EXAMPLE OF A DIRECTLY-COUPLED SYSTEM DESIGN

Specification

Design a directly-coupled pumping system (no batteries or power conditioning circuitry) for irrigation purposes near Melbourne (latitude 37.8°S). The only months of interest are December and January, during which period 2.5 million litres are required to be pumped. The relevant insolation data is given in Table H.2. The total pumping head is 8.8 m and this remains approximately constant during pumping and from season to season. There are no problems with the water source replenishment rate. Other months of the year are considered to be non-critical.

Table H.2. Average daily insolation data for a horizontal surface for the design months.

month______________________ S_____________ D____________ R

December 676 214 890

January___________________ 629___________ 210__________ 839

Design procedure

Everything necessary is specified above.

Step 2—Select tilt angle and adapt the solar insolation data

Since the design period is only for December and January, the tilt angle is simple to optimise, since we point the panels directly at the noon-time sun for these months. From Eqn. (1.7) in Chapter 1, we calculate that the declination of the sun for December is the maximum of 23.4°, while for mid-January it is not much different, at about 21.8°. Using the value of 22.8°, we get a noon-time solar altitude, as given by Eqn. (1.6) from Chapter 1, of 75°. Consequently, our tilt angle should be 15°. Similarly, from Eqn. (1.5) of Chapter 1, we correct the tilt angle to give the solar insolation incident on the solar panels, as shown in Table H.3.

Table H.3. Solar insolation incident on the solar panels, corrected for tilt angle.

 month а S D S15 R15 December 75.6 676 214 698 912 January 73.5 629 210 656 866

Step 3—Dislocation data manipulation

(a) Determine Is and Ic values

For December, using Eqns. (H.1) and (H.2),

Isl = 1.353 x 0.7AM0678 x 1.1 x sin(180°- 75.6° -15°)

where AM = 1/sin(75.6°), giving

Is1 = 103 mW/cm2 Ic1 = 9.4 mW/cm2.

For January (with a = 73.6)

Is2 = 102 mW/cm2 Ic2 = 9.3 mW/cm2.

(b) Calculate the mWh/cm2/day for sunny and cloudy days

Using Fig. H.2, we get the ‘sunny day’ mWh/cm2/day as

6.76 x Ni x Isi

with Nj = 1.4 for December and January and the Isi values as given in (c). Similarly, the ‘cloudy day’ mWh/cm2/day is given by

6.76 x Ni x Ia

for each month respectively.

(c) Determine the percentage of sunny and cloudy days

For December, using Fig. H.2 and Eqn. (H.3),

R15 = X x 6.76 x 1.4 x 103 + Y x 6.76 x 1.4 x 9.4

where R15 = 912, X and Y are the percentages of sunny and cloudy weather, respectively, and X + Y = 1 (or Y = 1 – X). This gives

912 = 975 x X + 89 x (1 – X)

and

X = 0.93, Y = 0.07.

In other words, during the days, it is sunny 93% of the time and cloudy only 7% of the time.

For January

866 = 695 x X2 + 89 x (1 – X2)

X2 = 0.89 Y2 = 0.11.

That is, it is sunny 89% of the time.

Neglecting the cloudy weather as being useless for pumping, the useful radiation incident on the solar panels is

RU1 = 0.93 x 912

= 848 mWh/cm2/day (December)

Ru2 = 0.89 x 866

= 771 mWh/cm2/day (January).

(d) Determine equivalent number of pumping hours (E) at light intensity Isa

Using Eqn. (H.4),

Isa = (103 x 31 X 0.98 + 102 x 31 x 0.89) /

(31 x 0.93 + 31 x 0.89)

= 103 mW/cm2.

Using Eqn. (H.5),

Ex = 6.76 x 1.4 x 103 / 103 = 9.5 hours E2 = 9.4 hours

and using Eqn. (H.6), the monthly number of such hours is Em1 = 9.5 x 31 x 0.93 = 274 hours Em2 = 9.4 x 31 x 0.89 = 259 hours.

Also, for the design period

Ey Em1 + Em2

= 533 hours (of equivalent sunshine at Isa).

Step 4—Pump selection

We require 2.5 million litres in total in an average year to be pumped throughout December and January combined. This gives a pumping rate (P) of

P = 2.5 x 106 / (53 3 x 60 x 60)

= 1.30 L/s.

In this instance, we do not need to be concerned about the fact that more water will be pumped in December than January, because this is simply the result of there being more sunny days on average in December, hence necessitating the increased water being pumped during that month.

Frictional losses in the pipes, since not otherwise stated, are assumed to effectively add 2% to the total head, which in this case gives a total of 9 m. Fig. H.4 gives the performance curves of a centrifugal pump well suited to a head of 9 m and operation at a pumping rate of 1.30 L/s.

Step 5—Select motor with compatible torque-speed characteristics

From Fig. H.4, the centrifugal pump operating at a head of 9 m and a pumping rate of 1.3 L/s will have a speed of about 2700 rpm. This corresponds to an input power (Pin) of about 230 W. Consequently, since the torque (t) is not given directly, it can be calculated from the power (Pin) and angular velocity (ю) as follows:

t = Pin / ю (H.15)

= 230 / (2rc x 2700 / 60)

= 0.814 Nm.

The performance curves of the DC motor in Fig. H.5 indicate that it will be a suitable match for the selected pump, giving about 76% efficiency at the design point for operation.

Step 6—Determine the load line for the subsystem, weighted by efficiencies

Fig. H.4 gives the necessary pump characteristics, although not in the most desirable form. However, from the supplied curves, the torque (t/) can be determined for each speed (N), via the input power to the pump as given by Eqn. (H.15) above. For each t/ value we then refer to Fig. 11.11 in Chapter 11 for our chosen motor.

The curves are provided, although it is far simpler and more accurate to use the provided equations, which can be rearranged to give

Finally, by including the pump and motor efficiencies (щ and qm, respectively) by reading the values off the curves, a table can be formed (Table H.4) that includes all the necessary information for the motor/pump load line, weighted by the subsystem efficiencies. Va in the final column is the voltage necessary from the array. Its values are 2% higher than the motor voltages (Vm), owing to the losses in the wiring.

Table H.4. Calculated values for determining the load line.

 N ю=лМ30 Pin pn nm Vsub Im %a Vm Va (rpm) (rad/s) (W) (Nm) (%) (%) (%) (A) (V) (V) 3200 335 400 1.19 75.5 41.0 31.0 9.20 55 56 3000 314 300 0.96 76.0 45.0 34.2 7.60 50 51 2800 293 250 0.85 75.5 47.5 36.0 6.75 46 47 2600 272 220 0.81 75.0 47.5 35.6 6.45 43 44 2400 251 170 0.68 74.0 45.0 33.3 5.50 40 41 2200 230 110 0.48 73.0 35.0 25.5 4.03 35 36 2120 1 222 89 0.40 73.0 21.4 15.6 3.44 34 35 2100 220 50 0.23 0 0 0 2.19 32 33

Note

1. At N = 2120, qp = (0.22 kg/s x (10.1 – 0.2) m x 9.8 m/s) / 100 J/s = 21.4%. Therefore, at a 9 m head on the same efficiency line, Pin = (0.22 kg/s x (9 – 0.2) x 9.8) / 0.214 = 89 W.

Step 7—Photovoltaic system sizing

The I-V curves for our PV panels are given in Fig. H.3. Our task is to configure the modules so as to match the PV output to the requirements of the motor/pump.

Voltage sizing

Since the maximum power point voltage remains roughly constant as light intensity changes, we need it to correspond to the voltage at which the subsystem achieves maximum efficiency. From Table H.4, this occurs for an array voltage of 47 V. From our standard normalised curves for commercial modules in Fig. H.3 and Table H.1, we find that, for three nominal 12 V modules connected in series, the maximum power point voltage at 45°C will be about 46 V, which is a little lower than ideal although quite acceptable. This selection therefore specifies the horizontal axis in Fig. H.3 with m = 3.

Current sizing

For peak subsystem performance, Table H.4 indicates that we require a motor current of 6.75 A. Since, however, we want this current when the light intensity is only

0. 80 x Isa, then our rated maximum power point current (Imp) at 100 mW/cm2 needs to be as given by Eqn. (H.14)

Imp = 6.75 x 100 / (0.80 x 103)

= 8.2 A.

We then allow for a derating factor (DR) of 0.74 and divide by 0.95 to convert to a short circuit current rating (Isc) of

Isc = 8.2 / (0.74 x 0.95)

= 11.6 A.

That is, this is the manufacturer’s rated current we need to ensure that, in the long term, we will still get a maximum power point current of 6.75 A when the light intensity is 0.80 x Isa,

We can now specify the vertical axis of Fig. H.3 for each month, with the short circuit current for the top curve in December being given by

Isc = (0.99 x 0.74 x 103 / 100 ) x 11.6 = 8.75 A.

This gives us the curves of Fig. H.5 for the month of December. Also from Table H.4 we can superimpose the subsystem load line onto the I-V curves to determine the operating points for each light intensity. This is also shown in Fig. H.5.

 Figure H.5. I-V curves corresponding to the daily insolation profile, as given by Fig. H.2 for the month of December, with the motor-pump load line from Table H.4 superimposed.

We are now in a position to determine the daily and monthly water pumped for each month and hence throughout the design period.

Notes

1. For a cloudy day, the peak light intensity of 9.4 mW/cm[1], as given by Eqn. (H.2), is insufficient to initiate or maintain any pumping.

3. If the calculated pumped water profile throughout the year does not match the demand profile, then the tilt angle needs to be optimised.

[1] The total water pumped throughout the design period should come to about 10-20% more than that specified, owing to the deliberate conservatism built into the design to allow for such things as the subsystem not meeting the manufacturer’s specifications, degradation etc. If the calculated volume is outside this range, then optimisation of the array size (current rating) needs to take place.

Updated: July 1, 2015 — 11:36 am