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HYDRO ELECTRIC POWER PLANT

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HYDRO ELECTRIC POWER PLANT. AGUS HARYANTO. Part II. Conservation of Energy. - PowerPoint PPT Presentation
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AGUS HARYANTO Part II
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Page 1: HYDRO ELECTRIC POWER PLANT

AGUS HARYANTO

Part II

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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.

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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

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Energy change = Energy at final state – Energy at initial stateEsystem = Efinal – Einitial = E2 – E1E = U + KE + PEU = U2 – U1 = m(u2 – u1)KE = KE2 – KE1 = 0.5 m(V2

2 – V12)

PE = PE2 – PE1 = mg(z2 – z1)

For stationary systems, KE = 0 and PE = 0), and E = U

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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

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

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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.

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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.

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Applying the energy balance on the system gives the final internal energy of the system is 400 kJ:

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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.

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

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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.

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= useful pumping power

= = Power rating x

= )( 12 PPolV

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inshaft

fluidmechfanmech W

E

,

,,

)(2

1 22, VmE fluidmech

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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

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outshaftinelgeneration WWQ ,,

motor

outshaftinel

WW

,

,

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1. Polusi Merusak lingkunan

dan kesehatan

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2. 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 (O3) : 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.

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3. 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

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4. GHG: CO2, CH4, H2O, NOx

Global Warming

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