Answer to Question #156865 in Mechanics | Relativity for mosts

Question

Water flowing along a horizontal pipe has a speed of 2.27 m/s and a pressure of 4.86 atm. Further along, the pipe narrows so that the cross-sectional area decreases by a factor of 6.44. What is the pressure in the narrow section?

Answer 1

$V_{1}=2.27m/s, P_{1}=4.86 atm$

cross-sectional area decreased by a factor of 6.44

$A_{1}/A_{2}=6.44$

$A_{1}V_{1}=A_{2}V_{2}$

$V_{2}=(A_{1}V_{1})/A_{2}=6.44\times2.27=14.6188m/s$

From Bernoulli's theorem,

$\Delta P=\rho (V_{1}^{2}-V_{2}^{2})/2$

$P_2-P_1 =1000\times(2.27^{2}-14.6188^{2})/2$

$P_2-P_1=-104278.2067P_a=-\frac{104278.2067}{101325}atm=-1.029atm$

$P_2=4.86-1.029=3.831atm$