- The overall efficiency of a hydroelectric power plant depends on the efficiency of the turbine (ηt) used to convert potential energy to mechanical energy, and the generator efficiency (ηe) in converting mechanical energy to electrical energy. - The electrical power generated (Pe) can be calculated as ρghQηoverall, where ρ is the density of water, g is gravity, h is the head, Q is discharge, and ηoverall is the product of turbine and generator efficiencies. - Key factors in siting a hydroelectric plant include available hydro potential (ρghV), reliability of water flow over time, and cost considerations regarding plant size and choice of generating units.

1. EE-417
POWER GENERATION
LECTURE # 10

15. THE OVERALL EFFICIENCY OF HYDROELECTRIC POWER PLANT

16. REVISION OF POWER/ENERGY FROM A DAM
• Hydro (Potential) Energy available in a Dam 𝑃. 𝐸ℎ= 𝑚𝑔ℎ = 𝛒Vgh ;
where 𝛒 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = 1000 (𝐾𝑔
𝑚3), V= 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑅𝑒𝑠𝑒𝑟𝑣𝑖𝑟𝑒 (𝑚3
), h= 𝐻𝑒𝑎𝑑 𝑜𝑓 𝐷𝑎𝑚 (m)
• Hydro-Power Available in a Dam (Joul or N-m)………… 𝑃ℎ =
𝛒Vgh
𝑡
= 𝛒gh∗Q
Where 𝑄 = 𝑉𝑜𝑙𝑢𝑚𝑒
𝑡 = 𝐷𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒 𝑜𝑓 𝑊𝑎𝑡𝑒𝑟 𝑓𝑟𝑜𝑚 𝑎 𝐷𝑎𝑚 (unit; 𝑚3
𝑠)
• Mechanical / Turbine Power available / Produced From a Hydroelectric Station (in Watts) ;……….
• 𝑃𝑚 = 𝛒gh∗Q∗𝜼𝑡 = 1000 ∗ 9.8h∗Q∗𝜼𝑡
Where 𝜼𝑡 = 𝑇𝑢𝑟𝑏𝑖𝑛𝑒 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦

17. .
• Mechanical/Turbine Power available / Produced From a Hydroelectric Station (in metric-
horsepower);
𝑃𝑚 =
𝛒hQ∗𝜼𝑡
75
Where 𝜼𝑡 = 𝑇𝑢𝑟𝑏𝑖𝑛𝑒 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦
• Electrical Power Generated From a Hydroelectric Station (in KW);
𝑃𝑒 =
736 ∗ 𝛒hQ∗𝜼𝑡 ∗ 𝜼𝑒
75 ∗ 1000
Where 𝜼𝑒 = 𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙/𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 of Generating units of Power House
𝜼𝑡 = 𝑇𝑢𝑟𝑏𝑖𝑛𝑒 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦
One metric horsepower as the power to raise a mass of 75 kilograms against the Earth's gravitational force over a distance of one metre in on1e second:[19] 75 kg × 9.80665 m/s2 × 1 m / 1 s = 735.49875 W
1 metric horsepower = 75 Kg*m/s = 736 W
1 metric horsepower = 0.736 KW
1 mechanical horsepower = 746 W
Note: In this formula for calculating the
power in metric horsepower there is no
“g” Note: In this formula
for calculating the
power in Watts there
is no “g” as
736
75
=
9.81 = 𝑔 hence
indirectly “g” is there

18. • Electrical Energy Generated From a Hydroelectric Station (in KWh);
𝐸𝑒 =
𝛒ghV∗𝜼𝑜𝑣𝑒𝑟𝑎𝑙𝑙
1000 ∗ 3600
Where V= 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑅𝑒𝑠𝑒𝑟𝑣𝑖𝑜𝑟
• Electrical Power Generated From a Hydroelectric Station (in KW);
𝑃𝑒 = 𝛒ghQ∗𝜼𝑜𝑣𝑒𝑟𝑎𝑙𝑙
Where ;𝜼𝑜𝑣𝑒𝑟𝑎𝑙𝑙 = 𝜼𝑡 ∗ 𝜼𝑒

19. PRACTICE PROBLEMS

25. CHOICE OF THE SITE FOR HYDROELECYTIC POWER STATION

28. HYDROGRAPH EXAMPLE
• .

29. FLOW DURATION CURVE

34. POWER DURATION CURVE

35. SIZE OF PLANT & CHOICE OF UNITS

38. CHOICE OF UNITS (EXAMPLE 8.1)

39. COST OF HYDROELECTRIC POWER PLANTS

45. PRACTICE PROBLM (TRY YOURSELF)