Question

Water flows steadily through a horizontal pipe of varying cross section. At one place the pressure 15 1.5 atm and the speed is 1.0 m/sec. Determine the pressure at another place where the speed is 10 m/sec.

A. 0.5 atm
B. 0.6 atm
C. 0.7 atm
D. 0.8 atm
E, 1.0 atm

Answer 1

D. 1 atm

Step-by-step explanation:

pressure P1 = 1.5 atm = 1.5 x 101325 = 151987.5 N/m^{2}

speed V = 1 m/s

speed U = 10 m/s

pressure P2 = ?

applying Bernoulli's equation we can find the pressure P2

`(1)/(2)ρv^(2)+P1=(1)/(2)ρu^(2)+P2`

`(1)/(2)ρv^(2)-(1)/(2)ρu^(2)+P1= P2`

`P2=(1)/(2)ρ(v^(2)-u^(2))+P1`

where ρ is the density of water = 1000 kg/m^{3}

`P2=(1)/(2)x1000x(1^(2)-10^(2))+151987.5`

`P2=(1)/(2)x1000x(-99)+151987.5`

P2 = 102487.7 N/m^{2}

P2 = 102487.7 / 101325 = 1.01 atm = 1 atm