HYDRO ELECTRIC POWER PLANT Part II AGUS HARYANTO

HYDRO ELECTRIC POWER PLANT Part II AGUS HARYANTO

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.

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

Energy Change, Esystem Energy change = Energy at final state – Energy at initial state Esystem = Efinal – Einitial = E 2 – E 1 E = U + KE + PE U = U 2 – U 1 = m(u 2 – u 1) KE = KE 2 – KE 1 = 0. 5 m(V 22 – V 12) PE = PE 2 – PE 1 = mg(z 2 – z 1) For stationary systems, KE = 0 and PE = 0), and E = U

Energy Balance: Clossed System For a closed system undergoing a cycle, the initial and final states are identical, thus Esystem = E 2 – E 1 = 0. The energy balance simplifies to: Ein – Eout = 0 or Ein = Eout 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

EXAMPLE 8: 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 k. J. During the cooling process, the fluid loses 500 k. J of heat, and the paddle wheel does 100 k. J of work on the fluid. Determine the final internal energy of the fluid. Neglect the energy stored in the paddle wheel.

Example 8: 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.

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

EXAMPLE 8: 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.

EXAMPLE 8: Solution

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

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.

Pump = useful pumping = power = = Power rating x

Fan

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

Motor and Generator

Energi dan Lingkungan 1. Polusi Merusak lingkunan dan kesehatan

Energi dan Lingkungan Smog (asap kota metropolitan) dengan ciri dark yellow or brown haze in stagnant air mass and hangs over on calm hot summer day. Komponen Smog: Ground Ozone (O 3) : menyebabkan iritasi mata, merusak paru-paru, dan merusak jaringan daun tanaman. CO : racun mematikan VOCs (Benzene, butane, …) Smog dapat dibawa angin melintasi perbatasan persoalan global. 2.

Energi dan Lingkungan 3. Acid Rain Fossil fuel contain sulfur SOx and NOx + Water + sunlight Sulfuric Acid + Nitric Acid The acid washed out by rain water Acid Rain

Energi dan Lingkungan 4. GHG: CO 2, CH 4, H 2 O, NOx Global Warming

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