According to \"U.S. Standard Atmosphere\" definition, temperature within the fir

Question

According to "U.S. Standard Atmosphere" definition, temperature within the first 11 km of the atmosphere drops linearly as T = T0 + Bz, where the temperature at the ground is T0 = 288 K and the temperature lapse rate is B = -0.0065 K/m. Considering that the pressure at the ground level (z = 0) is equal to p0 = 101,3 kPa and treating alias an ideal gas, obtain the pressure variation as a function of z within the first 11 km of the atmosphere. Use the equation derived in part (a) to calculate the atmospheric pressure at the peak of Mount Everest (z = 8848 m). How much percent error would there be in the previous calculation if we assume atmospheric air to be isothermal at ground temperature of 15 degree C, an average temperature of -14 degree C ?

Explanation / Answer

a)

T = T0 + Bz

We have, dP/dz = -g where density = P/RT

Thus, dP/P = -g/RT dz............(eqn 1)

dP/P = -g/R dz/(T0+Bz)

Integrating both the sides from z=0 to z=z,

ln P - ln Patm = -(g/BR) [ln(T0+Bz) - ln(T0)]

Or, ln (P/Patm) = -(g/BR) ln[(T0+Bz)/T0]

P/Patm = [1 + Bz/T0]^(-g/RB)

Thus, P = 101.3*[1 - 0.0065*z/288]^(-9.81/(287*(-0.0065)))

or, P = 101.3*[1 - 0.0065*z/288]^5.259

b)

For z = 8848 m, we get P = 101.3*[1 - 0.0065*8848/288]^5.259 = 31.39 kPa

c)

For isothermal air, integrating equation 1 from z = 0 to z = z yields,

ln P/Patm = -gz/RT

or, P = Patm exp(-gz/RT)

i) For T = 15 deg C = 273+15 K = 288 K,

P = 101.3 exp(-9.81*8848/287/288) = 35.44 kPa

% error = (35.44 - 31.39)/31.39*100 = 12.9%

ii) For T = -14 deg C = 273-14 K = 259 K we get,

P = 101.3 exp(-9.81*8848/287/259) = 31.51 kPa

% error = (31.51-31.39)/31.39*100 = 0.38 %

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