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Environmental Engineering · FE Reference Handbook section

Environmental Engineering
0 formulas
10 exam-style examples
~45 min
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Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

  • TYPICAL SUMMER DAY TYPICAL WINTER DAY
  • 50 SOLAR MODE 50 SOLAR MODE
  • Note: Correlation of solar generation to peaking power requirements
  • Spacing of Solar Flat-Plate Collectors to Avoid Shadowing

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

This section is conceptual; there are no equations to memorise.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1
Solar — solve for solar power output — Solar

solar power output from a photovoltaic array installation Given panel conversion efficiency (eta) = 0.1250; panel array area (A) = 1,813 m^2; solar irradiance (G) = 370.0 W/m^2, determine the solar power output (P) in W.

Given

  • panelconversionefficiency(eta)=0.1250panel conversion efficiency (eta) = 0.1250
  • panelarrayarea(A)=1,813m2panel array area (A) = 1,813 m^2
  • solarirradiance(G)=370.0W/m2solar irradiance (G) = 370.0 W/m^2

Find

solar power output (P), in W

Start with the thinking

  • The governing relation printed in this handbook section is Solar.
  • Everything except P is given, so isolate P symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Solar power output from a photovoltaic array depends on panel efficiency, array area, and incident solar irradiance.
Solar power output vs irradiance02505007501000300Low600Medium1000Highirradiance W/m^2

Figure 1 — schematic for Solar — solve for solar power output — Solar

Step-by-step solution

  1. Step 1 — State the governing relation:

    P=η×A×GP = \eta \times A \times G
  2. Step 2 — Rearrange symbolically for P:

    P=ηAGP = \eta A G
  3. Step 3 — List the givens: panel conversion efficiency (eta) = 0.1250, panel array area (A) = 1,813 m^2, solar irradiance (G) = 370.0 W/m^2.

  4. Step 4 — Substitute the given values:

    P=0.12501813370.0P = 0.1250 1813 370.0
  5. Step 5 — Evaluate:

    P=83851 WP = 83851\ \text{W}
  6. Step 6 — Check: returning P = 83,851 W to

    P=η×A×GP = \eta \times A \times G

    reproduces the given quantities, and both sides carry the same units.

Answer:
P=83851 WP = 83851\ \text{W}

Why the other options are there

  • 167,703 — kept a factor of two that cancels in the correct rearrangement.
  • 41,926 — dropped that same factor in the other direction.
  • 92,236 — rounded an intermediate value before the final step.

Reference: FE Handbook — Solar Energy

Example 2
Solar — solve for panel array area — Solar (2)

solar energy conversion calculation for a rooftop solar array Given panel conversion efficiency (eta) = 0.1740; solar irradiance (G) = 430.0 W/m^2; solar power output (P) = 3,166,540 W, determine the panel array area (A) in m^2.

Given

  • panelconversionefficiency(eta)=0.1740panel conversion efficiency (eta) = 0.1740
  • solarirradiance(G)=430.0W/m2solar irradiance (G) = 430.0 W/m^2
  • solarpoweroutput(P)=3,166,540Wsolar power output (P) = 3,166,540 W

Find

panel array area (A), in m^2

Start with the thinking

  • The governing relation printed in this handbook section is Solar.
  • Everything except A is given, so isolate A symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Solar power output from a photovoltaic array depends on panel efficiency, array area, and incident solar irradiance.
Solar power output vs irradiance02505007501000300Low600Medium1000Highirradiance W/m^2

Figure 2 — schematic for Solar — solve for panel array area — Solar (2)

Step-by-step solution

  1. Step 1 — State the governing relation:

    P=η×A×GP = \eta \times A \times G
  2. Step 2 — Rearrange symbolically for A:

    A=PηGA = \dfrac{P}{\eta G}
  3. Step 3 — List the givens: panel conversion efficiency (eta) = 0.1740, solar irradiance (G) = 430.0 W/m^2, solar power output (P) = 3,166,540 W.

  4. Step 4 — Substitute the given values:

    A=31665400.1740430.0A = \dfrac{3166540}{0.1740 430.0}
  5. Step 5 — Evaluate:

    A = 42322\ \text{m^2}
  6. Step 6 — Check: returning A = 42,322 m^2 to

    P=η×A×GP = \eta \times A \times G

    reproduces the given quantities, and both sides carry the same units.

Answer:
A = 42322\ \text{m^2}

Why the other options are there

  • 84,644 — kept a factor of two that cancels in the correct rearrangement.
  • 21,161 — dropped that same factor in the other direction.
  • 46,554 — rounded an intermediate value before the final step.

Reference: FE Handbook — Solar Energy

Example 3
Solar — solve for solar irradiance — Solar (3)

solar power generation as a function of irradiance and panel area Given panel conversion efficiency (eta) = 0.1650; panel array area (A) = 7,345 m^2; solar power output (P) = 1,437,200 W, determine the solar irradiance (G) in W/m^2.

Given

  • panelconversionefficiency(eta)=0.1650panel conversion efficiency (eta) = 0.1650
  • panelarrayarea(A)=7,345m2panel array area (A) = 7,345 m^2
  • solarpoweroutput(P)=1,437,200Wsolar power output (P) = 1,437,200 W

Find

solar irradiance (G), in W/m^2

Start with the thinking

  • The governing relation printed in this handbook section is Solar.
  • Everything except G is given, so isolate G symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Solar power output from a photovoltaic array depends on panel efficiency, array area, and incident solar irradiance.
Solar power output vs irradiance02505007501000300Low600Medium1000Highirradiance W/m^2

Figure 3 — schematic for Solar — solve for solar irradiance — Solar (3)

Step-by-step solution

  1. Step 1 — State the governing relation:

    P=η×A×GP = \eta \times A \times G
  2. Step 2 — Rearrange symbolically for G:

    G=PηAG = \dfrac{P}{\eta A}
  3. Step 3 — List the givens: panel conversion efficiency (eta) = 0.1650, panel array area (A) = 7,345 m^2, solar power output (P) = 1,437,200 W.

  4. Step 4 — Substitute the given values:

    G=14372000.16507345G = \dfrac{1437200}{0.1650 7345}
  5. Step 5 — Evaluate:

    G = 1186\ \text{W/m^2}
  6. Step 6 — Check: returning G = 1,186 W/m^2 to

    P=η×A×GP = \eta \times A \times G

    reproduces the given quantities, and both sides carry the same units.

Answer:
G = 1186\ \text{W/m^2}

Why the other options are there

  • 2,372 — kept a factor of two that cancels in the correct rearrangement.
  • 592.9 — dropped that same factor in the other direction.
  • 1,304 — rounded an intermediate value before the final step.

Reference: FE Handbook — Solar Energy

Example 4
Solar — solve for solar power output (case 2) — Solar (4)

solar power output from a photovoltaic array installation Given panel conversion efficiency (eta) = 0.2010; panel array area (A) = 6,245 m^2; solar irradiance (G) = 435.0 W/m^2, determine the solar power output (P) in W.

Given

  • panelconversionefficiency(eta)=0.2010panel conversion efficiency (eta) = 0.2010
  • panelarrayarea(A)=6,245m2panel array area (A) = 6,245 m^2
  • solarirradiance(G)=435.0W/m2solar irradiance (G) = 435.0 W/m^2

Find

solar power output (P), in W

Start with the thinking

  • The governing relation printed in this handbook section is Solar.
  • Everything except P is given, so isolate P symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Solar power output from a photovoltaic array depends on panel efficiency, array area, and incident solar irradiance.
Solar power output vs irradiance02505007501000300Low600Medium1000Highirradiance W/m^2

Figure 4 — schematic for Solar — solve for solar power output (case 2) — Solar (4)

Step-by-step solution

  1. Step 1 — State the governing relation:

    P=η×A×GP = \eta \times A \times G
  2. Step 2 — Rearrange symbolically for P:

    P=ηAGP = \eta A G
  3. Step 3 — List the givens: panel conversion efficiency (eta) = 0.2010, panel array area (A) = 6,245 m^2, solar irradiance (G) = 435.0 W/m^2.

  4. Step 4 — Substitute the given values:

    P=0.20106245435.0P = 0.2010 6245 435.0
  5. Step 5 — Evaluate:

    P=546032 WP = 546032\ \text{W}
  6. Step 6 — Check: returning P = 546,032 W to

    P=η×A×GP = \eta \times A \times G

    reproduces the given quantities, and both sides carry the same units.

Answer:
P=546032 WP = 546032\ \text{W}

Why the other options are there

  • 1,092,063 — kept a factor of two that cancels in the correct rearrangement.
  • 273,016 — dropped that same factor in the other direction.
  • 600,635 — rounded an intermediate value before the final step.

Reference: FE Handbook — Solar Energy

Example 5
Solar — solve for panel array area (case 2) — Solar (5)

solar energy conversion calculation for a rooftop solar array Given panel conversion efficiency (eta) = 0.2000; solar irradiance (G) = 475.0 W/m^2; solar power output (P) = 3,408,790 W, determine the panel array area (A) in m^2.

Given

  • panelconversionefficiency(eta)=0.2000panel conversion efficiency (eta) = 0.2000
  • solarirradiance(G)=475.0W/m2solar irradiance (G) = 475.0 W/m^2
  • solarpoweroutput(P)=3,408,790Wsolar power output (P) = 3,408,790 W

Find

panel array area (A), in m^2

Start with the thinking

  • The governing relation printed in this handbook section is Solar.
  • Everything except A is given, so isolate A symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Solar power output from a photovoltaic array depends on panel efficiency, array area, and incident solar irradiance.
Solar power output vs irradiance02505007501000300Low600Medium1000Highirradiance W/m^2

Figure 5 — schematic for Solar — solve for panel array area (case 2) — Solar (5)

Step-by-step solution

  1. Step 1 — State the governing relation:

    P=η×A×GP = \eta \times A \times G
  2. Step 2 — Rearrange symbolically for A:

    A=PηGA = \dfrac{P}{\eta G}
  3. Step 3 — List the givens: panel conversion efficiency (eta) = 0.2000, solar irradiance (G) = 475.0 W/m^2, solar power output (P) = 3,408,790 W.

  4. Step 4 — Substitute the given values:

    A=34087900.2000475.0A = \dfrac{3408790}{0.2000 475.0}
  5. Step 5 — Evaluate:

    A = 35882\ \text{m^2}
  6. Step 6 — Check: returning A = 35,882 m^2 to

    P=η×A×GP = \eta \times A \times G

    reproduces the given quantities, and both sides carry the same units.

Answer:
A = 35882\ \text{m^2}

Why the other options are there

  • 71,764 — kept a factor of two that cancels in the correct rearrangement.
  • 17,941 — dropped that same factor in the other direction.
  • 39,470 — rounded an intermediate value before the final step.

Reference: FE Handbook — Solar Energy

Example 6
Solar — solve for solar irradiance (case 2) — Solar (6)

solar power generation as a function of irradiance and panel area Given panel conversion efficiency (eta) = 0.1950; panel array area (A) = 17,028 m^2; solar power output (P) = 595,640 W, determine the solar irradiance (G) in W/m^2.

Given

  • panelconversionefficiency(eta)=0.1950panel conversion efficiency (eta) = 0.1950
  • panelarrayarea(A)=17,028m2panel array area (A) = 17,028 m^2
  • solarpoweroutput(P)=595,640Wsolar power output (P) = 595,640 W

Find

solar irradiance (G), in W/m^2

Start with the thinking

  • The governing relation printed in this handbook section is Solar.
  • Everything except G is given, so isolate G symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Solar power output from a photovoltaic array depends on panel efficiency, array area, and incident solar irradiance.
Solar power output vs irradiance02505007501000300Low600Medium1000Highirradiance W/m^2

Figure 6 — schematic for Solar — solve for solar irradiance (case 2) — Solar (6)

Step-by-step solution

  1. Step 1 — State the governing relation:

    P=η×A×GP = \eta \times A \times G
  2. Step 2 — Rearrange symbolically for G:

    G=PηAG = \dfrac{P}{\eta A}
  3. Step 3 — List the givens: panel conversion efficiency (eta) = 0.1950, panel array area (A) = 17,028 m^2, solar power output (P) = 595,640 W.

  4. Step 4 — Substitute the given values:

    G=5956400.195017028G = \dfrac{595640}{0.1950 17028}
  5. Step 5 — Evaluate:

    G = 179.4\ \text{W/m^2}
  6. Step 6 — Check: returning G = 179.4 W/m^2 to

    P=η×A×GP = \eta \times A \times G

    reproduces the given quantities, and both sides carry the same units.

Answer:
G = 179.4\ \text{W/m^2}

Why the other options are there

  • 358.8 — kept a factor of two that cancels in the correct rearrangement.
  • 89.6924 — dropped that same factor in the other direction.
  • 197.3 — rounded an intermediate value before the final step.

Reference: FE Handbook — Solar Energy

Example 7
Solar — solve for solar power output (case 3) — Solar (7)

solar power output from a photovoltaic array installation Given panel conversion efficiency (eta) = 0.1800; panel array area (A) = 15,967 m^2; solar irradiance (G) = 795.0 W/m^2, determine the solar power output (P) in W.

Given

  • panelconversionefficiency(eta)=0.1800panel conversion efficiency (eta) = 0.1800
  • panelarrayarea(A)=15,967m2panel array area (A) = 15,967 m^2
  • solarirradiance(G)=795.0W/m2solar irradiance (G) = 795.0 W/m^2

Find

solar power output (P), in W

Start with the thinking

  • The governing relation printed in this handbook section is Solar.
  • Everything except P is given, so isolate P symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Solar power output from a photovoltaic array depends on panel efficiency, array area, and incident solar irradiance.
Solar power output vs irradiance02505007501000300Low600Medium1000Highirradiance W/m^2

Figure 7 — schematic for Solar — solve for solar power output (case 3) — Solar (7)

Step-by-step solution

  1. Step 1 — State the governing relation:

    P=η×A×GP = \eta \times A \times G
  2. Step 2 — Rearrange symbolically for P:

    P=ηAGP = \eta A G
  3. Step 3 — List the givens: panel conversion efficiency (eta) = 0.1800, panel array area (A) = 15,967 m^2, solar irradiance (G) = 795.0 W/m^2.

  4. Step 4 — Substitute the given values:

    P=0.180015967795.0P = 0.1800 15967 795.0
  5. Step 5 — Evaluate:

    P=2284878 WP = 2284878\ \text{W}
  6. Step 6 — Check: returning P = 2,284,878 W to

    P=η×A×GP = \eta \times A \times G

    reproduces the given quantities, and both sides carry the same units.

Answer:
P=2284878 WP = 2284878\ \text{W}

Why the other options are there

  • 4,569,755 — kept a factor of two that cancels in the correct rearrangement.
  • 1,142,439 — dropped that same factor in the other direction.
  • 2,513,365 — rounded an intermediate value before the final step.

Reference: FE Handbook — Solar Energy

Example 8
Solar — solve for panel array area (case 3) — Solar (8)

solar energy conversion calculation for a rooftop solar array Given panel conversion efficiency (eta) = 0.2190; solar irradiance (G) = 500.0 W/m^2; solar power output (P) = 30,540 W, determine the panel array area (A) in m^2.

Given

  • panelconversionefficiency(eta)=0.2190panel conversion efficiency (eta) = 0.2190
  • solarirradiance(G)=500.0W/m2solar irradiance (G) = 500.0 W/m^2
  • solarpoweroutput(P)=30,540Wsolar power output (P) = 30,540 W

Find

panel array area (A), in m^2

Start with the thinking

  • The governing relation printed in this handbook section is Solar.
  • Everything except A is given, so isolate A symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Solar power output from a photovoltaic array depends on panel efficiency, array area, and incident solar irradiance.
Solar power output vs irradiance02505007501000300Low600Medium1000Highirradiance W/m^2

Figure 8 — schematic for Solar — solve for panel array area (case 3) — Solar (8)

Step-by-step solution

  1. Step 1 — State the governing relation:

    P=η×A×GP = \eta \times A \times G
  2. Step 2 — Rearrange symbolically for A:

    A=PηGA = \dfrac{P}{\eta G}
  3. Step 3 — List the givens: panel conversion efficiency (eta) = 0.2190, solar irradiance (G) = 500.0 W/m^2, solar power output (P) = 30,540 W.

  4. Step 4 — Substitute the given values:

    A=305400.2190500.0A = \dfrac{30540}{0.2190 500.0}
  5. Step 5 — Evaluate:

    A = 278.9\ \text{m^2}
  6. Step 6 — Check: returning A = 278.9 m^2 to

    P=η×A×GP = \eta \times A \times G

    reproduces the given quantities, and both sides carry the same units.

Answer:
A = 278.9\ \text{m^2}

Why the other options are there

  • 557.8 — kept a factor of two that cancels in the correct rearrangement.
  • 139.5 — dropped that same factor in the other direction.
  • 306.8 — rounded an intermediate value before the final step.

Reference: FE Handbook — Solar Energy

Example 9
Solar — solve for solar irradiance (case 3) — Solar (9)

solar power generation as a function of irradiance and panel area Given panel conversion efficiency (eta) = 0.1600; panel array area (A) = 399.0 m^2; solar power output (P) = 265,490 W, determine the solar irradiance (G) in W/m^2.

Given

  • panelconversionefficiency(eta)=0.1600panel conversion efficiency (eta) = 0.1600
  • panelarrayarea(A)=399.0m2panel array area (A) = 399.0 m^2
  • solarpoweroutput(P)=265,490Wsolar power output (P) = 265,490 W

Find

solar irradiance (G), in W/m^2

Start with the thinking

  • The governing relation printed in this handbook section is Solar.
  • Everything except G is given, so isolate G symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Solar power output from a photovoltaic array depends on panel efficiency, array area, and incident solar irradiance.
Solar power output vs irradiance02505007501000300Low600Medium1000Highirradiance W/m^2

Figure 9 — schematic for Solar — solve for solar irradiance (case 3) — Solar (9)

Step-by-step solution

  1. Step 1 — State the governing relation:

    P=η×A×GP = \eta \times A \times G
  2. Step 2 — Rearrange symbolically for G:

    G=PηAG = \dfrac{P}{\eta A}
  3. Step 3 — List the givens: panel conversion efficiency (eta) = 0.1600, panel array area (A) = 399.0 m^2, solar power output (P) = 265,490 W.

  4. Step 4 — Substitute the given values:

    G=2654900.1600399.0G = \dfrac{265490}{0.1600 399.0}
  5. Step 5 — Evaluate:

    G = 4159\ \text{W/m^2}
  6. Step 6 — Check: returning G = 4,159 W/m^2 to

    P=η×A×GP = \eta \times A \times G

    reproduces the given quantities, and both sides carry the same units.

Answer:
G = 4159\ \text{W/m^2}

Why the other options are there

  • 8,317 — kept a factor of two that cancels in the correct rearrangement.
  • 2,079 — dropped that same factor in the other direction.
  • 4,575 — rounded an intermediate value before the final step.

Reference: FE Handbook — Solar Energy

Example 10
Solar — solve for solar power output (case 4) — Solar (10)

solar power output from a photovoltaic array installation Given panel conversion efficiency (eta) = 0.1250; panel array area (A) = 18,879 m^2; solar irradiance (G) = 635.0 W/m^2, determine the solar power output (P) in W.

Given

  • panelconversionefficiency(eta)=0.1250panel conversion efficiency (eta) = 0.1250
  • panelarrayarea(A)=18,879m2panel array area (A) = 18,879 m^2
  • solarirradiance(G)=635.0W/m2solar irradiance (G) = 635.0 W/m^2

Find

solar power output (P), in W

Start with the thinking

  • The governing relation printed in this handbook section is Solar.
  • Everything except P is given, so isolate P symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Solar power output from a photovoltaic array depends on panel efficiency, array area, and incident solar irradiance.
Solar power output vs irradiance02505007501000300Low600Medium1000Highirradiance W/m^2

Figure 10 — schematic for Solar — solve for solar power output (case 4) — Solar (10)

Step-by-step solution

  1. Step 1 — State the governing relation:

    P=η×A×GP = \eta \times A \times G
  2. Step 2 — Rearrange symbolically for P:

    P=ηAGP = \eta A G
  3. Step 3 — List the givens: panel conversion efficiency (eta) = 0.1250, panel array area (A) = 18,879 m^2, solar irradiance (G) = 635.0 W/m^2.

  4. Step 4 — Substitute the given values:

    P=0.125018879635.0P = 0.1250 18879 635.0
  5. Step 5 — Evaluate:

    P=1498521 WP = 1498521\ \text{W}
  6. Step 6 — Check: returning P = 1,498,521 W to

    P=η×A×GP = \eta \times A \times G

    reproduces the given quantities, and both sides carry the same units.

Answer:
P=1498521 WP = 1498521\ \text{W}

Why the other options are there

  • 2,997,041 — kept a factor of two that cancels in the correct rearrangement.
  • 749,260 — dropped that same factor in the other direction.
  • 1,648,373 — rounded an intermediate value before the final step.

Reference: FE Handbook — Solar Energy

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