Pan Evaporation

Hydrology and Water Resources · FE Reference Handbook section

Hydrology and Water Resources

2 formulas

10 exam-style examples

~49 min

All Hydrology and Water Resources lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Pan Evaporation within Hydrology and Water Resources. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what pan evaporation describes physically and when it applies.
State every one of the 2 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: acre-in/hr ≈ cfs makes the rational formula work in US units.

Lecture

Why this section exists. Pan Evaporation is the part of Hydrology and Water Resources that lets you connect a watershed, aquifer or detention facility to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as a rainfall-runoff or well-drawdown calculation with one lookup. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. acre-in/hr ≈ cfs makes the rational formula work in US units. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Row of centrifugal pumps and valved steel piping inside a water pumping station. — Photo 1. Where this shows up in practice: pan evaporation.
Capstone Studio instructional photograph

Hydrology and Water Resources — Pan Evaporation: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a watershed, aquifer or detention facility. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 2 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

EL	Quantity produced by "EL = Pc Ep" — read its definition and unit from the handbook line directly above the equation.
where Pc	Quantity produced by "where Pc = pan coefficient (Pc ranges from 0.3 to 0.85; common Pc = 0.7)" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

The lake evaporation EL is related to the pan evaporation Ep by the expression:

Core formulas for this FE topic

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

Worked exam-style examples

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

Example 1

Monthly reservoir water balance and storage change — Pan Evaporation

A reservoir with a 80 acre surface receives 97 ac-ft of stream inflow in a month, plus 5.5 in of direct rainfall, while losing 2.0 in to evaporation and releasing 82 ac-ft to demand. Find the change in storage.

Given

I = 97 ac-ft
P = 5.5 in
E = 2.0 in
A_s = 80 ac
D = 82 ac-ft

Find

ΔS for the month

Start with the thinking

Depths over the water surface convert to volume with V = (depth in inches / 12) × area in acres.
The continuity budget is ΔS = inflows − outflows.

Step-by-step solution

Rain volume
$V_{P} = (P /12) A_{s}$
Substituting
$V_{P} = (5.5/12) \times 80 = 36.67 a c - f t$
Evaporation volume
$V_{E} = (E /12) A_{s}$
Substituting
$V_{E} = (2.0/12) \times 80 = 13.33 a c - f t$
Balance — ΔS = I + V_P − V_E − D
Substituting — ΔS = 97 + 36.67 − 13.33 − 82
Evaluate — ΔS = 38.33 ac-ft (storage gained)

Answer: ΔS ≈ 38.3 ac-ft

Why the other options are there

15.0 ac-ft (rain and evaporation ignored)
18.5 ac-ft (inches added as volumes)

Reference: FE Reference Handbook — Hydrology and Water Resources → Pan Evaporation

Example 2

Reservoir water budget with rainfall and evaporation — Pan Evaporation

A reservoir with a surface area of 847 ha receives 181 mm of rainfall and loses 167 mm to evaporation in a 30-day month while a stream delivers 0.05 m³/s. Compute the net storage change in volume and the corresponding change in water-surface elevation.

Given

A = 847 ha = 8,470,000 m²
P = 181 mm
E = 167 mm
Q_in = 0.05 m³/s for 30 days

Find

Net volume change and stage change

Start with the thinking

The water budget is simply inflow + precipitation − evaporation − outflow = ΔS.
Depths in millimetres must be converted to metres before multiplying by the surface area.

Step-by-step solution

Formula — ΔS = Q_in Δt + P·A − E·A
Stream inflow
$0.05 (86, 400) (30) = 129, 600 m^{3}$
Rainfall
$(181/1000) (8, 470, 000) = 1, 533, 070 m^{3}$
Evaporation
$(167/1000) (8, 470, 000) = 1, 414, 490 m^{3}$
Substituting — ΔS = 129,600 + 1,533,070 − 1,414,490 = 248,180 m³
Formula — Δz = ΔS/A
Substituting — Δz = 248,180/8,470,000 = 0.029 m

Answer: ΔS = 248,180 m³, raising the pool 0.03 m

Why the other options are there

118,580 m³ (stream inflow ignored)
14 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Pan Evaporation

Example 3

Monthly reservoir water balance and storage change — Pan Evaporation (2)

A reservoir with a 330.0 acre surface receives 85 ac-ft of stream inflow in a month, plus 3.0 in of direct rainfall, while losing 4.0 in to evaporation and releasing 65 ac-ft to demand. Find the change in storage.

Given

I = 85 ac-ft
P = 3.0 in
E = 4.0 in
A_s = 330.0 ac
D = 65 ac-ft

Find

ΔS for the month

Start with the thinking

Depths over the water surface convert to volume with V = (depth in inches / 12) × area in acres.
The continuity budget is ΔS = inflows − outflows.

Step-by-step solution

Rain volume
$V_{P} = (P /12) A_{s}$
Substituting
$V_{P} = (3.0/12) \times 330.0 = 82.50 a c - f t$
Evaporation volume
$V_{E} = (E /12) A_{s}$
Substituting
$V_{E} = (4.0/12) \times 330.0 = 110.0 a c - f t$
Balance — ΔS = I + V_P − V_E − D
Substituting — ΔS = 85 + 82.50 − 110.0 − 65
Evaluate — ΔS = -7.50 ac-ft (storage drawn down)

Answer: ΔS ≈ -7.5 ac-ft

Why the other options are there

20.0 ac-ft (rain and evaporation ignored)
19.0 ac-ft (inches added as volumes)

Reference: FE Reference Handbook — Hydrology and Water Resources → Pan Evaporation

Example 4

Reservoir water budget with rainfall and evaporation — Pan Evaporation (2)

A reservoir with a surface area of 260 ha receives 102 mm of rainfall and loses 96 mm to evaporation in a 30-day month while a stream delivers 0.15 m³/s. Compute the net storage change in volume and the corresponding change in water-surface elevation.

Given

A = 260 ha = 2,600,000 m²
P = 102 mm
E = 96 mm
Q_in = 0.15 m³/s for 30 days

Find

Net volume change and stage change

Start with the thinking

The water budget is simply inflow + precipitation − evaporation − outflow = ΔS.
Depths in millimetres must be converted to metres before multiplying by the surface area.

Step-by-step solution

Formula — ΔS = Q_in Δt + P·A − E·A
Stream inflow
$0.15 (86, 400) (30) = 388, 800 m^{3}$
Rainfall
$(102/1000) (2, 600, 000) = 265, 200 m^{3}$
Evaporation
$(96/1000) (2, 600, 000) = 249, 600 m^{3}$
Substituting — ΔS = 388,800 + 265,200 − 249,600 = 404,400 m³
Formula — Δz = ΔS/A
Substituting — Δz = 404,400/2,600,000 = 0.156 m

Answer: ΔS = 404,400 m³, raising the pool 0.16 m

Why the other options are there

15,600 m³ (stream inflow ignored)
6 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Pan Evaporation

Example 5

Monthly reservoir water balance and storage change — Pan Evaporation (3)

A reservoir with a 220.0 acre surface receives 44 ac-ft of stream inflow in a month, plus 6.0 in of direct rainfall, while losing 7.0 in to evaporation and releasing 17 ac-ft to demand. Find the change in storage.

Given

I = 44 ac-ft
P = 6.0 in
E = 7.0 in
A_s = 220.0 ac
D = 17 ac-ft

Find

ΔS for the month

Start with the thinking

Depths over the water surface convert to volume with V = (depth in inches / 12) × area in acres.
The continuity budget is ΔS = inflows − outflows.

Step-by-step solution

Rain volume
$V_{P} = (P /12) A_{s}$
Substituting
$V_{P} = (6.0/12) \times 220.0 = 110.0 a c - f t$
Evaporation volume
$V_{E} = (E /12) A_{s}$
Substituting
$V_{E} = (7.0/12) \times 220.0 = 128.3 a c - f t$
Balance — ΔS = I + V_P − V_E − D
Substituting — ΔS = 44 + 110.0 − 128.3 − 17
Evaluate — ΔS = 8.67 ac-ft (storage gained)

Answer: ΔS ≈ 8.7 ac-ft

Why the other options are there

27.0 ac-ft (rain and evaporation ignored)
26.0 ac-ft (inches added as volumes)

Reference: FE Reference Handbook — Hydrology and Water Resources → Pan Evaporation

Example 6

Reservoir water budget with rainfall and evaporation — Pan Evaporation (3)

A reservoir with a surface area of 440 ha receives 98 mm of rainfall and loses 93 mm to evaporation in a 30-day month while a stream delivers 0.50 m³/s. Compute the net storage change in volume and the corresponding change in water-surface elevation.

Given

A = 440 ha = 4,400,000 m²
P = 98 mm
E = 93 mm
Q_in = 0.50 m³/s for 30 days

Find

Net volume change and stage change

Start with the thinking

The water budget is simply inflow + precipitation − evaporation − outflow = ΔS.
Depths in millimetres must be converted to metres before multiplying by the surface area.

Step-by-step solution

Formula — ΔS = Q_in Δt + P·A − E·A
Stream inflow
$0.50 (86, 400) (30) = 1, 296, 000 m^{3}$
Rainfall
$(98/1000) (4, 400, 000) = 431, 200 m^{3}$
Evaporation
$(93/1000) (4, 400, 000) = 409, 200 m^{3}$
Substituting — ΔS = 1,296,000 + 431,200 − 409,200 = 1,318,000 m³
Formula — Δz = ΔS/A
Substituting — Δz = 1,318,000/4,400,000 = 0.300 m

Answer: ΔS = 1,318,000 m³, raising the pool 0.30 m

Why the other options are there

22,000 m³ (stream inflow ignored)
5 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Pan Evaporation

Example 7

Monthly reservoir water balance and storage change — Pan Evaporation (4)

A reservoir with a 40 acre surface receives 55 ac-ft of stream inflow in a month, plus 4.0 in of direct rainfall, while losing 4.0 in to evaporation and releasing 51 ac-ft to demand. Find the change in storage.

Given

I = 55 ac-ft
P = 4.0 in
E = 4.0 in
A_s = 40 ac
D = 51 ac-ft

Find

ΔS for the month

Start with the thinking

Depths over the water surface convert to volume with V = (depth in inches / 12) × area in acres.
The continuity budget is ΔS = inflows − outflows.

Step-by-step solution

Rain volume
$V_{P} = (P /12) A_{s}$
Substituting
$V_{P} = (4.0/12) \times 40 = 13.33 a c - f t$
Evaporation volume
$V_{E} = (E /12) A_{s}$
Substituting
$V_{E} = (4.0/12) \times 40 = 13.33 a c - f t$
Balance — ΔS = I + V_P − V_E − D
Substituting — ΔS = 55 + 13.33 − 13.33 − 51
Evaluate — ΔS = 4.00 ac-ft (storage gained)

Answer: ΔS ≈ 4.0 ac-ft

Why the other options are there

4.0 ac-ft (rain and evaporation ignored)
4.0 ac-ft (inches added as volumes)

Reference: FE Reference Handbook — Hydrology and Water Resources → Pan Evaporation

Example 8

Reservoir water budget with rainfall and evaporation — Pan Evaporation (4)

A reservoir with a surface area of 149 ha receives 130 mm of rainfall and loses 100 mm to evaporation in a 30-day month while a stream delivers 0.85 m³/s. Compute the net storage change in volume and the corresponding change in water-surface elevation.

Given

A = 149 ha = 1,490,000 m²
P = 130 mm
E = 100 mm
Q_in = 0.85 m³/s for 30 days

Find

Net volume change and stage change

Start with the thinking

The water budget is simply inflow + precipitation − evaporation − outflow = ΔS.
Depths in millimetres must be converted to metres before multiplying by the surface area.

Step-by-step solution

Formula — ΔS = Q_in Δt + P·A − E·A
Stream inflow
$0.85 (86, 400) (30) = 2, 203, 200 m^{3}$
Rainfall
$(130/1000) (1, 490, 000) = 193, 700 m^{3}$
Evaporation
$(100/1000) (1, 490, 000) = 149, 000 m^{3}$
Substituting — ΔS = 2,203,200 + 193,700 − 149,000 = 2,247,900 m³
Formula — Δz = ΔS/A
Substituting — Δz = 2,247,900/1,490,000 = 1.509 m

Answer: ΔS = 2,247,900 m³, raising the pool 1.51 m

Why the other options are there

44,700 m³ (stream inflow ignored)
30 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Pan Evaporation

Example 9

Monthly reservoir water balance and storage change — Pan Evaporation (5)

A reservoir with a 140.0 acre surface receives 111.0 ac-ft of stream inflow in a month, plus 5.0 in of direct rainfall, while losing 5.0 in to evaporation and releasing 23 ac-ft to demand. Find the change in storage.

Given

I = 111.0 ac-ft
P = 5.0 in
E = 5.0 in
A_s = 140.0 ac
D = 23 ac-ft

Find

ΔS for the month

Start with the thinking

Depths over the water surface convert to volume with V = (depth in inches / 12) × area in acres.
The continuity budget is ΔS = inflows − outflows.

Step-by-step solution

Rain volume
$V_{P} = (P /12) A_{s}$
Substituting
$V_{P} = (5.0/12) \times 140.0 = 58.33 a c - f t$
Evaporation volume
$V_{E} = (E /12) A_{s}$
Substituting
$V_{E} = (5.0/12) \times 140.0 = 58.33 a c - f t$
Balance — ΔS = I + V_P − V_E − D
Substituting — ΔS = 111.0 + 58.33 − 58.33 − 23
Evaluate — ΔS = 88.00 ac-ft (storage gained)

Answer: ΔS ≈ 88.0 ac-ft

Why the other options are there

88.0 ac-ft (rain and evaporation ignored)
88.0 ac-ft (inches added as volumes)

Reference: FE Reference Handbook — Hydrology and Water Resources → Pan Evaporation

Example 10

Reservoir water budget with rainfall and evaporation — Pan Evaporation (5)

A reservoir with a surface area of 208 ha receives 38 mm of rainfall and loses 161 mm to evaporation in a 30-day month while a stream delivers 1.05 m³/s. Compute the net storage change in volume and the corresponding change in water-surface elevation.

Given

A = 208 ha = 2,080,000 m²
P = 38 mm
E = 161 mm
Q_in = 1.05 m³/s for 30 days

Find

Net volume change and stage change

Start with the thinking

The water budget is simply inflow + precipitation − evaporation − outflow = ΔS.
Depths in millimetres must be converted to metres before multiplying by the surface area.

Step-by-step solution

Formula — ΔS = Q_in Δt + P·A − E·A
Stream inflow
$1.05 (86, 400) (30) = 2, 721, 600 m^{3}$
Rainfall
$(38/1000) (2, 080, 000) = 79, 040 m^{3}$
Evaporation
$(161/1000) (2, 080, 000) = 334, 880 m^{3}$
Substituting — ΔS = 2,721,600 + 79,040 − 334,880 = 2,465,760 m³
Formula — Δz = ΔS/A
Substituting — Δz = 2,465,760/2,080,000 = 1.185 m

Answer: ΔS = 2,465,760 m³, raising the pool 1.19 m

Why the other options are there

-255,840 m³ (stream inflow ignored)
-123.0 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Pan Evaporation

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a watershed, aquifer or detention facility, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Pan Evaporation contains 2 relations; you must be able to find this page in under 15 seconds.
Exam style: a rainfall-runoff or well-drawdown calculation with one lookup.
Unit rule: acre-in/hr ≈ cfs makes the rational formula work in US units.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

acre-in/hr ≈ cfs makes the rational formula work in US units
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.

Pan Evaporation

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

EL=PcEp

wherePc=pancoefficient(Pcrangesfrom0.3to0.85;commonPc=0.7)

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section