{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

PracticeTest01

# PracticeTest01 - EES 108 Earth and Atmosphere Practice...

This preview shows pages 1–5. Sign up to view the full content.

EES 108: Earth and Atmosphere Practice Test #1 For Tuesday September 27 Instructions You will have 75 minutes to complete the test. Answer all 20 multiple choice questions, which are worth 3 points each, and 5 out of the 6 short-answer questions, which are worth 10 points each, for a maximum possible total score of 110 points. The actual test will be quite similar to this one.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
EES 108 Practice Test # 1 Physical Constants Freezing point of water 273 K Boiling point of water 373 K Stefan-Boltzmann constant σ = 5 . 67 × 10 - 8 W / m 2 K 4 Planck constant h = 6 . 63 × 10 - 20 J s Average Distance from Earth to sun 1 AU = 1 . 50 × 10 11 m Average Distance from Mars to sun 1 . 52 AU Average Distance from Venus to sun 0 . 72 AU Radius of Earth 6 . 4 × 10 6 m Solar constant (solar flux at Earth) S Earth = 1370 W / m 2 Average albedo of the Earth A = 0 . 3 Average albedo of Venus A = 0 . 8 Average albedo of Mars A = 0 . 2 Greenhouse effect on Venus Δ T g = 510 K Greenhouse effect on Mars Δ T g = 6 K Effective radiating temperature of the Earth T e = 255 K Average surface temperature of the Earth T s = 288 K Atmospheric pressure at sea level P 0 = 1 . 013 bar Dry adiabatic lapse rate 10 C / km Normal tropospheric environmental lapse rate 6 . 5 C / km Dewpoint lapse rate L d = 2 C / km Average scale height of the atmosphere H 0 = 5 . 6km –2 of 18– Tuesday September 27
EES 108 Practice Test # 1 Equations Solar zenith angle zenith angle = latitude - declination (Positive latitude = northern, negative = southern) Solar elevation angle elevation angle = 90 - zenith angle Stefan-Boltzmann equation E = σT 4 Wien’s law λ max = 2898 μ m / K T Temperature conversion: Kelvin to Celsius T( C ) = T( K ) - 273 Fahrenheit to Celsius T( C ) = T( F ) - 32 1 . 8 Celsius to Fahrenheit T( F ) = 1 . 8 T( C ) + 32 Temperature lapse T(h) = T 0 - Lh , where T 0 is temperature at sea level, h is the height above sea level, and L is the lapse rate. Dew point lapse in unsaturated air T d (h) = T d, 0 - L d h , where T d (h) is the dew point at some height h above sea level, T d, 0 is dew point at sea level, h is the height above sea level, and L d is the dew point lapse rate. Lifting Condensation Level LCL = T - T d 8 C / km , where T = air temp and T d = dew point Barometric law P(h) = P 0 × 0 . 5 h/H 0 , where P(h) is the pressure at height h , P 0 is the pressure at sea level, and H 0 is the scale height of the atmosphere. Calculating fourth roots The fourth root ( 4 ) is the same as the square root of the square root. On your calculator, it may be easier to push the square-root ( ) key twice than to take the fourth root directly. Calculating fourth powers The fourth power is the same as the square of the square. On your calculator, if you have an x 2 key, it may be easier to push x 2 twice than to take the fourth power directly. –3 of 18– Tuesday September 27

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
EES 108 Practice Test # 1 Saturation specific humidity of water vapor Temp ( C ) Saturation spe- cific humidity ( g / kg ) - 40 0 . 10 - 30 0 . 30 - 20 0 . 75 - 10 2 . 0 0
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern