A hot-wire anemometer is a temperature-based instrument for measuring velocities

Question

A hot-wire anemometer is a temperature-based instrument for measuring velocities in a flowing gas. The key component of the device is a very fine wire (diameter = 0. 0004 in. , length = 0. 20 in. ) that is electrically heated. The wire is positioned so that the air stream passes over it in cross flow. The heating of the wire is accomplished by ohm ic dissipation of electric energy, in the same manner that the wires of a toaster are heated. The ohm ic heating of the wire is 0. 0253 W, and the wire temperature is 147 F. In the steady state, this heat is transferred to the air flow. The air that passes over the wire has a temperature of 73 F and a pressure of 726 mm Hg. Required results: The velocity of the air flow. Calculate the Reynolds number based on the wire diameter. If the velocity were to be doubled, and the wire and air stream temperatures were maintained as stated, what is the required ohm ic dissipation?

Explanation / Answer

a) D = 0.0004 in = 0.0004*0.0254 m = 1.016*10^-5 m

L = 0.2 in = 0.2*0.0254 m = 0.00508 m

T = 147 deg F = 63.9 deg C,

T0 = 73 deg F = 22.8 deg C.

Average temperature = (63.9+22.8)/2 = 43.4 deg C

Thermal conductivity of air at 43.4 deg C, k = 0.0271 W/mK, Prandtl no. Pr = 0.711 and kinematic viscosity = (17*10^-6) m^2/s

T-T0 = 63.9 - 22.8 = 41.1 deg C

Q = hA(T-T0) and A = DL

So, h = Q/[DL(T-T0)] = 0.0253/[*(1.016*10^-5)*0.00508*41.1] = 3798.3 W/m^2-C

Nusselt no Nu = hD/k = 3798.3*(1.016*10^-5)/0.0271 = 1.424

Using the relation Nu = 0.3 + 0.62*(Re^0.5)(Pr^0.33)/[1 + (0.4/Pr)^0.66]^0.25, we get

1.424 = 0.3 + 0.62*(Re^0.5)(0.711^0.33)/[1 + (0.4/0.711)^0.66]^0.25

This gives 1.424 = 0.3 + 0.48595*Re^0.5

So, Re = 5.35

Since Re = VD/

So, velocity V = 5.35*(17*10^-6)/(1.016*10^-5) = 8.95 m/s

So, V = 8.95 m/s or 8.95/0.3048 ft/s = 29.36 ft/s

b) As found above, Re = 5.35

c) When velocity is doubled, Re gets doubled. Re = 5.35*2 = 10.7

Nu = 0.3 + 0.48595*Re^0.5 yields that Nu = 1.89

h = Nu*k/D = 1.89*0.0271/(1.016*10^-5) = 5040 W/m^2-C

Q = h*(DL)*(T-T0)

Q = 5040*[*(1.016*10^-5)*0.00508*41.1] = 0.03357 W

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