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FIRST QUARTER EXERCISE
UNITS
In the first quarter exercises you will be reducing units in several of the problems. Some units you will need to write into basic units before reducing units including units of Pascals, Newtons, and Joules.
Pascals = N/m^2 = kgms^-2/m^2 = kgm^-1s^-2
Newtons = kgms^-2
Joules = Nm = (kgms^-2)*(m)= kgm^2s^-2
Dividing same units:
same units divided by same units with same power = no units
(e.g. m/m = no units, m^2/m^2 = no units)
same units divided by same units with different powers = subtraction of powers (power of numerator minus power of denominator)
(e.g. m^3/m^2 = m^1 = m)
(e.g. m^2/m^5 = m^-3)
Multiplying units:
Same units times same units = units powered to the addition of the exponents
(e.g. m^2 * m^2 = m^4)
(e.g. m^3 * m^7 = m^10)
(e.g. m^2 * m^-6 = m^-4)
a unit with no exponent is understood to have a power of 1
(e.g. m = m^1)
a unit with a zero exponent is equal to 1
(e.g. m^0 = 1)
(e.g. 45^0 = 1)
USING KELVINS
In the first quarter exercises #2, #11, #13, and #17 require units of Kelvins. A Kelvin is a basic SI unit. If units are in Fahrenheit or Celsius, the units in most cases need to be converted to Kelvins. When a temperature change is being dealt with, Celsius or Kelvins can be used because the increment of a degree Celsius is the same as a degree Kelvin. Fahrenheit will NEVER be used in any of the calculations in the 1st and 2nd quarter exercises. Convert F to C and then to K. If the equation asks for a temperature that is not related to a temperature change, always use Kelvins in the 1st and 2nd quarter exercises.
(e.g. #17 in 1st quarter exercise, 104 F = (104 - 32)*5/9 = 40 C = 40 C + 273 = 313 K).. numerator is 313 K
QUESTION 2 HINTS
1. density = Pressure / (Rd*T) = (N*m^-2) / (N*m*K^-1*kg^-1*K); Newtons cancel, Kelvins cancel; you are left with kg/m^3 which is density units; [note: A Pascal is a Newton per meter squared; A Joule is a Newton times a meter. Rd the gas constant has units of Joules per Kelvin per kg (JK^-1kg^-1]
m^-2/m = 1/m^3
2. Temperature will need to be in Kelvins
3. To convert millibars to Pascals, multiply millibars value times 100
4. For the gas constant, use Rd
QUESTION 3 HINTS
1) In data set 1, the molecular weight value of 6.576 comes from the multiplication of 32.00 times 0.2055. When a percent is changed to a decimal the decimal place is moves two spaces to the left.
2) The average molecular weight is the addition of all the individual molecular weights.
3) The percent change in molecular weight will be a small amount. It is found by:
( 1 - (avg set2/avg set1)) * 100%
QUESTION 9 HINTS
acceleration / constant pressure gas constant = g/Cp =
ms^-2 / Jkg^-1K^-1 = ms^-2 / Nmkg^-1K^-1; a Joule is a Newton times a meter
= ms^-2 / kgms^-2mkg^-1K^-1; acceleration in numerator cancels with acceleration in denominator; A Newton is a kgms^-2 (mass times acceleration)
= 1 / kgmkg^-1K^-1; mass cancels
= 1/mK^-1; move K to numerator (changes sign of power)
= K/m; since an increment of C is the same as K, it can also be expressed as:
= C/m; degrees Celsius per meter
Multiplying by 1000 changes it to C/km since there are 1000 m in a km
QUESTION 11 HINTS
1. The last term in the equation is Ln(p1/p2). Ln is the natural log. p1 is always the larger value. In this problem, p1 is 1000 and p2 is 500, therefore the value is:
LN(1000/500) = LN(2) = 0.693147181
2. Rd and g are simply constants
3. The virtual temperature must be in Kelvins, 0 C = 273 K
4. The hypsometric equation is one of the most important equations in meteorology. It's calculation produces all the height contours on analysis and graphical model images.
QUESTION 13 HINTS
1. A millibar is a hectopascal, therefore, 17 millibars is 1,700 Pascals
2. To convert Celsius to Kelvins, add 273 to Celsius value.
3. Density is the inverse of specific volume. The inverse button is the 1/x button on the calculator. Example, the inverse of 2 is 1/2. The units of specific volume are m^3/kg, the units of density is kg/m^3, notice these units are an inverse of one another. Once you have a numerical value for specific volume, put this value in the calculator and press the 1/x button, this will give you the numerical value of density. Another alternative is to rearrange the equation to e = density*Rv*T, then solve for density.
QUESTION 15 HINTS
1. Pressure = density * gravity * height
2. Units check: Pressure = kgm^-3*ms^-2*m = kgm^-1s^-2 = the basic units of the Pascal
3. When solving the pressure equation, pressure is in Pascals, change millibars to Pascals in part #2
4. Part 2 is solved the same way as Part 1, except in part 2 you are solving for h instead of P, P in part 2 is given to you in the problem. The density of water was already given in part 1 and g is a constant.
QUESTION 17 HINTS
1. The (1- Rd/Rv) term is a constant of 0.3774…; since Rd is 287 and Rv is 461
2. e/p = 30/1025 = 0.029268293 (this is a case where you keep pressure in mb because e has units of mb, if you convert numerator and denominator both to Pascals, you will get the same answer as when both are in units of mb)
3. Temperature is in Kelvins
4. The virtual temperature is always greater than the actual temperature (except in perfectly dry air)
5. Denominator is 1-(e/p)(0.3774)= a number that is a little less than 1.0 (exact number is critical)
6. e and P have same units, Rd and Rv have same units, therefore there are no units in denominator, numerator has units of K, therefore final answer will have units of K
7. Do not round numbers (let the digits go out as far as your calculator will let them) until you have the final answer. Rounding each number in the equation can result in significant error.
8. Keep final answer in Kelvin units
SECOND QUARTER EXERCISE
DIFFERENCE BETWEEN E AND Es
1. Es is the saturation vapor pressure and can be found by plugging temperature into the Clausius Clapeyron equation.
2. Es does not tell the meteorologist how much moisture is in the air (unless RH is 100%). Es is the vapor pressure that would result if the air was saturated with water vapor with respect to a certain temperature. Vapor pressure has units of millibars (mb).
3. What is vapor pressure? It is the pressure exerted by water vapor alone. Suppose the amount of water vapor in the air is 3% of the total pressure. If the total air pressure outside was 1000 mb, the contribution of pressure from water vapor would be 1000*0.03 = 30 mb
4. E (actual vapor pressure) does indicate the actual amount of moisture in the air. E is found by plugging dewpoint into the Clausius Clapeyron equation. The terms vapor pressure and actual vapor pressure mean the same.
5. If the saturation vapor pressure is 35 mb and the actual vapor pressure is 25 mb, then the RH = 25/35*100% = 71.4%
6. When the RH is 100%, Es = E
7. In the second half of the course you will learn how to find E on the Skew-T diagram
8. Remember, Es is the vapor pressure relative to the temperature and E is the vapor pressure relative to the dewpoint temperature. Since the RH=100% when the dewpoint is reached, plugging the dewpoint in for T in the CC equation yields the actual vapor pressure. Plugging the temperature into the CC equation yields Es. E and Es will have different values when the air is NOT saturated.
MIXING RATIO UNITS
Mixing ratio is defined as the mass of water vapor in a parcel of air compared to the weight of dry air in the parcel. Suppose the mass of water vapor in a parcel of air was 6 grams and the weight of the rest of the air in that parcel was 500 grams. The mixing ratio would be 6 grams per 500 grams = 6 grams/500 grams = 0.012 grams per grams. Notice the units are the same in the numerator (top number) and the denominator (bottom number). Thus, the units cancel. BUT, we keep the units instead of canceling them for comparative purposes.
The units of mixing ratio are often expressed as grams per kilogram (grams of water vapor per kilogram of dry air), because there is much more mass in the air from Nitrogen and Oxygen than there is from water vapor. 0.012 grams of water vapor per gram of dry air becomes, 0.012*1000 = 12 grams of water vapor per kilogram of dry air. 12 is an easier number to deal with than 0.012. This is why units are often expressed as g/kg.
When you solve equation #19 on the second quarter exercise you will notice that there are no units! the millibars in the numerator cancel with the millibars in the denominator. This actually makes sense because mixing ratio really does not have units (they cancel out) because mixing ratio is a RATIO. Relative humidity is also a ratio (changed to a percentage). Just keep in mind that the units will be in grams per gram. Once you solve the equation in #19, that number needs to be multiplied by 1000 to get the mixing ratio in grams per kilogram.
LATENT HEAT VALUES
On page 46 of the text are the values of latent heat for various states of water vapor. Notice they are in units of cal/g. Below is the conversion of units from cal/g to Joules/kg. There are 1000 grams in a kilogram and 4.186 Joules in a calorie.
Lv(latent heat of vaporization) = 597.3 cal/g
597.3cal/g*(1000g/kg)*(4.186J/cal) = 2.5*10^6 J/kg
Ls(latent heat of sublimation) = 677.0 cal/g
677.0cal/g*(1000g/kg)*(4.186J/cal) = 2.83*10^6 J/kg
Lm (latent heat of melting) = 79.7 cal/g
79.7cal/g*(1000g/kg)*(4.186J/cal) = 3.33*10^5 J/kg
Note: Ls = Lv + Lm
2.83*10^6 J/kg = 2.5*10^6 J/kg + 3.33*10^5 J/kg
LFC, EL and Tdd
1) It is operational practice to locate the LFC and EL that correlates to the largest region of CAPE on the sounding. The LFC is the pressure level at the bottom of the largest area of CAPE and the EL is the pressure level at the top of the largest area of CAPE. On a sounding with fairly large CAPE, the EL will be near the tropopause, the MPL (maximum parcel level) may be above the tropopause due to updraft momentum.
2) The dewpoint depression is always a positive number or 0. It must be positive or 0 because the formula is the temperature minus the dewpoint and the temperature will always be greater than or equal to the dewpoint. Dewpoint depression is understood to be a temperature difference. A Tdd of 15 C, has nothing to do with an actual temperature of 15 C, it is understood to be a temperature difference (depression). If the temperature is -15 C and the dewpoint is -25 C, the Tdd is 10 C. If the temperature is +10 C and the dewpoint is -5 C, the Tdd is 15 C.
VIRTUAL TEMPERATURE INFORMATION
The virtual temperature is defined as the temperature dry air must be in order to have the same density as air that does have water vapor in it.
Take two parcels of air that are the same temperature. Suppose one has no water vapor and the other one does. The parcel of air with moisture will be less dense than the one without moisture because the H20 vapor molecule weights less than diatomic Oxygen or diatomic Nitrogen.
Warming the air will cause the density of air to decrease.
Thus, if the dry air parcel is warmed it will begin to approach the same density as the parcel that had moisture.
When the dry air has reached the same density as the moist air, then the virtual temperature (Tv) has been reached.
Because moist air is less dense than dry air at the same temperature, the virtual temperature is always greater than the actual temperature.
The virtual temperature is generally just a couple of degrees or a partial degree more than the actual temperature because a slight temperature change can change the density of air more rapidly than a moisture change.
Virtual temperature is used in the place of actual temperature in thermodynamic equations because virtual temperature does not result in errors from density changes caused by water vapor.
QUESTION 1 HINTS
The first law of thermodynamics states energy can neither be created or destroyed. All the energy in the universe is accounted for. Energy can change forms but can not be created or destroyed. Energy can exist as heat or work. Both heat and work have units of Joules. The equation given in lecture 4 which mathematically describes heat and work is dq - dw = du.
1. In problem, du is 0, therefore dq = dw, dw = Pressure*volume change
2. The basic units of pressure is the Pascal kgm^-1s^-2, the units of volume are m^3, a Pascal times volume yields units of kgm^2s^-2... this is the basic units of a Joule
3. A Joules is a Newton times a meter = Nm = kgms^-2m = kgm^2s^-2
4. Again, The formula you use for this problem is dq = P * dV
QUESTION 3 HINTS
Poisson's equation is the solution for potential temperature.
1. Potential temperature, THETA= T*(p1/p2)^0.286
2. Potential temperature is defined as the temperature a parcel of air will have if brought to the 1000 millibar level, therefore, p1= 1000mb
3. Example problem, what is the potential temperature of air at 700 mb at a temp of 273 K.
THETA= 273(1000/700)^0.286 = 302 K;
4. You should have a button on your calculator that says y^x, this button is used for solving exponents. example, what is 3^5?, to find this press 3, then the y^x button, then 5, answer is 243
5. When solving Poisson's equation, temperature must be in Kelvins
6. You may hear THETA referenced to in forecast discussions. Air parcels tend to flow along constant THETA surfaces. Constant THETA surfaces are constant "density" (more accurately potential temperature) surfaces. When air lifts over a warm front, the air lifts to attain constant THETA. Air in cool sector of warm front is more dense, therefore, less dense air will lift over more dense air. THETA can also be used to determine if air at different levels of the atmosphere has a greater or smaller THETA than air at another level in the atmosphere. This helps a forecaster determine which regions aloft are experiencing WAA and CAA.
EXAMPLE PROBLEM USING POISSON'S EQUATION
Using Poisson's equation, what is the potential temperature at 500 mb when the initial temperature of the parcel is -10.0C (263.0 K) ); T must be in Kelvin units anytime solving this equation.
Poisson's equation, THETA = T*(1000/p)^0.286
THETA = 263(1000mb/500mb)^0.286 =
THETA = 263*2^0.286 (263 times 2 to the 0.286 power) =
THETA = 320.7 K
notes: units of pressure cancel leaving degrees K as units; performing powers requires the use of the y to the x button on the calculator (y^x)
i.e. 5 to the third power = 5^3 = press 5 then press y^x key then press 3 then enter = 125
QUESTION 16 EXAMPLE PROBLEM
Example problem for solving Clausius-Clapeyron equation
What is the saturation vapor pressure when the temperature is 50 F?
1. The equation is LN(es/6.11) = (L/Rv)(1/273 - 1/T); L is 2.453*10^6 J/kg, Rv is 461 J/kg/K; L/Rv = 5,321.04 K
2. Equation can be rewritten as
LN(es/6.11) = 5,321.04K(1/273 - 1/T)
3. T is given as 50 F, Temperature must be in Kelvins, therefore 50 F = 10 C = 283 K; equation becomes
LN(es/6.11) = 5,321.04K(1/273 - 1/283)
4. We can now reduce (1/273 - 1/283)to one number; it reduces to 0.000129435 1/K
5. Equation becomes
LN(es/6.11) = 5,321.04(0.000129435); units eliminated since K*1/K = 1), final units are set to millibars
LN(es/6.11) = 0.688727527
6. taking the exponent of a LN cancels it, taking the exponent of each side of the equation yields
es/6.11 = e^0.688727527 = 1.991180197
**I want to point out that the exponential function (e) and the vapor pressure (e) have the same abbreviation but they are totally separate items. e^0.688727527 above has nothing to do with vapor pressure. It is the exponent of that number.
7. Multiply both sides of equation by 6.11
es = 6.11(1.991180197) = 12.166 mb
8. es is the maximum amount of vapor pressure that can exist in the atmosphere at a certain temperature. Vapor pressure increases exponentially as temperature increases.
9. This equation is explained in section 4.5 of the book
QUESTION 17 EXAMPLE PROBLEM
Example problem, What is the percentage of water vapor in the air using a surface pressure of 997.2 mb? Given: the vapor pressure is 17 mb.
Ratio = 17 mb vapor pressure / 997.2 mb total pressure =
17mb/997.2mb = 0.01705
expressed as a percent = 0.01705*100% = 1.705% (1.705% of the total air pressure is from water vapor)
**Assume vapor pressure equals saturation vapor pressure for #17
QUESTION 18 HINTS
1. Find e and es using Clausius-Clapeyron equation for both 50 F and 104 F, divide es at 50 F by es at 104 F. The relative humidity is the vapor pressure divided by the saturation vapor pressure. RH = e/es*100%
note- e is the vapor pressure at the dewpoint while es is the vapor pressure at the temperature. They are both found by the same method of solving the C-C equation.
2. e, the vapor pressure, is the maximum amount of moisture that can be in the atmosphere when the temperature is equal to the dewpoint
3. es, the saturation vapor pressure, is the maximum amount of moisture that can be in the atmosphere at the actual air temperature.
4. es will always be greater than or equal to e
5. You already found es at 104 F in #16, now just find e at 50 F and then divide e by es and multiply by 100% to get RH
6. Your final answer will have a low relative humidity, which makes sense because the dewpoint depression between 104 and 50 is very large
7. Always remember that the C-C (Clausius-Clapeyron) equation requires temperature to be in units of Kelvins (whether you are plugging in the actual temperature or the dewpoint temperature for the temperature term)
QUESTION 19 EXAMPLE PROBLEM
Example problem: What is the mixing ratio when the vapor pressure is 7 millibars at 1000 mb?
mixing ratio = (0.622e)/(P - e)
*note: denominator (P - e) will sometimes be written as just P because P is much larger than e, the difference in answer will be small if P is used instead of (p - e). The difference in which formula to use becomes more significant at higher vapor pressures.
mixing ratio = (0.622*7mb) / (1000mb - 7mb)
= 4.354mb / 993mb = 0.00438 GRAMS of water vapor per GRAM of dry air
= 0.00438*1000 = 4.38 GRAMS of water vapor per KILOGRAM of dry air
**For the second part of #19 assume that the RH=100%, thus mixing ratio equals saturation mixing ratio. Plug answer to #16 directly into mixing ratio equation. Thus you are finding the saturation mixing ratio since saturation vapor pressure is being plugged into equation.
**For #20, the value of the saturation mixing ratio is meant to relate to the dewpoint. Use the 9 g/kg (0.009 kg/kg) value to plug directly into virtual temperature equation.
QUESTION 21 HINTS
a. The dewpoint represents the temperature at which saturation will occur. es represents maximum vapor pressure for actual temperature. Use dewpoint in Clausius-Clapeyron equation (for T) to find e for Td.
b. Use formula from #19 (w = (0.622*e) / P - e)) ; e is value you found by solving for dewpoint (51 F) in Clausius-Clapeyron equation. Final answer units will be in grams per kilogram
c. Tv = T*(1 + 0.61w); w was found in part b but must be converted to units of kg per kg; Temperature (T) must be in Kelvins, 72 F = ? Kelvins
d. Divide e at 51 F by es at 72 F, RH = e/es*100%
QUESTION 23 HINTS AND EXAMPLE PROBLEMS
1. Tv is the symbol for the virtual temperature; formula: Tv = T / (1 - 0.378*e/p); this is simplified form of equation given in #17 of first quarter exercise
2. T= 60 F = ? K
3. p = 1006 mb
4. Since e and p have same units, the units cancel, whether you use Pascals or millibars, the division will yield the same result for e/p. keep units in mb since e will have units of mb.
5. Find e by plugging and chugging given temperature (60 F = ? K) into Clausius-Clapeyron equation. Multiple the e value by 0.65 to get e with RH of 65%. Remember RH = 100e/es
6. Plug and chug temperature, 0.65e (the new e calculated in step 5), and pressure value into virtual temperature equation given in step 1. Your answer will be in Kelvins
7. Virtual temperature is ALWAYS greater than the actual temperature in air that has moisture. The virtual temperature is usually within a few degrees or less of the actual temperature. Adding moisture to the air changes the density of the air (causes it to be less dense). Virtual temperature was developed to account for density differences between air with different amounts of moisture. The thickness equation (hypsometric equation) uses Tv to account for thickness changes from moisture.
8. Alternative method: you can also use formula Tv = T*(1 + 0.61*w) where w = (0.622*e) / (P - e). Once you find e, then w can be found, then w can be plugged into Tv equation. Both methods will result in the same answer.
Example problems:
Given Es= 23.1 mb, What is e if the RH is 70%?
RH (relative humidity) = e/Es*100% (vapor pressure divided by saturation vapor pressure times 100%)
70% = 0.70 (changed from percent to decimal)
RH (as ratio) = 0.70 = e/Es; Es = 23.1 mb
Therefore: 0.70 = e/23.1 mb (solve for e)
e = 0.70*23.1 mb = 16.1 mb
What is Tv at 1000 mb and 293.0 K given RH is 70% and Es= 23.1mb?
e was found to be 16.1 mb
Tv = T/(1 - 0.378e/p)
The 0.378 came from (1 - Rd/Rv) = (1 - 287/461) = 0.378
see equation in #17 of 1st quarter exercise
Tv = 293.0 / (1 - 0.378*(16.1/1000)) = 293/0.9939142 = 294.8 K
QUESTION 24 HINTS
See page 81 in text
dT/dz = -g/R
The negative sign has nothing to do with gravity. The equation could just as easily be written as
-dT/dz = g/R or dT/dz = g/-R
The negative sign is there (in front of the g/R) term because the temperature lapse rate with height dT/dz is negative (in the homogeneous atmosphere, temperature decreases with height).
As an example in earth's atmosphere, suppose the temperature is 20 C at the surface and 0 C at 4,000 meters. The change in temperature with height is -20 C/4 km = -5 C/km (temperature decreases by ascending in the atmosphere).
By convention, distance is positive when moving away from the earth's surface, therefore, when temperature decreases with height the change in temperature with height must be negative.
The value of gravity only changes slightly from the bottom to the top of the atmosphere. For all practical purposes gravity can be treated as a constant in the atmosphere of +9.8m/s^2. g is always positive, the negative sign in the equation is the result of dT being negative.
QUESTION 25 EXAMPLE PROBLEM AND HINTS
note** to find e in #25 to begin with, you must first find es (by plugging T into C-C equation), second find Ws using the formula, Ws = (0.622es) / (P - es), then third find RH using, RH = 100*W/Ws (W is given in original problem), then fourth find e using formula RH = 100*e/es. Once you have e, then you can plug e into C-C equation to find dewpoint.
Example problem of finding e when given dewpoint
What is the dewpoint given the vapor pressure is 15.8 mb?
CC equation:
LN(Es/6.11) = L/Rv(1/273 - 1/T)
note: plugging T (temperature in CC equation yields Es)
LN(e/6.11) = L/Rv(1/273 - 1/Td)
note: plugging Td (dewpoint) in CC equation yields e
Since we are given vapor pressure in the original question we use the formula below
LN(e/6.11) = L/Rv(1/273 - 1/Td)
we are given e is 15.8 mb, that leaves us with only one unknown which is Td
rearrange CC equation using algebra to solve for Td
LN(15.8/6.11) = L/Rv(1/273 - 1/Td)
(Rv*LN(15.8/6.11))/L = (1/273 - 1/Td)
(Rv*LN(15.8/6.11))/L - (1/273) = -1/Td
Td = -1/(Rv*LN(15.8/6.11))/L - (1/273))
Rv/L = 0.000187933 (inverse of L/Rv)
LN(15.8/6.11) = LN(2.585924714) = 0.950083167
Td = -1/(0.000187933*0.950083167 - (1/273))
Td = -1/(-0.003484452) = 287 K = 14 C
More explanation of finding Td when given e:
LN(15.8/6.11) = L/Rv(1/273 - 1/Td)
Easy way to reduce above equation is to immediately plug numbers in for all variables that are unknown. This can be done in one continuous step on a scientific calculator.
Here are steps: 15.8 divided by 6.11 and then take natural log of result. Divide this new number by 5,321.04 (which is L/Rv). Subtract 1/273 from the newest new number. Next take the inverse of number and cancel the negative sign. Convert number to Celsius from Kelvin.
LN(15.8/6.11) = L/Rv(1/273 - 1/Td)
(LN(15.8/6.11) / 5,321.04) - 1/273 = -1/Td
-0.003484452 = -1/Td
negative sign cancels
0.003484452 = 1/Td
Td*0.003484452 = 1
Td = 1 / 0.003484452
Td = 287 K = 14 C
Easy way is to take inverse of 0.003484452. When this number is in calculator, press the 1/X button on calculator. It will convert to 287 with just that one press of a button.