Evapotranspiration Rates for Grasses

Hydrology and Water Resources · FE Reference Handbook section

Hydrology and Water Resources

0 formulas

10 exam-style examples

~45 min

All Hydrology and Water Resources lectures

Learning objectives

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

This chapter section covers Evapotranspiration Rates for Grasses 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 evapotranspiration rates for grasses describes physically and when it applies.
State every one of the 0 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. Evapotranspiration Rates for Grasses 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: evapotranspiration rates for grasses.
Capstone Studio instructional photograph

Hydrology and Water Resources — Evapotranspiration Rates for Grasses: 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 0 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.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

Grass-Reference Evapotranspiration Rates
Range (mm/day) Classification
0–2.5 Low
2.5–5.0 Moderate
5.0–7.5 High
>7.5 Very High
Source: Allen et al. (1998)
Chin, David, Water-Resources Engineering, 2nd ed., Pearson, 2006, p. 542.

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

Reservoir water budget with rainfall and evaporation — Evapotranspiration Rates for Grasses

A reservoir with a surface area of 110 ha receives 31 mm of rainfall and loses 158 mm to evaporation in a 30-day month while a stream delivers 1.45 m³/s. Compute the net storage change in volume and the corresponding change in water-surface elevation.

Given

A = 110 ha = 1,100,000 m²
P = 31 mm
E = 158 mm
Q_in = 1.45 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
$1.45 (86, 400) (30) = 3, 758, 400 m^{3}$
Rainfall
$(31/1000) (1, 100, 000) = 34, 100 m^{3}$
Evaporation
$(158/1000) (1, 100, 000) = 173, 800 m^{3}$
Substituting — ΔS = 3,758,400 + 34,100 − 173,800 = 3,618,700 m³
Formula — Δz = ΔS/A
Substituting — Δz = 3,618,700/1,100,000 = 3.290 m

Answer: ΔS = 3,618,700 m³, raising the pool 3.29 m

Why the other options are there

-139,700 m³ (stream inflow ignored)
-127.0 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Evapotranspiration Rates for Grasses

Example 2

Reservoir water budget with rainfall and evaporation — Evapotranspiration Rates for Grasses (2)

A reservoir with a surface area of 256 ha receives 30 mm of rainfall and loses 57 mm to evaporation in a 30-day month while a stream delivers 0.80 m³/s. Compute the net storage change in volume and the corresponding change in water-surface elevation.

Given

A = 256 ha = 2,560,000 m²
P = 30 mm
E = 57 mm
Q_in = 0.80 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
$0.80 (86, 400) (30) = 2, 073, 600 m^{3}$
Rainfall
$(30/1000) (2, 560, 000) = 76, 800 m^{3}$
Evaporation
$(57/1000) (2, 560, 000) = 145, 920 m^{3}$
Substituting — ΔS = 2,073,600 + 76,800 − 145,920 = 2,004,480 m³
Formula — Δz = ΔS/A
Substituting — Δz = 2,004,480/2,560,000 = 0.783 m

Answer: ΔS = 2,004,480 m³, raising the pool 0.78 m

Why the other options are there

-69,120 m³ (stream inflow ignored)
-27 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Evapotranspiration Rates for Grasses

Example 3

Reservoir water budget with rainfall and evaporation — Evapotranspiration Rates for Grasses (3)

A reservoir with a surface area of 601 ha receives 53 mm of rainfall and loses 107 mm to evaporation in a 30-day month while a stream delivers 0.95 m³/s. Compute the net storage change in volume and the corresponding change in water-surface elevation.

Given

A = 601 ha = 6,010,000 m²
P = 53 mm
E = 107 mm
Q_in = 0.95 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
$0.95 (86, 400) (30) = 2, 462, 400 m^{3}$
Rainfall
$(53/1000) (6, 010, 000) = 318, 530 m^{3}$
Evaporation
$(107/1000) (6, 010, 000) = 643, 070 m^{3}$
Substituting — ΔS = 2,462,400 + 318,530 − 643,070 = 2,137,860 m³
Formula — Δz = ΔS/A
Substituting — Δz = 2,137,860/6,010,000 = 0.356 m

Answer: ΔS = 2,137,860 m³, raising the pool 0.36 m

Why the other options are there

-324,540 m³ (stream inflow ignored)
-54 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Evapotranspiration Rates for Grasses

Example 4

Reservoir water budget with rainfall and evaporation — Evapotranspiration Rates for Grasses (4)

A reservoir with a surface area of 643 ha receives 156 mm of rainfall and loses 94 mm to evaporation in a 30-day month while a stream delivers 0.30 m³/s. Compute the net storage change in volume and the corresponding change in water-surface elevation.

Given

A = 643 ha = 6,430,000 m²
P = 156 mm
E = 94 mm
Q_in = 0.30 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
$0.30 (86, 400) (30) = 777, 600 m^{3}$
Rainfall
$(156/1000) (6, 430, 000) = 1, 003, 080 m^{3}$
Evaporation
$(94/1000) (6, 430, 000) = 604, 420 m^{3}$
Substituting — ΔS = 777,600 + 1,003,080 − 604,420 = 1,176,260 m³
Formula — Δz = ΔS/A
Substituting — Δz = 1,176,260/6,430,000 = 0.183 m

Answer: ΔS = 1,176,260 m³, raising the pool 0.18 m

Why the other options are there

398,660 m³ (stream inflow ignored)
62 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Evapotranspiration Rates for Grasses

Example 5

Reservoir water budget with rainfall and evaporation — Evapotranspiration Rates for Grasses (5)

A reservoir with a surface area of 832 ha receives 153 mm of rainfall and loses 177 mm to evaporation in a 30-day month while a stream delivers 0.20 m³/s. Compute the net storage change in volume and the corresponding change in water-surface elevation.

Given

A = 832 ha = 8,320,000 m²
P = 153 mm
E = 177 mm
Q_in = 0.20 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
$0.20 (86, 400) (30) = 518, 400 m^{3}$
Rainfall
$(153/1000) (8, 320, 000) = 1, 272, 960 m^{3}$
Evaporation
$(177/1000) (8, 320, 000) = 1, 472, 640 m^{3}$
Substituting — ΔS = 518,400 + 1,272,960 − 1,472,640 = 318,720 m³
Formula — Δz = ΔS/A
Substituting — Δz = 318,720/8,320,000 = 0.038 m

Answer: ΔS = 318,720 m³, raising the pool 0.04 m

Why the other options are there

-199,680 m³ (stream inflow ignored)
-24 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Evapotranspiration Rates for Grasses

Example 6

Reservoir water budget with rainfall and evaporation — Evapotranspiration Rates for Grasses (6)

A reservoir with a surface area of 895 ha receives 163 mm of rainfall and loses 135 mm to evaporation in a 30-day month while a stream delivers 1.50 m³/s. Compute the net storage change in volume and the corresponding change in water-surface elevation.

Given

A = 895 ha = 8,950,000 m²
P = 163 mm
E = 135 mm
Q_in = 1.50 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
$1.50 (86, 400) (30) = 3, 888, 000 m^{3}$
Rainfall
$(163/1000) (8, 950, 000) = 1, 458, 850 m^{3}$
Evaporation
$(135/1000) (8, 950, 000) = 1, 208, 250 m^{3}$
Substituting — ΔS = 3,888,000 + 1,458,850 − 1,208,250 = 4,138,600 m³
Formula — Δz = ΔS/A
Substituting — Δz = 4,138,600/8,950,000 = 0.462 m

Answer: ΔS = 4,138,600 m³, raising the pool 0.46 m

Why the other options are there

250,600 m³ (stream inflow ignored)
28 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Evapotranspiration Rates for Grasses

Example 7

Reservoir water budget with rainfall and evaporation — Evapotranspiration Rates for Grasses (7)

A reservoir with a surface area of 350 ha receives 120 mm of rainfall and loses 140 mm to evaporation in a 30-day month while a stream delivers 0.45 m³/s. Compute the net storage change in volume and the corresponding change in water-surface elevation.

Given

A = 350 ha = 3,500,000 m²
P = 120 mm
E = 140 mm
Q_in = 0.45 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
$0.45 (86, 400) (30) = 1, 166, 400 m^{3}$
Rainfall
$(120/1000) (3, 500, 000) = 420, 000 m^{3}$
Evaporation
$(140/1000) (3, 500, 000) = 490, 000 m^{3}$
Substituting — ΔS = 1,166,400 + 420,000 − 490,000 = 1,096,400 m³
Formula — Δz = ΔS/A
Substituting — Δz = 1,096,400/3,500,000 = 0.313 m

Answer: ΔS = 1,096,400 m³, raising the pool 0.31 m

Why the other options are there

-70,000 m³ (stream inflow ignored)
-20 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Evapotranspiration Rates for Grasses

Example 8

Reservoir water budget with rainfall and evaporation — Evapotranspiration Rates for Grasses (8)

A reservoir with a surface area of 164 ha receives 175 mm of rainfall and loses 88 mm to evaporation in a 30-day month while a stream delivers 1.40 m³/s. Compute the net storage change in volume and the corresponding change in water-surface elevation.

Given

A = 164 ha = 1,640,000 m²
P = 175 mm
E = 88 mm
Q_in = 1.40 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
$1.40 (86, 400) (30) = 3, 628, 800 m^{3}$
Rainfall
$(175/1000) (1, 640, 000) = 287, 000 m^{3}$
Evaporation
$(88/1000) (1, 640, 000) = 144, 320 m^{3}$
Substituting — ΔS = 3,628,800 + 287,000 − 144,320 = 3,771,480 m³
Formula — Δz = ΔS/A
Substituting — Δz = 3,771,480/1,640,000 = 2.300 m

Answer: ΔS = 3,771,480 m³, raising the pool 2.30 m

Why the other options are there

142,680 m³ (stream inflow ignored)
87 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Evapotranspiration Rates for Grasses

Example 9

Reservoir water budget with rainfall and evaporation — Evapotranspiration Rates for Grasses (9)

A reservoir with a surface area of 824 ha receives 30 mm of rainfall and loses 75 mm to evaporation in a 30-day month while a stream delivers 1.40 m³/s. Compute the net storage change in volume and the corresponding change in water-surface elevation.

Given

A = 824 ha = 8,240,000 m²
P = 30 mm
E = 75 mm
Q_in = 1.40 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
$1.40 (86, 400) (30) = 3, 628, 800 m^{3}$
Rainfall
$(30/1000) (8, 240, 000) = 247, 200 m^{3}$
Evaporation
$(75/1000) (8, 240, 000) = 618, 000 m^{3}$
Substituting — ΔS = 3,628,800 + 247,200 − 618,000 = 3,258,000 m³
Formula — Δz = ΔS/A
Substituting — Δz = 3,258,000/8,240,000 = 0.395 m

Answer: ΔS = 3,258,000 m³, raising the pool 0.40 m

Why the other options are there

-370,800 m³ (stream inflow ignored)
-45 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Evapotranspiration Rates for Grasses

Example 10

Reservoir water budget with rainfall and evaporation — Evapotranspiration Rates for Grasses (10)

A reservoir with a surface area of 485 ha receives 75 mm of rainfall and loses 81 mm to evaporation in a 30-day month while a stream delivers 0.65 m³/s. Compute the net storage change in volume and the corresponding change in water-surface elevation.

Given

A = 485 ha = 4,850,000 m²
P = 75 mm
E = 81 mm
Q_in = 0.65 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
$0.65 (86, 400) (30) = 1, 684, 800 m^{3}$
Rainfall
$(75/1000) (4, 850, 000) = 363, 750 m^{3}$
Evaporation
$(81/1000) (4, 850, 000) = 392, 850 m^{3}$
Substituting — ΔS = 1,684,800 + 363,750 − 392,850 = 1,655,700 m³
Formula — Δz = ΔS/A
Substituting — Δz = 1,655,700/4,850,000 = 0.341 m

Answer: ΔS = 1,655,700 m³, raising the pool 0.34 m

Why the other options are there

-29,100 m³ (stream inflow ignored)
-6 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Evapotranspiration Rates for Grasses

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

Evapotranspiration Rates for Grasses contains 0 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.