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
Ohmic dissipation of electric energy, in the same manner that the wires of a toaster are heated. The Ohmic 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:
(a) The velocity of the air flow.
(b) Calculate the Reynolds number based on the wire diameter.
(c) If the velocity were to be doubled, and the wire and air stream temperatures were maintained as stated, what is the required Ohmic dissipation?
To find the velocity of the air flow, we can use the heat transfer equation and solve for the velocity. The heat transferred to the air flow can be calculated by the equation:
Q = m * Cp * ΔT
where Q is the heat transfer rate, m is the mass flow rate, Cp is the specific heat capacity of air, and ΔT is the temperature difference between the wire and the air. We can rearrange this equation to solve for the mass flow rate:
m = Q / (Cp * ΔT)
The mass flow rate can also be expressed as:
m = ρ * A * V
where ρ is the density of air, A is the cross-sectional area of the flow, and V is the velocity of the air flow.
Since the wire is electrically heated, the power dissipated by the wire (0.0253 W) is equal to the heat transferred to the air flow. Therefore, we can substitute this value for Q:
0.0253 W = (ρ * A * V) * Cp * ΔT
Now, let's calculate each value:
ρ = 726 mm Hg / (0.3704 lb/ft^3) = 1960 kg/m^3 (converting pressure to density)
A = π * (0.0004 in./12 ft * 2)^2 / 4 = 0.000000051 m^2 (converting diameter to radius and then calculating the area)
Cp = 0.24 BTU/lb·F * 1055 J/BTU / 0.1 kg/g = 2.52 kJ/kg·C (converting specific heat capacity to SI units)
ΔT = (147 - 73) F * 5/9 = 36.11 °C (converting temperature difference to °C)
Now we can substitute these values into the equation and solve for V:
0.0253 W = (1960 kg/m^3 * 0.000000051 m^2 * V) * 2.52 kJ/kg·C * 36.11 °C
Simplifying the equation:
0.0253 W = 4.5386 * 10^-12 V
Solving for V:
V = 0.0253 W / (4.5386 * 10^-12) = 5.57 m/s (rounded to two decimal places)
Therefore, the velocity of the air flow is approximately 5.57 m/s.
To calculate the Reynolds number based on the wire diameter, we can use the equation:
Re = (ρ * V * D) / μ
where Re is the Reynolds number, ρ is the density of air, V is the velocity of the air flow, D is the wire diameter, and μ is the dynamic viscosity of air.
Let's calculate each value:
μ = 0.0000159 lb/(ft·s) * 47.8803 kg/(lb·s) = 0.000763 kg/(m·s) (converting dynamic viscosity to SI units)
D = 0.0004 in. * 0.0254 m/in. = 0.00001016 m (converting diameter to meters)
Substituting these values into the equation:
Re = (1960 kg/m^3 * 5.57 m/s * 0.00001016 m) / 0.000763 kg/(m·s)
Simplifying the equation:
Re = 260
Therefore, the Reynolds number based on the wire diameter is 260.
If the velocity were to be doubled while maintaining the wire and air stream temperatures, we can use the same heat transfer equation to find the new required Ohmic dissipation.
0.0253 W = (ρ * A * V_new) * Cp * ΔT
We already know the values for ρ, A, Cp, and ΔT. Let's solve for V_new:
0.0253 W = (1960 kg/m^3 * 0.000000051 m^2 * V_new) * 2.52 kJ/kg·C * 36.11 °C
Simplifying the equation:
0.0253 W = 4.5386 * 10^-12 V_new
Solving for V_new:
V_new = 0.0253 W / (4.5386 * 10^-12) = 5.57 m/s
Since the velocity was doubled, the required Ohmic dissipation remains the same at 0.0253 W.