Question

A flat-plate solar collector is used to heat water by having water flow through tubes attached at the back of the thin solar absorber plate. The absorber plate has a surface area of 2 m2 with emissivity and absorptivity of 0.9. The surface temperature of the absorber is 35°C, and solar radiation is incident on the absorber at 500 W/m2 with a surrounding temperature of 0°C. Convection heat transfer coefficient at the absorber surface is 5 W/m2 ∙K, while the ambient temperature is 25°C. Net heat rate absorbed by the solar collector heats the water from an inlet temperature (Tin) to an outlet temperature (Tout). If the water flow rate is 5 g/s with a specific heat of 4.2 kJ/kg∙K, determine the temperature rise of the water.

Answer 1

Answer:

dT = 21.35 C

Explanation:

Given:

- Area of the absorber plate A = 2 m^2

- e= a = 0.9

- T_s = 35 C

- q''_solar = 500 W/m^2

- T_surr = 0 C

- h = 5 W/m^2 K

- T_amb = 25 C

- flow(m) = 5 g/s

- c_p = 4.2 KJ/kgK

Find:

- The temperature rise of water.

Solution:

- Using Energy balance on the plate:

                                   E_in - E_out = q_in

- Find E_in and E_out:

                                   E_in = a*q''_solar

                            E_out = q"_convec + q"_rad

                   E_out = h*(T_s - T_amb) + e*σ*(T_s^4 - T_surr^4)

- Compute E_in and E_out:

                            E_in = 0.9*(500) = 450 W/m^2

         E_out = 5*(35 - 25) + 0.9*5.67*10^-8*(308^4 - 273^4)

                                E_out = 225.778 W/m^2

- Hence,

                                  E_in - E_out = q_in

                                 450 - 225.778 = q_in

                                  q_in = 224 W/m^2

- Assuming thickness of plate and pipe through which water flows as negligible, then:

                                flow(m)*c_p*(dT) = q_in*A

                                dT = q_in*A / flow(m)*c_p

- plug values in:

                                    dT = 224*2 / 5*4.2

                                    dT = 21.35 C