Solar Thermal Collectors: Optical and Thermal Efficiency, Flat-Plate, Evacuated-Tube, Air-Heater and Concentrating Collectors

This chapter, the second of two on solar thermal systems in the Engineering Sciences (XE) paper, answers the remainder of the third topic of Section XE8.1 (Energy Resources and Conversion Technologies) of XE8 Energy Science, one of the nine optional sections, any two of which a candidate chooses in the exam: “Solar thermal systems: … optical efficiency, thermal efficiency, energy conversation [conservation] for solar thermal collectors, flat plate collectors, evacuated tube collectors, solar air heaters, concentrating collectors.” It teaches the collector energy balance, optical and thermal efficiency, the Hottel-Whillier-Bliss equation with the heat-removal factor, flat-plate and evacuated-tube collectors, solar air heaters, and concentrating collectors (concentration ratio, acceptance angle, tracking). The numericals are useful gain, stagnation temperature, heat-removal factor, efficiency-line interpretation, air-heater area and concentrator efficiency.

1. Energy balance, optical and thermal efficiency

A solar thermal collector absorbs radiation, loses part of it as heat to the surroundings and carries the rest away in a fluid. Per unit aperture area, the radiation absorbed by the absorber is S = G (τα), where G is the incident irradiance on the collector plane and (τα) is the transmittance-absorptance product: the fraction of G that passes the cover (τ) and is absorbed by the plate (α). The energy balance on the absorber is

Q_u = A_c [S − U_L (T_p − T_a)],

with Q_u the useful heat gain, A_c the collector area, U_L the overall heat-loss coefficient (top, bottom and edge losses, referred to the plate), T_p the mean plate temperature and T_a the ambient temperature. The useful gain is also what the fluid receives, Q_u = ṁ c_p (T_o − T_i). The instantaneous thermal efficiency is η = Q_u/(A_c G). Two limits matter: the optical efficiency, the efficiency when there is no loss (T_p = T_a), which is η₀ = (τα) (or F_R(τα) with the heat-removal factor below); and the stagnation state, when no heat is extracted (Q_u = 0), where the plate settles at T_p = T_a + S/U_L.

Worked example: G = 800 W/m², (τα) = 0.85, U_L = 6 W/m²K, T_a = 20 °C. S = 0.85 × 800 = 680 W/m². Stagnation temperature = 20 + 680/6 = 133.3 °C. If the plate is held at 60 °C, Q_u/A_c = 680 − 6 × 40 = 440 W/m², so η = 440/800 = 0.55. Lower losses (U_L) or a higher (τα) raise the efficiency; a higher plate temperature lowers it.

⚠️ Exam trap
Stagnation uses the absorbed radiation S and the loss coefficient only; the heat-removal factor F_R does not appear, because with no flow there is no heat removal. Also, U_L is not constant: it rises with plate temperature, which is why efficiency curves bend slightly at high temperatures.

2. The Hottel-Whillier-Bliss equation

The mean plate temperature is hard to measure, so the performance is written with the fluid inlet temperature T_i, which is known. The Hottel-Whillier-Bliss (HWB) equation is Q_u = A_c F_R [S − U_L (T_i − T_a)]. The heat-removal factor F_R is the ratio of the actual useful gain to the gain that would occur if the whole absorber were at the inlet temperature. For a collector of efficiency factor F′ and a mass flow ṁ of fluid with specific heat c_p,

F_R = [ṁ c_p/(A_c U_L)] [1 − exp(−A_c U_L F′/(ṁ c_p))].

The collector efficiency factor F′ is the ratio of the actual gain to the gain if the absorber were at the local fluid temperature; it depends on how well heat gets from the plate to the fluid. For a tube-and-sheet absorber the fin efficiency is F = tanh(mW)/(mW) with m = √(U_L/(kδ)), where W is the half-width of plate between tubes, k its conductivity and δ its thickness: a thinner plate, lower k or wider tube spacing lowers F and F′. As ṁ → ∞, F_R → F′ (the upper limit), so F_R ≤ F′ ≤ 1.

Dividing by A_c G gives the efficiency line: η = F_R(τα) − F_R U_L (T_i − T_a)/G. Plotted against (T_i − T_a)/G, it is a straight line (if U_L is constant) with intercept F_R(τα) (the optical efficiency) and slope −F_R U_L; it meets the axis at (T_i − T_a)/G = (τα)/U_L, the stagnation point. Worked example: A_c = 2 m², F′ = 0.90, U_L = 6 W/m²K, ṁ = 0.03 kg/s water with c_p = 4180 J/kgK. ṁ c_p = 125.4 W/K; the exponent is 0.90 × 6 × 2/125.4 = 0.0861; F_R = (125.4/12)(1 − e^(−0.0861)) = 10.45 × 0.0825 = 0.86. With (τα) = 0.85, G = 800 W/m², T_i = 50 °C and T_a = 20 °C and this F_R = 0.86 in the HWB equation: Q_u/A_c = 0.86 × (680 − 6 × 30) = 430 W/m², so η = 0.54. (A poorer flow giving F_R = 0.80 would cut this to 0.80 × 500 = 400 W/m², η = 0.50.)

🎯 Why it matters
F_R falls as the flow rate falls (the fluid heats up more, the plate runs hotter, losses rise), which is why a low flow gives a hotter outlet but a lower efficiency. The line shows the trade: a collector with a high intercept but a steep slope suits low temperatures; a low slope suits high temperatures or low sunshine.

3. Flat-plate and evacuated-tube collectors

A flat-plate collector has a dark absorber plate with fluid passages (tubes or channels), one or two transparent covers (glass), insulation at the back and edges, and a weatherproof casing. It uses both beam and diffuse radiation, needs no tracking and is fixed at a tilt near the latitude. The cover transmits short-wave sunlight but is nearly opaque to the long-wave radiation emitted by the hot plate (the greenhouse effect) and it suppresses convective loss from the plate; more covers cut the loss but also cut the transmittance τ, so one or two are typical. A selective surface, high absorptance α for solar radiation and low emittance ε in the infrared, reduces radiative loss.

An evacuated-tube collector places the absorber inside a sealed glass tube (often two concentric tubes with the space between them evacuated, the absorber coating on the inner tube). The vacuum removes gas conduction and convection between absorber and cover, so U_L is much smaller. The heat is taken out by fluid flowing through the tube or, commonly, by a heat pipe in each tube. Consequences: the efficiency line is much flatter, so the collector keeps its efficiency at high T_i − T_a (for process heat, or in cold weather) and at low irradiance; the round tubes let the absorber face the sun over a wide range of angles; and a broken tube can be replaced individually. The cost per unit area is higher, and the intercept is not necessarily higher than a good flat plate, so at small T_i − T_a a flat plate can perform as well.

FeatureFlat plateEvacuated tube
Loss coefficient U_LHigher; convection in the air gapMuch lower; vacuum removes convection and gas conduction
Efficiency lineSteeperFlatter
Best useLow-temperature heating in mild conditionsHigher temperatures, cold or cloudy conditions
Cost per unit areaLowerHigher

4. Solar air heaters

In a solar air heater the working fluid is air, which is free of freezing and corrosion and needs no leak-tight plumbing, so it suits space heating and crop drying. Air has low density, low specific heat and low thermal conductivity, so the heat-transfer coefficient between plate and air is small (the plate runs hot and U_L is high) and a large volume must be moved. The design problem is to raise the plate-to-air heat transfer without a large fan-power penalty.

  • Extended surfaces: fins, corrugated or v-corrugated absorbers, and a matrix or packed bed raise the area for convection.
  • Artificial roughness (ribs on the absorber surface) breaks the laminar sub-layer and raises the heat-transfer coefficient, at the cost of higher pressure drop.
  • Flow arrangement: air passing behind the absorber, or both sides, or a double pass under a second cover, lengthens contact and reduces loss.
  • Higher flow rate raises efficiency (lower plate temperature) but lowers outlet temperature and raises fan power.

The same HWB form applies. Worked example: the air flow of 0.04 kg/s with c_p = 1005 J/kgK is heated by 25 K, so Q_u = 0.04 × 1005 × 25 = 1005 W. With a thermal efficiency of 0.45 and G = 800 W/m² the collector area is A_c = Q_u/(η G) = 1005/(0.45 × 800) = 2.79 m².

5. Concentrating collectors: concentration ratio, acceptance angle, tracking

A concentrating collector uses mirrors or lenses to focus radiation from a large aperture area A_a onto a small receiver (absorber) area A_r. The geometric concentration ratio is C = A_a/A_r. The heat-loss area is the small receiver area, so the loss per unit aperture is U_L (T_r − T_a)/C; high temperatures become practicable. The price: it responds to beam radiation only (diffuse light cannot be focused), and it must track the sun. Worked example: a trough with aperture width 2.0 m and a tubular receiver of diameter 0.05 m has, per metre length, A_a = 2.0 m² and A_r = π × 0.05 = 0.157 m², so C = 12.7.

The acceptance angle 2θ_a is the range of directions of incoming beam (in the plane considered) over which all rays still reach the receiver without moving the concentrator. The sun subtends about 0.53° (half angle about 0.267°), so θ_a cannot be less than that. Thermodynamics (conservation of étendue) limits the ideal concentration: C_max = 1/sin θ_a for a line-focus (two-dimensional) concentrator and C_max = 1/sin²θ_a for a point-focus (three-dimensional) one. For θ_a = 2°, 1/sin 2° = 28.7. For the sun's own half-angle these limits are about 215 (line focus) and about 46 000 (point focus). Higher C therefore means a smaller acceptance angle and more precise tracking.

TypeFocus and trackingRemarks
Parabolic troughLine focus; one-axis trackingModerate C; receiver tube along the focal line
Linear FresnelLine focus; flat mirror strips tracking in one axisCheaper structure, lower optical efficiency
Parabolic dishPoint focus; two-axis trackingHighest C and temperature
Central receiver (heliostat field)Point focus; each mirror tracks in two axesLarge scale, high temperature
Compound parabolic concentrator (CPC)Non-imaging; wide acceptance, little or seasonal adjustmentLow C near 1/sin θ_a; also collects some diffuse light

Performance. With an optical efficiency η₀ (reflectance × transmittance × absorptance × the fraction of rays intercepted), the efficiency based on the beam on the aperture is η = η₀ − U_L (T_r − T_a)/(C G_b). Worked example: η₀ = 0.70, U_L = 8 W/m²K on the receiver area, T_r − T_a = 200 K, C = 20, G_b = 800 W/m²: loss term = 8 × 200/(20 × 800) = 0.10, so η = 0.60. At the same T_r − T_a a concentrator with C = 1 would lose 8 × 200/800 = 2.0, which is more than the whole input: only a high C works at this temperature.

⚠️ Exam trap
Concentrator efficiency is quoted on the beam, not the global, radiation, and on aperture area. Do not use the receiver area in the denominator of η, and do not apply C = A_a/A_r to the flat plate (where C = 1).

Key takeaways

  • Energy balance: Q_u = A_c[S − U_L(T_p − T_a)] with S = G(τα); optical efficiency is (τα) (or F_R(τα)); stagnation is T_p = T_a + S/U_L, with no F_R.
  • HWB: Q_u = A_c F_R[S − U_L(T_i − T_a)]; efficiency η = F_R(τα) − F_R U_L (T_i − T_a)/G, a line with intercept F_R(τα) and slope −F_R U_L.
  • F_R = [ṁc_p/(A_cU_L)][1 − exp(−A_cU_LF′/(ṁc_p))] ≤ F′ ≤ 1; F_R rises with flow rate and tends to F′.
  • Flat plates use cover, selective surface and insulation and need no tracking; evacuated tubes remove convection with a vacuum, giving low U_L and a flat efficiency line at high T_i − T_a.
  • Air heaters suffer from poor air heat transfer; fins, corrugations, roughness and packed beds help at the cost of pressure drop and fan power.
  • Concentrators: C = A_a/A_r; C_max = 1/sin θ_a (line) or 1/sin²θ_a (point); higher C needs tighter tracking; η = η₀ − U_L(T_r − T_a)/(C G_b); only beam radiation is used.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. In a flat-plate collector the transmittance-absorptance product (τα) represents:

    1. the fraction of incident radiation that passes the cover and is absorbed by the plate
    2. the fraction of the absorbed heat that is finally carried away by the working fluid in the tubes
    3. the ratio of the radiative heat loss to the convective heat loss from the hot plate
    4. the fraction of the incident radiation that is reflected by the plate and the cover taken together
    Show answer

    Answer: A — the fraction of incident radiation that passes the cover and is absorbed by the plate

    (τα) is the product of the cover transmittance and the plate absorptance, which is the fraction of the incident radiation that ends up absorbed. The fraction taken away by the fluid is the efficiency, not (τα).

  2. A collector's efficiency is plotted against (T_i − T_a)/G. For the HWB model with constant U_L, the intercept and the slope of the straight line are respectively:

    1. (τα) and −U_L
    2. F_R and −F_R(τα)
    3. F_R(τα) and −F_R U_L
    4. F_R U_L and −F_R(τα)
    Show answer

    Answer: C — F_R(τα) and −F_R U_L

    Dividing Q_u = A_c F_R[G(τα) − U_L(T_i − T_a)] by A_c G gives η = F_R(τα) − F_R U_L (T_i − T_a)/G. The intercept is the optical efficiency F_R(τα) and the slope is −F_R U_L; (τα) and −U_L would be the values without the heat-removal factor.

  3. The glass cover of a flat-plate collector improves its performance mainly because it:

    1. raises the absorptance of the absorber plate for visible light by acting as an anti-reflection layer on the plate
    2. concentrates the sunlight onto the absorber plate by refraction, like a lens, so that more energy arrives
    3. cools the plate by conduction into the glass so that the plate stays at a much lower temperature than it would uncovered
    4. transmits solar short-wave radiation but blocks the long-wave emission of the hot plate and suppresses convective loss
    Show answer

    Answer: D — transmits solar short-wave radiation but blocks the long-wave emission of the hot plate and suppresses convective loss

    Glass is largely transparent to short-wave sunlight and opaque to the long-wave infrared from the plate (a greenhouse effect), and the still air gap lowers convective loss. It does not raise α, concentrate light or cool the plate.

  4. A selective absorber coating is one with:

    1. a low solar absorptance and a high infrared emittance, to reject heat
    2. a high solar absorptance and a low infrared emittance, to reduce radiative loss
    3. equal absorptance and emittance at all wavelengths, as for an ideal black body
    4. a high reflectance in the visible range, to avoid overheating
    Show answer

    Answer: B — a high solar absorptance and a low infrared emittance, to reduce radiative loss

    A selective surface absorbs strongly in the solar band (high α) yet emits weakly in the infrared where a warm plate radiates (low ε), which reduces radiative loss. An ideal black body has α = ε and is not selective.

  5. Compared with a flat-plate collector, an evacuated-tube collector typically keeps a higher efficiency at large (T_i − T_a)/G because:

    1. its absorber has a much higher solar absorptance than any flat-plate absorber at all operating temperatures
    2. the vacuum nearly eliminates convective and conductive loss, so the loss coefficient U_L is much smaller
    3. the round tubes concentrate the beam radiation by a factor of several onto the absorber inside
    4. the working fluid inside the tube always has a higher specific heat than the fluid in a flat plate
    Show answer

    Answer: B — the vacuum nearly eliminates convective and conductive loss, so the loss coefficient U_L is much smaller

    The vacuum suppresses gas conduction and convection, giving a low U_L and hence a flatter efficiency line. It does not raise α or concentrate light, and the fluid is not the reason.

  6. A solar air heater usually needs fins, corrugations or artificial roughness on the absorber mainly because:

    1. air has a low heat-transfer coefficient and low heat capacity, so plate-to-air transfer must be enhanced
    2. air freezes at low winter temperatures inside the duct and the fins prevent ice from forming on the plate
    3. fins increase the transmittance of the glass cover for solar radiation by reducing reflection losses
    4. the air would otherwise corrode the metal absorber quickly and the fins serve as a protective surface layer
    Show answer

    Answer: A — air has a low heat-transfer coefficient and low heat capacity, so plate-to-air transfer must be enhanced

    Air's low density, conductivity and specific heat give a small heat-transfer coefficient, so extended surfaces and roughness are used, at the cost of pressure drop. Air does not freeze or corrode, and fins do not change cover transmittance.

  7. An ideal two-dimensional (line-focus) concentrator has an acceptance half-angle θ_a = 2°. Its maximum possible geometric concentration ratio, C = 1/sin θ_a, is closest to:

    1. 14.3
    2. 57.3
    3. 28.7
    4. 820
    Show answer

    Answer: C — 28.7

    sin 2° = 0.034899, so C = 1/0.034899 = 28.65, about 28.7. The value 57.3 is double the correct result (it comes from using 1° instead of the stated 2°), 14.3 comes from using 4°, and 820 is 1/sin²θ_a, the point-focus limit.

  8. When the design concentration ratio of a tracking concentrator is increased, which change is unavoidable?

    1. The acceptance angle must increase, so tracking can be less accurate
    2. The collector can use more of the diffuse radiation from the sky
    3. The acceptance angle must decrease, so tracking must be more accurate
    4. The receiver temperature must fall because the receiver area is smaller
    Show answer

    Answer: C — The acceptance angle must decrease, so tracking must be more accurate

    Since C_max = 1/sin θ_a (or 1/sin²θ_a), a larger C means a smaller acceptance angle and tighter tracking. A larger C also lowers loss per unit aperture, so the attainable temperature rises.

  9. A concentrator has optical efficiency 0.70, receiver loss coefficient 8 W/m²K (on receiver area), concentration ratio 20, beam irradiance 800 W/m² and a receiver 200 K above ambient. Using η = η₀ − U_L(T_r − T_a)/(C G_b), the efficiency is:

    1. 0.60
    2. 0.50
    3. 0.68
    4. 0.10
    Show answer

    Answer: A — 0.60

    Loss term = 8 × 200/(20 × 800) = 1600/16000 = 0.10, so η = 0.70 − 0.10 = 0.60. The value 0.10 is the loss term alone, and 0.68 would need a loss term of only 0.02, which is a factor of five too small.

  10. Which statements about the Hottel-Whillier-Bliss model are correct?

    1. The heat-removal factor F_R is the ratio of the actual useful gain to the gain if the whole absorber were at the fluid inlet temperature
    2. Increasing the mass flow rate raises F_R towards the efficiency factor F′
    3. F_R can exceed F′ if the flow rate is high enough
    4. The HWB equation in terms of F_R uses the mean plate temperature, not the inlet temperature
    Show answer

    Answer: A — The heat-removal factor F_R is the ratio of the actual useful gain to the gain if the whole absorber were at the fluid inlet temperature; B — Increasing the mass flow rate raises F_R towards the efficiency factor F′

    F_R is defined relative to the inlet temperature and it rises with flow towards the upper limit F′, which it cannot exceed. The point of F_R is to write the equation with the known inlet temperature, so the statement about the mean plate temperature is wrong.

  11. Which statements about concentrating collectors are correct?

    1. They respond mainly to beam radiation, since diffuse light cannot be focused
    2. For a given optical design, a higher concentration ratio means a smaller acceptance angle
    3. Heat loss per unit aperture area falls as the concentration ratio rises, because the loss area is only the receiver
    4. The concentration ratio can be made as large as desired for a fixed acceptance angle
    Show answer

    Answer: A — They respond mainly to beam radiation, since diffuse light cannot be focused; B — For a given optical design, a higher concentration ratio means a smaller acceptance angle; C — Heat loss per unit aperture area falls as the concentration ratio rises, because the loss area is only the receiver

    The first three are standard results. The last is false because étendue conservation limits C to 1/sin θ_a (line) or 1/sin²θ_a (point) for a given acceptance angle.

  12. Which statements about flat-plate and evacuated-tube collectors are correct?

    1. The efficiency of a flat-plate collector falls as (T_i − T_a)/G rises
    2. With no fluid flow the absorber reaches a stagnation temperature T_a + G(τα)/U_L
    3. In an evacuated tube most of the heat loss is by convection across the vacuum
    4. A flat-plate collector needs two-axis tracking to work
    Show answer

    Answer: A — The efficiency of a flat-plate collector falls as (T_i − T_a)/G rises; B — With no fluid flow the absorber reaches a stagnation temperature T_a + G(τα)/U_L

    The line has negative slope, and stagnation follows from Q_u = 0. A vacuum cannot support convection, and flat plates are normally fixed, so the other two are false.

  13. A flat-plate collector has F_R = 0.80, (τα) = 0.85 and U_L = 6.0 W/m²K. The irradiance on the collector plane is 800 W/m², the fluid inlet temperature is 50 °C and the ambient temperature is 20 °C. Using the Hottel-Whillier-Bliss equation, find the useful heat gain per square metre of collector, in W/m².

    Numerical answer — type the value.

    Show answer

    Answer: 400

    Q_u/A_c = F_R[G(τα) − U_L(T_i − T_a)] = 0.80 × [800 × 0.85 − 6.0 × 30] = 0.80 × (680 − 180) = 400 W/m². Python check: 400.0.

  14. A collector has absorbed flux S = G(τα) = 680 W/m² and loss coefficient U_L = 6.0 W/m²K, and the ambient temperature is 20 °C. With no fluid flow (no heat removal), find the stagnation temperature of the absorber in °C, to one decimal place.

    Numerical answer — type the value.

    Show answer

    Answer: 133.3

    With Q_u = 0, S = U_L(T_p − T_a), so T_p = T_a + S/U_L = 20 + 680/6.0 = 20 + 113.33 = 133.3 °C.

  15. A collector of area 2.0 m² has efficiency factor F′ = 0.90 and loss coefficient U_L = 6.0 W/m²K. Water flows at 0.030 kg/s with c_p = 4180 J/kgK. Using F_R = [ṁc_p/(A_cU_L)][1 − exp(−A_cU_LF′/(ṁc_p))], find F_R to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 0.86

    ṁc_p = 0.030 × 4180 = 125.4 W/K. A_cU_L = 12.0 W/K. Exponent = 12.0 × 0.90/125.4 = 0.08612. F_R = (125.4/12.0)(1 − e^(−0.08612)) = 10.45 × 0.08251 = 0.862, which is 0.86. Python check: 0.8623.

  16. A collector has the efficiency line η = 0.78 − 5.0 (T_i − T_a)/G, with G in W/m² and temperatures in K or °C (a difference). At G = 900 W/m², by how many kelvin must the inlet temperature exceed the ambient temperature for the efficiency to fall to 0.40? Give the answer to one decimal place.

    Numerical answer — type the value.

    Show answer

    Answer: 68.4

    0.40 = 0.78 − 5.0 ΔT/900 gives 5.0 ΔT/900 = 0.38, so ΔT = 0.38 × 900/5.0 = 68.4 K. Python check: 68.4.

  17. An air flow of 0.040 kg/s with c_p = 1005 J/kgK is heated by 25 K in a solar air heater whose thermal efficiency is 0.45 under an irradiance of 800 W/m² on the collector plane. Find the required collector area in m², to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 2.79

    Q_u = ṁc_pΔT = 0.040 × 1005 × 25 = 1005 W. A_c = Q_u/(ηG) = 1005/(0.45 × 800) = 1005/360 = 2.792 m², or 2.79 m². Python check: 2.7917.