Question:medium
Water of density 1000 kg/m3 flows at a steady rate in a converging duct of circular cross-section. Let the areas of sections 1 and 2 be 100 cm2 and 50 cm2 respectively. The pressure and average velocity at section 1 are 110 kPa and 1 m/s.
If the centres of the two sections are at the same elevation, then the pressure at section 2 is ______ kPa (answer in one decimal place).
Water of density 1000 kg/m3 flows at a steady rate in a converging duct of circular cross-section. Let the areas of sections 1 and 2 be 100 cm2 and 50 cm2 respectively. The pressure and average velocity at section 1 are 110 kPa and 1 m/s.
If the centres of the two sections are at the same elevation, then the pressure at section 2 is ______ kPa (answer in one decimal place).
If the centres of the two sections are at the same elevation, then the pressure at section 2 is ______ kPa (answer in one decimal place).
Show Hint
Use continuity to get the velocity at section 2, then apply Bernoulli's equation between the two sections.
Updated On: Jul 28, 2026
Show Solution
Correct Answer: 108.5
Solution and Explanation
Step 1: Get $V_2$ from the continuity equation.
The volume flow rate stays constant along the duct, so $Q = A_1 V_1 = A_2 V_2$.
Using $A_1 = 0.01\ \text{m}^2$, $A_2 = 0.005\ \text{m}^2$, $V_1 = 1\ \text{m/s}$, we get $Q = 0.01\ \text{m}^3/\text{s}$, so $V_2 = Q/A_2 = 0.01/0.005 = 2\ \text{m/s}$.
Step 2: Write Bernoulli's equation in head form.
$\dfrac{P_1}{\rho g} + \dfrac{V_1^2}{2g} + z_1 = \dfrac{P_2}{\rho g} + \dfrac{V_2^2}{2g} + z_2$, and since $z_1 = z_2$, the elevation heads cancel out.
Step 3: Solve for the pressure difference directly.
$P_1 - P_2 = \dfrac{1}{2}\rho(V_2^2 - V_1^2) = \dfrac{1}{2}(1000)(4-1) = 1500\ \text{Pa}$.
So $P_2 = P_1 - 1500 = 110000 - 1500 = 108500\ \text{Pa}$, the pressure drops because the duct narrows and speeds up the flow.
Final Answer:
The pressure at section 2 comes out to 108.5 kPa.
\[ \boxed{108.5} \]
The volume flow rate stays constant along the duct, so $Q = A_1 V_1 = A_2 V_2$.
Using $A_1 = 0.01\ \text{m}^2$, $A_2 = 0.005\ \text{m}^2$, $V_1 = 1\ \text{m/s}$, we get $Q = 0.01\ \text{m}^3/\text{s}$, so $V_2 = Q/A_2 = 0.01/0.005 = 2\ \text{m/s}$.
Step 2: Write Bernoulli's equation in head form.
$\dfrac{P_1}{\rho g} + \dfrac{V_1^2}{2g} + z_1 = \dfrac{P_2}{\rho g} + \dfrac{V_2^2}{2g} + z_2$, and since $z_1 = z_2$, the elevation heads cancel out.
Step 3: Solve for the pressure difference directly.
$P_1 - P_2 = \dfrac{1}{2}\rho(V_2^2 - V_1^2) = \dfrac{1}{2}(1000)(4-1) = 1500\ \text{Pa}$.
So $P_2 = P_1 - 1500 = 110000 - 1500 = 108500\ \text{Pa}$, the pressure drops because the duct narrows and speeds up the flow.
Final Answer:
The pressure at section 2 comes out to 108.5 kPa.
\[ \boxed{108.5} \]
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