How is the outlet temperature found?

From Qu = m cp (To - Ti): rise = 1126/(0.03 x 4186) = 8.97 K, so To ≈ 39 °C

What does (Ti - Ta) = 135 K represent?

The point where useful heat is zero (stagnation)

Practice · 6 marks

Flat plate collector useful heat and outlet temperature

Short answer

Using Qu = Ac[FR(τα)GT - FR UL (Ti - Ta)]: Qu = 2[0.76×800 - 4.5×10] = 1126 W; efficiency = 1126/(2×800) = 70.4%; outlet temperature To = 30 + 1126/(0.03×4186) ≈ 39.0 °C; stagnation occurs at Ti - Ta = (0.76×800)/4.5 = 135.1 K.

A flat plate collector of area 2 m² has $F_R(\tau\alpha) = 0.76$ and $F_R U_L = 4.5$ W/m²K. The solar radiation incident on the collector plane is 800 W/m², the ambient temperature is 20 °C and water enters at 30 °C with a flow rate of 0.03 kg/s ( $c_p = 4186$ J/kgK). Calculate (a) the useful heat gain, (b) the collector efficiency, (c) the outlet water temperature, and (d) the maximum value of $(T_i - T_a)$ at which the collector still gives useful heat at this radiation.

Hottel-Whillier-Bliss equation

Q_u = A_c\left[F_R(\tau\alpha)\,G_T - F_R U_L\,(T_i - T_a)\right]

(a) Useful heat

\begin{aligned} Q_u &= 2\,[0.76(800) - 4.5(30-20)] \\ &= 2\,[608 - 45] = 2 \times 563 = 1126\ \text{W} \end{aligned}

(b) Efficiency

\eta = \frac{Q_u}{A_c G_T} = \frac{1126}{2 \times 800} = 0.704 \;(70.4\ \%)

(c) Outlet temperature

\begin{aligned} Q_u &= \dot m\,c_p\,(T_o - T_i) \\ T_o - T_i &= \frac{1126}{0.03 \times 4186} = 8.97\ \text{K} \\ T_o &= 30 + 8.97 = 38.97\ \text{°C} \end{aligned}

(d) Stagnation condition

Useful heat becomes zero when $F_R(\tau\alpha)G_T = F_R U_L (T_i - T_a)$ :

T_i - T_a = \frac{0.76 \times 800}{4.5} = 135.1\ \text{K}

Answer: $Q_u = 1126$ W; $\eta = 70.4\ \%$ ; $T_o = 39.0$ °C; useful heat is zero when $T_i - T_a = 135$ K (the collector is then at stagnation).

✓ Answer checked by Claude Opus 5.5 (high effort)

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