Water flows at a rate of 2.0 $\times$ 10^-3 m³/s through a horizontal pipe of 2.5-cm² cross-sectional area at a pressure of 150 kPa. The flow is constricted at a valve having an effective cross-sectional area of 1.5 cm². The pressure at the valve is A) 57 kPa. B) 93 kPa. C) 160 kPa. D) 210 kPa.

Question

Quizplus · Accepted Answer

The pressure at the valve can be found using Bernoulli's equation, which in this case simplifies to considering the conservation of energy between two points in the flow. The equation relating the pressures, velocities, and areas at two points (1 and 2) in a horizontal flow is given by $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$, where $P$ is the pressure, $\rho$ is the density of the fluid (not given but not needed due to the ratios), $v$ is the velocity of the fluid, and the subscripts 1 and 2 refer to the conditions at the two points respectively. Given that the flow rate $Q = vA$, and is constant, we can express $v_2$ in terms of $v_1$ as $v_2 = v_1 \frac{A_1}{A_2}$. Substituting the given areas and solving for $P_2$ using the given initial pressure $P_1 = 150$ kPa and the fact that the flow rate through both areas is the same, we find that the pressure at the valve (point 2) is lower than at point 1 due to the constriction increasing the velocity of the fluid, which according to Bernoulli's principle, decreases the pressure. Calculating this difference using the provided areas and initial conditions leads to the conclusion that option B) 93 kPa is the correct pressure at the valve, as it reflects the pressure drop due to the increased velocity in the constricted area.

Water Flows at a Rate of 2 ×\times× 10-3 M3/s Through a Horizontal Pipe of 2

Water Flows at a Rate of 2 $\times$ 10^-3 M³/s Through a Horizontal Pipe of 2