Chapter_2 (2)

pdf

School

University of Minnesota-Twin Cities *

*We aren’t endorsed by this school

Course

4501

Subject

Civil Engineering

Date

Oct 30, 2023

Type

pdf

Pages

52

Uploaded by MajorNarwhalPerson799

Report
CEGE 4501 Hydrologic Design Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj Department of Civil, Environmental, and Geo-Engineering September 22, 2021
Outline Introduction Atmospheric Composition and States Equation of State for Air Hydrostatic Equilibrium and Pressure Profiles Quantifying Atmospheric Water First Law of Thermodynamics Adiabatic Expansion and Stability Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj
Introduction I Thermodynamics a branch of physics and deals with conservation and conversion of various forms of energy and, specifically, the relationship between heat energy and its effects on changes in properties of materials. A thermodynamic system is a definite quantity of matter that can exchange energy with its surrounding environment by performing mechanical work or by transferring heat across its boundary with the environment.The state variables that define the earth’s atmosphere as a thermodynamic system are temperature, pressure and density . In the atmosphere, the water vapor affects the state variables, so it can be thought of as a fourth state variable. The state variables are related by the equation of state . Why do we need to know about thermodynamics for hydrology? The hydrologic cycle is the result of mechanical works done by a system similar to a steam engine at the size of the planet Earth. The main source of heat is the energy received by the Sun. The differential heating of earth is the main source of atmospheric pressure gradient from equators towards polar regions as mentioned in Chapter 1. In the atmosphere, water is transported and transformed (phase change) by two hydrologic fluxes: - Precipitation: A flux of water that reaches the earth’s surface, as a result of the conversion of atmospheric water vapor to liquid or solid water. - Evapotranspiration: conversion of surface liquid water to water vapor through combination of direct evaporation from the soil/water surfaces and transpiration due to the vegetation metabolism. In order to understand these fluxes, we must first understand how the vertical profile of key atmospheric state variables change in space and time. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 1
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Atmospheric Composition and States I Atmospheric Composition Nitrogen N 2 and Oxygen O 2 constitute approximately 98% of the mass and volume of the homosphere (first 100 km of earth’s atmosphere). Concentration of most of the atmospheric gases is constant except water that, on average, constitutes 0-4% of the volume of atmosphere. Figure 1: Main gaseous constituents of air and their relative percentages with respect to dry air (left; Curry and Webster, 1999). Atmospheric moisture profiles at different latitudes (right) from the tropical ∼ 23 S-N, middle latitudes ∼ 23-66 S-N to polar regions > 66 S-N (Oort 1983). Atmospheric Moisture Clouds contain suspended water vapor, liquid water and ice particles . Hydrometeors are those particles that become large enough to fall to the earth surface. Warmer atmosphere can hold more water ; therefore, atmospheric moisture is higher near the surface and equator compared to higher elevations and polar regions. Water vapor is the most variable constituent in the Earth’s atmosphere and plays a critical role in transfer of energy from equators to polar regions. Why? Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 2
Atmospheric Composition and States II Atmospheric Pressure and Density Gravitational force is the main force contributing to atmospheric pressure. The earth’s atmospheric mass per unit area at the surface is around 10 4 [kg m -2 ]. The air pressure at earth surface is thus around 10 5 Pascal [Pa] or [N m -2 ]. Approximately 90% of the atmospheric weight lies below 15 [km]. Common pressure units are: 1 [bar] = 10 5 [Pa] 1 [ mb ] = 10 2 [Pa] = 1 [ hPa ] 1 [atm] = 1.01325 [bar] = 1.01325 × 10 5 [Pa]. Atmospheric pressure and density are tightly coupled, decaying almost exponentially as a function of atmospheric height. Density of air has an average value of 1.3 [kg m -3 ] near the earth’s surface. Figure 2: Atmospheric pressure and density decreases with height almost exponentially due to the compressibility of air (Curry and Webster, 1999). Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 3
Atmospheric Composition and States III Atmospheric Temperature Atmospheric lapse rate is the rate of temperature change as function of elevation, defined as follows: Γ = - dT dz , As shown below, typical lapse rates is approximately linear in different nominal layers of the atmosphere and vary depending on latitude. For a dry atmosphere the lapse rate is 9.8 [ ◦ C km -1 ], while the global average for moist air is around 6.5 [ ◦ C km -1 ]. Figure 3: Typical vertical temperature profiles in nominal layers of the atmosphere (left) and variations in temperature profiles with latitude (right, credit: The COMET Program). Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 4
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Atmospheric Composition and States IV Temperature inversion may occur when Γ < 0 and thus the air becomes colder near the surface. Γ = 0 is referred to an isothermal atmosphere where the temperature does change with height. During an inversion, atmospheric circulation is often stable and suppressed as heavier and colder air parcels remain below lighter and warmer air aloft. When Γ > 0, it is very likely that the atmosphere is unstable , meaning that warmer and lighter air parcels near the earth surface begin to rise and generate wind and atmospheric circulation. Atmospheric stability is a function of density profile throughout the atmospheric column, which is not only related to the air temperature profile but also moisture content of the atmosphere. Unstable Neutral Stable ρ , T z z T ρ z ρ T ρ , T ρ , T ρ T Figure 4: Atmospheric stability is a function of its density profile, which is related to the temperature (laps rate) and moisture profiles. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 5
Equation of State for Air I Equation of state for Dry Air How can we compute pressure in air? Roughly speaking, an ideal gas is a hypothetical gas whose molecules occupy a negligible volume compared to the entire space that the gas occupies and whose molecular movements are approximately independent of one another. For an ideal gas, the states are related to each other as follows, known as the ideal gas law : PV = nR * T , P: Pressure [pa] V: Volume [m 3 ] n: Number of moles of the gas R * : Universal gas constant, 8.314 [J mole -1 K -1 ] T: Temperature [K]. The mole is a unit of measurement for amount of a substance. It is defined as the amount of a substance that contains as many atoms as there are in 12 grams of pure Carbon-12, which is 6.02214 × 10 23 [# atoms/mole], known as the Avogadro constant. Dividing both sides by the mass ( m ) of the gas molecules to obtain the intensive form of the ideal gas law, we get P V m = n m R * T . Given that ν = V m = 1 ρ is the specific volume , M = m / n represents the molar mass of the gas and R = R * / M denotes the specific gas constant [J kg -1 K -1 ], we have P ν = RT , which can be organized as P = ρ RT , where ρ is the gas density [kg m -3 ]. Air is a mixture of gases and thus we need to generalize the ideal gas law to a mixture of gas molecules. Let us first focus on dry air where the moisture content is zero. Pressure of a mixture of gasses is explained by the Dalton’s law of partial pressures. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 6
Equation of State for Air II The Dalton’s Law states that the total pressure of a mixture of ideal gasses is equal to the sum of the partial pressures of each individual gas. The ideal gas law for j th gas is P j = ρ j R j T , which can be expressed as follows, considering ρ j = m j / V : P j V = m j R j T . Figure 5: Illustration of Dalton’s Law for partial pressures (credit: eTAP). Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 7
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Equation of State for Air III Using Dalton’s law of superposition, one can obtain V Σ j P j = T Σ j m j R j and thus Σ j P j V Σ j m j = T Σ j m j R j Σ j m j . Given that P = Σ j P j , and the specific volume ν = V / Σ m j , we have P ν = T Σ j m j R j Σ j m j = T m 1 R 1 + m 2 R 2 + . . . Σ j m j . Therefore, specific gas constant of a mixture of gases is the weighted average of the specific gas constants of each gas, where the weights are the ratio of the mass of each gas to the mass of the mixture. As a result, one can simply compute the specific gas constant of the dry air ( R d ) as follows: R d ≈ 0 . 75 R nitrogen + 0 . 23 R oxygen + · · · = 0 . 75 × 296 . 8 + 0 . 23 × 259 . 8 + · · · = 287 . 14 [J kg -1 K -1 ] , which is identical to R d = R * / M d , where M d = 28 . 96 [gram mole -1 ] is the molar mass of the dry air. Thus, for a dry air parcel we have, P = ρ R d T . The above equation is the equation of state for dry air . Now the question is – how can we modify this equation for a moist air parcel ? which is necessary for the study of hydrologic fluxes in the atmosphere. We need to note that the water vapor is another gas that can change the pressure as it alters both density and specific gas constant of the air. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 8
Equation of State for Air IV Equation of State for Moist Air Let us assume that the water vapor is not near condensation , and write the ideal gas law only for the water vapor: e = ρ v R v T , where e is the water vapor partial pressure , ρ v ≈ 0 . 804 [kg m -3 ] and R v = R * / M v = 461 . 51 [J kg -1 K -1 ] is the water vapor specific gas constant—given that the water vapor molar mass is M v = 18 [grams mole -1 ]. In a mixture of dry air and water vapor, using the Dalton’s Law, we have, P = P d + e = ( ρ d R d + ρ v R v ) T . Previously, we showed that the specific gas constant of a gas mixture can be calculated via weighted averaging as follows: R = m d R d + m v R v m d + m v = m d m d + m v R d + m v m d + m v R v . From this, we define an important variable called specific humidity as follows q v = m v m d + m v , which expresses the moisture content of the atmosphere in terms of mass of water to the mass of air such as [grams of water/grams of air] . Therefore, the specific gas constant for the moist air can be written as follows: R = (1 - q v ) R d + q v R v . (1) Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 9
Equation of State for Air V At the same time, we know that R * = R v M v = R d M d and thus we have R v = R d M d M v = R d - 1 , (2) where = M v / M d = 18 / 28 . 96 = 0 . 622. From Eq. 1 and 2, we have R = (1 - q v ) R d + q v R d - 1 = R d (1 + 0 . 608 q v ) and thus combining this equation and ideal gas law P = ρ RT , we have P = ρ R d (1 + 0 . 608 q v ) T = ρ R d T v where T v = (1 + 0 . 608 q v ) T is called virtual temperature , which is a hypothetical temperature that a dry air parcel should have to represent the same density and pressure of moist air . Therefore, the ideal gas law for moist air can be stated as follows: P = ρ R d T v . The specific humidity rarely exceeds 0.02 [gram of water/ gram of air], so T v rarely exceeds the normal temperature by more than 2-3 ◦ K. However, we should note that changes in humidity could have important implications on air density and thus atmospheric stability. Note that a moist air is always less dense than its dry air counterpart. Therefore, when air parcels near the earth surface pick moisture, they become lighter and ascend to higher levels. Now we have the moist air equation of state, and can compute the pressure in a moist atmosphere. The new question is – how do the state variables change throughout the atmospheric depth? To answer this question, we will introduce the concepts of hydrostatic balance and later on will discuss the first law of thermodynamics to better understand the process of condensation and precipitation. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 10
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Hydrostatic Equilibrium and Pressure Profiles I Hydrostatic Equilibrium We know from basics of fluid mechanics that the difference in fluid pressure ( dp ) because of a change of depth ( dz ), can be explained as follows: dP = - ρ ( z ) g dz , where ρ ( z ) is the fluid density as a function of elevation from a datum and g represents the gravitational acceleration. The negative sign encodes the fact that the pressure decreases when z increases. The above equation is often referred to as the hydrostatic equilibrium , which is valid when the fluid is at rest or the flow velocity at each point is constant over time. Figure 6: Hydrostatic pressure decreases with altitude. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 11
Hydrostatic Equilibrium and Pressure Profiles II One can integrate the hydrostatic equation as follows: z 0 - dP = z 0 ρ ( z ) gdz . Because air is compressible and density decreases as a function of atmospheric height, the above integration is not straightforward and we need a model to explain the changes in atmospheric density at different elevations. To further expand the hydrostatic equilibrium, the following two assumptions can be made: 1. Constant Temperature for an Isothermal Atmosphere Given that dP = - ρ gdz and from the ideal gas law ρ = P R d Tv , we have P 2 P 1 dP P = z 2 z 1 - g R d T v dz (3) for a finite thickness Δ z = z 2 - z 1 and constant T v , one can obtain: ln P 2 - ln P 1 = - g R r T v ( z 2 - z 1 ) ⇒ P 2 = P 1 exp - g R d T v ( z 2 - z 1 ) which is known as the hypsometric equation that explains atmospheric pressure versus height in an isothermal atmosphere . This equation can be used to calculate the pressure profile in a piecewise manner, if the temperature profile varies slowly in height. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 12
Hydrostatic Equilibrium and Pressure Profiles III 2. Constant Temperature Lapse Rate Let us assume that the lapse rate of the virtual temperature is denoted by T v = T v 0 - Γ z , where Γ = - dT / dz is the ambient temperature lapse rate. Integrating both sides of Eq. 13, from ( P 0 , T v 0 ) to ( P , T v ), one can obtain the following expression for changes of the pressure profile: ln P P 0 = g R d Γ ln T v 0 - Γ z T v 0 ⇒ P = P 0 T v T v 0 g R d Γ . It is important to note that even assuming a linear temperature profile, the pressure profile is non-linear. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 13
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Quantifying Atmospheric Water I Air has limited capacity to hold water and becomes eventually saturated depending on its temperature. A warmer air can hold more water vapor . It can be shown that the water vapor pressure at saturation condition e s is only a function of air temperature T . This relationship is explained through the known Clausius-Clapeyron equation as follows: Clausius-Clapeyron equation e s ( T ) = 611 exp - L v R v 1 T - 1 273 . 15 [Pa] where T is in Kelvin. e s ( T ) = 611 exp 17 . 27 T 237 . 3 + T [Pa] where T is in Celsius. L v : Latent heat of vaporization = 2.5x10 6 [J kg -1 ], R v : Water vapor specific gas constant = 461 [J kg -1 K -1 ], T : Air temperature [K or ◦ C depending on the equation]. As discussed before, the water vapor pressure ( e ) represents partial pressure of water vapor in atmosphere P = e + P d , where P is the total air pressure and P d denotes the dry air pressure. Typical values of the partial water vapor pressure near the earth surface varies from 10 to 30 hPa. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 14
Quantifying Atmospheric Water II Specific Humidity: is the ratio of the mass of water vapor to the total mass of dry air and water vapor. q v = m v m v + m d , where m v and m d are the masses of water vapor and dry air in a unit volume of air. Therefore, the unit of q v is [kg of water/ kg of air]. Using the ideal gas law, P = ρ RT , the specific humidity can be further expanded as follows: q v = e P - (1 - ) e , where = R d / R v = 0 . 622, e is the water vapor pressure, and P is the total air pressure. Mixing Ratio: is the ratio of the mass of the water vapor to the mass of dry air. w v = m v m d = e P - e , where the unit is [kg of water/ kg of dry air]. For the case where the water vapor pressure e P , the following approximation holds: q v ∼ = w v ∼ = e P . Note: We can see from the Clausius-Clapeyron that the saturated water vapor pressure is only a function of temperature ; however, the saturated specific humidity or mixing ratio are also a function of total air pressure. Relative Humidity: is the ratio of water vapor pressure to the saturated water vapor pressure RH = e e s ∼ = q v q s ∼ = w v w s . Clearly 0 ≤ RH ≤ 100 and shows strong diurnal variability as it is a function of e s ( T ). Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 15
Quantifying Atmospheric Water III Precipitable Water: is the total mass of water in the atmospheric column per area from surface to top of the atmosphere, W v = ∞ 0 ρ v ( z ) dz , [kg m -2 ] , where ρ v ( z ) is the density of the water vapor at elevation z . We can present the total precipitable water (TPW) in terms of the depth of liquid water as h TPW = W v /ρ l , where ρ l is the density of liquid water. Obviously, one can further expand the calculation of TPW in terms of the vertical profile of the atmospheric pressure using the hydrostatic pressure equation, dp = - ρ a ( z ) gdz , as follows: W v = 1 g p 0 0 ρ v ρ a dp ∼ = 1 g p 0 0 q v dp ∼ = 1 g p 0 0 w v dp , where ρ a is the air density and we assumed that the pressure is zero when z → ∞ and is p 0 at the surface z = 0. Note that, we changed the integration variable from elevation to pressure. Dew Point Temperature: is the temperature to which the air must be cooled to reach its saturation level. Therefore, when the relative humidity is low the due point temperature is way colder than the actual temperature and vice versa. Vapor Pressure Deficit: is the difference between the saturated and actual water vapor pressure VPD= e s - e , which is zero when the air is saturated. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 16
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Quantifying Atmospheric Water IV T e s (T) T dew T a e s (T a ) e s (T dew ) Figure 7: The relationship between the dew point and air temperature through their corresponding saturated water vapor pressure values. The blue curve shows the Clausius-Clapeyron equation. Dry and Wet bulb Temperature: The dry bulb temperature of air is the usual temperature measured by a thermometer. The wet bulb temperature is the temperature measured by a thermometer, while it is covered by a wet cotton cloth. The wet bulb temperature is the temperature that an air parcel would have if it is brought to its saturation level by evaporative cooling. In other words, it is the lowest temperature that can be reached by evaporating water into the air while the internal energy of the air is used for the evaporation. Thus, the wet bulb temperature will always be less than or equal to the actual temperature and greater than the dew point temperature. When relative humidity increases the difference between the dry and wet bulb temperature decreases and vice versa. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 17
First Law of Thermodynamics I Work ( dW ) When a force vector -→ F acts and moves a body of mass with a displacement vector -→ dx , the work being done is equal to the following inner product: dW = -→ F · -→ dx = | -→ F || -→ dx | cos( θ ) , where θ is the angle between the force and displacement vectors and |·| refers the tha magnitude of the vector. As a result one can conclude that any expansion or contraction of an air parcel can be explained as work, dW = F dx = PA dx = P dV , where P is pressure and A is the parcel area. In other words, when an air parcel expands, it works on the surrounding environment ( dW > 0) and when it contracts the surrounding environment works on the parcel ( dW < 0). Figure 8: Schematic of expansion work (right; Credit: Boundless.com). Air parcels expand due to decreased surrounding pressure and thus work on their environment as they rise due to orographic (a), frontal lifting (b), low-level convergence (c), buoyant rising of warm air (d), and turbulent mixing (e) (from Curry and Webster, 1999). Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 18
First Law of Thermodynamics II Heat ( dQ ) Heat is the transfer of thermal energy between the system and the environment. Empirically, it has been found that heat is: dQ = m c dT (extensive) ⇒ dq = c dT (intensive) where m is mass [kg], c is the specific heat capacity [J kg -1 K -1 ], and dT is the change in temperature [K]. The specific heat of air is measured for two scenarios: 1) constant pressure ( c p ) with typical values of 1004 [J kg -1 K -1 ] and constant volume ( c v ) with typical values 717 [J kg -1 K -1 ] for dry air. Why is there a difference in values of specific heat? Another important observation about specific heat values is the following proven relationship: R d = c p - c v = 1004 - 717 = 287 [J kg -1 K -1 ] Figure 9: When the volume is constant (left) all added heat dq increases the temperature. However, when the pressure is constant (right), the added heat not only increases the temperature but also expands the volume. As a result, when the pressure is constant, we need more heat to increase the temperature of the system and thus c p > c v (credit, Wallace and Hobbs, 2006). Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 19
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
First Law of Thermodynamics III Internal Energy ( dU ) The change in internal energy of a thermodynamic system, dU , consists of the change in microscopic kinetic and potential energy. The kinetic energy from random molecular movements is called sensible energy , which we can be felt and measured through changes in temperature. The potential energy are associated with the forces that bind molecules together , and, in our case, it refers to the latent energy associated with the phase change of system material (e.g., water vapor in an air parcel). Based on the Joule’s Law, the internal energy of an ideal gas is only dependent on its temperature. This allows us to create the following relationship for an air parcel with constant volume: du = c v dT (intensive form) First Law of Thermodynamics The first law states that the heat added to a system must be equal to the sum of changes in internal energy (microscopic effect) and the work done “by” the system “or” on the system (macroscopic effect): dQ = dU + dW (extensive) × 1 m -→ dq = du + dw (intensive) . Note: The sign convention for work is positive for work done on the environment by the system ( air expansion; dW > 0 ) and negative for work done on the system by the environment ( air contraction; dW < 0 ) . Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 20
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
First Law of Thermodynamics IV First Law for an Air Parcel Our focus is on a parcel of air in the atmosphere, as a closed thermodynamic system that does not exchange mass with the environment. Therefore, we want to put the first law in a form that makes sense for the atmospheric state variables such as temperature and pressure that are readily measurable . The first step is to substitute expansion work into the first law and our definition of internal energy for an ideal gas: dq = c v dT + Pd ν. (intensive) Now using the product rule from calculus, we can substitute for the work term as follows: dq = du + dw = c v dT + d ( P ν ) - ν dP . Then, using the ideal gas law ( P ν = RT ) for the middle term, we get, dq = c v dT + d ( RT ) - ν dP const . R -→ dq = c v dT + RdT - ν dP Finally, use the fact that c v + R = c p to get the first law in the form of: dq = c p dT - ν dP where dh = c p dT is called enthalpy . The above expression is known as the enthalpy form of the first law. The enthalpy, can be thought as the sensible heat . This form allows us to analyze the heat exchange between the air parcels and the environment in terms of changes in pressure and temperature , which are easily measurable–compared to density ν = 1 /ρ . Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 21
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Adiabatic Expansion and Stability I Dry Adiabatic Expansion In order to understand how water and heat are transferred in the atmosphere, we must understand what happens to an air parcel as it moves vertically . We know that if air rises via a lifting mechanism such as orographic lifting, its temperature and pressure will decrease as internal energy is reduced because the work done on the environment by the parcel. Let us assume that the time scale of these changes is short enough that no heat is exchanged with the environment ( dq = 0) and no water phase change occurs during the expansion, which is known as the dry adiabatic expansion . This means that the enthalpy form of the first law reduces to: c p dT = ν dP Given the ideal gas law P ν = R d T , let us further expand the above equation as follows: c p dT v = R d T v dP P ⇒ c p R d dT v T v = dP P T 2 T 1 c p R d dT v T v = P 2 P 1 dP P Therefore, c p R d ln T v 2 T v 1 = ln P 2 P 1 and thus we obtain the Poisson’s equation for adiabatic expansion as follows: T v 2 T v 1 = P 2 P 1 R d cp . Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 22
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Adiabatic Expansion and Stability II Adiabatic Lapse Rate: Finally, the adiabatic assumption allows estimating the lapse rates . Using the enthalpy form of the first law, c p dT = ν dP , together with hydrostatic equilibrium ( ν = - gdz / dP ), one can obtain c p dT = - gdz . Thus, the dry and moist adiabatic temperature lapse rates are: Γ dry = - dT dz = g c pd ≈ 9 . 8 [K km -1 ] Γ moist = - dT dz = g c p = - g c pd (1 + 0 . 87 q v ) In the above derivation, knowing that c p = q v c pv + (1 - q v ) c pd where c pd = 1004 and c pv = 1864 [J kg -1 K -1 ], we can conclude that c p = c pd (1 + 0 . 87 q v ). Observations show that, on average, the moist atmospheric lapse rate is around 6.5 [K km -1 ]. It is also customary to use the dry adiabatic lapse rate to roughly infer atmospheric stability as follows: Γ < Γ d ⇒ (statically stable) Γ = Γ d ⇒ (statically neutral) Γ > Γ d ⇒ (statically unstable) We need to note that the atmospheric stability must be determined based on the density profile, which is largely a function of temperature and to a lesser degree a function of the air moisture content. Therefore, the use of adiabatic laps rate is only an approximation when there is not instability cased by wind shear forces. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 23
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Cloud and Precipitation I To this point, we have been dealing primarily with water vapor and its interaction with the atmospheric state variables and have ignored condensation . Now we will discuss condensation of water vapor in terms of hydrologic flux of precipitation beginning with a few definitions: Precipitation: is the conversion of atmospheric water to liquid or solid water that reaches the earth’s surface. Clouds: are collections of small water particles that are either in vapor, liquid (water droplets), or solid (ice crystals) phases. Hydrometeors: are body of liquid or solid water such as raindrops and snow flakes that are heavy enough to fall to the earth surface. Note that in addition to atmospheric gases, water in liquid or solid phases, the earth’s atmosphere contains aerosol particles , which are suspended tiny solid particles with a size in the order of 0.1-10 μ m (e.g., dust, sea-slat, soil particles, by products of combustion processes, etc). Cloud Formation As we discussed, when a parcel of air rises adiabatically, the temperature decreases as internal energy is converted to work on the environment through expansion. When temperature of a rising air parcel drops its water holding capacity reduces , which can be explained via the Clausius-Clapeyron equation. Therefore, the air parcel can no longer hold the amount of moisture it could hold at the earth surface with a warmer temperature. As a result, condensation happens, which leads to the formation of clouds and perhaps precipitation . Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 24
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Cloud and Precipitation II After condensation happens, the temperature lapse rate is significantly reduced as the latent heat released counteracts a portion of the loss of eternal energy to expansion work. We can derive the saturated adiabatic lapse rate from the first law as: Γ s = Γ d 1 + L lv cp dqvs dT As is evident, dq vs / dT > 0 ⇒ Γ s < Γ d . The normal rates of saturated lapse rate may vary between 3–9.8 [K Km -1 ] . The low bound refers to the condition of a lower warm atmosphere and upper bound represents the condition of an upper cold atmosphere. Note: Γ s is different than Γ moist , since the latter is just a minor correction to Γ d for water vapor and not saturated conditions. height Temperature Cloud Base d s d Figure 10: Dry (Γ d ) versus saturated (Γ s ) lapse rate. Saturated lapse rate varies nonlinearly with height and is typically smaller than the dry lapse rate. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 25
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Cloud and Precipitation III Intro to Cloud Microphysics It turns out that the water vapor molecules do not easily join together under saturation condition to form liquid water droplets —due to the high surface tension and their polar structure. Water vapor molecules ( D ∼ 0 . 3 nm ) need some condensation sites to join together efficiently and form the clouds. These condensation sites are provided by the atmospheric aerosols ( D ∼ 0 . 1 - 10 μ m ). These aerosols are necessary ingredients for cloud formation and are called cloud-condensation nuclei or CCN. It turns out that without CCNs, the air can be in a super-saturation condition ( RH > 100%) without any condensation and thus clouds cannot be formed. However,in the presence of CCNs, it is likely that water molecules stick to the CCNs surfaces and form the cloud droplets. However, we have seen a lot of non-precipitating clouds! Basically, when condensation continues and moisture is sufficient, the cloud droplet size starts to grow by a diffusional growth mechanism until they are large enough to fall. However, diffusional growth is not fast enough to create precipitation at the time scales we have observed, as short as ∼ 30 minutes. It turns out that the main growth mechanism is due to atmospheric turbulence . In effect, because of turbulent motions in atmosphere, the probability of collision and thus coalescence of the cloud droplets increases significantly, which eventually gives rise to formation of larger raindrops ( D ∼ 0.5–1 mm ) that can counteract the updraft drag forces and fall to the ground surface. The figures on the next slide give an illustration of the complex microphysical processes. An in-depth look at these processes is beyond the scope of this course and can be pursued in a course on the topic of cloud microphysics. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 26
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Cloud and Precipitation IV Figure 11: Schematic of precipitation micro-physical formation in a warm cloud and the concept of warm vs cold cloud (right). Drop Size Distribution (DSD) The random process of cloud and precipitation formation leads to a wide range of raindrop sizes . Obviously, deterministic explanation of motions of each drop is extremely complex. As a result, probability distributions are often employed to explain the drop size distribution (DSD) and precipitation microphysical properties. Marshall and Palmer (1947) suggested that the rainfall drop size distribution can be explained by the family of exponential probability density functions (i.e., f X ( x ) = λ exp ( - λ x ) for x ≥ 0 and λ > 0). Basically they counted the number of raindrops of size [ D , D + dD ] in a unit volume N D [# m -3 mm -1 ] as show in Figure 12 . Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 27
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Cloud and Precipitation V Figure 12: Schematic of drop size distribution in a unit volume (left). Distribution of number versus diameter for raindrops recorded at Ottawa, summer 1946. Curves A is for rate of rainfall 1.0 mm hr - 1 , curves B, C, D, for 2.8, 6.3, 23.0 mm hr - 1 , N D dD is the number of drops per cubic meter, of diameter between D and D + dD mm (from Marshall and Palmer, 1948). Specifically, Marshall and Palmer suggested the following dropsize distribution function N D = N 0 e - Λ D , where N D [# m -3 mm -1 ] is the number of raindrops with unit diameter in a unit volume of air, N 0 = 0 . 08 [cm - 4 ] is an experimental fitting parameter, Λ = 1 / D and D denotes the mean of the drop sizes. Note: When a random variable x has a probability distribution of f x ( x ), the mean of x is + ∞ -∞ g ( x ) f X ( x ) dx and mean of g ( x ) is + ∞ -∞ g ( x ) f X ( x ) dx . Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 28
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Cloud and Precipitation VI Applications of the Marshall-Palmer DSD: Total number of raindrops per unit volume [# m -3 ]: ∞ 0 N D dD = ∞ 0 N 0 e - Λ D dD = N 0 Λ Total liquid water content [kg m -3 ]: ∞ 0 ρ w π D 3 6 N D dD = ∞ 0 ρ w π D 3 6 N 0 e - Λ D dD = ρ w π N 0 Λ 4 Cloud-base precipitation rate [m s -1 ] : ∞ 0 π D 3 6 V t N D dD = ∞ 0 π D 3 6 α D N 0 e - Λ D dD = 4 π N 0 α Λ 5 , where V t = α D is the average falling velocity of raindrops at the base of the cloud. To understand how the above integrals are computed, we need to recall that ∞ 0 x n e - ax dx = n ! a n +1 with n = 1 , 2 , 3 , . . . and a > 0. Note that in computation of the cloud-base precipitation rate, we have simply assumed that the rainfall terminal velocity V t is a linear function of the drop diameter V t = α D , where α [s -1 ] is a proportionality constant. It should be emphasized that this linear assumption is not always true and experimentation often signifies a power law relationship V = α D β between the terminal velocity and the raindrop diameter. However, the above linear assumption can be considered as a first order approximation, which as we will see later on, can provide insightful information for retrieval of precipitation rate from active ground-based radar observations. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 29
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Cloud and Precipitation VII Precipitation Types Precipitation can be in liquid or frozen forms. Liquid form Drizzle : is a liquid precipitation with the average drop size of less than 0.5 [mm], generated by stratiform clouds with surface rainfall rate of < 1 [mm hr -1 ]. Rain : is a liquid precipitation with and average dropsize > 0 . 5 [mm] and a rate of more than 1 [mm hr -1 ]. Frozen form Freezing Drizzle/Rain : is the rainfall that falls through a subzero temperature layer of atmosphere near the earth surface and becomes supercooled . The supercooled raindrops become frozen immediately upon the first impact with the ground surface. The resulting ice is called glaze , which can grow to a thickness of several centimeters. Sleet : is a mixture of rain and partly melted snow. Snow : is small ice particle with average specific gravity of ρ snow /ρ water = 0 . 1. Ice Pellets : are small but translucent balls of ice with D < 5 mm, which are smaller than hail. Hail : is solid precipitation D=5–125 mm, with specific gravity of ρ snow /ρ water = 0 . 8. Graupel : is a soft hail or pellet with D = 2 - 5 mm and specific gravity 0.2–0.3. The graupels are opaque. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 30
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Cloud and Precipitation VIII Figure 13: Frozen precipitation types including, ice pellets (left), hail (middle), and graupel (right). Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 31
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Precipitation Mechanisms I As mentioned previously, there are five main lifting mechanisms that may give rise to precipitation: 1. Orographic lifting: Orographic lifting occurs when flow of moist air is lifted because of the presence of a topographic feature . When the moist air parcel ascends over mountains, condensation occurs, which often gives rise to a substantial precipitation on the windward side of the mountain range. On the leeward side of the mountain, rainfall is usually absent as the moisture content significantly drops. This process often creates a very dry area on the leeward of the mountain called the rain shadow . Figure 14: Orographic precipitation and rain shadows (Credit: NASA). Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 32
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Precipitation Mechanisms II 2. Frontal lifting : Warm Fronts: Warm air mass moves faster than the cold air mass. Typically the slope between the cold and warm air masses is mild (1:100 to 1:400) and lifting is gentle. This lifting mechanism produces mild precipitation over an area that may cover 300-500 km ahead of the frontal surface. Cold Fronts: Cold air mass moves faster than the warm air mass. Typically the slope between the cold and warm air masses is sharper than the warm front (1:25 to 1:100) and lifting is stronger. This lifting mechanism often produces heavy precipitation over an area that may cover 80-100 km ahead of the frontal surface. Stationary Fronts: is formed when the relative velocity is almost equal for the cold and warm air masses and neither of them can take over the other one. Figure 15: Frontal systems. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 33
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Precipitation Mechanisms III 3. Low level convergence: The low-level convergence occurs when two large-scale air masses at synoptic-scale ( > 1000 km) of similar temperature merge and are forced upward as shown in the schematic below. Figure 16: Schematic of low-level convergence (Credit: LakeErieWX) Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 34
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Precipitation Mechanisms IV 4. Buoyant rising: During a warm summer day, the air near the surface is heated by a warm earth surface , which is typically accompanied by an increased evaporation from the surface. This causes the air parcels near the surface to be less dense than the cooler air above, due to increased temperature and water vapor, causing the air to buoyantly rise. Figure 17: Buoyant lifting gives rise to cloud formation and perhaps precipitation. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 35
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Precipitation Mechanisms V 5. Mechanical lifting: Due to local turbulent mixing that picks up and takes the surface moisture to higher atmospheric elevations. As explained, this process accelerates the condensation and often produces short and scattered summertime precipitation. Figure 18: A schematic of mechanical lifting which may lead to condensation and precipitation. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 36
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Precipitation Measurements and Analysis I There are three ways to measure precipitation: 1) rain gauges, 2) ground-based weather radars, and 3) satellites. Rain Gauges Rain gauges are ground-based point measurements that are often categorized as follows: Non-recording gauges : Non-recording type rain gauge is the most common type of rain gauge used by meteorological departments round the world. It consists of a cylindrical vessel 127 [mm] in diameter with a base enlarged to 210 [mm] diameter. At its top section, funnel . This funnel shank is inserted in the neck of a receiving bottle. Figure 19: A non-recording type rain gauge. A receiving bottle has capacity of 100mm and during heavy rainfall, the amount of rain may exceeds the capacity. Therefore, frequent reading might be necessary during heavy storms . Water contained in this receiving bottle is measured by a graduated measuring glass with an accuracy up to 0.1 [mm]. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 37
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Precipitation Measurements and Analysis II Recording gauges : Automatic recording devices that provide high temporal resolution measurements of precipitation. Weighting gauges: Weighing bucket type rain gauge is one of the most common self-recording rain gauge. It consists of a receiver bucket supported by a mechanism to record the weight of water. The movement of bucket due to its increasing weight is transmitted to a pen, which gives a plot of accumulated (increased) rainfall in time. Figure 20: A weighting rain gauge. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 38
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Precipitation Measurements and Analysis III Float gauges : The working of this type of rain gauge is similar to weighing bucket rain gauge. A funnel receives the water which is collected in a rectangular container. A float is provided at the bottom of container, and this float raises as the water level rises in the container. Its movement being recorded by a pen to obtain an estimate of the collected rainfall mass . When water rises, this float reaches to the top floating in water, then siphon comes into operation and releases the water outwards through the connecting pipe, thus all water in box is drained out. Figure 21: A floating rain gauge. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 39
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Precipitation Measurements and Analysis IV Tipping bucket gauges : Tipping bucket rain gauges are adopted by the US weather bureau. It has 30cm diameter sharp edged receiver and at the end of the receiver there is a funnel. Pair of buckets are pivoted under this funnel in such a manner that when one bucket receives 0.25 mm of rainfall, it tips discharging its rainfall into the container, bringing the other bucket under the funnel and a pen records each tipping. Figure 22: A tipping bucket rain gauge. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 40
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Precipitation Measurements and Analysis V Optical rain gauges create a radiation field and record the altered intensity of the field as rain or snow passes through it. There are algorithms that calculate a precipitation rate based on the recorded irregularities in the radiation field. These sensors are a superior tool but are primarily used in research currently. Figure 23: An optical rain rain gauges. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 41
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Precipitation Measurements and Analysis VI Spatial Analysis of Rain Gauges Given that rain gauges are point measurements , there are different methods for interpolating these measurements to a spatial field that would be representative of precipitation over a watershed. These methods will be covered during the recitation of this course, but a brief overview is as follows: Figure 24: Construction of Thiessen polygons. The dashed lines connect nearby gauges, the solid lines are perpendicular bisectors of those lines, and the solid curved lines show the boundary of the watershed. Points in each subregion are the nearest neighbors to the gauge close to the center of each polygon. Arithmetic Mean Method : This method simply takes the mean of all the gauges within the spatial area of interest, meaning that all gauges have the same weight . Using Figure 24 as an example, the average precipitation is calculated as follows: R = Σ 7 i =1 R i 7 where R i is the precipitation amount or rate at each gauge and the denominator is the number of gauges. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 42
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Precipitation Measurements and Analysis VII Thiessen Polygon Method : This method geometrically calculates the area that is nearest to each rain gauge and assumes that the precipitation is constant over that area. The areal weighted average is then calculated as follows: R = Σ 7 i =1 A i R i Σ 7 i =1 A i where A i and R i are the area and precipitation rate within each polygon. Inverse Weighted Distance Interpolation: This method allows to interpolate the precipitation at each i th point on the watershed by taking a weighted average based on how far the point is from j th rain gauge: R i = Σ N j =1 w ij R j where w ij = 1 / d ij Σ j 1 / d ij . This interpolation uses the inverse of the distance of each point on the watershed to the rain gauge to determine the areal average of the precipitation over the watershed. If one wants to reduce the weights of the rain gauges that are far from analysis point on the watershed the weights can be obtained by squaring the distances as w ij = 1 / d 2 ij Σ j 1 / d 2 ij . Radar Measurement of Precipitation During the World War II, military radars noticed background clutters and noise due to weather phenomena such as precipitation. This observation became the main motivation for developing weather radar since early 50s. Ground-based precipitation radars have become operational across the United State since late 80’s. The Next Generation Weather Radar (NEXRAD) is a network of 160 high-resolution S-band (2-4 GHz) Doppler weather radars operated by the National Weather Service (NWS). Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 43
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Precipitation Measurements and Analysis VIII Figure 25: A NEXRAD radar (left), its reflectivity field in dBZ (middle) and the network of radar stations across the United States. A weather radar is equipped with a transmitter that generates fluxes of electromagnetic energy in microwave bands and a receiver that captures the backscattered waves by the raindrops. The power and pattern of the returned echoes captured by the receiver determine the type and intensity of precipitation. Typically due to the atmospheric attenuation, the amount of received energy flux is much smaller ( ∼ 10 - 9 watt) than the transmitted energy flux ( ∼ 10 6 watt). The returned power to the receiver is often expressed as reflectivity ( Z ). It turns out that Z is proportional to the sixth power of the diameter of the raindrops as Z i ∝ D 6 i and thus Z = ∞ 0 D 6 N D dD , Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 44
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Precipitation Measurements and Analysis IX where N D : [# m -3 mm -1 ], N D dD : [# m -3 ], and Z: [mm 6 m -3 ]. Reflectivity has a a wide dynamic range from 10 - 5 for drizzle to 10 7 for heavy rainfalls. Due to this wide dynamic range, reflectivity is often expressed in decibels of reflectivity as follows: dBZ = 10 log 10 Z 1 mm 6 m - 3 Theoretical and experimental evidence suggests that Z = aR b , where Z : [mm 6 m -3 ] and R : [mm hr -1 ] with typical values 200 ≤ a ≤ 600 and 1 . 0 ≤ b ≤ 2 . 0. The a and b are often obtained empirically through experimentation; even though, some theoretical derivations are also possible. The default values in NEXRAD WSR-88D are: Z = 200 R 1 . 6 stratiform storms (updraft < 1 m/s, lower troposphere is stably stratified) Z = 250 R 1 . 2 tropical convective storms (updraft > 1 m/s, lower troposphere is unstable) Z = 300 R 1 . 4 a compromise between stratiform and convective storms Figure 26: NEXRAD scans the entire vertical structure of atmosphere in 5 to 6 minutes at spatial resolution 1-4 km. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 45
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Precipitation Measurements and Analysis X An example of theoretically derived a and b can be obtained using the Marshal-Palmer dropsize distribution and assuming that the terminal velocity of raindrops is V t = α D . As discussed before, the cloud-based rainfall rate can be derived as follows: R = 4 π N 0 α 1 Λ 5 → Λ = 1 ¯ D = (4 π N 0 α ) 1 / 5 R - 1 / 5 . From the reflectivity integral, we have: Z = ∞ 0 D 6 N 0 e - Λ D dD = N 0 6! Λ 7 , plugging Λ from the first to the second equation, we get Z = 6! N 0 (4 π N 0 α ) 7 / 5 R 7 / 5 = a R b . Figure 27: Limitations of ground-based radar precipitation monitoring. Figure 28: Cone of silence in ground-based radars. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 46
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Precipitation Measurements and Analysis XI To visualize and process the radar data across the U.S., NOAA has offers a weather and climate toolkit (WCT) for free at https://www.ncdc.noaa.gov/wct/ . The data can be ordered through NEXRAD data archive and inventory at https://www.ncdc.noaa.gov/nexradinv/ . Figure 29: Visualizing the Base reflectivity image of a storm over Minnesota on 06/08/2017 at 12:04 am using the NOAA-WCT. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 47
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Precipitation Measurements and Analysis XII Temporal rain gauge analysis We are often interested to understand and quantify time series of precipitation. A hyetograph is a plot of rainfall depth [mm] or intensity [mm/hr] as a function of time, typically shown as a histogram. Given the rainfall measurement techniques, we assume the intensity is constant over a measurement period. Additionally, we can aggregate temporal data to determine monthly and yearly precipitation totals that can be used to see large-scale climatic changes. Figure 30: Sample of a rainfall hyetograph. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 48
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Precipitation Measurements and Analysis XIII Figure 31: United States and Minnesota 30-yr Annual Precipitation (Credit: PRISM) Figure 32: Global mean monthly rain rate using multi-satellite precipitation and rain gauge data from the Global Precipitation Climatology Project (GPCP) project. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 49
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Satellite Precipitation Measurement NASA launched a satellite called Global Precipitation Measurement (GPM) satellite in 2014 to map global changes of rain and snowfall using active and passive measurements. Physical Hydrology, Chapter 2: Atmospheric Water and Precipitation Ardeshir Ebtehaj 50
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help

Related Documents

CIVL 402 Syllabus 2023.pdf
PS-Business-Practices-Related-Professions.pdf
Assignment 1.pdf
Assignment 2.pdf
Assignment 3.pdf
Chapter_3 (1).pdf
Chapter_1 (1).pdf
Chapter_4 (1).pdf
Untitled document (6).pdf
CE643_Spring2023_Week6_GS (6).pdf
CIVE 208 Fall 2023 Full Course Outline.pdf
syllabus.pdf
Recommended textbooks for you
Text book image
Engineering Fundamentals: An Introduction to Engi...
Civil Engineering
ISBN:9781305084766
Author:Saeed Moaveni
Publisher:Cengage Learning
SEE MORE TEXTBOOKS
Recommended textbooks for you
Text book image
Engineering Fundamentals: An Introduction to Engi...
Civil Engineering
ISBN:9781305084766
Author:Saeed Moaveni
Publisher:Cengage Learning
SEE MORE TEXTBOOKS

Browse Popular Homework Q&A

Q: Taylor series and interval of convergencea. Use the definition of a Taylor/Maclaurin series to find…
Q: 3 Which of the following is the domain of the function f (x) = x ³ ? O All Real Numbers O [0, ∞) Ⓒ…
Q: QUESTION 1 A common treatment for Bipolar Disoder(s) is Electric Convulsion Therapy (ECT) True False
Q: A combination of four charges and four Gaussian surfaces are displayed in the figure. The charges…
Q: You are considering buying stock A. If the economy grows rapidly, you may earn 20 percent on the…
Q: mud silt sand conglomerate trough crossbedding ripple marks planar lam 50 પીપી 3. Which environments…
Q: Sharon Incorporated is headquartered in State X and owns 100 percent of Carol Corporation, Josey…
Q: What mass of oxygen is needed for the complete combustion of 3.70×10−3 g of methane?
Q: Find the derivative of the function. y = [x + (x + sin2(x))7]³
Q: A student conducted an experiment in a Styrofoam cup calorimeter that contained 100.0 g of water at…
Q: Find the flux of F=−zi+yj−xk out of a sphere of radius 4 centered at the origin. Hint: Your answer…
Q: Find the distance from the point (-7,-8,5) to the line ä(t) =(2, 5, 3)t + (-1, 3, 4).
Q: The following densities of water were obtained by Student A: 1.09g/mL, 1.14g/mL, and…
Q: Journalize the entries for costs transferred from Milling to Sifting and the costs transferred from…
Q: Solve the LP problem. If no optimal solution exists, indicate whether the feasible region is empty…
Q: Evaluate the following indefinite integral. et 1-1+ di dt 1 + et Note: Don't forget your C! And be…
Q: which 2 vertebrate clades suffered the greatest extinctions?
Q: Construct a 4-bit Johnson counter using J-K flip-flops. (See Figure 12-12 for a Johnson counter.)…
Q: 2. Cold packs take advantage of the fact that dissolving NH4NO3 in water is an endothermic process.…
Q: Required information [The following information applies to the questions displayed below.] Sammy's…
Q: Tamara owns interest in two different entities: a partnership and an S-Corp.  Her basis in the…
Q: 3. Inverse trigonometric outcomes For this section, recall: du I Ja du -1 = sin-¹ ()+C, /du- -tan-¹…
Q: Ice sheet is another name for mountain glaciers. O True O False
Q: What are the amount and character of Kimberly's recognized gain or loss on the distribution? What…
Q: Find the volume of a soild generated by roatating the region bounded by f(x)= x^3 and g(x)=2-x^2 and…