Example $6\mathrm{.}3.$ Air at $1\mathrm{.}7$atm gauge and $15C$ enters a horizontal $75-mm$ steel pipe

that is $70m$ long. The velocity at the entrance of the pipe is $60\frac{m}{s}.$ Assuming

isothermal flow, what is the pressure at the discharge end of the line?

Solution

Use Eq$.(6\mathrm{.}52).$ The quantities needed are

$D=0\mathrm{.}075m,r_{\text{I}}=\frac{0\mathrm{.}075}{4}=0\mathrm{.}01875m$

$=0\mathrm{.}0174cP($ Appendix $8)=1\mathrm{.}74{10}^{-5}k\frac{g}{m}-s$

${}_E=\frac{29}{22\mathrm{.}4}\frac{2\mathrm{.}7}{1}\frac{273}{288}=3\mathrm{.}31k\frac{g}{m^3}$

Then

$G=gra{d}_{}=603\mathrm{.}31=198\mathrm{.}6k\frac{g}{m^2}=s$

$N_{Re}=0\mathrm{.}075\frac{198\mathrm{.}6}{1\mathrm{.}74{10}^{-5}}=8\mathrm{.}56{10}^5$

$\frac{k}{d}=0\mathrm{.}00015\frac{0\mathrm{.}3048}{0\mathrm{.}075}=0\mathrm{.}00061$

$f=0\mathrm{.}0044(Fig\mathrm{.}5\mathrm{.}9)$

Let $\frac{p_a+p_2}{2}\frac{?}{b}=ar(p),$ so

$(p_a-p_b{)}^2=(p_a-p_b)(p_a+p_b)=2p(p_a-p_b)$

From Eq$.(6\mathrm{.}52),$ omitting $g_c$

$p_a-p_b=\frac{RT{G}^2}{(\frac{?}{\text{b}\text{a}\text{r}}(p))M}(\frac{fL}{2r_B}+ln(\frac{p_a}{p_b}))$

where $p_a=2\mathrm{.}7$atm

$M=29$

$L=70m$

$R=82\mathrm{.}056{10}^{-3}{m}^{3}at\frac{m}{k}g$ mol$-K(Appendix2)$

$T=15+273=288K$

By trial, $\frac{?}{\text{b}\text{a}\text{r}}(p)$ is found to be $1\mathrm{.}982$atm and $p_b=1\mathrm{.}264$atm.

Check: Since $1$atm$=101,325\frac{N}{m^2}=101,325k\frac{g}{m}-s^2[$Eqs$.(1\mathrm{.}14)$ and $(1\mathrm{.}15)].$

$p_b=2\mathrm{.}7-\frac{82\mathrm{.}056{10}^{-3}288{198\mathrm{.}6}^2}{1\mathrm{.}98229101,325}(\frac{0\mathrm{.}004470}{20\mathrm{.}01875}+ln(\frac{2\mathrm{.}7}{1\mathrm{.}264}))$

$=1\mathrm{.}264$atm $absor0\mathrm{.}264$atm gauge

The average pressure is

tilde$(p)=\frac{27+1\mathrm{.}264}{2}=1\mathrm{.}992$atm

in this example from mcabe smith book how did we find p bar and pb