A common procedure for measuring the velocity of an air stream involves insertion of an electrically heated wire (called a hot-wire anemometer) into the air flow, with the axis of the wire oriented perpendicular to the flow direction. The electrical energy dissipated in the wire is assumed to be transferred to the air by forced convection. Hence, for a prescribed electrical power, the temperature of the wire depends on the convection coefficient, which, in turn, depends on the velocity of the air. Consider a wire of length L - 20 mm and diameter D = 0.5 mm, for which a calibration of the form. V = 6.25 times 10-5 h2, has been determined. The velocity V and the convection coefficient h have units of m/s and W/m2 K, respectively. In an application involving air at a temperature of Tx = 25 degree C, the surface temperature of the anemometer is maintained at Ts = 75 degree C with a voltage drop of 5 V and an electric current of 0.1 A. What is the velocity of the air?

Question

Engineering

Bongo

5 September, 12:04

A common procedure for measuring the velocity of an air stream involves insertion of an electrically heated wire (called a hot-wire anemometer) into the air flow, with the axis of the wire oriented perpendicular to the flow direction. The electrical energy dissipated in the wire is assumed to be transferred to the air by forced convection. Hence, for a prescribed electrical power, the temperature of the wire depends on the convection coefficient, which, in turn, depends on the velocity of the air. Consider a wire of length L - 20 mm and diameter D = 0.5 mm, for which a calibration of the form. V = 6.25 times 10-5 h2, has been determined. The velocity V and the convection coefficient h have units of m/s and W/m2 K, respectively. In an application involving air at a temperature of Tx = 25 degree C, the surface temperature of the anemometer is maintained at Ts = 75 degree C with a voltage drop of 5 V and an electric current of 0.1 A. What is the velocity of the air?

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Answers (1)

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Henry Morris · Answer 1

V = 6.33 m/s

Explanation:

Given:

- The length of the wire L = 0.02 m

- The diameter of the wire D = 0.0005 m

- The calibration expression V = 0.0000625*h^2

- Environment temperature T_inf = 298 K

- Surface temperature T_s = 348 K

- The voltage drop dV = 5 V

- The electric current I = 0.1 A

Find:

- the velocity of Air

Solution:

- Calculate the surface area of the wire:

A = pi*D*L

A = pi * (0.0005) * (0.02) = 0.00003142 m^2

- The rate of energy in the wire P:

P = I*dV = 0.1*5 = 0.5 W

- Apply Newton's Law of Cooling:

P = h*A * (T_s - T_inf)

h = P / A * (T_s - T_inf)

Plug in the values:

h = 0.5 / 0.00003142 * (348 - 298)

h = 318.27 W / m^2K

- Using the calibration relationship given, compute the velocity of air:

V = 6.25*10^-5 * h^2

V = 6.25*10^-5 * (318.27) ^2

V = 6.33 m/s