---
title: "Flat plate collector useful heat and outlet temperature"
subject: "Energy Resources"
chapter: 2
key: "semester_6/energy_resources:flat-plate-collector-efficiency-numerical"
source: https://www.bctengineer.com/q/semester_6/energy_resources/flat-plate-collector-efficiency-numerical
site: BCT Engineering
---

# Flat plate collector useful heat and outlet temperature

Energy Resources · Chapter 2: Solar energy

- Asked in: year not given, 6 marks
- Video: [Solar Thermal Course, Flat Plate Solar Collector #Question 1](https://www.youtube.com/watch?v=eS2nvyZEtxo) — Engineering Topics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/energy_resources/flat-plate-collector-efficiency-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_6/energy_resources/chapter-2

## Question

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.

## 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.

## Answer

#### 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).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With FR(τα)=0.76, G=800, FRUL=4.5, Ti-Ta=10, Ac=2, what is Qu?
   A. 1216 W
   B. 563 W
   C. 1280 W
   D. 1126 W
   Answer: D. Qu = 2[608 - 45] = 1126 W. (Key point: Useful heat)

2. How is efficiency calculated and what is it here?
   A. Qu/GT = 563%
   B. FR(τα) = 76%
   C. Qu/(Ac GT) = 70.4%
   D. Qu/(m cp) = 70.4%
   Answer: C. 1126/(2 x 800) = 0.704. (Key point: Efficiency)

3. How is the outlet temperature found?
   A. To = Ti + Qu/Ac = 30 + 563
   B. From Qu = m cp (To - Ti): rise = 1126/(0.03 x 4186) = 8.97 K, so To ≈ 39 °C
   C. To = Ta + 8.97 = 28.97 °C
   D. To = Ti x efficiency = 21 °C
   Answer: B. The rise is applied to the inlet temperature of 30 °C. (Key point: Outlet temperature)

4. What does (Ti - Ta) = 135 K represent?
   A. The point where useful heat is zero (stagnation)
   B. Maximum efficiency
   C. The temperature of the sun
   D. The point where Qu is maximum
   Answer: A. FR(τα)GT = FRUL(Ti - Ta), so absorbed heat equals losses. (Key point: Stagnation condition)

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