Suppose the temperature of the atmosphere decreased from a surface virtual temperature of 15.8°C with a lapse rate of 7.4 K/km. Find the average virtual temperature of a layer extending from the surface to 7250 m.
ii. Find the scale height of an atmosphere with an average virtual temperature of −11°C.
iii. Using your result from (ii), find the pressure at a height of 7250 m above MSL in air with the above mean virtual temperature and a MSL pressure of 1025 hPa.
Please help me, 10 points if you do?
Maybe for 10 bucks
Reply:i) 7.4 x 7.250 = 53.65
15.8 - 53.65 = -37.85
ii) 15.8 - x = -11, x = 16.8
7.4 x height = 16.8
height = 2270 km
iii) 1025*2270 = pressure*7250
pressure = 321 hPa
Reply:ownage
Subscribe to:
Post Comments (Atom)
No comments:
Post a Comment