Water flows through a horizontal pipe of cross-sectional area $10.0 \mathrm {~cm} ^ { 2 }$ at a pressure of 0.250 atm
with a flow rate is 1.00 L/s. At a valve, the effective cross-sectional area of the pipe is reduced to $5.00 \mathrm {~cm} ^ { 2 }$ What is the pressure at the valve? The density of water is $1000 \mathrm {~kg} / \mathrm { m } ^ { 3 }$, and treat it as an ideal incompressible fluid.
A) 0.157 atm
B) 0.110 atm
C) 0.200 atm
D) 7700 pa
E) 0.235 atm

Question

Water flows through a horizontal pipe of cross-sectional area $10.0 \mathrm {~cm} ^ { 2 }$ at a pressure of 0.250 atm
with a flow rate is  1.00 L/s. At a valve, the effective cross-sectional area of the pipe is reduced to $5.00 \mathrm {~cm} ^ { 2 }$ What is the pressure at the valve? The density of water is $1000 \mathrm {~kg} / \mathrm { m } ^ { 3 }$, and treat it as an  ideal incompressible fluid.
A) 0.157 atm
B) 0.110 atm
C) 0.200 atm
D) 7700 pa
E) 0.235 atm

Quizplus · Accepted Answer

To solve this problem, we can use the Bernoulli's equation and the equation of continuity. The equation of continuity states that $A_1v_1 = A_2v_2$, where $A_1$ and $A_2$ are the cross-sectional areas of the pipe before and at the valve, respectively, and $v_1$ and $v_2$ are the velocities of the fluid at these points. Given $A_1 = 10.0 \, \text{cm}^2 = 1.0 \times 10^{-3} \, \text{m}^2$ and $A_2 = 5.00 \, \text{cm}^2 = 0.5 \times 10^{-3} \, \text{m}^2$, and a flow rate of $1.00 \, \text{L/s} = 1.00 \times 10^{-3} \, \text{m}^3/\text{s}$, we can find the velocities $v_1$ and $v_2$.$$v_1 = \frac{Q}{A_1} = \frac{1.00 \times 10^{-3}}{1.0 \times 10^{-3}} = 1.00 \, \text{m/s}$$$$v_2 = \frac{Q}{A_2} = \frac{1.00 \times 10^{-3}}{0.5 \times 10^{-3}} = 2.00 \, \text{m/s}$$Bernoulli's equation in the form $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$ allows us to solve for $P_2$, the pressure at the valve. Given $P_1 = 0.250 \, \text{atm} = 0.250 \times 101325 \, \text{Pa} = 25331.25 \, \text{Pa}$, and $\rho = 1000 \, \text{kg/m}^3$, we can substitute the values:$$25331.25 + \frac{1}{2}(1000)(1.00)^2 = P_2 + \frac{1}{2}(1000)(2.00)^2$$$$25331.25 + 500 = P_2 + 2000$$$$P_2 = 25331.25 - 1500 = 23831.25 \, \text{Pa}$$To convert $P_2$ back to atm:$$P_2 = \frac{23831.25 \, \text{Pa}}{101325 \, \text{Pa/atm}} \approx 0.235 \, \text{atm}$$Therefore, the pressure at the valve is approximately 0.235 atm.

Water Flows Through a Horizontal Pipe of Cross-Sectional Area 10.0 cm210.0 \mathrm {~cm} ^ { 2 }10.0 cm2

Water Flows Through a Horizontal Pipe of Cross-Sectional Area $10.0 \mathrm {~cm} ^ { 2 }$