Question

a parking lot te storm sewer. rop in a 100-ft " sure difference is due to elevation effects, and how much is uue tu frictional effects? 850 8.50 WIT. A 40-m-long, 12-mm-diameter pipe with a friction fac- tor of 0.020 is used to siphon 30 °C water from a tank as shown in with a diameter essure drop per 10 m lost to the fric- cosity of 0.007 ermine the head 30 m rizontal pipe at ft of pipe. De- -mm-diameter m/s and exits bove the outlet r will air enter if the average ugh an 8-in.- versus pressure Figure P8.50

Answer 1

Objective: To determine the maximum value of h allowed.

.

Car ONS VAN Tom 7m 70) datum RBETE SSNAHMEN

Solution :

At 30°C the vapor pressure of water P, = 4.2343 kPa

Assume that minimum pressure is at the top of the hose : P3 = P

Applying energy equation between (1) and (3) :

$\therefore \frac{P_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{P_3}{\rho g}+\frac{V_3^2}{2g}+z_3+f\frac{L_{1-3}}{D}\frac{V^2_3}{2g}$

$\textup{As we know, }P_1=P_{atm}=101.325 \, \, kPa\, \, (abs), z_1=3\, \, m,\, \,L_{1-3}=10\,\, m$

$V_1=0, \, \, V_3=V,\, \,z_3=7\, \, m\, \, \textup{and }\, \, P_3=P_v$

72 101.325 x 103 (995.92)(9.81) 1 +0+3= 4.2343 x 103 + (995.92) (9.81) 2(9.81)+7 10 +7+(0.02) 0.012 1 V2 2(9.81)

$\therefore V=2.5679\, \, m/s\, \, \, \, \, \textup{(Velocity inside the pipe.)}$

.

$\textup{To obtain h applying energy equation between (1) and (2)} :$

$\therefore \frac{P_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+z_2+f\frac{L_{1-2}}{D}\frac{V^2_2}{2g}$

$\textup{As we know, }P_1=P_2=P_{atm}=0 \, \, kPa\, \, (gage), z_1=3\, \, m,\, \,L_{1-2}=40\,\, m$

$V_1=0, \, \, V_2=V,\,\, \textup{ and } \,z_2=-h\, \, m\, \,$

$\therefore0+0+3=0+\frac{(2.5679)^2}{2(9.81)}-h+(0.02)\frac{40}{0.012}\left (\frac{(2.5679)^2}{2(9.81)} \right )$

$\therefore h=19.7\, \, m$

.

$\textup{Let us check our assumption that minimum pressure occurs at (3) :}$

$\textup{From figure : }\frac{L}{z_x}=\frac{L_{2-3}}{h+7}$

$\therefore \frac{L}{z_x}=\frac{30}{19.7+7}$

$\therefore L=1.122 z_x$

$\textup{Applying energy equation between (x) and (2) :}$

$\therefore \frac{P_x}{\rho g}+\frac{V_x^2}{2g}+z_x=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+z_2+f\frac{L_{2-x}}{D}\frac{V^2_2}{2g}$

$\textup{As we know, }P_2=P_{atm}=0\, \, (gage), V_2=V_x=V,\, \, z_2=0(\textup{ (2) as datum})$

$\therefore \frac{P_x}{\rho g}+\frac{V^2}{2g}+z_x=0+\frac{V^2}{2g}+0+f\frac{1.122z_x}{D}\frac{V^2}{2g}$

$\therefore \frac{P_x}{\rho g}=(0.02)\frac{1.122z_x}{0.012}\left (\frac{2.5679^2}{2(9.81)} \right )-z_x$

$\therefore P_x=-0.3716z_x\times (995.9)(9.81)$

$\therefore P_x=-3630.6z_x$

$\textup{Hence, As the value of }z_x\textup{ increases, }P_x\textup{ decreases .}$

$\textup{Therefore, the assumed minimum pressure occurs at (3) is correct .}$

.

${\color{DarkBlue} \therefore h=19.7\, \, m}\, \, \, \, \, \, \, \, \, \, \, \, \, \textbf{(Answer)}$