A small (area = 3.0 km2) agricultural catchment on average receives 950 mm of pr

Question

A small (area = 3.0 km2) agricultural catchment on average receives 950 mm of precipitation. This agricultural catchment is drained by a stream, and a continuous record of stream discharge (runoff) is available. The total amount of surface-water runoff for the year, determined from the stream discharge record, is 1.1x106 m3. i) What is volume of water (in m3) evapotranspired for the year (assume no change in water stored in the catchment)? (5 points) i) What is the depth of water (in mm) evapotranspired for the year (agaitn assuming no change in water stored in the catchment)? (5 points) ii) What is the runoff ratio for the catchment? (5 points)

Explanation / Answer

Answer:

Area= 3 sqkm = 3 * 106 m; precipitation = 950 mm = 950*10-2 m ; Stream discharge = 1.1x106 m3

i) From, the water-budget balance equation,

P= E + Q + S where E,Q,S are Evapotransporation, run-off and change in storage.

By the question, change in storage S is negligible.

Then, E = P – Q = (950*10-2 * 3 * 106 ) – 1.1x106 m3 = 27.4x106 m3 (Ans)

ii) Depth of water evapotransporation = Volume/catchment area= 27.4* 102 * 106 m3 /3*106 mm = 913.33 mm (Ans)

iii) Run-off ratio = run-off / Precipitation = 1.1x106 m3/ 950*10-2 * 3 * 106 m3 = 0.039 (Ans)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.