1. **State the problem:** We have a parabolic trough concentrator (PTC) with given dimensions and properties. We want to calculate the exit fluid temperature after heating. 2. **Given data:** - Width of aperture, $W = 5$ m - Length of aperture, $L = 18$ m - Absorbed radiation per unit area, $C$ W/m$^2$ (unknown constant) - Diameter of receiver tube, $d = 53$ mm = 0.053 m - Fluid inlet temperature, $T_i = 81^6$C - Fluid flow rate, $\dot{m} = 0.06$ kg/s - Specific heat capacity, $C_p = 4.1$ kJ/kg$^6$C = 4100 J/kg$^6$C - Overall loss coefficient, $U_L = 3.5$ W/m$^2$$^6$C - Heat removal factor, $F_R = 0.91$ - Ambient temperature, $T_a = 23^6$C 3. **Formula used:** The useful heat gain by the fluid is given by: $$ Q_u = A_c F_R [C - U_L (T_i - T_a)] $$ where $A_c$ is the collector aperture area. The heat gained by the fluid is also: $$ Q_u = \dot{m} C_p (T_o - T_i) $$ where $T_o$ is the exit fluid temperature. 4. **Calculate aperture area:** $$ A_c = W \times L = 5 \times 18 = 90 \text{ m}^2 $$ 5. **Equate heat gained expressions and solve for $T_o$:** $$ \dot{m} C_p (T_o - T_i) = A_c F_R [C - U_L (T_i - T_a)] $$ $$ T_o = T_i + \frac{A_c F_R [C - U_L (T_i - T_a)]}{\dot{m} C_p} $$ 6. **Substitute known values:** $$ T_o = 81 + \frac{90 \times 0.91 \times [C - 3.5 \times (81 - 23)]}{0.06 \times 4100} $$ 7. **Simplify the temperature difference term:** $$ 81 - 23 = 58 $$ $$ U_L (T_i - T_a) = 3.5 \times 58 = 203 $$ 8. **Final expression for exit temperature:** $$ T_o = 81 + \frac{90 \times 0.91 \times (C - 203)}{246} $$ **Note:** To find a numerical value for $T_o$, the absorbed radiation per unit area $C$ must be known. **Summary:** The exit fluid temperature is given by $$ T_o = 81 + \frac{81.9 (C - 203)}{246} $$ where $C$ is the absorbed radiation per unit area in W/m$^2$.