Example

Ex$\mathrm{.}1$: Calculate the overall loss coefficient for a flat$-$plate collector with two glass covers. The following data is given: $($a$)$ size of the absorber plate: $1\mathrm{.}90m0\mathrm{.}90m,($b$)$ spacing between plate and first glass cover: $5cm,($c$)$ spacing between first and second glass cover: $5cm,($d$)$ plate emissivity : $0\mathrm{.}90,($e$)$ glass cover emissivity: $0\mathrm{.}85,($f$)$ collector tilt: $23,($g$)$ Mean plate temperature: $73C,($h$)$ Ambient air temperature: $25C,($i$)$ wind speed : $2\mathrm{.}7\frac{m}{s},($j$)$ Back insulation thickness: $10cm,($k$)$ side insulation thickness: $5cm,($I$)$ thermal conductivity of insulation: $0\mathrm{.}07\frac{W}{m}-K.$ Use the appropriate correlation from the correlation Table$-1.$ The properties of air is given in Table$-2.$

Table$-1$ Correlation table

$1$atmT,$ka,\frac{m^2}{s}m,m\frac{i^2}{s}4\mathrm{.}158{10}^{-4}8\mathrm{.}036{10}^{-6}1\mathrm{.}252{10}^{-5}1\mathrm{.}356{10}^{-5}1\mathrm{.}465{10}^{-5}8\mathrm{.}636{10}^{-4}1\mathrm{.}189{10}^{-6}1\mathrm{.}474{10}^{-1}1\mathrm{.}527{10}^{-5}1\mathrm{.}579{10}^{-5}3\mathrm{.}013{10}^{-6}5\mathrm{.}837{10}^{-5}9\mathrm{.}319{10}^{-6}1\mathrm{.}008{10}^{-5}1\mathrm{.}087{10}^{-5}1\mathrm{.}578{10}^{-5}1\mathrm{.}696{10}^{-5}1\mathrm{.}818{10}^{-5}1\mathrm{.}880{10}^{-5}1\mathrm{.}944{10}^{-5}1\mathrm{.}630{10}^{-5}1\mathrm{.}680{10}^{-5}1\mathrm{.}729{10}^{-5}1\mathrm{.}754{10}^{-5}1\mathrm{.}778{10}^{-5}1\mathrm{.}169{10}^{-5}1\mathrm{.}252{10}^{-5}1\mathrm{.}338{10}^{-5}1\mathrm{.}382{10}^{-5}1\mathrm{.}426{10}^{-5}2\mathrm{.}009{10}^{-5}2\mathrm{.}074{10}^{-5}2\mathrm{.}141{10}^{-5}2\mathrm{.}208{10}^{-5}2\mathrm{.}277{10}^{-5}1\mathrm{.}802{10}^{-5}1\mathrm{.}825{10}^{-5}1\mathrm{.}849{10}^{-5}1\mathrm{.}872{10}^{-5}1\mathrm{.}895{10}^{-5}1\mathrm{.}470{10}^{-5}1\mathrm{.}516{10}^{-5}1\mathrm{.}562{10}^{-5}1\mathrm{.}608{10}^{-5}1\mathrm{.}655{10}^{-5}$

Top loss coefficient:

$\frac{q_t}{A_p}=h_{p-c_1}(T_{pm}-T_{c1})+\frac{(T_{pm}^4-T_{cl}^4)}{(\frac{1}{{}_p}+\frac{1}{{}_c}-1)}$

$\frac{q_t}{A_p}=h_{c1-c2}(T_{c1}-T_{c2})+\frac{(T_{c1}^4-T_{c2}^4)}{(\frac{1}{{}_c}+\frac{1}{{}_c}-1)}$

$\frac{q_t}{A_p}=h_w(T_{c2}-T_{sky})+{}_c(T_{c2}^4-T_{sky}^4)$

Example

Ex$\mathrm{.}1$: Calculate the overall loss coefficient for a flat$-$plate collector with two glass covers. The following data is given: $($a$)$ size of the absorber plate: $1\mathrm{.}90m0\mathrm{.}90m,($b$)$ spacing between plate and first glass cover: $5cm,($c$)$ spacing between first and second glass cover: $5cm,($d$)$ plate emissivity : $0\mathrm{.}90,($e$)$ glass cover emissivity: $0\mathrm{.}85,($f$)$ collector tilt: $23,($g$)$ Mean plate temperature: $73C,($h$)$ Ambient air temperature: $25C,($i$)$ wind speed : $2\mathrm{.}7\frac{m}{s},($j$)$ Back insulation thickness: $10cm,($k$)$ side insulation thickness: $5cm,($I$)$ thermal conductivity of insulation: $0\mathrm{.}07\frac{W}{m}-K.$ Use the appropriate correlation from the correlation Table$-1.$ The properties of air is given in Table$-2.$

Table$-1$ Correlation table

$1$atmT,$ka,\frac{m^2}{s}{x}_n\frac{m^2}{5}4\mathrm{.}158{10}^{-4}8\mathrm{.}036{10}^{-6}1\mathrm{.}252{10}^{-5}1\mathrm{.}356{10}^{-5}1\mathrm{.}465{10}^{-5}8\mathrm{.}636{10}^{-4}1\mathrm{.}189{10}^{-6}1\mathrm{.}474{10}^{-5}1\mathrm{.}527{10}^{-5}1\mathrm{.}579{10}^{-5}3\mathrm{.}013{10}^{-4}5\mathrm{.}837{10}^{-5}\frac{?}{?}$