Calculating Pressure Outside a Box: Bernoulli's Equation

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SUMMARY

The discussion focuses on calculating the pressure outside an open box using Bernoulli's Equation. Given the atmospheric pressure of 101325 Pa, air velocity of 45 m/s, and air density of 1.3 kg/m³, the calculation yields an external pressure of 100008.75 Pa. The application of Bernoulli's principle confirms that the terms represent energy density, validating the computation. The conversation emphasizes the importance of clearly defining the problem context when applying fluid dynamics principles.

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  • Understanding of Bernoulli's Equation
  • Knowledge of fluid dynamics concepts
  • Familiarity with pressure and density units
  • Basic algebra for solving equations
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Homework Statement


Calculate pressure outside the box- ATM pressure is 101325 Pa, air moving outside the box is 45 m/s, air density is 1.3 kg/m^3

Homework Equations


P + 1/2(roe)(v)^2 = Patm

The Attempt at a Solution


P + 1/2 (1.3kg/m^3)(45m/s)^2 = 101325 Pa
P + 1316.25 Pa = 101325 Pa
P= 100008.75 Pa
 
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I guess the problem is an open box with static air inside and air blowing over the open end, in which case your computation is right.
But you need to define problems more fully.
 
Bernoulli's equation is an express of energy conservation. All terms have energy density dimensions and your calculation is correct.
 

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