Question 1 of 4: Flat-Plate Solar Collector Performance Test
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams December 2013 — 04-Agric-A3, Heat Engineering (3 hours, open book). Four questions constitute a complete exam paper; each is of equal value (25 points) and all require calculation.
Reference texts: Çengel & Ghajar, Heat and Mass Transfer: Fundamentals and Applications (conduction, solar-collector performance testing); J.P. Holman, Heat Transfer (natural-convection correlations, boundary-layer thickness); R.F. Barron, Cryogenic Heat Transfer (concentric-cylinder radiation shields).
Problem 1: Flat-Plate Solar Collector Performance Test (25 points)
Test-chart scatter, 13 points of η (%) vs. (Tin−Tamb)/I
range 0.06–0.47 h·ft²·°F/Btu, η 33–78%
Incident irradiation, I
200 Btu/h·ft²
Ambient temperature, Tamb
30°F
Inlet water temperature, Tin
60°F
Find. FR; UL; the useful energy delivery rate at the stated test condition; and the stagnation (no-flow) collector temperature.
Approach. The standard flat-plate-collector efficiency line, $\eta = F_R(\tau\alpha) - F_RU_L\left(\dfrac{T_{in}-T_{amb}}{I}\right)$, is a straight line in $\eta$ vs. $x=(T_{in}-T_{amb})/I$: a least-squares fit of the 13 digitized test points gives the intercept $F_R(\tau\alpha)$ and slope magnitude $F_RU_L$ directly. Divide the intercept by the known $\tau\alpha$ to isolate $F_R$, then use $\eta$ and $I$ for the useful-energy rate and set $\eta=0$ for the stagnation temperature.
Least-squares fit through the 13 digitized test points (η% vs. (Tin−Tamb)/I): intercept FR(τα) = 0.787 at x = 0, slope magnitude FRUL = 0.933 Btu/h·ft²·°F.
Fit the performance line to the scattered test data. A least-squares regression of the 13 plotted (x, η) points gives
$$\eta(\%) = 78.70 - 93.25\,x \quad\Rightarrow\quad F_R(\tau\alpha)=0.787,\ \ F_RU_L = 0.9325\ \text{Btu/h}\cdot\text{ft}^2\cdot{}^{\circ}\text{F}$$
Part (a) — heat removal factor FR. The intercept of the efficiency line is $F_R(\tau\alpha)$, and $\tau\alpha = 0.90\times0.92 = 0.828$, so
$$F_R = \frac{F_R(\tau\alpha)}{\tau\alpha} = \frac{0.787}{0.828} = \boxed{0.950}$$
Part (b) — overall loss conductance UL. The slope magnitude of the line is $F_RU_L$, so dividing by the just-found $F_R$,
$$U_L = \frac{F_RU_L}{F_R} = \frac{0.9325}{0.950} = \boxed{0.981\ \text{Btu/h}\cdot\text{ft}^2\cdot{}^{\circ}\text{F}}$$
Part (c) — useful energy delivery rate. At the stated test condition,
$$x = \frac{T_{in}-T_{amb}}{I} = \frac{60-30}{200} = 0.15\ \text{h}\cdot\text{ft}^2\cdot{}^{\circ}\text{F/Btu}$$
Reading the efficiency line at this x (equivalently, computing it directly from the intercept and slope already found),
$$\eta = F_R(\tau\alpha) - F_RU_L\,x = 0.787 - 0.9325(0.15) = 0.647$$
$$q_u = \eta\,I = 0.647\times 200\ \text{Btu/h}\cdot\text{ft}^2 = \boxed{129.4\ \text{Btu/h}\cdot\text{ft}^2}$$
Part (d) — stagnation temperature (η = 0). At zero flow the collector heats up until losses exactly balance absorbed radiation, i.e. $\eta=0$:
$$0 = F_R(\tau\alpha) - F_RU_L\left(\frac{T_{stag}-T_{amb}}{I}\right)
\;\Rightarrow\; T_{stag} = T_{amb} + \frac{F_R(\tau\alpha)}{F_RU_L}\,I$$
$$T_{stag} = 30 + \frac{0.787}{0.9325}(200) = 30 + 168.8 = \boxed{198.8\,{}^{\circ}\text{F}}$$
