More Questions from Hydraulics and Fluid Mechanics

What is the specific weight (unit weight) of water in SI units, to the nearest standard value?

Mechanical Engineering Hydraulics and Fluid Mechanics Difficulty: Easy
Choose an option
  • A
    9.81 kN/m^3
  • B
    981 kN/m^3
  • C
    9.81 N/m^3
  • D
    9810 kN/m^3

Answer

Correct Answer: 9.81 kN/m^3

Explanation

Introduction:Specific weight (unit weight) is weight per unit volume. For water at standard conditions, engineers commonly use a convenient rounded value for design and quick checks.

Given Data / Assumptions:

  • Standard reference conditions near 4°C to 20°C.
  • Acceleration due to gravity g ≈ 9.81 m/s^2.

Concept / Approach:Specific weight gamma = rho * g. For water, rho ≈ 1000 kg/m^3. Therefore gamma ≈ 1000 * 9.81 N/m^3 = 9810 N/m^3 = 9.81 kN/m^3. This is the widely accepted design value in SI.

Step-by-Step Solution:1) Take rho_w ≈ 1000 kg/m^3.2) Multiply by g: 1000 * 9.81 = 9810 N/m^3.3) Convert to kN: 9810 N/m^3 = 9.81 kN/m^3.

Verification / Alternative check:Using more precise densities changes only the third decimal place of gamma, not the standard rounded 9.81 kN/m^3 used for hand calculations.

Why Other Options Are Wrong:

  • 981 kN/m^3: Off by a factor of 100; implies rho ≈ 100,000 kg/m^3.
  • 9.81 N/m^3: Off by a factor of 1000; would imply extremely low density.
  • 9810 kN/m^3: Off by a factor of 1000; numerically equals 9.81 MN/m^3, not water.

Common Pitfalls:

  • Mixing N/m^3 and kN/m^3 units.
  • Using mass density (kg/m^3) directly as specific weight without multiplying by g.

Final Answer:9.81 kN/m^3

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