91Ó°ÊÓ

To estimate the steady-state surface temperature, we'll balance the heat generated in the heating element with the convective heat loss to the air: \( q_{\mathrm{electrical}} = hA(T_\mathrm{s} - T_\mathrm{\infty}) \) Where \(q_{\mathrm{electrical}}\) is the rate of electrical energy dissipation, \(h\) is the convective heat transfer coefficient, \(A\) is the heater surface area, \(T_\mathrm{s}\) is the heater surface temperature, and \(T_\mathrm{\infty}\) is the air temperature. In this problem, we are given the rate of electrical energy dissipation per unit length, so we can write the above equation as: \( q'_{\mathrm{electrical}} = hA'(T_\mathrm{s} - T_\mathrm{\infty}) \) First, we need to find the convective heat transfer coefficient and the surface area of the heater per unit length.

To calculate the convective heat transfer coefficient, we'll need to find the Nusselt number (Nu). For a long cylinder in cross flow, we'll use the correlation for Nu: \( \mathrm{Nu} = \frac{hD}{k_\mathrm{air}} \) Where \(D\) is the diameter of the heater and \(k_\mathrm{air}\) is the thermal conductivity of air. We also need to find the following properties for air at a film temperature of \(T_\mathrm{f} = (T_\mathrm{s} + T_\mathrm{\infty}) / 2\): - Reynolds number, \(Re = \frac{\rho_\mathrm{air}VD}{\mu_\mathrm{air}}\) - Prandtl number, \(Pr = \frac{\mu_\mathrm{air}c_{p_\mathrm{air}}}{k_\mathrm{air}}\) We can estimate these properties using air property tables at a temperature of 27°C. After calculating Re and Pr, we can use the Zukauskas correlation to find Nu: \( \mathrm{Nu} = 0.3 + \frac{0.62 \times Re^{0.5} \times Pr^{1/3}}{[1 + (0.4 / Pr)^{2/3}]^{1/4}} \times [1 + (Re / 282000)^{5/8}]^{4/5} \) Finally, we can use the expression for Nu to find the convective heat transfer coefficient, \(h\).

We can now calculate the surface area of the heater per unit length, \(A'\), as follows: \( A' = \pi D \)

Using the convective heat transfer coefficient and surface area per unit length, we can calculate the steady-state surface temperature using the heat balance equation found in Step 1: \( T_\mathrm{s} = T_\mathrm{\infty} + \frac{q'_{\mathrm{electrical}}}{hA'} \) This will give us the steady-state surface temperature of the heater, neglecting radiation.

To estimate the time required to reach within 10°C of the steady-state surface temperature, we'll use the lumped capacitance method for transient heat transfer. This is appropriate because the heater's geometry has a small ratio of volume to surface area, and its thermal conductivity is high. The temperature difference as a function of time is given by: \( \frac{T_\mathrm{s}(t) - T_\mathrm{\infty}}{T_\mathrm{s,ss} - T_\mathrm{\infty}} = \exp{\left(-\frac{hA'c_bt}{k_ac_bV'}\right)} \) Where \(T_\mathrm{s}(t)\) is the heater surface temperature at time \(t\), \(T_\mathrm{s,ss}\) is the steady-state surface temperature, \(c_b\) is the specific heat of the heater, \(k_a\) is the heater's thermal conductivity, and \(V'\) is the volume of the heater per unit length. We can find the volume of the heater per unit length as follows: \( V' = \frac{\pi D^2}{4} \) Putting all values in the above temperature difference equation and solving for the time when \(T_\mathrm{s}(t) = T_\mathrm{s,ss} - 10^{\circ} \mathrm{C}\), we can obtain the time required for the surface temperature to come within 10°C of its steady-state value.