HYDRO ELECTRIC POWER PLANT

1. HYDRO ELECTRIC POWER PLANT AGUS HARYANTO Part II

2. Conservation of Energy The conservation of energy principle can be expressed as follows: The net change (increase or decrease) in the total energy of the system during a process is equal to the difference between the total energy entering and the total energy leaving the system during that process.

3. Energy Balance Esystem = Ein - Eout = Qin - Qout + Win - Wout + Emass,in - Emass,out Noting that energy can be transferred in the forms of heat, work, and mass, the energy balance can be written more explicitly as:

4. Energy Change, Esystem Energy change = Energy at final state – Energy at initial state Esystem = Efinal– Einitial = E2– E1 E = U + KE + PE U = U2 – U1 = m(u2 – u1) KE = KE2 – KE1 = 0.5 m(V22 – V12) PE = PE2 – PE1 = mg(z2 – z1) For stationary systems, KE = 0 and PE = 0), and E = U

5. Energy Balance: Clossed System • Noting that a closed system does not involve any mass flow across its boundaries, the energy balance for a cycle can be expressed in terms of heat and work interactions as: Wnet,out = Qnet,in For a closed system undergoing a cycle, the initial and final states are identical, thus Esystem = E2 – E1 = 0. The energy balance simplifies to: Ein – Eout = 0 or Ein = Eout

6. EXAMPLE8: Hot Fluid Cooling A rigid tank contains a hot fluid that is cooled while being stirred by a paddle wheel. Initially, the internal energy of the fluid is 800 kJ. During the cooling process, the fluid loses 500 kJ of heat, and the paddle wheel does 100 kJ of work on the fluid. Determine the final internal energy of the fluid. Neglect the energy stored in the paddle wheel.

7. Example8: Solution Assumptions : The tank is stationary and thus the kinetic and potential energy changes are zero, KE = PE = 0. Therefore, E = U. Energy stored in the paddle wheel is negligible.

8. Example8: Solution (cont’d) Applying the energy balance on the system gives the final internal energy of the system is 400 kJ:

9. EXAMPLE8: Air Acceleration by Fan A fan that consumes 20 W of electric power when operating is claimed to discharge air from a ventilated room at a rate of 0.25 kg/s at a discharge velocity of 8 m/s (Fig. 2–48). Determine if this claim is reasonable.

10. EXAMPLE8: Solution

11. EFISIENSI KONVERSI ENERGI Performance = efficiency, is expressed in desired output by the required input

12. Efficiencies of Mechanical and Electrical Devices A pump or a fan receives shaft work (from an electric motor) and transfers it to the fluid as mechanical energy (less frictional losses). A turbine, converts the mechanical energy of a fluid to shaft work.

13. Pump = = = Power rating x  = useful pumping power

14. Fan

15. Turbine is the rate of decrease in the mechanical energy of the fluid, which is equivalent to the mechanical power extracted from the fluid by the turbine

16. Motor and Generator

18. EnergidanLingkungan Polusi Merusak lingkunan dan kesehatan

19. EnergidanLingkungan Smog (asapkota metropolitan) denganciri dark yellow or brown haze in stagnant air mass and hangs over on calm hot summer day. Komponen Smog: Ground Ozone (O3) : menyebabkaniritasimata, merusakparu-paru, danmerusakjaringandauntanaman. CO : racunmematikan VOCs (Benzene, butane, …) Smog dapatdibawaanginmelintasiperbatasan persoalan global.

20. EnergidanLingkungan Acid Rain Fossil fuel contain sulfur  SOx SOx and NOx + Water + sunlight  Sulfuric Acid + Nitric Acid The acid washed out by rain water  Acid Rain

21. EnergidanLingkungan GHG: CO2, CH4, H2O, NOx  Global Warming

22. PR