An alternative, less empirical approach (though still full of brazen assumptions) is the following:

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1 ERTH 500: More Notes on Final Project: Dr. Dave Dempsey Earth Systems II Modeling the Dept. of (Spring 2016) Cenozoic Icehouse Earth Earth & Climate Sciences More notes on Upslope/Monsoon Precipitation One approach to this might be to try to estimate (a) contemporary precipitation rates in the Himalaya and (b) precipitation rates over flat terrain, using the Amazon Basin as a reference for the latter, at today s global average temperature. Then assume a linear relationship between precipitation rate and mountain height (or, equivalently, slope), and scale the constant of proportionality by a function of temperature. (The ability of air to hold water vapor depends increases with increasing temperature, so as the planet warms or cools, the precipitation rates should vary as the temperature varies.) An alternative, less empirical approach (though still full of brazen assumptions) is the following: Make the following assumptions: o Air coming from the Bay of Bengal as part of the Indian monsoon is saturated with water vapor. o When the slope of the Himalaya Mts. is heated by the sun, a pressure difference develops along the slope that pushes air up the slope. (Note: This is a fact, not an assumption.) The speed of air along this slope depends on the slope (greater slope means faster upslope winds). Schumann (1990) present model simulations of winds created by heating a slope at a range of slopes; we ll use those results to develop a rough relationship between slope and upslope wind speed. o Only air within a layer next to the sloping surface participates in the upslope flow. Typical values of the depth of that layer we ll take from a text by David Whiteman, Mountain Meteorology: Fundamentals and Applications, excerpts of which are available online. o The temperature of the air arriving at the base of the slope is the same as the temperature it had coming off of the Bay of Bengal. We ll assume that this temperature is warmer then the global average air temperature by an amount equal to the climatological average for the time of year when the summer monsoon blows minus the global average temperature, and we ll assume that this has been true for millions of years. (Need to look up the climatological average surface air temperature over the northern Bay of Bengal in early summer. The global average surface temperature today is about 16 C.)

2 o Air arriving at the base of the Himalaya Mts. rises along the slope to the summit, but no farther. [This is not a very good assumption often the air becomes unstable and rises vertically to great heights in thunderstorms, but we re not in a position to try to model that behavior.] o As the saturated air rises along the slope, it encounters lower pressure, expands, and cools as a result. As it cools, it is able to hold less water vapor, so it s capacity to hold vapor drops below the amount it actually has in it, and the excess water vapor condenses to form clouds. [Note: this is not an assumption: this actually happens.] We ll assume that the pressure at the base of the mountains is roughly typical of sea-level pressures (around 10 5 Pascals). We ll also assume that the atmosphere is approximately hydrostatic (so that the upward net force due to pressure roughly balances the downward force on air due to gravity). Furthermore, we ll assume that the temperature profile in the air surrounding the Himalaya has a constant environmental lapse rate. (A lapse rate is the decrease in temperature per unit vertical distance. The environmental lapse rate is the change in air temperature per unit vertical distance that a balloon would measure as it rises through the atmosphere.) We ll assume that the environmental lapse rate (Γ e ) is equal to the global average value, 6.5 C/km. With these assumptions, we can combine the hydrostatic equation and the gas law for air and derive a relationship between altitude (z) and pressure (p): Γ e R d p g z(p) = ( T(0) Γ e ) 1 p(0) (1) We can solve Eq.(1) for pressure as a function of altitude, too: (2) ( ) Γ ez ( ) p(z) = p(0) T 0 T 0 g Γ e R d where p(0) is the pressure at altitude z = 0 (sea level); T(0) is the temperature (in Kelvins) at altitude z = 0; g is the acceleration of gravity (9.81 m/s 2 ); and R d is the gas constant for dry air (287 J/kg/K).

3 o A flux is the rate at which any quantity, Q, passes through a unit of surface area. (It has dimensions of Q/time/area.) It is possible to show that the flux of the mass of air through a unit of surface area at the base of the Himalaya Mts. is ρv, where ρ is the air density and v is the wind speed, which we ll take to be the upslope wind speed. We can estimate the density of air arriving at the bottom of the slope, ρ 0 ( ), using the ideal gas law: ( ( )R d ) (3) ρ( 0) = p( 0) T 0 (where T is the temperature in Kelvins and p is pressure in Pascals. The density will be in kg/m 3.) o The total rate at which the mass of air arriving at the bottom of the slope moves up the slope will be the air mass flux times the surface area through which the air moves. That surface area will be the depth of the slope flow times the length of the Himalaya Mts., l Hence, the rate at which air mass moves up the slope is: Rate of flow of air mass up the slope = area x flux = ( l mts h slopeflow ) ρ( 0)v The saturation water vapor mixing ratio of air (q s ) is the mass of water vapor in saturated air per unit mass of dry air. It is related to the temperature and pressure of the air as follows: (4) q s ( T, p) = e s (T ) ( p e s (T )) where e s (T ) is the pressure that the water vapor exerts (the saturation vapor pressure), which depends only on the air temperature. The relationship between the saturation vapor pressure and temperature is given by the Clausius-Clapeyron equation: (5) e s (T ) = e s (T ref )exp L H 1 R v T ref 1 T where L H is the latent heat of water vapor (2.6x10 6 J/kg); R v is the gas constant for water vapor (461 J/kg/, and T ref is a reference temperature at which the saturation vapor pressure is known. (At the freezing point of water, K, the saturation vapor pressure is 611 Pascals.)

4 The amount of water vapor that condenses out of the air rising up the slope will depend on the change in the temperature and pressure of the air from the based of the mountains to the summit. The change in the saturation mixing ratio, Δq s, gives us the change in water vapor mass per unit mass of dry air: (6) Δq s = q s T ( 0), p 0 ( ( )) q s T z top ( ( ), p( z top )) We can get the pressure at the top of the mountains from Eq.(2). All we need to know is the temperature of the rising air when it reaches the top of the mountains. This will depend on how much it cools in response to the decrease in pressure as is rises and on how much latent heat is released into the air as water vapor condenses in it. Equations for this exist, but they are complicated, so we ll make approximations to them. In particular, we ll (a) read values of temperature of saturated, rising air at six pressure levels from 10 5 Pascals (near sea level) to 0.5x10 5 Pascals (around 6000 meters altitude) for each of three starting temperatures (5 C, 20 C, and 35 C) at 10 5 Pascals, from a pseudoadiabatic chart (such as P_diagram#/media/File:Teph2.svg); (b) calculate the altitude at each of the six pressure levels using Eq.(1); (c) estimate the saturated air parcel s lapse rate at three levels using the six pairs of temperature and pressure values; fit the three lapse rate estimates to a quadratic function of altitude, giving us three quadratic coefficients; do the same at each of the three sea-level temperatures, and fit each of the three quadratic altitude cofficients to a quadratic function of temperature; and finally, integrate the lapse rate in its doubly quadratic form between sea level and the top of the mountains, to get an estimate of the temperature of saturated air starting at the base of the mountains and rising to the top.) The results are as follows: (7) T ( z top ) = T 0 ( ) z top ( ( ) + Cb T ( 0) 2 2 ) z top ( ( ) + Cc T ( 0) 2 3 ) z top ( ) + Aa + Ba T ( 0) + Ca T ( 0) Ab + Bb T Ac + Bc T 0 where: Aa = C/km Ba = /km Ca = /(km* C) Ab = -3.50E-04 C/(km*meter) Bb = 1.40E-04/(km*meter) Cb = -3.30E-06/( C*km*meter) Ac = 4.75E-07 C/km*meter 2 ) Bc = -4.63E-08/(km*meter 2 ) Cc = 8.98E-10/( C*km* meter 2 )

5 We ll assume that all of the water that condenses to form clouds will also precipitation onto the mountain slope. In that case, the rate at which water falls onto the mountain slope will be: Rate at which (8) water mass = Δq s T, p ( ) l mts h slopeflow precipitates onto the slope ( ) ρ 0 ( )v The precipitation rate in depth per unit time can be calculated from this using the area onto which the water precipitates and the density of water. Scale the precipitation rate that we calculate by the fraction of the year that the monsoon flow occurs and the fraction of the day when the slopes are heated enough to cause air to rise up them.