Answer:
Explanation:
Rate of change of pressure with respect to heigh
[tex]\frac{dp}{dh}[/tex] = k p ( given )
[tex]\frac{dp}{p}[/tex] = k dh
Integrating on both sides
∫[tex]\frac{dp}{p}[/tex] = ∫ k dh
lnp = kh + c , c is a constant
when h = 1000 m , p = 89.1 k Pa
ln 89.1 = 1000 k + c -------------- (1)
when h = 0 , p = 103.6 k Pa
ln 103.6 = 0 x k + c
c = ln 103.6
Putting it in quation (1)
ln 89.1 = 1000 k + ln103.6
ln [tex]\frac{89.1}{103.6}[/tex] = 1000k
k = [tex]\frac{-0.15}{1000}[/tex]
= - .15 x 10⁻³
lnp = - .15 x 10⁻³ h + ln 103.6
when h = 3000
lnp = - .15 x 10⁻³ x 3000 + ln 103.6
= - .45 +4.64
= 4.2
p = 66.68 k Pa
b )
when h = 6358
lnp = - .15 x 10⁻³ x 6358 + ln 103.6
= - .9537 + 4.64
= 3.6863
39.9 k Pa.
