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