Question 2 of 4: Flat-Plate Solar Collector Performance Test
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 04-Agric-A3 Heat Engineering, National Exams
May 2013 — a three-hour open-book examination; any non-communicating
calculator is permitted. The cover page states that four (4) questions constitute a
complete exam paper and that only the first four as they appear in the answer book are
marked, that each question is of equal value, and that all questions require calculation.
All four printed problems are worked here.
Reference texts. Y.A. Çengel & A.J. Ghajar, Heat and Mass
Transfer: Fundamentals and Applications, 5th ed. (conduction, flat-plate solar
collectors); J.P. Holman, Heat Transfer, 10th ed. (natural-convection
correlations and boundary-layer thickness); R.F. Barron, Cryogenic Heat Transfer,
2nd ed. (radiation shields for cryogenic lines).
Problem 2: Flat-Plate Solar Collector Performance Test (25 points)
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. $x=(T_{in}-T_{amb})/I$: its intercept is $F_R(\tau\alpha)$ and its slope magnitude
is $F_RU_L$. Read both off the plotted line, 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.
Reconstructed test line from the printed performance chart: intercept FR(τα) = 0.80 at x = 0, slope
magnitude FRUL = 1.04 Btu/h·ft²·°F (dashed segment
is the graph's own extrapolation back to x = 0).
Check: the intercept and slope used here were read from the printed chart against its own gridlines (checked against the "50" y-axis label and the x-axis tick labels) and by fitting a line through the plotted test points, giving FR(τα) = 0.80 and FRUL =
1.04 Btu/h·ft²·°F with R² = 0.998 — these are the values
used throughout.
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
$$F_R = \frac{F_R(\tau\alpha)}{\tau\alpha} = \frac{0.80}{0.828} = \boxed{0.966}$$
Part (b) — overall loss conductance UL. The slope
magnitude of the line is $F_RU_L$, so dividing by the just-found $F_R$,
$$U_L = \frac{F_RU_L}{F_R} = \frac{1.04}{0.966} = \boxed{1.08\ \text{Btu/h}\cdot
\text{ft}^2\cdot{}^{\circ}\text{F}}$$
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 (or computing it directly from the intercept and
slope already found),
$$\eta = F_R(\tau\alpha) - F_RU_L\,x = 0.80 - 1.04(0.15) = 0.644$$
$$q_u = \eta\,I = 0.644\times 200\ \text{Btu/h}\cdot\text{ft}^2 =
\boxed{128.8\ \text{Btu/h}\cdot\text{ft}^2}$$
Part (d) — stagnation temperature (η = 0). At zero flow the
collector heats up until losses exactly balance absorbed radiation, i.e. $\eta=0$:
$$0 = F_R(\tau\alpha) - F_RU_L\left(\frac{T_{stag}-T_{amb}}{I}\right)
\;\Rightarrow\; T_{stag} = T_{amb} + \frac{F_R(\tau\alpha)}{F_RU_L}\,I$$
$$T_{stag} = 30 + \frac{0.80}{1.04}(200) = 30 + 153.8 =
\boxed{183.8\,{}^{\circ}\text{F}}$$
