Solutions to Chapter 10: Hint: In the next exercises only estimations are done. There are different ways to the solution therefore the presented solutions are to be considered only as example solutions. However the alternative solutions should yield similar results. Exercise 10.1: Potential Estimation for Pitched Roofs a) From Figure 2.7: The Saarland is situated around Saarbrücken in South-West Germany: H ′ ≈ 1075 kWh m2 ⋅ a Theoretical Potential: WTheo = A ⋅ H ′ = 2750 km 2 ⋅ 1075 kWh kWh = 2.956 ⋅ 1012 ≈ 3 ⋅ 1012 kWh/a 2 a m ⋅a b) Roof surfaces: APitch ≈ 0.3 % ⋅ 2750 km 2 = 7.71 km 2 Optical energy on pitched surface (incl. inclination losses of 15 %, see Table 10.1): WOptical_Pitch ≈ 1200 c) kWh H′ kWh 1075 ⋅ 0.85 ⋅ ⋅ APitch ≈ 1200 2 ⋅ 0.85 ⋅ ⋅ 7.71 km 2 ≈ 8.45 TWh/a 2 H 1000 m ⋅a m ⋅a W ⋅ 0.2 = 1.54 GWp m2 kWh ⋅ 0.18 = 1.52 TWh/a = 8.45 ⋅ 109 a Power: PSTC = APitch ⋅ ESTC ⋅ η Module = 7.71 km 2 ⋅ 1000 Electrical Energy: WElec_Pitch = WOptical_Pitch ⋅ ηSystem K. Mertens Textbook PV Solutions to Exercises Exercise 10.2: Potential Estimation for Free Areas a) From Figure 2.7: The Saarland (around Saarbrücken) has a latitude of about 49 °, Smallest noonday solar altitude in winter with Equation (2.6): γ S_Min = 66.6 ° − ϕ = 66.6 ° − 49 ° = 17.6 ° Area utilization factor with Equations (9.4) and (9.5): f Util = sin(γ S ) b sin(17.6 °) = = = 49.6 % ≈ 50 % d Min sin(γ S + β ) sin(17.6 ° + 20 °) b) Module area: AModule = f Util ⋅ AFree_Area = 0.5 ⋅ 10 000 m 2 = 5000 m 2 PV-Power: PSTC = AModule ⋅ ESTC ⋅ η Modul e = 5000 m 2 ⋅ 1000 W ⋅ 0.2 = 1 MWp m2 Electrical Energy with Equation (10.4): WElectrical_PV = AModule ⋅ f Util ⋅ η Module ⋅ ESTC ⋅ ⇒ WElectrical_PV = 10 000 m 2 ⋅ 0.5 ⋅ 0.2 ⋅ 1000 H′ ⋅ YF H kWh W 1075 = 967 500 kWh /a ⋅ ⋅ 900 2 kWp ⋅ a m 1000 Alternative way of solution: Electrical Energy: WElectrical = WOptical ⋅ηSystem ⇒ WElectrical = 1200 c) kWh H ′ kWh 1075 ⋅ ⋅ AModule ⋅ηSystem ≈ 1200 2 ⋅ ⋅ 5000 m 2 ⋅ 0.18 ≈ 1.2 Mio. kWh/a 2 m ⋅a H m ⋅ a 1000 According to Section 10.1.4 energy corn yields an electrical energy of about 17 000 kWh/a. Thus photovoltaics on the same area yields about the 57- to 70-fold of electrical energy! K. Mertens Textbook PV Solutions to Exercises
