡
𝑁𝑃𝑉 =
𝑃𝑝𝑒𝑎𝑘/𝑝𝑣 ∗ 𝜂𝑃𝑉 ∗ 𝜂𝐶 ∗ 𝜂𝐵 ∗ 𝑇
Where,

t- Working time per day of the load

Ppeak/pv - peak power output of a single PV panel

PC - controller self-consumption power

𝜼PV - efficiency of PV module (including dust loss, temperature influence loss, deteriorate loss)

𝜼C - Efficiency of controller

𝜼B - Efficiency of battery

T- Minimum sunshine time of the year

 7. Determine required number of MPPT modules (NMPPT)
𝑁𝑃𝑉 ∗ 𝑃𝑝𝑒𝑎𝑘/𝑝𝑣
𝑁𝑀𝑃𝑃𝑇 =
𝑃𝑀𝑃𝑃𝑇/𝑚𝑜𝑑𝑢𝑙𝑒

Where,
𝑃𝑀𝑃𝑃𝑇/𝑚𝑜𝑑𝑢𝑙𝑒 - Chosen a single MPPT module capacity in watt

9. Determine number of PV panels to be connected per MPPT module (NPV/MPPT)

𝑁𝑃𝑉
𝑁𝑃𝑉/𝑀𝑃𝑃𝑇 =
𝑁𝑀𝑃𝑃𝑇
10. Determine possible range for number of PV panels to be connected in series per
MPPT module i.e., (NS_min) and (NS_max) and chose NS accordingly.
𝑉𝑀𝑃𝑃𝑇_𝑚𝑖𝑛 𝑉𝑀𝑃𝑃𝑇_𝑚𝑎𝑥
𝑁𝑆_𝑚𝑖𝑛 = < 𝑁𝑆 < 𝑁𝑆_𝑚𝑎𝑥 =
𝑉𝑃𝑉_𝑚𝑎𝑥 𝑉𝑃𝑉_𝑚𝑎𝑥
11. Determine number of PV panels (NP) to be connected in parallel per MPPT module
as of step 10.
𝑁𝑃𝑉/𝑀𝑃𝑃𝑇
𝑁𝑃 =
𝑁𝑆
12. Determine the number of junction boxes or combiners required i.e., NJB and check
that minimum input capacity of junction boxes are equal to NP.
𝑁𝐽𝐵 = 𝑁𝑀𝑃𝑃𝑇

13. Determine the breaker capacity required

𝐶𝐵𝑟𝑒𝑎𝑘𝑒𝑟 = 1.2 ∗ 𝐼𝑃𝑉_𝑚𝑎𝑥 ∗ 𝑁𝑃
14. Determine the cable requirement (length, quantity, cross-sectional area,.. from each
of the junction boxes (combiners) to controller.
Example:
 Suppose a newly built BTS site located at latitude 6.927000 and longitude
37.83000 has network elements of total power consumption 3000 watts expected
to work for about 24 hours and the site loads are to be powered by DC power
harvested from pure solar. On hand solar battery of 1000Ah/2V is desired to
supply the loads for 72 hours and assume that recommended DoD by
manufacturer is 80%. Each of the MPPT solar module capacities is 3000 watt
and use controller self consumption 76-watt, controller efficiency=0.98, PV
efficiency= 0.94, and battery efficiency= 0.85. Size and configure accordingly.
Note: Use the information provided in tables 1, 2 & 3.
Table: - Monthly average solar insolation values of the site

Item Specification

Maximum Power *(Pmax) 200W

Power Tolerance 0~+5W

Rated Voltage at Pmax* 36.7 V

Rated Current at Pmax* 5.45A

Open Circuit Voltage* 45.5V

Short Circuit Current* 5.81A

Module Efficiency 15.7%

Cells Quantity 72 in series

Cell Type Mono-crystalline Cell

Cell Size 125 x 125 mm

Weight 16kg

Dimensions 1580(L) x 808(W) x 46(H) mm

Solution
1. Collect data from national metrology & determine average solar insolation
(kWh/m2/day). At STC), solar radiation intensity of PV modules is 1kWh/m 2 or
1000 Wh/m2.

2. Determine average sunshine time of the year (T) (Note: Minimum sunshine time of
the year is to be chosen to make the design more accurate). Example: let average solar
insolation of the site is 4.6 kWh/m2/day, and then average sunshine time of the day (T)
will be
4.6𝑘𝑊ℎ/𝑚2⁄
𝑑𝑎𝑦
𝑇= 2
= 4.6ℎ/𝑑𝑎𝑦
1𝑘𝑊ℎ/𝑚

3 Sum up all the DC load wattages (PLoad) including future expansion if any
𝑃𝐿𝑜𝑎𝑑 =3000 watts (given)
4. Choose load battery backup time (user defined), T backup
Tbackup=72 hour (given)
5. Determine load battery capacity required, C (Ah)
𝑷𝑳𝒐𝒂𝒅 ∗𝑻𝒃𝒂𝒄𝒌𝒖𝒑
C= 𝑽𝒏 ∗𝑫𝒐𝑫

=3000w*72h/ 48v*.8
= 216000AVh/38.4h
=5625Ah
6. Then, determine required number of battery strings/groups (NBB)
𝑪
𝑵𝑩𝑩 =
𝑪𝒉𝒐𝒔𝒆𝒏 𝑩𝒂𝒕𝒕𝒆𝒓𝒚 𝑪𝒂𝒑𝒂𝒄𝒊𝒕𝒚 (𝑨𝒉)
= 5625Ah/1000Ah
=5.625
Since we have 1000Ah/2V on hand, we need 6*1000Ah/2V battery
 8. Determine required number of PV modules (NPV)
(𝑃𝐿𝑜𝑎𝑑 +𝑃𝐶 ) ∗ 𝑡
𝑁𝑃𝑉 =
𝑃𝑝𝑒𝑎𝑘/𝑝𝑣 ∗ 𝜂𝑃𝑉 ∗ 𝜂𝐶 ∗ 𝜂𝐵 ∗ 𝑇
Where,

t- Working time per day of the load

Ppeak/pv - peak power output of a single PV panel

PC - controller self-consumption power

𝜼PV - efficiency of PV module (including dust loss, temperature influence loss, deteriorate loss)

𝜼C - Efficiency of controller

𝜼B - Efficiency of battery

T- Minimum sunshine time of the year

(𝟑𝟎𝟎𝟎 + 𝟕𝟔) ∗ 𝟐𝟒
𝑵𝑷𝑽 =
𝟐𝟎𝟎 ∗ 𝟎. 𝟗𝟖 ∗ 𝟎. 𝟗𝟒 ∗ 𝟎. 𝟖𝟓 ∗ 𝟒. 𝟔
𝟕𝟑𝟖𝟐𝟒
= 𝟕𝟐𝟎.𝟒

=102.5
9. Determine required number of MPPT modules (NMPPT)
𝑁𝑃𝑉 ∗ 𝑃𝑝𝑒𝑎𝑘/𝑝𝑣
𝑁𝑀𝑃𝑃𝑇 =
𝑃𝑀𝑃𝑃𝑇/𝑚𝑜𝑑𝑢𝑙𝑒

Where,
𝑃𝑀𝑃𝑃𝑇/𝑚𝑜𝑑𝑢𝑙𝑒 - Chosen a single MPPT module capacity in watt
102 ∗ 200
𝑁𝑀𝑃𝑃𝑇 =
3000
20400
= 3000

=6.8
We need to configure 7*3000w MPPT module to safely deploy all PV modules
9. Determine number of PV panels to be connected per MPPT module (NPV/MPPT)
𝑁𝑃𝑉
𝑁𝑃𝑉/𝑀𝑃𝑃𝑇 =
𝑁𝑀𝑃𝑃𝑇
102
𝑁𝑃𝑉/𝑀𝑃𝑃𝑇 =
7
= 14.57
Since each MPPT module can control their connected pv module individually, we configure 15 PV
module for 6 MPPT and 12 PV modules for the remaining 1 MPPT module

10. Determine possible range for number of PV panels to be connected in series per
MPPT module i.e., (NS_min) and (NS_max) and chose NS accordingly.
𝑉𝑀𝑃𝑃𝑇_𝑚𝑖𝑛 𝑉𝑀𝑃𝑃𝑇_𝑚𝑎𝑥
𝑁𝑆_𝑚𝑖𝑛 = < 𝑁𝑆 < 𝑁𝑆_𝑚𝑎𝑥 =
𝑉𝑃𝑉_𝑚𝑎𝑥 𝑉𝑃𝑉_𝑚𝑎𝑥
58 150
𝑁𝑆_𝑚𝑖𝑛 = < 𝑁𝑆 < 𝑁𝑆_𝑚𝑎𝑥 =
36.7 36.9
𝑁𝑆_𝑚𝑖𝑛 = 1.58 < 𝑁𝑆 < 𝑁𝑆_𝑚𝑎𝑥 = 4.08

As we can notice from the equation result, the minimum number of PV modules that can
be connected in series is 2 and the maximum is 4. We choose three module in series
connection to use all of the 102 PV modules
11. Determine number of PV panels (NP) to be connected in parallel per MPPT module
as of step 10.
𝑁𝑃𝑉/𝑀𝑃𝑃𝑇
𝑁𝑃 =
𝑁𝑆
= 15
𝑁𝑃 =
3
=5
12. Determine the number of junction boxes or combiners required i.e., NJB and check
that minimum input capacity of junction boxes are equal to NP.
𝑁𝐽𝐵 = 𝑁𝑀𝑃𝑃𝑇

𝑁𝐽𝐵 = 7
13. Determine the breaker capacity required
𝐶𝐵𝑟𝑒𝑎𝑘𝑒𝑟 = 1.2 ∗ 𝐼𝑃𝑉_𝑚𝑎𝑥 ∗ 𝑁𝑃
𝐶𝐵𝑟𝑒𝑎𝑘𝑒𝑟 = 1.2 ∗ 5.45 ∗ 5𝑃
=32.7A
14. Determine the cross-sectional area of the cable from each of the junction boxes
(combiners) to controller. Assume the longest distance b/n the JB and controller is 30m.
𝐼∗𝐿
𝑆𝑡 =
𝜌 ∗ 𝛥𝑉
St= (5*5.45*30)/ (57*1.7)
=817.5/96.9
= 8.43 mm2
S= 1.2*8.43mm2
=10.12mm2
We can configure 10mm2 cable for connection b/n JB and controller

Exercise: DC Power System Sizing

Suppose DC powered loads with maximum power consumption of 50000 watts
are planned to be deployed in new core site of first floor (1F). Consider that
primary DC distribution rack is to be installed besides two parallel rectifier racks
so that to be connected via DC bus bar and two secondary DC distribution racks
(DCDR-1 & DCDR-2) are to be located at 1F of equipment room and 50% of load
is to be supplied from each of the racks. Assume that each of the rectifier module
capacities to be configured is 100 A, and on hand 3000Ah/2V load battery is
intended to supply for 3 hours and its manufacturer recommended equalize
voltage and charging coefficient are, Vequalize=56.4 & KC=0.15, respectively. The
battery cable loop required per group is 20m in length and the load cable loop
required i.e., from primary DC distribution rack to secondary DC distribution rack
1 (DCDR-1) is 30m in length. Note that secondary DC distribution racks are to be
supplied via two redundant input cables to make them secure and the input
cables are to be connected from different breakers of primary DC distribution
rack. Use table below showing Discharge Time & Discharge Time Factor relation
to plan & design the DC power system requirement.
1. Determine the battery capacity required
2. Configure rectifier system (No. of modules, propose rectifier rack and secondary
DC distribution rack capacities).
Note: Use N+2 rectifier configuration scenario
3 Determine CSA of the battery DC power cable
4 Determine CSA of battery working ground & its cabinet earthing cable
5 Determine CSA of the secondary DC distribution rack 1 (DCDR-1) DC power cable
6 Determine CSA of the secondary DC distribution rack 1 (DCDR-1) working ground
cable & its earthing cable
7 Conclude the rectifier system working ground and its earthing cables
8 Determine the breaker capacity needed for the secondary DC distribution rack 1
connection from primary DC distribution rack.

Solution

1 Determine max. load current of the DC-powered load (Imax) in the unit of A
𝑃𝑚𝑎𝑥
𝐼𝑚𝑎𝑥 = ⁄𝑉
𝑛

Imax=50000w/48v
=1041.67A
2 Choose load battery backup time (user defined), Tbackup
=3 hours

3 Determine battery capacity (C)

C= 𝐴 ∗ 𝐾𝑇 ∗ 𝐼𝑚𝑎𝑥 ∗ 𝑇𝑏𝑎𝑐𝑘𝑢𝑝
Where,

A- Life time compensating time =1.25

K T - Discharge time factors which depends on Tbackup

Vn - Nominal system voltage, (48V)

C=1.25*1.34*1041.67A*3h
=52234.39 Ah
Since we have 1000Ah/2V on hand, we need 6*1000Ah/2v battery to handle the load for three hours

4 Calculate total battery charging current (Ic)

𝐼𝑐 = 𝐾𝑐 ∗ C
Where,

𝑲𝒄 - battery charging coefficient

C- Total battery capacity

Ic=0.15/h*(6*1000Ah)
=900A

5. Determine the maximum output current (IOut) required

𝐼𝑂𝑢𝑡 = 𝐼𝑚𝑎𝑥 + 𝐼𝑐
= 1041.67A +900A
=1941.67
6. Determine quantity of rectifier modules required (N+1)
𝐼𝑂𝑢𝑡
𝑁𝑚𝑜𝑑𝑢𝑙𝑒𝑠 = ⁄𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 (𝐴)
/𝑚𝑜𝑑𝑢𝑙𝑒

=1941.67/ 100
=19.42
We need 20 rectifier module to charge battery and handle the site load, Using the N+2 scenario we need to configure the
site with 22*100A rectifier modules

Note after obtaining the above information, we proceed to size cable and
protection equipment as follow
A) Determine CSA of the battery DC power cable
The maximum battery charging current is calculated as follow. Since we use 6*1000Ah/2V battery, we consider the cable
size that is enough to handle the charging current of one bank

Ic=0.15/h*(1000Ah)
=150A
St =(150A*20)/57*1
=52mm2
 S=1.2*52mm2
=63mm2
Since we have no 63mm2 cable on store, choose 70mm2 for battery
B) Determine CSA of battery working ground & its cabinet earthing cable
The working ground cable of battery is 70mm2 and the earthing cable of battery
rack/cabinet is 35mm2

C) Determine CSA of the secondary DC distribution rack 1 (DCDR-1) DC power cable

Note:- it is given that each DC distribution rack handle half of the load equipment,
so the cable b/n primary and secondary distribution rack is required to handle the
load current of 25000W load equipment.
𝑃𝑚𝑎𝑥
𝐼𝑚𝑎𝑥 = ⁄𝑉
𝑛

Imax=25000w/48v
=520.83A
There fore St=(520.83A*30)/57*1.7
=15625/96.9
=161.25mm2
S=1.2*161.25mm2
=193.5mm2
Since we have no 193.5mm2 cable on warehouse, choose 240mm2
D) Determine CSA of the secondary DC distribution rack 1 (DCDR-1) working ground
cable & its earthing cable
The CSA of the secondary DC distribution Rack 1 (DCDR-1) is 240mm2 and is
earthing cable CSA is 120mm2
E) Conclude the rectifier system working ground and its earthing cables
It is mentioned that the two parallel rectifier rack connected to the primary DC
distribution implying that one rectifier is to handle 25000w. Therefore the CSA of its
earthing cable is 120 mm2 like DCDR-1 for both rectifier rack
F) Determine the breaker capacity needed for the secondary DC distribution rack 1
connection from primary DC distribution rack.
C=1.2*Imax
= 1.2*520.83A

= 625A