The wind-chill index is the rate of surface-convection (and surface-radiation) heat transfer per unit area D from a person with a surface temperature Ts in a crossing wind [1]. This is depicted in Figure Ex. 6.10(a). The heat loss is characterized by the ambient air velocity uf,∞(m/s) and

Question

The wind-chill index is the rate of surface-convection (and surface-radiation) heat transfer per unit area [latex]\left\langle q_{ku}\right\rangle _{D} [/latex] from a person with a surface temperature [latex]T_{s}[/latex] in a crossing wind [1]. This is depicted in Figure (a). The heat loss is characterized by the ambient air velocity [latex]u_{f,∞}(m/s)[/latex] and temperature [latex]T_{f,∞}(^{\circ }C)[/latex]. The body is approximated as a long cylinder. The average Nusselt number [/latex]\left\langle Nu\right\rangle _{D}[/latex] is determined empirically as a function of the wind velocity. The correlation used is slightly different than that listed in Table.
The equivalent wind-chill temperature (also called the wind-chill factor) is
the apparent ambient temperature [latex] (T_{f,∞})_{app} [/latex] that results in the same heat loss but under a calm wind condition (the calm wind velocity is considered to be 6.4 km/hr), [latex]\left\langle Nu\right\rangle _{D,calm}[/latex]

(a) Draw the thermal circuit diagram.
(b) Determine the wind-chill index [latex]\left\langle q_{ku}\right\rangle _{D} [/latex].
(c) Determine the wind-chill factor [latex](T_{f,∞})_{app}[/latex] .
[latex]D = 35 cm, u_{f,∞} = 10 km/hr, T_{s} = 33^{\circ }C, and T_{f,∞} = 10^{\circ }C[/latex].
Evaluate the air properties from Table at T = 300 K.

Accepted Answer

(a) The thermal circuit diagram is shown in Figure (b). The energy equation is the surface energy equation [latex] riangledown .q\equiv \lim _{\Delta V ightarrow 0}\frac{\int_{\Delta A}{(q.s_{n})dA} }{\Delta V} =\lim _{\Delta V ightarrow 0}\frac{Q\mid_{\Delta A}}{\Delta V} [/latex], which reduces to [latex]Q\mid _{A} = \left\langle Q_{ku} ight angle _{D} = \dot{S}_{r}[/latex].
(b) The surface-convection heat flux is given by [latex]\left\langle Q_{ku} ight angle _{L( or D)}=\frac{T_{s}-T_{f,\infty }}{\left\langle R_{ku} ight angle_{L(or D)} } =A_{ku}\frac{k_{f}}{L(or D)}\left\langle Nu ight angle _{L(or D)}(T_{s}-T_{f,\infty }) ,Re_{L(or D)}=\frac{u_{f,\infty }L(or D)}{v_{f}}[/latex] as

[latex]\left\langle q_{ku} ight angle _{D} =\frac{\left\langle Q_{ku} ight angle _{D}}{A_{ku}} =\left\langle Nu ight angle _{ D}\frac{k_{f}}{ D}(T_{s}-T_{f,\infty }) [/latex]

In order to evaluate [latex]\left\langle Nu ight angle _{D}[/latex] from Table, we need to first determine the Reynolds number given by [latex]\left\langle Q_{ku} ight angle _{L( or D)}=\frac{T_{s}-T_{f,\infty }}{\left\langle R_{ku} ight angle_{L(or D)} } =A_{ku}\frac{k_{f}}{L(or D)}\left\langle Nu ight angle _{L(or D)}(T_{s}-T_{f,\infty }) ,Re_{L(or D)}=\frac{u_{f,\infty }L(or D)}{v_{f}}[/latex],[latex] Re_{D} = u_{f,∞}D/ν_{f}[/latex] . From Table, we have for T = 300 K

[latex]ν_{f} = 15.66 × 10^{−6} m^{2}/s[/latex] Table
[latex]k_{f} = 0.0267 W/m-K[/latex] Table
Pr = 0.69 Table

Then the Reynolds number is

[latex]Re_{D}=\frac{u_{f,\infty }D}{v_{f}}=\frac{\frac{10 imes 10^{3}}{3,600} (m/s) imes 0.35(m)}{15.66 × 10^{−6} (m^{2}/s)} = 6.208 × 10^{4} [/latex]

From Table, we have

[latex]\left\langle Nu ight angle _{D} = a_{1}Re^{a2}_{ D} Pr^{1/3} , a_{1} = 0.027 , a_{2} = 0.805 [/latex]

[latex]\left\langle Nu ight angle _{D} =0.027(6.208 × 10^{4} ) ^{0.805}(0.69)^{1/3} = 172.2[/latex]

For [latex]\left\langle q_{ku} ight angle _{D}[/latex], we have

[latex] \left\langle q_{ku} ight angle _{D}=172.2 × \frac{0.06267(W/m-K)}{0.35(m)} (33 − 10)(^{\circ }C) [/latex]

[latex] = 302.1 W/m^{2} [/latex] wind-chill index.

(c) The wind-chill factor is defined as

[latex] \left\langle q_{ku} ight angle _{D}=\frac{k_{f}}{D}\left\langle Nu ight angle _{D,calm}[T_{s}-\left(T_{f,\infty } ight) _{app}] [/latex]

or

[latex]\left(T_{f,\infty } ight) _{app}=T_{s}-\frac{\left\langle q_{ku} ight angle_{D}D }{k_{f}\left\langle Nu ight angle _{D,calm}}[/latex]

where [latex] \left\langle Nu ight angle _{D,calm}[/latex] is for [latex]u_{f,∞} = 6.4 km/hr[/latex]. We need to determine [latex] \left\langle Nu ight angle _{D,calm}[/latex] and first we calculate [latex]Re_{D}[/latex], i.e.,

[latex]Re_{D}=\frac{\frac{6.4 imes 10^{3}}{3,600} (m/s) × 0.35(m)}{15.66 × 10^{−6} (m^{2}/s)} = 3.974 × 10^{4} \simeq 4 × 10^{4} [/latex]

From Table, this is in the same range of [latex]Re_{D}[/latex] as above, i.e.,

[latex]\left\langle Nu ight angle _{D} = 0.027(3.947 × 10^{4} )^{0.805}(0.69)^{1/3} = 120.2 [/latex]

Then

[latex] (T_{f,∞})_{app} = 33(^{\circ}C) −\frac{302.1(W/m^{2} ) × 0.35(m)}{0.0267(W/m-K) × 120.2} = 33(^{\circ }C) − 32.94(^{\circ }C) [/latex]

[latex]= 0.06183^{\circ}C [/latex] wind-chill factor (apparent ambient temperature).

Question 6.10: The wind-chill index is the rate of surface-convection (and ...

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