a) Define laminar flow and turbulent flow of liquid. [2] b) Water flows steadily through a horizontal pipe of non uniform cross section. If the pressure of water is 4 10^4\ N\,m^ 2 at a point where…

a) Laminar Flow and Turbulent Flow

Laminar Flow

• Definition: Laminar flow is a smooth, orderly, and predictable flow of a liquid in which the fluid moves in parallel layers or laminae, with no disruption between them.
• Characteristics:
  • Fluid particles move in straight or smooth curved paths.
  • Velocity of the fluid at any point remains constant over time.
  • Low Reynolds number ($Re<2000$).
  • Minimal energy loss due to friction.
  • Common in viscous fluids (e.g., honey, oil) or slow-moving liquids in narrow pipes.

Turbulent Flow

• Definition: Turbulent flow is an irregular, chaotic, and unpredictable flow of a liquid in which the fluid moves in random eddies and swirls, causing mixing.
• Characteristics:
  • Fluid particles move erratically in all directions.
  • Velocity at any point fluctuates rapidly over time.
  • High Reynolds number ($Re>4000$).
  • Significant energy loss due to increased friction and mixing.
  • Common in fast-moving liquids (e.g., rivers, blood flow in arteries).

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b) Calculation of Pressure at Reduced Cross-Section

Given Data

• Pressure at point 1, $P_1=4\times 10^{4}N/{\mathrm{m}}^2$
• Velocity at point 1, $v_1=2m/s$
• Cross-sectional area at point 1, $A_1=200\mathrm{c}{\mathrm{m}}^2=200\times 10^{-4}{\mathrm{m}}^2=0.02{\mathrm{m}}^2$
• Cross-sectional area at point 2, $A_2=50\mathrm{c}{\mathrm{m}}^2=50\times 10^{-4}{\mathrm{m}}^2=0.005{\mathrm{m}}^2$
• Density of water, $\rho =1000kg/{\mathrm{m}}^3$

Assumptions

• The pipe is horizontal (no change in height, $h_1=h_2$).
• Flow is steady and incompressible (water density remains constant).
• No viscosity or energy loss (ideal fluid).

Step 1: Apply the Continuity Equation

The continuity equation for incompressible flow states: $$A_1{v}_1=A_2{v}_2$$ Solving for $v_2$: $$v_2=\frac{A_1{v}_1}{A_2}=\frac{0.02\times 2}{0.005}=8m/s$$

Step 2: Apply Bernoulli’s Equation

Bernoulli’s equation for horizontal flow ($h_1=h_2$) is: $$P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2$$ Rearranging to solve for $P_2$: $$P_2=P_1+\frac{1}{2}\rho (v_1^2-v_2^2)$$ Substitute the known values: $$P_2=4\times 10^4+\frac{1}{2}\times 1000\times (2^2-8^2)$$ $$P_2=4\times 10^4+500\times (4-64)$$ $$P_2=4\times 10^4+500\times (-60)$$ $$P_2=4\times 10^4-30000$$ $$P_2=1\times 10^{4}N/{\mathrm{m}}^2$$

Final Answer

The pressure at the point where the cross-section reduces to $50\mathrm{c}{\mathrm{m}}^2$ is $1\times 10^{4}N/{\mathrm{m}}^2$.