--- title: "Free water evaporation by Penman method, June" subject: "Engineering Hydrology" chapter: 3 key: "semester_5/engineering_hydrology:penman-june-33n" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/penman-june-33n site: BCT Engineering --- # Free water evaporation by Penman method, June Engineering Hydrology · Chapter 3: Hydrological Losses - Asked in: 2071 Shrawan, 8 marks - Video: [Penman’s Equation | Concept on Equation With Solved Question](https://www.youtube.com/watch?v=1pOXvsUAlEY) — Hydropower - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/penman-june-33n - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3 ## Question Calculate the free water surface evaporation in June using the Penman method from an area whose latitude is approximately 33$^\circ$N. The available data include air temperature = 30$^\circ$C, wind speed at 2 m height = 10 km/h, relative humidity = 60%, mean observed sun shine hours = 12 and reflection coefficient = 0.05. ## Short answer Free water surface evaporation = 9.2 mm/day (about 275 mm in June). ## Answer Free-water evaporation is obtained from Penman with $r = 0.05$ for a water surface. **Penman equation (potential evapotranspiration):** $$ PET = \frac{A\,H_n + E_a\,\gamma}{A + \gamma} $$ with $$ H_n = H_a(1-r)\left(a + b\frac{n}{N}\right) - \sigma T_a^4\,(0.56 - 0.092\sqrt{e_a})\left(0.1 + 0.9\frac{n}{N}\right) $$ $$ E_a = 0.35\left(1 + \frac{u_2}{160}\right)(e_w - e_a), \qquad a = 0.29\cos\varphi,\ b = 0.52 $$ where $H_n$ is the net radiation in mm of evaporable water per day, $E_a$ the aerodynamic term in mm/day, $u_2$ the wind speed at 2 m in km/day, and $e_w, e_a$ in mm of Hg. #### Data and constants used - $T = 30\,^\circ$C $\Rightarrow T_a = 303$ K; RH $= 60\%$; $n = 12$ h; $N = 14.2$ h; $u_2 = 240$ km/day - $H_a = 16.9$ mm/day; $r = 0.05$; $\varphi = 33.00^\circ$; $e_w = 31.82$ mm Hg; $A = 1.83$ mm Hg/$^\circ$C; $\gamma = 0.49$ mm Hg/$^\circ$C; $\sigma = 2.01\times10^{-9}$ mm/day/K$^4$ - **Assumed values (not given):** $e_w$ at $30^\circ$C $= 31.82$ mm Hg and slope $A = 1.83$ mm Hg/$^\circ$C; for $33^\circ$N in June, $H_a \approx 16.9$ mm/day and possible sunshine $N \approx 14.2$ h (standard tables / solar geometry); wind $10\ \text{km/h} = 240$ km/day; $\gamma = 0.49$. #### Step 1: Actual vapour pressure $$ e_a = \frac{RH}{100}\,e_w = 0.60\times31.82 = 19.09\ \text{mm Hg} $$ #### Step 2: Net radiation $H_n$ $a = 0.29\cos(33.00^\circ) = 0.2432$, and $n/N = 0.8451$ $$ \begin{aligned} H_a(1-r)\left(a + b\tfrac{n}{N}\right) &= 16.9\times0.950\times(0.2432 + 0.52\times0.8451) = 10.960\\ \sigma T_a^4 &= 2.01\times10^{-9}\times(303)^4 = 16.942\\ \text{Back radiation} &= 16.942\times(0.56 - 0.092\sqrt{19.09})\times(0.1 + 0.9\times0.8451) = 2.304\\ H_n &= 10.960 - 2.304 = 8.656\ \text{mm/day} \end{aligned} $$ #### Step 3: Aerodynamic term $$ E_a = 0.35\left(1 + \frac{240}{160}\right)(31.82 - 19.09) = 11.137\ \text{mm/day} $$ #### Step 4: PET $$ PET = \frac{1.83\times8.656 + 11.137\times0.49}{1.83 + 0.49} = \frac{15.841 + 5.457}{2.320} = 9.18\ \text{mm/day} $$ **Answer: Free water surface evaporation = 9.2 mm/day, so about 275 mm in June (30 days).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why r = 0.05 here? A. Snow B. Green crop C. Bare soil D. Water surface reflects little Answer: D. For free-water evaporation the albedo of water is 0.05. (Key point: Free water r) 2. Wind 10 km/h becomes A. 10 km/day B. 240 km/day C. 100 km/day D. 2.4 km/day Answer: B. 10 x 24 = 240 km/day. (Key point: Wind unit) 3. Ea = 0.35(1 + 240/160)(ew - ea) with ew - ea = 12.73 is about A. 1.1 mm/day B. 4.5 mm/day C. 22 mm/day D. 11.1 mm/day Answer: D. 0.35 x 2.5 x 12.73 = 11.14. (Key point: Aerodynamic term) 4. Free-water evaporation is about A. 9.2 mm/day (275 mm in June) B. 30 mm/day C. 2 mm/day D. 92 mm/day Answer: A. PET = 9.18 mm/day x 30 days = about 275 mm. 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