# Step-by-Step Solution: Measuring Air Velocity Let's solve each sub-part step by step: --- ## **1) Total Energy Input to the Wire (in terms of Q)** - **Given:** Electric energy input per unit length of wire = \( Q \) (W/m) Length of the wire = \( L \) (m) - **Total energy input to the wire:** The total energy input, for the entire length, is: \[ \text{Total energy input} = QL \] --- ## **2) Total Energy Lost from the Wire by Convection** - **Convective heat transfer equation:** \[ \text{Heat loss by convection} = hA(T_s - T_\infty) \] Where: - \( h \) = convective heat transfer coefficient (W/m²·K) - \( A \) = surface area for heat transfer (for a wire: \( A = \pi D L \)) - \( T_s \) = surface temperature of wire (°C or K) - \( T_\infty \) = ambient air temperature (°C or K) - **For a cylinder in cross flow, \( h \) can be found using the Nusselt number, \( Nu \):** \[ Nu = \frac{hD}{k} \] Where \( k \) is the thermal conductivity of air. - **Empirical correlation for Nusselt number:** \[ Nu = B \left( \frac{u_\infty D}{\nu} \right)^n Pr^{1/3} \] Where: - \( B, n \) = empirical constants (from literature) - \( u_\infty \) = air velocity - \( D \) = wire diameter - \( \nu \) = kinematic viscosity of air - \( Pr \) = Prandtl number of air - **So,** \[ h = \frac{Nu \cdot k}{D} = \frac{k}{D} B \left( \frac{u_\infty D}{\nu} \right)^n Pr^{1/3} \] - **Total heat loss by convection:** \[ QL = h \cdot \pi D L (T_s - T_\infty) \] Substitute for \( h \): \[ QL = \left[ \frac{k}{D} B \left( \frac{u_\infty D}{\nu} \right)^n Pr^{1/3} \right] \pi D L (T_s - T_\infty) \] Simplify: \[ QL = \pi D L (T_s - T_\infty) \frac{k}{D} B \left( \frac{u_\infty D}{\nu} \right)^n Pr^{1/3} \] \[ QL = \pi L (T_s - T_\infty) k B \left( \frac{u_\infty D}{\nu} \right)^n Pr^{1/3} \] --- ## **3) Expression for Air Velocity \( u_\infty \) (Steady State)** - **From the above, for unit length (\( L = 1 \)):** \[ Q = \pi (T_s - T_\infty) k B \left( \frac{u_\infty D}{\nu} \right)^n Pr^{1/3} \] - **Solve for \( u_\infty \):** \[ Q = \pi (T_s - T_\infty) k B D^n u_\infty^n \nu^{-n} Pr^{1/3} \] \[ u_\infty^n = \frac{Q}{\pi (T_s - T_\infty) k B (D/\nu)^n Pr^{1/3}} \] \[ u_\infty^n = \frac{Q}{\pi (T_s - T_\infty) k B} \cdot \frac{\nu^n}{D^n Pr^{1/3}} \] \[ u_\infty^n = \frac{Q}{\pi (T_s - T_\infty) k B (D/\nu)^n Pr^{1/3}} \] --- ## **4) Numerical Example** - **Given:** - Air temperature, \( T_\infty = 30^\circ C \) - Wire temperature, \( T_s = 80^\circ C \) - Heat input per length, \( Q = 5 \) W/m - Wire diameter, \( D = .15 \) mm \( = .00015 \) m - **Find:** Air velocity, \( u_\infty \) (in m/s) **Assume (for air at 30°C):** - \( k \approx .026 \) W/m·K - \( \nu \approx 1.6 \times 10^{-5} \) m²/s - \( Pr \approx .7 \) - Use typical empirical values for a cylinder: - \( B \approx .47 \) - \( n \approx .5 \) (for turbulent flow) Now, plug values into the derived formula: \[ u_\infty^n = \frac{Q}{\pi (T_s - T_\infty) k B (D/\nu)^n Pr^{1/3}} \] Substitute: - \( Q = 5 \) W/m - \( T_s - T_\infty = 80 - 30 = 50 \) K - \( k = .026 \) W/m·K - \( B = .47 \) - \( D = .00015 \) m - \( \nu = 1.6 \times 10^{-5} \) m²/s - \( Pr = .7 \) - \( n = .5 \) Calculate \( (D/\nu)^n \): \[ \frac{D}{\nu} = \frac{.00015}{1.6 \times 10^{-5}} = 9.375 \] \[ (D/\nu)^n = (9.375)^{.5} = \sqrt{9.375} \approx 3.062 \] \[ Pr^{1/3} = (.7)^{1/3} \approx .887 \] Now, compute denominator: \[ \pi (T_s - T_\infty) k B (D/\nu)^n Pr^{1/3} \] \[ = \pi \times 50 \times .026 \times .47 \times 3.062 \times .887 \] Let's compute step by step: - \( \pi \times 50 = 157.08 \) - \( 157.08 \times .026 = 4.084 \) - \( 4.084 \times .47 = 1.920 \) - \( 1.920 \times 3.062 = 5.882 \) - \( 5.882 \times .887 = 5.218 \) So, denominator ≈ **5.22** Now, solve for \( u_\infty^n \): \[ u_\infty^{.5} = \frac{Q}{5.22} = \frac{5}{5.22} = .958 \] \[ u_\infty = (.958)^2 = .918~\text{m/s} \] --- ## **Final Answer** ### **Air velocity, \( u_\infty \), is approximately:** \[ \boxed{.92~\text{m/s}} \] --- ### **Summary Table** | Subpart | Formula/Result | |---------|------------------------------------------------------------------------------------| | 1 | Total energy input: \( QL \) | | 2 | Energy lost by convection: \( \pi D L (T_s - T_\infty) \frac{k}{D} B \left( \frac{u_\infty D}{\nu} \right)^n Pr^{1/3} \) | | 3 | \( u_\infty^n = \frac{Q}{\pi (T_s - T_\infty) k B (D/\nu)^n Pr^{1/3}} \) | | 4 | \( u_\infty \approx .92~\text{m/s} \) | --- Let me know if you want to see the calculation with other values or further explanation!