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## Main Question or Discussion Point

Let us assume we have a cylindrical wind tunnel having a 0.5 m diameter fed by an electric fan. The cross-sectional area of the wind tunnel would be A1 = (PI/4) D1^2 = 0.196349541 m2. Let us suppose the motor driving the fan has a power rating of 1,500 W. At this stage, let us assume that the electric motor is consuming its maximum power rating, and that the measured airflow velocity is V1 = 20 m/s, just at the output of the wind tunnel. The kinetic power P1 of the airflow just at the output of the wind tunnel can be calculated as by applying the formula P1 = (1/2) (density of air) (A1) (V1^3) = 962.11 W. If we viewed the wind tunnel plus the fan and the electric motor as a system (System 1), we could say that the efficiency of this system 1 would be System 1 Efficiency = 100 P1/1500 = 64.14%.

At this stage, let us consider a new system (System 2), formed by system 1 plus a 2:1 ratio convergent nozzle attached to the output of system 1. Now the output of system 2 would have a cross-sectional area A2 given by A2 = 0.09817477 m2. If now the electric motor is ran at its maximum power rating as before, can we expect the velocity of the airflow just at the output of the convergent nozzle to be V2 = 40 m/s?

If that were the case, the kinetic power P2 of of the airflow just at the output of the convergent nozzle would be P2 = (1/2) (density of air) (A2) (V2^3) = 3,848.45 W. Notice that now the power of the output airflow would be greater than the input power to the system (1, 500 W), which would apparently be a violation of first law of thermodynamics. Moreover, the efficiency of system 2 would be System 2 Efficiency = 100 P2/1500 = 256.56%, which apparently constitute a violation of second law of thermodynamics also.

Is there a flaw in this analysis?

At this stage, let us consider a new system (System 2), formed by system 1 plus a 2:1 ratio convergent nozzle attached to the output of system 1. Now the output of system 2 would have a cross-sectional area A2 given by A2 = 0.09817477 m2. If now the electric motor is ran at its maximum power rating as before, can we expect the velocity of the airflow just at the output of the convergent nozzle to be V2 = 40 m/s?

If that were the case, the kinetic power P2 of of the airflow just at the output of the convergent nozzle would be P2 = (1/2) (density of air) (A2) (V2^3) = 3,848.45 W. Notice that now the power of the output airflow would be greater than the input power to the system (1, 500 W), which would apparently be a violation of first law of thermodynamics. Moreover, the efficiency of system 2 would be System 2 Efficiency = 100 P2/1500 = 256.56%, which apparently constitute a violation of second law of thermodynamics also.

Is there a flaw in this analysis?