Lecture 3 – Practice Problems in Renewable Energy Engineering

Lecture 3 – Practice Problems in Renewable Energy Engineering

Course: Solar, Wind and Biomass Energy Systems

The third lecture focuses on applying the nine solar‑radiation parameters introduced in Lectures 1 and 2. All calculations are performed for horizontal surfaces first and then converted to tilted surfaces.

1. Problem Set Overview

1. Flat‑plate collector
2. Tilt (β) = 30°
3. Latitude (φ) = 19° 7′ N → 19.28°
4. Longitude (λ) = 72° 51′ E → 72.85°
5. Surface azimuth (γ) = 0° (facing due south)

Task:
- Compute the angle of incidence (θ) of the beam radiation on April 1 at 10 h local apparent time (LAT).
- Show how to obtain LAT from Indian Standard Time (IST, based on 82.50° E).

1. Radiation on the same site
2. Average sunshine hours (S̅) = 7.2 h
3. Elevation = 14 m

Task:
- Determine the monthly average of daily and hourly global and diffuse radiation on a horizontal plane and on the tilted surface (β = 30°).

2. Key Solar‑Geometry Definitions

The angle of incidence is required to convert the beam flux I_b to a flux normal to the surface:

\[\nI_{b,\text{normal}} = I_b \cos\theta\n\]

The general expression for cos θ is

\[\n\cos\theta = \sin\phi\sin\delta\cos\beta\n + \cos\phi\cos\delta\cos\omega\cos\beta\n + \cos\phi\sin\delta\sin\beta\cos\gamma\n - \sin\phi\cos\delta\cos\omega\sin\beta\cos\gamma\n + \cos\delta\sin\omega\sin\beta\sin\gamma\n\]

For a surface facing due south (γ = 0°) the formula simplifies to a function of φ, β, δ and ω only.

3. Angle of Incidence on April 1

• Day number N = 91 (31 Jan + 28 Feb + 31 Mar).
• Declination

\[\n\delta = 23.45^\circ \sin!\left(\frac{360^\circ}{365}(284+N)\right)\n = 23.45^\circ \sin!\left(\frac{360^\circ}{365}(284+91)\right)\n = 4.02^\circ\n\]

• Hour angle at 10 h LAT

\[\n\omega = (10 - 12)\times15^\circ = -30^\circ\n\]

• Substituting φ = 19.28°, β = 30°, γ = 0°, δ = 4.02°, ω = –30° into the simplified cosine‑θ expression gives

\[\n\cos\theta = 0.835 \;\;\Rightarrow\;\; \theta = 33.29^\circ\n\]

The same result is obtained using the alternative formulation based on the zenith angle (θ_z) and solar azimuth (γ_s).

4. Sunrise, Sunset and Day Length

For a horizontal surface (β = 0°) the sunrise/sunset condition is θ_z = 90°, yielding

\[\n\cos\omega_s = -\tan\phi\,\tan\delta\n\]

\[\n\omega_s = \cos^{-1}(-\tan\phi\,\tan\delta)\n\]

For the given location (φ = 19.28°, δ = 4.02°):

\[\n\omega_s = \pm 91.40^\circ\n\]

Day length

\[\nS_{\max}= \frac{2}{15}\,\omega_s = 12.19\ \text{h}\n\]

For a tilted surface (β = 30°) the sunrise/sunset hour angle becomes

\[\n\omega_{st}= \cos^{-1}!\bigl(-\tan(\phi-\beta)\,\tan\delta\bigr)\n\]

Using the same φ, β and δ gives

\[\n\omega_{st}= \pm 89.23^\circ\n\]

During the summer period (March 21 – September 22) the horizontal‑surface formula over‑predicts ω_s; the tilted‑surface formula must be used. In the winter period (September 22 – March 21) the horizontal formula is adequate.

5. Local Apparent Time (LAT)

LAT is obtained from standard time (IST) by

\[\n\text{LAT}= \text{ST} + 4(\lambda_{\text{std}}-\lambda) + \text{EOT}\n\]

• Standard time (ST) = 14 h (IST)
• Standard meridian λ_std = 82.50° E
• Location longitude λ = 72.85° E
• Equation of Time (EOT) for N = 91

\[\nB = \frac{360^\circ}{365}\,(N-81) = 90^\circ\n\]

\[\n\text{EOT}= 229.18 + 0.00075B - 0.001868\cos B - 0.032077\sin B\n -0.014615\cos2B -0.04089\sin2B = -4.4\ \text{min}\n\]

Putting the numbers together

\[\n\text{LAT}= 14\ \text{h} - 4\,(82.50-72.85) - 0.073\ \text{h}\n = 13\ \text{h}\ 17\ \text{min}\n\]

6. Monthly Average Daily Global Radiation (H̅_g)

The correlation (Modi et al., 1979)

\[\n\frac{H_g}{H_0}= a + b\frac{S̅}{S_{\max}}\n\]

with

• a = 0.31, b = 0.43
• S̅ = 7.2 h, S_max = 12.44 h (computed for April 15)

First, compute the extraterrestrial daily radiation (H_0) for the representative day April 15 (N = 105).

• Declination δ = 19.42°
• Sunrise/sunset hour angle ω_s = 93.32° → ω_s = 1.628 rad
• (S_{\max}= \frac{2}{15}\,ω_s = 12.44) h

Using the standard extraterrestrial radiation formula (solar constant (I_{sc}=1367) W m⁻²) and converting to kJ m⁻² day⁻¹ gives

\[\nH_0 = 37\,957\ \text{kJ m}^{-2}\,\text{day}^{-1}\n\]

Now

\[\n\frac{S̅}{S_{\max}} = \frac{7.2}{12.44}=0.579\n\]

\[\nH_g = H_0\bigl(a + b\frac{S̅}{S_{\max}}\bigr)\n = 37\,957\,(0.31 + 0.43\times0.579)\n = 21\,213\ \text{kJ m}^{-2}\,\text{day}^{-1}\n\]

Alternative a‑b values (Gopinathan, 1995)

Using elevation (E_L = 0.014) km, the fitted constants become

• a ≈ 0.3666
• b ≈ 0.3511

which yields

\[\nH_g = 21\,627\ \text{kJ m}^{-2}\,\text{day}^{-1}\n\]

Both results are close; either set may be used when the required parameters are known.

7. Monthly Average Daily Diffuse Radiation (H̅_d)

Three empirical correlations are presented.

1. Modi et al.

\[\nH_d = H_g\,(1.411 - 1.696\,\frac{H_g}{H_0})\n\]

Result: 9 825 kJ m⁻² day⁻¹

1. Garg & Garg (1988)

\[\nH_d = H_g\,(0.8677 - 0.7365\,\frac{S̅}{S_{\max}})\n\]

Result: 9 364 kJ m⁻² day⁻¹

1. Gopinathan & Soler (1995)

\[\nH_d = H_g\bigl[0.87813 - 0.33280\,\frac{H_g}{H_0}\n - 0.53039\,\frac{S̅}{S_{\max}}\bigr]\n\]

Result: 8 171 kJ m⁻² day⁻¹

The first two correlations give similar values (≈9.5 kJ m⁻² day⁻¹) and are preferred for further calculations.

8. Monthly Average Hourly Global Radiation (I̅_g)

Hourly radiation is expressed in kJ m⁻² h⁻¹. The steps are:

1. Select the representative hour – for the 9–10 h interval the midpoint is 9.30 h, giving hour angle

\[\n\omega = (12 - 9.30)\times15^\circ = 37.5^\circ\n\]

1. Extraterrestrial hourly radiation (I_0) (using the same April 15 data)

\[\nI_0 = 3\,871\ \text{kJ m}^{-2}\,\text{h}^{-1}\n\]

1. Determine coefficients a, b (Collares‑Pereira & Rabi, 1992)

\[\na = 0.409 + 0.5016\sin(\omega_s-60^\circ) = 0.6845\n\] \[\nb = 0.6609 - 0.4767\sin(\omega_s-60^\circ) = 0.3990\n\]

1. Correction factor f_c (Gueymard, 1986)

\[\nf_c = 1 - \frac{\sin\omega_s - \omega_s\cos\omega_s}\n {\sin\omega_s - \omega_s\cos\omega_s}\n \approx 0.9924\n\]

1. Compute I̅_g

\[\nI_g = I_0\Bigl[a + b\cos\omega\Bigr]\,f_c\n = 3\,871\,(0.6845 + 0.3990\cos37.5^\circ)\times0.9924\n \approx 2\,182\ \text{kJ m}^{-2}\,\text{h}^{-1}\n\]

This value represents the monthly average hourly global radiation for the chosen hour interval.

9. Summary of Results

10. References Cited in the Lecture

• Sukhatme & Nayak – fundamental solar‑geometry formulas.
• Modi, G., et al. (1979) – a, b constants for global radiation correlation.
• Klein (1997) – recommendation that (H_0) for a representative day equals the monthly average ( \overline{H}_0).
• Gopinathan & Soler (1995) – alternative a, b formulations incorporating elevation.
• Collares‑Pereira & Rabi (1992) – hourly global radiation correlation.
• Gueymard (1986) – correction factor (f_c) for hourly radiation.

End of Lecture 3 summary.

The lecture demonstrates how solar‑geometry fundamentals are applied to determine the angle of incidence for a tilted collector and to convert standard time to local apparent time. It distinguishes between sunrise and sunset calculations for horizontal versus tilted surfaces, highlighting seasonal differences in the appropriate formula. Empirical correlations are used to estimate monthly average global and diffuse radiation on both horizontal and inclined planes, with alternative coefficient sets reflecting site elevation. Finally, the method extends to hourly global radiation estimation by selecting a representative hour and applying correction factors, providing a comprehensive workflow for renewable energy engineering calculations.