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August 20, 1999, 20:06 
a VERY SIMPLE problem

#1 
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Can any of you experts out there solve my simple problem.
I need to know how to find the energy of air flowing through an imaginary cross sectional area. For example, what is the energy of a volume of air flowing through a 20m by 50m area at 5m/s? I'd be grateful for any help. 

August 20, 1999, 23:04 
Re: a VERY SIMPLE problem

#2 
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More information is required. But you can easily calculate by yourself because it's really simple as you said.
Energy = (density * vel * vel / 2 ) * Vol. (1) Density is the function of temperature and density of air at room temperature(25C) = 1.177 kg/metercubic. (2) To define volume, depth must be defined. Let the depth be unity(1 meter), then volume= 20 * 50 * 1 = 1000 metercubic So, energy of the air volume with 5 m/s velocity is 14712.5 (Joule). (3) For your application with different temperature, you can find density of air from any thermodynamic text. But simply, you can use following equation. density of air at temperature T(degree C) : air density = (28.8/22.4)*273/(273+T) Sincerely, 

August 23, 1999, 08:02 
Re: a VERY SIMPLE problem

#3 
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Adds...
Without the need of any depth you can estimate the Power (energy/timeunit) of the airstream by: Power=massflow*velocity^2/2 massflow=area*velocity*density hence: P=0.5*A*V^3*\rho Regards Jonas 

August 23, 1999, 10:39 
Re: a VERY SIMPLE problem

#4 
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The energy per unit mass is the enthalpy of the stream plus its kinetic energy:
E = h + v^2/2 Since air is an ideal gas, the enthalpy is a function of temperature only and can be found by: h = Cp * T Cp = 1.005 kJ/kgK (specific heat for air, assumed constant) To find the energy transferred across the area per unit time (power) you need to multiply the result above by the mass flowrate: P = E * mdot mdot = rho * A * v where rho is density of the air and is found from the ideal gas equation of state: rho = P / (R * T) R is the gas constant for air (look it up since I forgot) A is crossectional area v is velocity T is temperature (in Kelvin) Have fun, Peter 

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