- #1
kallistos
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Homework Statement
At ground level in Colorado Springs, Colorado, the atmospheric pressure and temperature are 83.2 kPa and 25 C. Calculate the pressure on Pike’s Peak at an elevation of 2690 m above the city assuming an adiabatic atmosphere.
Known
Patm = 83200 Pa
T0 = 25° C
z-z0 = 2690 m
ρair at 25 C = 1.19 kg/m3
γ = 1.4
R = 286.9 J/(kg ⋅ K)
(Assume z0=0 for math work)
(γ is the ratio of specific heats)
Unknown
Pf = ?
Homework Equations
dP/dz = -ρg
P = ρRT
T=T0 - mz
Remember, temperature in the ideal gas equation must be absolute!
For an adiabatic atmosphere:
P/ρ^k = constant
The Attempt at a Solution
Calling the constant 'c':
P = c*ρ^k
ln (P) = ln (c*ρ^k)
ln (P) = ln (c) + k ln (ρ)
Substituting in ideal gas expression where ρ is:
ln (P) = ln (c) + k ln (P/RT)
ln (P) = ln (c) + k ln (P) - k (ln (R) + ln (T))
Consolidating all constants into some constant, C, and segregating temperature from pressure:
(1-k) ln (P) = -k * ln (T) + C
Now, differentiating with respect to P on the left and T on the right, we get:
(1-k) dP/P = -k * dT/T
dP/P = -k/(1-k)dT/T
dP/P = k/(k-1)dT/T
Integrating the left from P0 to P and the right from T0 to T:
ln (P) - ln (P0) = k/(k-1) (ln(T) - ln(T0))
ln (P/P0) = k/(k-1) ln (T/T0)
P = P0(T/T0)^(k/(k-1))
According to my professor, this equation and the adiabatic relationship, P/ρ^k = constant, are all I need to solve the problem. And I don't see how, given that we aren't given the final temperature at the new elevation. My instinct was to plug in T=T0 - mz, since we have the initial temperature, look up 'm' for dry adiabatic atmospheric conditions, and solve for 'p' that way. However, my professor says there is a way to solve for T without looking up the value for 'm'. He would prefer an answer this way as well. Any ideas?
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NOTE 1: Logarithmic rules:
ln (ab) = ln (a) + ln (b)
ln (a/b) = ln (a) - ln (b)
ln (a^k) = k ln (a)
NOTE 2: Differentiation rule:
d/dx(ln(x)) = 1/x