Consider a 1.2-m-high and 2-m-wide glass window with a thickness of 6 mm, thermal conductivity k = 0.78W / m .K and emissivity  $\epsilon$ = 0.9 The room and the walls that face the window are maintained at 25°C , and the average temperature of the inner surface of the window is measured to be 5^° C If the temperature of the outdoors is - 5^° C determine (a) the convection heat transfer coefficient on the inner surface of the window, (b) the rate of total heat transfer through the window. [Cengel HT; Problem 9-27; 5^th Edition]

Method:1

Radiation heat transfer between room and glass,

${\stackrel{°}{Q}}_{ratiation} = \epsilon {{\rm A}}_s \sigma \left(T_{room} 4 - T_{glass,in }4\right)$

Convective heat transfer between room and glass,

${\stackrel{°}{Q}}_{convection} = h{{\rm A}}_s \sigma \left(T_{room} -T_{glass,in }\right)$

Total Heat transfer, $\stackrel{°}{Q} = {\stackrel{°}{Q}}_{convection} + {\stackrel{°}{Q}}_{ratiation}$

Taking the outer surface temperature of glass window, as -5°C,

(b)

the rate of total heat transfer through the window, 

$\stackrel{°}{Q}=\frac{∆T}{R_{glass}}=\frac{∆T}{{\displaystyle \frac{L}{k{A}_s}}}=\frac{5-(-5)}{{\displaystyle \frac{0.006}{(0.78)(1.2)\left(2\right)}}}$

= 3120 W

(a)

Now,  ${\stackrel{°}{Q}}_{convection} = \stackrel{°}{Q}- {\stackrel{°}{Q}}_{ratiation}$

$\Rightarrow h{A}_s (T_{room} - T_{glass,in}) = \stackrel{.}{Q} – \epsilon {{\rm A}}_s \sigma \left(T_{room} - T_{glass,in} 4\right)$

$\Rightarrow h (1.2\times 2)(25-5)=3120-0.9(1.2\times 2)\times 5.67\times {10}^{-8} ({298}^4-{278}^4)$

$\Rightarrow h=60.12W/m^{2}K$

Method:2

The properties of air at 1 atm and the film temperature of $\frac{(T_s+T_s)}{2}=\frac{(5+25)}{2}={15}^{°}C$ are (Table A-15)

k = 0.02476W / $m^{°}C$

$v=1.47\times {10}^{-5}{m}^2/s$

Pr = 0.7323

$\beta =\frac{1}{T_f}=\frac{1}{(15+273) K}=0.003472K^{-1}$

(a)

The characteristic length in this case is the height of the window,  $L_{\epsilon }$ = L = 1.2m Then,

$R_a=\frac{g\beta (T_{\infty }-T_s) L_c^3}{v^2} Pr =\frac{\left({\displaystyle \frac{9.81m}{s^2}}\right)(0.003472 K^{-1})(25 – 5 K) {(1.2 m)}^3}{\left(1.470 \times {\displaystyle \frac{10-5 m^2}{s}}\right)}(0.7323) = 3.989 \times {10}^9$

$Nu ={\left\{0.825 +\frac{0.387R{a}^{{\displaystyle \frac{1}{6}}}}{{\left[{\left(1+{\displaystyle \frac{0.492}{Pr}}\right)}^{{\displaystyle \frac{9}{16}}}\right]}^{{\displaystyle \frac{8}{27}}}}\right\}}^2= {\left\{0.825 +\frac{0.387{(3.989 \times {10}^9)}^{{\displaystyle \frac{1}{6}}}}{{\left[{\left(1+{\displaystyle \frac{0.492}{0.7323}}\right)}^{{\displaystyle \frac{9}{16}}}\right]}^{{\displaystyle \frac{8}{27}}}}\right\}}^2$

$h =\frac{k}{L}Nu = \frac{0.02476 W/m .°C}{1.2 m}(189.7)=3.915 W/m^2. °C$

$A_s = (1.2 m) (2 m) = 2.4 m^2$

(b)

The sum of the natural convection and radiation heat transfer from the room to the window is,

${\stackrel{·}{Q}}_{coevection} = h{A}_s (T_{room} - T_{glass,in}) = (3.915 W/m^2.{ }^{a}C) (2.4 m^2) (25-5)°C = 187.9 W$

${\stackrel{·}{Q}}_{raticition} = \epsilon {{\rm A}}_s \sigma (T_{room} 4 -T_{glass,in} 4)$

= $(0.9) (2.4 m^2) (5.67 \times {10}^{-8} W/m^2. K^4) [{(25+273 K)}^4- {(5+273 K)}^4] = 234.3 W$

${\stackrel{·}{Q}}_{boal} = {\stackrel{.}{Q}}_{convection} + {\stackrel{.}{Q}}_{ratiation}$

= 187.9 + 234.3 = 422.2 W