At a height of 1000 m the pressure is 82,000 N/m2
, find the pressure at zero altitude when the
temperature is 13 0C. In addition, find the temperature and density at 1000 m. Assume adiabatic
conditions. Use Ȣ = 1.2 and R = 287 J/Kg/K.
Use: p2=p1(1-g(z2-z1)/RT1 (y-1//y)^y(y-1)
Answers
Answer:
Explanation:
Taking ‘g’ as constant, Eq.(2.5) can be integrated between two altitudes h1 and
h2. Taking h1 as sea level and h2 as the desired altitude (h), the integration gives
the following equation, the intermediate steps are left as an exercise.
(p/p0) = (T/T0)
(g/λR) (2.6)
where T is the temperature at the desired altitude (h) given by Eq.(2.4).
Equation (2.6) gives the variation of pressure with altitude.
The variation of density with altitude can be obtained using Eq.(2.6) and
the equation of state. The resulting variation of density with temperature in the
troposphere is given by:
(ρ/ρ0) = (T/T0)
(g/λR)-1 (2.7)
Thus, both the pressure and density variations are obtained once the
temperature variation is known.
As per the ISA, R = 287.05287 m2
sec-2 K and g = 9.80665 m/s2
.
Using these and λ = 0.0065 K/m in the troposphere yields (g/Rλ) as 5.25588.
Thus, in the troposphere, the pressure and density variations are :
(p/p0) = (T/T0)
5.25588 (2.8)
(ρ/ρ0) = (T/T0)
4.25588 (2.9)
Note: T= 288.15 - 0.0065 h; h in m and T in K.
In order to obtain the variations of properties in the lower stratosphere (11
to 20 km altitude), the previous analysis needs to be carried-out afresh with λ = 0
i.e., ‘T’ having a constant value equal to the temperature at 11 km (T = 216.65 K).
From this analysis the pressure and density variations in the lower stratosphere
are obtained as :
(p / p11) = (ρ / ρ11) = exp { -g (h - 11000) / RT11 } (2.10)
where p11, ρ11 and T11 are the pressure, density and temperature respectively at
11 km altitude.
In the middle stratosphere (20 to 32 km altitude), it can be shown that (note in
this case λ = -0.001 K / m):
(p / p20) = (T / T20)
- 34.1632 (2.11)
(ρ / ρ20) = (T/ T20)
- 35.1632