The roof of a refrigerated truck compartment is of composite construction, consisting of a layer of foamed urethane insulation (t2 = 50 mm, ki= 0.026 W/m-k) sandwiched between aluminum alloy panels (t1=5mm and Kp=180 W/m · K). The length and width of the roof are L = 10m and W = 3.5 m , respectively, and the temperature of the inner Ts, i = -100 C Consider conditions for which the truck is moving at a speed of V = 105 km/h, the air temperature is T? = 32°C, and the solar irradiation is Gs = 750 W/m2 • Turbulent flow be assumed over the entire length of the roof. (a) For equivalent values of the solar absorptivity and the emissivity of the outer surface (?s = ? = 0.5), estimate the average temperature Ts. o of the outer surface. What is the corresponding heat load imposed on the refrigeration system?(b) A special finish (?s = 0.15, ?= 0.8) may be applied to the outer surface. What effect would such an application have on the surface temperature and the heat load?(c) If, with ?s = ? = 0.5, the roof is not insulated (t2 = 0), what are the corresponding values of the surface temperature and the heat load?Thermal Paint

(a) = 33.8°C   , q = 797 W

(b) = 27.1°C    , q = 675.3 W

(c) = -9.9°C    , q =90,630 W

Comment: Use the special surface finish reduces the solar input, while increasing radiation emission from the surface.

Explanation:

(a)

From an energy balance from the outer surface

Where, 2.78 x 10⁻⁵ m².k/w

            1.923 m².k/w

            [2.92 m/s x 10 m] / [15.89 x 10⁻⁶ m²/s]

                                   = 1.84 x 10⁴

            |h| = (K/L)0.037

            substituting parameters,

            |h| = 0.0263/10*(0.037)*[(184x10⁷)^(4/5)]*(0.707)^1/3

            |h| =562Wm⁻²K⁻¹

Hence,

0.5(950) + 56.2(305-)-0.5*5.67x10⁻⁸ = ( - 268) / (5.56x10⁻⁵ + 1.923)

Solving, we obtain

= 306.8K

    = 33.8°C

Hence, the heat load is

q = (W.L)q" = (3.5 x 10)

q = 797 W

(b) With the special surface finish

= 301.1K = 27.1°C

Similarly, q = 675.3 W

(c) Without the insulation (t₂ = 0) and with (∝s = E = 0.5)

= 263.1k = -9.9°C

Similarly, q = 90,630 W

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explanation: