## To T [S L T Ta L

AUlF'

tn cp

The energy gain by the air stream is then given by

tn W

tn cp

Using the equation for the heat removal factor given by Eq. (3.58), Eq. (3.75) gives

Since S = (Ta)Gt, Eq. (3.76) is essentially the same as Eq. (3.60).

Example 3.6

Estimate the outlet air temperature and efficiency of the collector shown in Figure 3.30 for the following collector specifications:

Total insolation, Gt = 890 W/m2

Emissivity of absorber plate, ep = 0.92.

Emissivity of back plate, eb = 0.92. Mass flow rate of air = 0.06 kg/s. Inlet air temperature, T = 50°C = 323 K.

### Solution

Here we need to start by assuming values for Tp and Tb. To save time, the correct values are selected; but in an actual situation, the solution needs to be found by iteration. The values assumed are Tp = 340 K and Tb = 334 K (these need to be within 10 K). From these two temperatures, the mean air temperature can be determined from

from which t = l(Tp + Tb)(T¡ + T2) = 3|(340 + 334)(3402 + 3342) = 33?R

The radiation heat transfer coefficient from the absorber to the back plate is given by h r, p-g2

From Tm, ^ the following properties of air can be obtained from Appendix 5:

^ = 2.051 X 10^5 kg/m-s k = 0.029 W/m-K cp = 1008J/kg-K

From fluid mechanics the hydraulic diameter of the air channel is given by D _ 41 Flow cross-sectional area

 = 4 = 2i 2W

The Reynolds number is given by pVD _ mD 0.06 X 0.03

4875.5

Therefore, the flow is turbulent, for which the following equation applies: Nu = 0.0158(Re)08. Since Nu = (hcD)/k, the convection heat transfer coefficient is given by h = h

0.029

0.0158(Re)

0.03

0.0158(4875.5)08 = 13.625WIm2-K

18.4

0.739

AcUlF'

323 + 351K

0.06 X 1007

Therefore, the average air temperature is ^(351 + 323) = 337 K, which is the same as the value assumed before. If there is a difference in the two mean values, an iteration is required. This kind of problem requires just one iteration to find the correct solution by using the assumed values, which give the new mean temperature.