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HYDRO ELECTRIC POWER PLANT

HYDRO ELECTRIC POWER PLANT. AGUS HARYANTO. OBJECTIVES:. Introduce concept of energy and its various forms. Discuss the nature of internal energy. Define concept of heat and terminology associated Define concept of work and forms of mechanical work. Define energy conversion efficiencies.

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HYDRO ELECTRIC POWER PLANT

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  1. HYDRO ELECTRIC POWER PLANT AGUS HARYANTO

  2. OBJECTIVES: Introduce concept of energy and its various forms. Discuss the nature of internal energy. Define concept of heat and terminology associated Define concept of work and forms of mechanical work. Define energy conversion efficiencies. Discuss relation of energy conversion and environment.

  3. Recall: ENERGY SISTEM TERMODINAMIKA BENTUK ENERGI: 1. Energi Kinetik (KE)  2. Energi Potensial (PE)  PE = mgh 3. Energi dakhil atau Internal Energy (U) ENERGI TOTAL: E = U + KE + PE e = u + ke + pe (per satuan massa)

  4. Thermodynamics Concern Thermodynamics deals only with the change of the total energy (E). Thus E of a system can be assigned to zero (E = 0) at some reference point. Thechange in total energy (E) of a system is independent of the reference point selected. For stationary systems, the E is identical to the change of internal energy U.

  5. Macroscopic vs. Microscopic Energy The macroscopicforms of energy are those a system possesses as a whole with respect to some outside reference frame, such as kinetic and potential energies. The microscopicforms of energy are those related to the molecular structure of a system and the degree of the molecular activity, and they are independent of outside reference frames. The sum of all the microscopic forms of energy is called the internal energy of a system and is denoted byU.

  6. More on Internal Energy SENSIBLE and LATENT energy CHEMICAL energy NUCLEAR energy The internal energy of a system is the sum of all forms of the microscopic energies.

  7. More on Internal Energy: Sensible Energy The various forms of microscopic energies that make up sensible energy The portion of the internal energy of a system associated with the kinetic energies of the molecules is called the sensible energy

  8. More on Internal Energy: Latent Energy The internal energy associated with the phase of a system is called the latent energy. The amount of energy absorbed or released during a phase-change process is called the latent heat coefficient. At 1 atm, the latent heat coefficientof water vaporization is 2256.5 kJ/kg.

  9. More on Internal Energy: Chemical and Nuclear Energy The internal energy associated with the atomic bonds in a molecule is called chemical energy. The tremendous amount of energy associated with the strong bonds within the nucleus of the atom itself is called nuclear energy.

  10. Energy Transfer: Heat vs. Works *) 1,2 for Clossed System; 1,2,3 for Open System Energy crosses the boundaries in the form of: Heat Work Mass flow*

  11. HEAT (Q) Heat : the form of energy that is transferred between two systems (or a system and its surroundings) by virtue of a temperature difference

  12. WORK (W) Mechanics: work is the energy transfer associated with a force acting through a distance (W = F.s). Thermodynamics: work is an energy interaction that is not caused by a temperature difference between a system and its surroundings.

  13. Sign Convention + - - + Qin = + (positive) Qout = - (negative) Win = - (negative) Wout = + (positive)

  14. Note on HEAT and WORK Both are recognized at the boundaries of a system as they cross the boundaries. That is, both heat and work are boundary phenomena. Systems possess energy, but not heat or work. Both are associated with a process, not a state. Unlike properties, heat or work has no meaning at a state. Both are path functions (i.e., their magnitudes depend on the path followed during a process as well as the end states), and not pointfunctions.

  15. Path vs. Point Functions Path functions have inexact differentials designated by (Q or W)NOT dQ or dW. Properties are point functions (i.e., they depend on the state only, and not on how a system reaches that state), and they have exact differentials designated by d. A small change in volume, for example, is represented by dV.

  16. Path vs. Point Functions Properties are point functions Heat and Work are path functions

  17. Example1 A candle is burning in a well-insulated room. Taking the room (the air plus the candle) as the system, determine (a) if there is any heat transfer during this burning process and (b) if there is any change in the internal energy of the system.

  18. Example1: Solution (a) The interior surfaces of the room form the system boundary. As pointed out earlier, heat is recognized as it crosses the boundaries. Since the room is well insulated, we have an adiabatic system and no heat will pass through the boundaries. Therefore, Q = 0 for this process. (b) The internal energy involves energies that exist in various forms. During the process just described, part of the chemical energy is converted to sensible energy. Since there is no increase or decrease in the total internal energy of the system, U = 0 for this process.

  19. Example2 A potato initially at room temperature (25°C) is being baked in an oven that is maintained at 200°C, as shown in Fig. 2–21. Is there any heat transfer during this baking process?

  20. Example2: Solution This is not a well-defined problem since the system is not specified. Let us assume that we are observing the potato, which will be our system. Then the skin of the potato can be viewed as the system boundary. Part of the energy in the oven will pass through the skin to the potato. Since the driving force for this energy transfer is a temperature difference, this is a heat transfer process. Note: if the system is the oven, then Q = 0

  21. Example2 Electrical Work: Wel = V.I.t = I.R.I.t A well-insulated electric oven is being heated through its heating element. If the entire oven, including the heating element, is taken to be the system, determine whether this is a heat or work interaction. How if the system is taken as only the air in the oven without the heating element.

  22. Example3: Solution 1st Case The energy content of the oven obviously increases during this process, as evidenced by a rise in temperature. This energy transfer to the oven is not caused by a temperature difference between the oven and the surrounding air. Instead, it is caused by electrons crossing the system boundary and thus doing work. Therefore, this is a work interaction.

  23. Example3: Solution 2nd Case This time, no electrons will be crossing the system boundary at any point. Instead, the energy generated in the interior of the heating element will be transferred to the air around it as a result of the temperature difference between the heating element and the air in the oven. Therefore, this is a heat transfer process.

  24. MECHANICAL FORMS OF WORK Kinetical Work Wk = F.s Wb = P.A.ds = P.dV

  25. Example: Wb = 0 karena dV = 0 Sebuah tangki kokoh berisi udara pada 500 kPa dan 150oC. Akibat pertukaran panas dengan lingkungannya, suhu dan tekanan di dalam tangki berturut-turut turun menjadi 65oC dan 400 kPa. Tentukan kerja lapisan batas selama proses ini.

  26. Shaft Work Shaft Work Wsh = 2..n.  = torsi = F.r DayaPoros:

  27. Example4 Determine the power transmitted through the shaft of a car when the torque applied is 200 N.m and the shaft rotates at a rate of 4000 revolutions per minute (rpm).

  28. Example4: Solution The shaft power is determined directly from: = = 83.8 kW (112 HP)

  29. Spring Work Spring Work Wsp = 0.5 k (x12 – x22) k = spring constant (kN/m) F = kx

  30. Work by Elastic Bars n = normal stress n = F/A

  31. Acceleration & Grafitational Work Wa = 0.5 m.(V22-V12) Wg = m.g.z = m.g. h

  32. Example5 Consider a 1200-kg car cruising steadily on a level road at 90 km/h. Now the car starts climbing a hill that is sloped 30° from the horizontal (Fig. 2–35). If the velocity of the car is to remain constant during climbing, determine the additional power that must be delivered by the engine.

  33. Example5: Solution The additional power required is simply the work that needs to be done per unit time to raise the elevation of the car, which is equal to the change in the potential energy of the car per unit time:

  34. Example6 Determine the power required to accelerate a 900-kg car shown in Fig. 2–36 from rest to a velocity of 80 km/h in 20 s on a level road.

  35. Example6: Solution

  36. 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.

  37. 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:

  38. 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

  39. 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

  40. 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.

  41. 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.

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

  43. 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.

  44. EXAMPLE8: Solution

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

  46. 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.

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

  48. Fan

  49. 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

  50. Motor and Generator

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