How Much Rain Falls on the Kathmandu Valley?

Start smaller than a valley. Lay a square meter of plastic sheet flat on a roof and let 1 mm of rain fall on it. How much water is on the sheet?

The water is a very flat box: 1 m long, 1 m wide and 1 mm deep. Its volume is length × width × depth, but only once all three are in the same unit. In meters it is 1 × 1 × 0.001 = 0.001 m³. A cubic meter is 1,000 liters, so the sheet holds exactly 1 liter.

This is worth remembering: 1 mm of rain on 1 m² is 1 liter. It is why rain is measured as a depth, not a volume. A gauge with a mouth of any size, a cup or a bucket, collects the same depth, and the depth tells you how much fell on every square meter around it.

Now let it grow:

Rain	On	Water
1 mm	1 m²	1 liter
1 mm	a roof of 60 m²	60 liters
1 mm	a hectare (100 m × 100 m)	10,000 liters
1 mm	a square kilometer (1,000 m × 1,000 m)	10,00,000 liters, ten lakh
1 mm	the valley, 587 km²	58,70,00,000 liters, about 59 crore

A square kilometer is a million square meters, so 1 mm on it is a million liters, about 167 of the 6,000-liter tankers that carry water round the valley. Every millimeter on the whole valley is about 59 crore liters: more than the valley uses in a day (43 crore), from a drizzle you would hardly notice.

In the widget you turned 239.7 mm into meters, 587 km² into square meters, and cubic meters into liters. Each step multiplied by a number that changed the unit but not the amount. That number is a conversion factor: a fraction whose top and bottom are the same quantity in two units, so it equals 1.

$$\frac{1 \text{ m}}{1{,}000 \text{ mm}} = 1, \qquad \frac{1{,}000{,}000 \text{ m}^2}{1 \text{ km}^2} = 1, \qquad \frac{1{,}000 \text{ L}}{1 \text{ m}^3} = 1.$$

Multiplying by 1 changes nothing, so you may multiply by these as often as you like. Write the units as if they were letters, and they cancel like letters:

$$239.7 \text{ mm} \times \frac{1 \text{ m}}{1{,}000 \text{ mm}} \times 587 \text{ km}^2 \times \frac{1{,}000{,}000 \text{ m}^2}{1 \text{ km}^2} \times \frac{1{,}000 \text{ L}}{1 \text{ m}^3}$$

The mm on top cancels the mm below, and km² cancels km². What is left is m × m² × L/m³, and m × m² = m³ cancels the m³ below, leaving liters. Doing arithmetic on the units alongside the numbers is called dimensional analysis. It is the most useful check there is: if the units left at the end are not the ones you wanted, the working is wrong.

$$0.2397 \times 587{,}000{,}000 \times 1{,}000 = 140{,}703{,}900{,}000 \text{ L}.$$

About 1.4 × 10¹¹ liters, or 1.4 × 10⁸ m³ (Grade 8’s scientific notation keeps the zeros in order). That is 2.3 crore tankers; 330 days of the water the valley wants; the reservoir at Kulekhani, just south of the valley, filled more than one and a half times.

Why areas and volumes trip people up

A square meter is not 100 cm². It is 100 cm long and 100 cm wide: 10,000 cm². A cubic meter is 100 × 100 × 100 = 10⁶ cm³. When a length is multiplied by k, an area is multiplied by k² and a volume by k³: the factor comes in once for each length.

$$1 \text{ km}^2 = (1{,}000 \text{ m})^2 = 1{,}000^2 \text{ m}^2 = 10^6 \text{ m}^2.$$

So on a map of scale 1 : 50,000, a centimeter stands for half a kilometer, but a square centimeter for a quarter of a square kilometer.

The widget gave 140,703,900 m³, a nine-figure number. Is the valley’s water really known to the last cubic meter? Look at what went in.

The rain was not the same everywhere. The airport’s 239.7 mm is one gauge. On the same morning the gauge at Budhanilkantha, in the north of the valley, had 187.3 mm, and the one at Chapagaun, in Lalitpur to the south, had 323 mm. Nine gauges in the valley recorded the most rain in 54 years, and they did not agree. The airport’s number is one point standing in for hundreds of square kilometers.

The area depends on where you draw the edge. One study of the valley’s watershed, the land whose streams leave through the gorge at Chobhar, gives 587 km²; an area often given for the valley is 665 km². The three districts of Kathmandu, Lalitpur and Bhaktapur together cover 934 km², but they reach over the hills beyond the valley’s rim.

So the honest answer is a range. The least it could be, the lowest gauge on the smallest area:

$$0.1873 \text{ m} \times 587 \times 10^6 \text{ m}^2 \approx 1.1 \times 10^8 \text{ m}^3,$$

and the most, the highest gauge on the largest area: 0.323 m × 665 × 10⁶ m² ≈ 2.1 × 10⁸ m³. Between about 260 and 500 days of the valley’s water. Writing 140,703,900 m³ claims a precision the measurements don’t have: the last six figures are noise. Between about 1.1 and 2.1 × 10⁸ m³ is what the data support; the airport’s 1.4 × 10⁸ m³ is one value inside that range.

This is a rule of every measured answer: report it to a level of accuracy that matches the measurements behind it. A good habit is to count significant figures, the digits from the first one that isn’t zero: 239.7 has four, 587 has three. An answer worked out by multiplying shouldn’t have more significant figures than the roughest number that went into it, and fewer when, as here, a number is less certain than its digits suggest.

How the uncertainties combine

The area is somewhere between 587 and 665 km²: about 626 km², give or take 6%. The rain, from the gauges, is somewhere between 187 and 323 mm: about 255 mm, give or take 27%. When two numbers are multiplied, their percentage uncertainties roughly add: 6% + 27% ≈ 33%. So the volume is about 0.255 m × 626 km² ≈ 1.6 × 10⁸ m³, give or take a third: 1.1 to 2.1 × 10⁸ m³, the range found above. The rain is the weak link: more gauges would help far more than a better map.

A record night is rare. In an ordinary year, Tribhuvan airport collects 1,502 mm of rain (the average of 1991 to 2020), most of it in four months:

Month	Jan	Feb	Mar	Apr	May	Jun	Jul	Aug	Sep	Oct	Nov	Dec
Rain (mm)	15.2	23.6	38.0	62.9	128.8	241.4	384.1	342.7	207.8	43.3	6.4	7.8

June to September bring 1,176 mm, 78% of the year’s rain. November brings 6.4 mm.

Over 587 km², a year’s 1.502 m is

$$1.502 \text{ m} \times 587 \times 10^{6} \text{ m}^2 \approx 8.8 \times 10^{8} \text{ m}^3.$$

The valley wants 430 million liters a day, 0.43 × 10⁶ m³, which over 365 days is about 1.6 × 10⁸ m³. So in a normal year the rain on the valley is between five and six times the water its people want. Shared among the three districts’ 2,996,341 people (the 2021 census), the rain comes to about 800 liters per person per day, against a demand of about 144.

Yet the valley is famous for its water shortage: in 2016 at least 200,000 taps in the valley gave water once in 9 to 13 days. Where does the rain go?

A study that ran three different computer models of the valley’s water (Thapa and others, 2017) found the answer. Of each year’s rain, 55 to 59% runs off into the streams and leaves the valley down the Bagmati, and 41 to 48% evaporates from the ground and from the leaves of plants. Almost none is kept: the water stored in the ground changed by 35 mm a year or less, up or down.

$$0.865 \text{ m} \times 587 \times 10^{6} \text{ m}^2 \approx 5.1 \times 10^{8} \text{ m}^3 \text{ a year runs out of the valley.}$$

That is over three times the water the valley wants, flowing away, most of it in the monsoon. The problem is not the amount. It is the timing: the rain comes when it is least needed, and there is nowhere to keep it. To keep one year’s water, 1.6 × 10⁸ m³, from the monsoon would take about two reservoirs the size of Kulekhani’s as it was built (85.3 million m³), with not a drop lost; nothing that size fits in the crowded valley, which is why it brings water in from outside (the Melamchi tunnel, its first phase planned for 170 million liters a day) and pumps it from under the ground.

The same problem appears on every house: a roof catches rain, a tank keeps it. Most of the rain on a roof reaches the gutter: engineers’ tables give roofs a runoff coefficient of 0.75 to 0.95, the fraction that runs off. Take 0.8.

A roof of 60 m² in a normal year catches

$$60 \text{ m}^2 \times 1.502 \text{ m} \times 0.8 \approx 72 \text{ m}^3 = 72{,}000 \text{ L}.$$

A family using 150 liters a day uses 150 × 365 = 54,750 L in a year. The roof catches more than the family uses. Does that mean they never need the tap?

With a 5,000-liter tank, which is already a big one for a house, the family is dry on 165 days of the year, while 35,000 liters spill over the top in the monsoon. To be never dry they would need a tank of about 22,500 liters, nearly four tankers’ worth.

Why so big? In April the 60 m² roof catches about 100 L a day; in July, 595 L; in November, 10 L. Take a family using 100 liters a day: from October to March the roof catches less each day than they use, so for six months the tank only empties. At the start of October it must hold everything those months fall short by:

$$182 \text{ days} \times 100 \text{ L} - 6{,}446 \text{ L caught} \approx 11{,}750 \text{ L}.$$

The day-by-day model behind the widget needs 11,900 L (on its slider, 12,000). The size of the tank isn’t set by the year’s total at all, only by the longest dry stretch.

On the night of the record, it was not only the amount of rain that flooded the valley but how fast it came. A total tells you how much; a rate tells you how fast, in a unit like millimeters per hour or cubic meters per second.

Suppose all of that night’s 0.2397 m on 587 km² had run off within the same 24 hours. The average flow out of the valley would have been

$$\frac{1.407 \times 10^8 \text{ m}^3}{24 \times 3600 \text{ s}} \approx 1{,}600 \text{ m}^3/\text{s},$$

1,600 tonnes of water a second, through one gorge at Chobhar. It didn’t all run off, and not in one day, but the size of it explains the flood.

How engineers size a drain

An engineer designing a drain for a street or a school yard needs the largest flow it must carry. For small areas they use the rational method:

$$Q = \frac{C \, I \, A}{360},$$

with Q the peak flow in m³/s, C the runoff coefficient, I the rain’s intensity in mm per hour, and A the area in hectares. For a neighborhood of 10 hectares, mostly roofs and roads (C = 0.8), in a cloudburst of 50 mm an hour:

$$Q = \frac{0.8 \times 50 \times 10}{360} \approx 1.1 \text{ m}^3/\text{s},$$

about 1,100 liters a second: the drain must carry that, or the road becomes the drain.

The 360 is not a law of nature. It is a conversion factor in disguise, and you can find it yourself.

Where the rule breaks

The rational method assumes the rain falls evenly over the whole area, steadily, and that nothing is stored on the way. That is close to true for a school yard. It is false for a valley: on 28 September Chapagaun had 1.7 times Budhanilkantha’s rain, fields and ponds held water back, and water from the far rim took hours to arrive. So the Texas manual limits the method to about 80 hectares; for a river basin, hydrologists use models that follow the water across the land and through time. The arithmetic doesn’t change; the assumptions stop being true.

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