Question-and-Answer Resource for the Building Energy Modeling Community
Get s tarted with the Help page
Ask Your Question
2

what percentage of battery loss and other losses are counted in eQUEST for modelling of SPV plant ? (please reply with source)

asked 2014-11-17 04:29:21 -0500

Ashok Dhayal gravatar image

updated 2015-07-11 14:50:00 -0500

I am modelling a Solar PV plant on eQUEST, I need to know various losses this software considers for calculation of outputs and where would I get part load efficiency curve of inverter which is used as default?

edit retag flag offensive close merge delete

2 Answers

Sort by » oldest newest most voted
5

answered 2014-11-17 05:33:25 -0500

updated 2014-11-17 05:43:29 -0500

Ashok,

You're a little out of luck as far as getting proper sources for this, at least in terms for something clearly written and directly quotable. I've looked into the DOE-2 Engineers Manual version 2.1A (here) and there is nothing about photovoltaic generators. I suppose it was added later, after 1993.

Though, looking at eQuest and the help files, you can already understand what is going on. I've looked at the source code of DOE22 as well, especially in PLT.LIS that is in the folder SRC48r, and you could do the same.

Anyway, in DOE2 you'll enter the characteristics of your PV Modules, especially the performance data. That's basically going to play on the efficiency of your modules. Then, you'll create your generator. A number of parameters come into play in the calculation routine. The array tilt for example is used to determine the electric capacity (direct and diffuse solar, angle of incidence, air mass, and ambient temperature are the parameters for that).

As far as modeling the "other losses", what you really want to be looking at is the EIR and the EIR-FPLR curves:


ELEC-INPUT-RATIO

specifies the ratio of direct-current input power to alternating-current output power. This number is the inverse of the full-load efficiency, and as such is always greater than 1.0.


EIR-FPLR

accepts the U-name of a curve that modifies the nominal electric input ratio as a function of the hourly part load ratio. This use of this curve is identical to part-load curves documented elsewhere.


I guess you might also play with the auxiliary power, but I think the main goal of this was to include say the power used for a sun tracking system.

Choose the ELEC-INPUT-RATIO to include all your regular losses such as wiring losses, potential step up transformer losses, etc. Choose the EIR-FPLR to match your inverter characteristics.

The EIR-FPLR that is included by default in eQuest is called "Inverter EIR-fPLR" and it's easy to see its coefficients. For your convenience, it's a Quadratic curve (Z = a + bX + cX²) with coefficients (a, b, c) = (0.01677774, 1.03766644, -0.05444432).

Now, this is not what you are used to see. What you are used to see is the efficiency of the inverter. The default eQuest inverter looks like this: Defaut eQuest Inverter Efficiency Curve

If I assume an ELEC-INPUT-RATIO of 1.075 (= 93% efficient, I'm voluntarily taking a fairly low number in order to be able to see the difference), the Total Efficiency of the system would look like: eQuest Total Efficiency Curve

I hope this clarifies things.

edit flag offensive delete link more

Comments

hey Julien many thanks for response, (At 50% load) Z = a + bX + cX², Z = 0.01677774 + 1.03766644 * 0.5 - 0.05444432 * 0.5 * 0.5 = 0.522, Is this okay ?. According to curve, the value of Z should be 0.96

Ashok Dhayal gravatar image Ashok Dhayal  ( 2014-11-25 03:20:39 -0500 )edit

@Ashok Dhayal: Correct, EIR(PLR=50%) = 0.522. The curve is the efficiency... that's not the same. I got Actual EIR by multiplying EIR with full-load EIR I chose (1.075*0.522=0.561) and then Total efficiency by doing PLR/Actual EIR=89.1%

Julien Marrec gravatar image Julien Marrec  ( 2014-12-01 03:41:05 -0500 )edit
2

answered 2014-11-17 15:32:17 -0500

In particular for Solar PV the industry is advancing so quickly that a set of default values made at some unknown time in the past may not be accurate for a modern PV system.

If it is possible to substitute your own performance from a current product that might be the preferred path.

edit flag offensive delete link more

Comments

Thanks David Eldridge

Ashok Dhayal gravatar image Ashok Dhayal  ( 2014-11-25 03:35:28 -0500 )edit

Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

Add Answer

 

Question Tools

1 follower

Stats

Asked: 2014-11-17 04:29:21 -0500

Seen: 289 times

Last updated: Nov 17 '14