Question

Both please

13.3 Estimate the amount of solar heating that is lost as a result of using one or two tempered 1/8 in thick glass plates covering a flat-plate absorber. The transmission coefficient for radiation is 2% for wavelengths greater than 3 m and 80% for wavelengths ranging from 0.2 to 3 um. 13.4 The amount of solar energy captured in a given application is expressed in terms of how much thermal energy is delivered relative to the total demand or load. This amount is often referred to as the production function (Q). Using the Kreith and Kreider (1978) method of analysis, a simplified empirical model of a solar building in Boston can be developed to relate Qs parametrically to the load in GJ (L), the amount of storage in m3 (S), and the collector surface area in m2 (A) as: 1/20 100 G Q, =L| 0.84 In If the collector area is increased by 10%, what happens to Q? If the amount of storage is increased by the same amount, what happens to Q.? What do you conclude from this comparison?

Answer 1

a) The solar heat is produced by Infrared radiation which covers the spectrum form 700nm to 1mm. For wavelength range 200nm to 3000nm, the transmission coefficient is 80% and for wavelengths greater than 3000nm, transmission coefficient is 2%.

The 700nm to 3000nm band forms %age of total IR band:

$\frac{(3000-700)nm}{1mm-700nm} \approx \frac{2300 \times 10^{-9}m}{10^{-3}m}=0.23 \times 10^{-2}$

Thus, 0.23% of IR band has transmission coefficient 80%.

The rest 99.77% has transmission coefficient 2%. So, the total transmission:

$0.23 \times 0.8 +99.77 \times 0.02=0.184+1.9954=2.18$

So, the total transmission is 2.18%. This much amount of heat escapes through glass plates.

b) The change in $Q_s$ if any of its variable changes :

$dQ_s=\frac{\partial Q_s}{\partial A_c}dA_c+\frac{\partial Q_s}{\partial S}dS$

$Q_s=L\left [ 0.8+ln\left \{ \left ( A_C/L \right )^{2/3} \left ( S/L \right )^{1/2} \right \} \right ]$

It can be further simplified using properties of logarithmic functions:

$Q_s=L\left [ 0.8+\frac{2}{3}ln\left ( A_c/L \right )+\frac{1}{2}ln(S/L) \right ]$

Calculating the partial derivatives:

$\frac{\partial Q_s}{\partial A_c}=\frac{2}{3} \frac{1}{A_c/L} \times 1/L=\frac{2}{3A_c}$

$\frac{\partial Q_s}{\partial S}=\frac{1}{2} \frac{1}{S/L} \times 1/L=\frac{1}{2S}$

Substituting the values, we get

$dQ_s=\frac{2}{3A_c}dA_c+\frac{1}{2 S}dS$

If collector area is increased by 10%, then

$dA_c=0.1$

Then the values of $Q_s$ changes by:

$\frac{2}{3} \times 0.1=0.07$

That it, 7%.

If storage is increased by 10%, then change in $Q_s$ :

$\frac{1}{2} \times 0.1=0.05$ i.e. 5%