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--- opinion:solar:sizing.walkthrough [2022/05/27 14:36]
frater_secessus [panel wattage]
+++ opinion:solar:sizing.walkthrough [2022/05/27 15:09] (current)
frater_secessus [total daily power requirement]
@@ Line 11: / Line 11: @@
   * how much power we will require - every single watt
   * where we will camp, and what time of year we will be there
-  * avoiding shade and so the panels can actually work
+  * avoiding shade so the panels can actually work
 ===== daily power requirements =====
-This part is the least fun and the most important;  we must account for every watt we intend to "spend" off-grid.
+This part is the least fun and the most important;  we must account for every watt we intend to "spend" off-grid.  Some people prefer pencil-and-paper, but doing the work on a spreadsheet (M$ Excel, freeware [[https://www.libreoffice.org/|LibreOffice]] Calc, or free-to-use [[https://drive.google.com/|Google Docs]]) can make the work easier and more organized.  You can also rapidly alter it to model new build ideas.
 The goal is to come up with a daily target in **watt-hours** (**Wh**).  There will be times we have to work in other units but Wh is the main thing.((Wh is independent of voltage))  **Watt-hours are calculated** by multiplying **the load's draw in watts** by **the hours it is expected to run**.    If we run a 4w device for 6 hours that is 24Wh (4W x 6h = 24Wh).
@@ Line 71: / Line 71: @@
 We will need **2,084Wh every day** to run our intended loads.
+==== day/night modeling ====
+You may find it useful to make day/night sections for both AC and DC.  This will yield more accurate numbers if you run many of your loads in daylight instead of at night.  It could save you from oversizing your battery bank.
+Note that very big [[electrical:12v:loads|loads]] may require bigger banks to get the current throughput required.  If you have a 120A load but the lithium is only rated to 100A discharge you will need ≥120Ah capacity regardless how small your actual Wh use is.
@@ Line 113: / Line 120: @@
 The good news is that different locations get roughly predictable average amounts of sunlight during specified months, and [[https://web.archive.org/web/20180811212703/http://www.solarinsolation.org/wp-content/uploads/2012/01/Solar_insolation.jpg|scientists have tabulated this data]]. The data is expressed in terms of hours of Full Sun Equivalent and assumes flat-mounted panels.  You can think of this as "hours of laboratory-perfect conditions".    The actual number of hours of sunlight in the day don't matter, nor do the average climatological conditions.  4 hours of FSE in Phoenix in summer might be 6.5 hours by the clock, while the same 4 hours of FSE in Anchorage might be 14 hours by the clock.  No matter.  The math works.
-Example:  if we need 2,317Wh and will be in an area with 4 hours of FSE in March, in theory we could get by with
-In practice 100% yield is not normal.  After various losses MPPT-equipped systems will get more like 85%, and PWM systems around 72%.((again, if you have observed your own system and know what it does then use the actual number)). For the purposes of this article we will call this //system efficiency//.  Examples:
-  * theoretical - 1,200Wh harvest
-  * MPPT - 1,200Wh x 0.85 = 1,020Wh harvest
-  * PWM - 1,200Wh x 0.72 = 864Wh harvest
 ==== where and when we are camping ====
-The actual hours of FSE will depend on where we are and month of the year.  If we want the solar to Just Work((by itself, no help from [[electrical:12v:alt_and_solar|other charging sources]])) we have to size it for the worst average yield we will experience.   Let's consider these very different scenarios:
+The actual hours of FSE will depend on where we are and month of the year.  If we want the solar to Just Work((by itself, no help from [[electrical:12v:alt_and_solar|other charging sources]])) **we have to size it for the worst average yield we will experience**.   Let's consider these very different scenarios:
   * vacation-camping with the family in Montana only in the summer
@@ Line 138: / Line 137: @@
-We can work backward from FSE and required Wh to get our required panel:  Wh required / hours of FSE / system efficiency = panel wattage.  Assuming an MPPT-equipped system:
+We can work backward from FSE and required Wh to get our required panel:  Wh required / hours of FSE / system efficiency((I will stipulate .85 for MPPT and .70 for PWM)) = panel wattage required.  Assuming an MPPT-equipped system:
   * **Montana in July** - 2,316Wh((daily charging requirement)) / 6.44 FSE / 0.85 system efficiency = **423w of panel**

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