Question 3.SP.1: Compute the atmospheric pressure at elevation 20,000 ft, con......

From Appendix A, Table A.3, the conditions of the standard atmosphere at sea level are T_{1}=59.0^{\circ} \mathrm{F}, p_{1}=14.70 \mathrm{psia}, \gamma_{1}=0.07648 \mathrm{lb} / \mathrm{ft}^{3}, where subscript 1 indicates conditions at our reference elevation, sea level.

(a) Constant density

From Sec. 3.2: \frac{d p}{d z}=-\gamma ; \quad d p=-\gamma d z ; \quad \int_{p_{1}}^{p} d p=-\gamma \int_{z_{1}}^{z} d z

so

and    p=14.70(144)-0.07648(20,000)=587 \mathrm{lb} / \mathrm{ft}^{2} \mathrm{abs}=4.08 psia

(b) Isothermal

From Sec. 2.7: p v= constant; so \frac{p}{\gamma}=\frac{p_{1}}{\gamma_{1}} if g is constant

Eq. (3.2) :

Integrating, \quad \int_{p_{1}}^{p} \frac{d p}{p}=\ln \frac{p}{p_{1}}=-\frac{\gamma_{1}}{p_{1}} \int_{z_{1}}^{z} d z=-\frac{\gamma_{1}}{p_{1}}\left(z-z_{1}\right)

Thus

(c) Isentropic

From Sec. 2.7: \quad p v^{1.4}=\frac{p}{\rho^{1.4}}= constant; \quad so \quad \frac{p}{\gamma^{1.4}}= constant =\frac{p}{\gamma_{1}^{1.4}}

Eq. (3.2): \quad \frac{d p}{d z}=-\gamma, \quad where \gamma=\gamma_{1}\left(\frac{p}{p_{1}}\right)^{1 / 1.4}=\gamma_{1}\left(\frac{p}{p_{1}}\right)^{0.714}

so      d p=-\gamma_{1}\left(\frac{p}{p_{1}}\right)^{0.714} d z

\begin{aligned}& \text { Integrating: } \\& \qquad \begin{aligned}\int_{p_{1}}^{p} p^{-0.714} d p & =-\gamma_{1} p_{1}^{-0.714} \int_{z_{1}}^{z} d z \\p^{0.286}-p_{1}^{0.286} & =-0.286 \gamma_{1} p_{1}^{-0.714}\left(z-z_{1}\right) \\p^{0.286} & =(14.70 \times 144)^{0.286}-0.286(0.07648)(14.70 \times 144)^{-0.714}(20,000) \\p & =942 \mathrm{lb} / \mathrm{ft}^{2} \mathrm{abs}=6.54 \mathrm{psia} \end{aligned}\end{aligned}

(d) Temperature decreasing linearly with elevation

For the standard lapse rate (Fig. 2.2): T=a+b z, where a=59.00+459.67=518.67^{\circ} \mathrm{R} and b=-0.003560^{\circ} \mathrm{R} / \mathrm{ft}

Eqs. (3.2) and (2.4): \quad \frac{d p}{d z}=-\rho g ; \quad \rho=\frac{p}{R T}

Combining to eliminate \rho, which varies, rearranging, and substituting for T,

Integrating:

i.e.,          \quad \frac{p_{2}}{p_{1}}=\left(\frac{a+b z_{2}}{a+b z_{1}}\right)^{-g / R b}

Here        \frac{-g}{R b}=\frac{-32.174}{1716(-0.003560)}=5.27

and, from Table A.3: p_{1}=14.696 psia when z_{1}=0.

Thus \quad \frac{p_{2}}{14.696}=\left(\frac{518.67-0.003560 \times 20,000}{518.67+0}\right)^{5.27}=0.459

p_{2}=14.696(0.459)=6.75 psia \quad

 \frac{d p}{d z}=-\gamma      (3.2)

 \frac{p}{\rho}=p v=R T          (2.4)