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Some revision questions for midterm exam

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- 1. Mid-Term Review Topics 1) Energy Analysis of Closed System (Lecture 4, Chap. 4 of textbook). 2) Mass and Energy Analysis Of Control Volumes (Lecture 5, Chap. 5 of textbook). 3) Second Law of Thermodynamics (Lecture 6, Chap. 6 of textbook). 1
- 2. 2 1) Energy Analysis of Closed System Know how to interpret condition at final or initial state, given the nature of the process; e.g. constant pressure process, isothermal process, process inside a closed rigid tank, process inside a piston-cylinder device. Know how to obtain boundary work wb for the constant pressure process, the isothermal process and the polytropic process. Know and understand the various forms of the energy balance equation for a closed system.
- 3. 3 Examples (Energy Analysis of Closed System) 1. An insulated piston–cylinder device contains 5 L of saturated liquid water at a constant pressure of 175 kPa. Water is stirred by a paddle wheel while a current of 8 A flows for 45 min through a resistor placed in the water. If one-half of the liquid is evaporated during this constant pressure process and the paddle-wheel work amounts to 400 kJ, determine the voltage of the source. Also, show the process on a P-v diagram.
- 4. 4 Solution We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as since ΔU + Wb = ΔH during a constant pressure quasi- equilibrium process.
- 5. 5 The properties of water are Substituting,
- 6. 6 2. A mass of 5 kg of saturated liquid–vapor mixture of water is contained in a piston–cylinder device at 150 kPa. Initially, half of the water is in the liquid phase and the rest is in the vapor phase. Heat is now transferred to the water, and the piston, which is resting on a set of stops, starts moving when the pressure inside reaches 300 kPa. Heat transfer continues until the total volume increases by 20 percent. Determine (a) the initial and final temperatures, (b) the mass of liquid water when the piston first starts moving, and (c) the work done during this process. Also, show the process on a P-v diagram.
- 7. 7 Solution Initially the system is a saturated mixture at 150 kPa pressure, and thus the initial temperature is Ti = Tsat@150 kPa = 111.35oC The specific volumes at 150 kPa are vf = 0.001053 m3/kg and vg = 1.1594 m3/kg; thus total initial volume is V1 = mf vf + mg vg = (2.5 kg)( 0.001053 m3/kg) + (2.5 kg)(1.1594 m3/kg) = 2.901 m3 Then the total and specific volumes at the final state are V3 = 1.2 V1 = 1.2 2.901 = 3.481 m3 v3 = V3/m = 3.481 m3 / 5 kg = 0.69627 m3/k Thus at the final state, P3 = 300 kPa, v3 = 0.69627 m3/kg.
- 8. 8 At 300 kPa, vg = 0.60582 m3/kg. Hence since v3 > vg@300 KPa, state 3 is superheated vapour. Interpolating from Table A-6 (b) When the piston first starts moving, P2 = 300 kPa and V2 = V1 = 2.901 m3. The specific volume at this state is V2 = V2/m = 2.901 m3 / 5 kg = 0.5802 m3/kg which is less than vg = 0.60582 m3/kg at 300 kPa. Thus the quality at state 2 is Thus mass of the liquid at state 2 = (5 kg)(1 0.9576) = 0.212 kg. 3 3 150 0.69627 0.63402 0.69627 0.63402 150 200 150 187.8 C 200 150 0.71643 0.63402 0.71643 0.63402 T T 2 @300 kPa 2 @300 kPa 0.5802 0.001073 0.9576 0.60582 0.001073 f fg v v x v
- 9. 9 (c) No work is done during process 1-2 since V1 = V2. The pressure remains constant during process 2-3 and the work done during this process is 3 3 2 3 2 300 kPa 3.481 2.901 m 206.4 kPa m 174 kJbW P V V
- 10. 10 Example Steam enters a turbine operating at steady state with a mass flow rate of 4600 kg/h. The turbine develops a power output of 1000 kW. At the inlet, the pressure is 60 bar, the temperature is 400C, and the velocity is 10 m/s. At the exit, the pressure is 0.1 bar, the quality is 0.9 (90%), and the velocity is 50 m/s. Calculate the rate of heat transfer between the turbine and surroundings, in kW.
- 11. 11 (assuming net heat transfer out of the system and pe = 0.)
- 12. 12 From the tables and using the usual procedures we obtain h1 = 3178.3 kJ/kg and h2 = 2344.7 kJ/kg. Also Substituting, Thus heat is transferred out of the system at a rate of 63.6 kW.
- 13. 13 Expansion or compression process where pressure is not constant - Examples
- 14. 14 FIGURE 4–10 Schematic and P-V diagram for Example 4–4.
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- 22. 22 A water heater in steady operation. Mass balance Energy balance 2) Mass and Energy Analysis Of Control Volumes
- 23. 23 Turbines and Compressors Turbine drives the electric generator In steam, gas, or hydroelectric power plants. As the fluid passes through the turbine, work is done against the blades, which are attached to the shaft. As a result, the shaft rotates, and the turbine produces work. Compressors, as well as pumps and fans, are devices used to increase the pressure of a fluid. Work is supplied to these devices from an external source through a rotating shaft. A fan increases the pressure of a gas slightly and is mainly used to mobilize a gas. A compressor is capable of compressing the gas to very high pressures. Pumps work very much like compressors except that they handle liquids instead of gases. Energy balance for the compressor in this figure:
- 24. 24 Example 1 The power output of an adiabatic steam turbine is 5 MW, and the inlet and the exit conditions of the steam are as indicated in the diagram. (a) Compare the magnitudes of h, ke, and pe. (b) Determine the work done per unit mass of the steam flowing through the turbine. (c) Calculate the mass flow rate of the steam.
- 25. 25 Solution We take the turbine as the system. This is a control volume since mass crosses the system boundary during the process. We observe that there is only one inlet and one exit and thus (a) At the inlet, steam is in a superheated vapor state, and its enthalpy is At the turbine exit, we have a saturated liquid–vapor mixture at 15 kPa. The enthalpy at this state is .21 mm
- 26. 26 Then
- 27. 27 Single inlet and outlet, subs. 1 inlet, 2 outlet
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- 29. 29 A steam turbine has an inlet of 2 kg/s steam at 1000 kPa, 350 C and velocity of 15 m/s. The exit is at 100 kPa, x = 1 and very low velocity. Determine the power output of the turbine in kW. Solution Properties of water Example 2
- 30. 30 The energy balance equation for the steady flow turbine is Neglecting change in PE and given that V2 0, we then have Or Substituting values, we have
- 31. 31 At 2 kg/s steam flow rate, the power output
- 32. 32 A well-insulated turbine operating at steady conditions develops 23 MW of power for a steam flow rate of 40 kg/s. The steam enters at 360°C with a velocity of 35 m /s and exits as saturated vapor at 0.06 bar with a velocity of 120 m /s. Neglecting potential energy effects, determine the inlet pressure, in bar. Solution Properties of water Example 3
- 33. 33 Neglecting change in PE and heat transfer (as turbine is well insulated), the energy balance equation for the steady flow process on a unit mass basis is, Thus Substituting values, = 3147.6 kJ/kg
- 34. 34 Hence Checking Table A-4, we find hg@360 C = 2481.6 kJ/kg. Since h1 is higher, we have superheated vapour state at 1. (Note: Steam at turbine inlet is always superheated.) Interpolating from Table A-6 (see below), we find P1 = 2.6 MPa or 26 bar.
- 35. 35 Refrigerant-134a enters an adiabatic compressor as saturated vapor at 24°C and leaves at 0.8 MPa and 60°C. The mass flow rate of the refrigerant is 1.2 kg/s. Determine (a) the power input to the compressor and (b) the volume flow rate of the refrigerant at the compressor inlet. Solution From the refrigerant tables Example 4
- 36. 36 The energy balance for this steady-flow system can be expressed in the rate form as
- 37. 37 Air enters the compressor of a gas-turbine plant at ambient conditions of 100 kPa and 25 C with a low velocity and exits at 1 MPa and 347 C with a velocity of 90 m/s. The compressor is cooled at a rate of 1500 kJ/min, and the power input to the compressor is 250 kW. Determine the mass flow rate of air through the compressor. Example 5
- 38. 38 Solution The energy balance for this steady-flow system can be expressed in the rate form as Substituting, the mass flow rate is determined to be
- 39. 39 Throttling valves Throttling valves are any kind of flow-restricting devices that cause a significant pressure drop in the fluid. What is the difference between a turbine and a throttling valve? The pressure drop in the fluid is often accompanied by a large drop in temperature, and for that reason throttling devices are commonly used in refrigeration and air-conditioning applications. The temperature of an ideal gas does not change during a throttling (h = constant) process since h = h(T). During a throttling process, the enthalpy of a fluid remains constant. But internal and flow energies may be converted to each other. Energy balance
- 40. 40 Example – Expansion of R-134a in a Refrigerator Refrigerant-134a enters the capillary tube of a refrigerator as saturated liquid at 0.8 MPa and is throttled to a pressure of 0.12 MPa. Determine the quality of the refrigerant at the final state and the temperature drop during this process.
- 41. 41 Solution A capillary tube is a simple flow-restricting device that is commonly used in refrigeration applications to cause a large pressure drop in the refrigerant. Flow through a capillary tube is a throttling process; thus, the enthalpy of the refrigerant remains constant. Since hf < h2 < hg, the refrigerant exists as a saturated mixture at the exit. The quality of this mixture is
- 42. 42 Since the exit state is a saturated mixture at 0.12 MPa, the exit temperature must be the saturation temperature at this pressure, which is 22.32°C. Then the temperature change for this process is
- 43. Part of the heat received by a heat engine is converted to work, while the rest is rejected to a sink. The devices that convert heat to work. 1. They receive heat from a high- temperature source (solar energy, oil furnace, nuclear reactor, etc.). 2. They convert part of this heat to work (usually in the form of a rotating shaft.) 3. They reject the remaining waste heat to a low-temperature sink (the atmosphere, rivers, etc.). 4. They operate on a cycle. Heat engines and other cyclic devices usually involve a fluid to and from which heat is transferred while undergoing a cycle. This fluid is called the working fluid. Work can be converted to heat directly and completely, but converting heat to work requires the use of heat engines. 43 3) Second Law of Thermodynamics Heat Engines
- 44. 44 Thermal efficiency Some heat engines perform better than others (convert more of the heat they receive to work). Schematic of a heat engine. Even the most efficient heat engines reject almost one-half of the energy they receive as waste heat.
- 45. 45 Example 1 Heat is transferred to a heat engine from a furnace at a rate of 80 MW. If the rate of waste heat rejection to a nearby river is 50 MW, determine the net power output and the thermal efficiency for this heat engine.
- 46. 46 Solution A schematic of the heat engine is shown in the previous slide. The furnace serves as the high-temperature reservoir for this heat engine and the river as the low-temperature reservoir. The given quantities can be expressed as
- 47. 47 Example 2 A car engine produces 136 hp on the output shaft. The thermal efficiency of the engine is 30%. Find the rate of heat rejected to the ambient and the rate of fuel consumption if the engine uses a fuel with a heating value of 35,000 kJ/kg. Note: 1 hp = 745.7 W = 0.7457 kW
- 49. 49 An automobile engine consumes fuel at a rate of 28 L/h and delivers 60 kW of power to the wheels. If the fuel has a heating value of 44,000 kJ/kg and a density of 0.8 kg/L, determine the thermal efficiency of this engine. Example 3
- 50. 50 Solution The mass consumption rate of the fuel is The rate of heat supply to the car is The thermal efficiency of the car is then
- 51. 51 REFRIGERATORS AND HEAT PUMPS • The transfer of heat from a low- temperature medium to a high- temperature one requires special devices called refrigerators. • Refrigerators, like heat engines, are cyclic devices. • The working fluid used in the refrigeration cycle is called a refrigerant. • The most frequently used refrigeration cycle is the vapor- compression refrigeration cycle. Basic components of a refrigeration system and typical operating conditions. In a household refrigerator, the freezer compartment where heat is absorbed by the refrigerant serves as the evaporator, and the coils usually behind the refrigerator where heat is dissipated to the kitchen air serve as the condenser.
- 52. 52 Coefficient of Performance The objective of a refrigerator is to remove QL from the cooled space. The efficiency of a refrigerator is expressed in terms of the coefficient of performance (COP). The objective of a refrigerator is to remove heat (QL) from the refrigerated space.
- 53. 53 Heat Pumps The objective of a heat pump is to supply heat QH into the warmer space. The work supplied to a heat pump is used to extract energy from the cold outdoors and carry it into the warm indoors. for fixed values of QL and QH
- 54. 54 The food compartment of a refrigerator, shown at right, is maintained at 4°C by removing heat from it at a rate of 360 kJ/min. If the required power input to the refrigerator is 2 kW, determine (a) the COP of the refrigerator and (b) the rate of heat rejection to the room that houses the refrigerator. Example 1
- 55. 55 Solution (a) The coefficient of performance of the refrigerator is That is, 3 kJ of heat is removed from the refrigerated space for each kJ of work supplied. (b) The rate at which heat is rejected to the room that houses the refrigerator is determined from the conservation of energy relation:
- 56. 56 A heat pump is used to meet the heating requirements of a house and maintain it at 20°C. On a day when the outdoor air temperature drops to 2°C, the house is estimated to lose heat at a rate of 80,000 kJ/h. If the heat pump under these conditions has a COP of 2.5, determine (a) the power consumed by the heat pump and (b) the rate at which heat is absorbed from the cold outdoor air. Example 2
- 57. 57 (a) The power consumed by this heat pump is determined from the definition of the coefficient of performance to be (b) The house is losing heat at a rate of 80,000 kJ/h. If the house is to be maintained at a constant temperature of 20°C, the heat pump must deliver heat to the house at the same rate, that is, at a rate of 80,000 kJ/h. Then the rate of heat transfer from the outdoor becomes
- 58. 58 When a man returns to his well-sealed house, he finds that the house is at 32°C. He turns on the air conditioner, which cools the entire house to 20°C in 15 min. If the COP of the air-conditioning system is 2.5, determine the power drawn by the air conditioner. Assume the entire mass within the house is equivalent to 800 kg of air for which cv = 0.72 kJ/kg · °C and cp = 1.0 kJ/kg · °C. Solution We first have to find the amount of heat that need to be removed. Since the house is well-sealed, we can assume a constant volume cooling process so that This heat is removed in 15 minutes. Thus the average rate of heat removal from the house is Example 3
- 59. 59 Using the definition of the coefficient of performance, the power input to the air-conditioner is determined to be
- 60. 60 Refrigerant-134a enters the condenser of a residential heat pump at 800 kPa and 35°C at a rate of 0.018 kg/s and leaves at 800 kPa as a saturated liquid. If the compressor consumes 1.2 kW of power, determine (a) the COP of the heat pump and (b) the rate of heat absorption from the outside air. Example 4
- 61. 61 Solution (a) The enthalpies of R-134a at the condenser inlet and exit are An energy balance on the condenser gives the heat rejected in the condenser The COP of the heat pump is
- 62. 62 (b) The rate of heat absorbed from the outside air
- 63. 63 THE CARNOT HEAT ENGINE The Carnot heat engine is the most efficient of all heat engines operating between the same high- and low- temperature reservoirs. No heat engine can have a higher efficiency than a reversible heat engine operating between the same high- and low- temperature reservoirs. Any heat engine Carnot heat engine
- 64. 64 A Carnot heat engine receives 500 kJ of heat per cycle from a high-temperature source at 652°C and rejects heat to a low- temperature sink at 30°C. Determine (a) the thermal efficiency of this Carnot engine and (b) the amount of heat rejected to the sink per cycle. Example
- 65. 65 (a) The Carnot heat engine is a reversible heat engine, and so its efficiency is given by (b) The amount of heat rejected QL by this reversible heat engine is Solution
- 66. 66 THE CARNOT REFRIGERATOR AND HEAT PUMP No refrigerator can have a higher COP than a reversible refrigerator operating between the same temperature limits. Any refrigerator or heat pump Carnot refrigerator or heat pump
- 67. 67 Example 1 An inventor claims to have developed a refrigerator that maintains the refrigerated space at 2°C while operating in a room where the temperature is 24°C and that has a COP of 13.5. Is this claim reasonable? Solution The maximum COP is the COP of a reversible refrigerator operating between these temperature limits. Since the COPR,rev is lower than the claimed COP, the claim cannot be justified. 512 1227324273 1 1 1 . // COPCOP max,Rrev,R LH TT
- 68. 68 Example 2 A heat pump is to be used to heat a house during winter. The house is to be maintained at 21°C at all times. The house is estimated to be losing heat at a rate of 135,000 kJ/h when the outside temperature drops to 5°C. Determine the minimum power required to drive this heat pump.
- 69. 69 Solution The heat pump must supply heat to the house at a rate of QH = 135,000 kJ/h = 37.5 kW. The power requirements are minimum when a reversible heat pump is used to do the job. The COP of a reversible heat pump operating between the house and the outside air is The required power input to this reversible heat pump is then 311 2127352731 1 1 1 . // COP rev,HP HL TT kW. . kW. COPHP 323 311 537H ,innet Q W
- 70. 70 THE CARNOT CYCLE Reversible Isothermal Expansion (process 1-2, TH = constant) Reversible Adiabatic Expansion (process 2-3, temperature drops from TH to TL) Reversible Isothermal Compression (process 3-4, TL = constant) Reversible Adiabatic Compression (process 4-1, temperature rises from TL to TH) Most efficient cycles are reversible cycles. Most familiar reversible cycle is the Carnot cycle Execution of the Carnot cycle in a closed system.
- 71. 71 P-V diagram of the Carnot cycle. P-V diagram of the reversed Carnot cycle. The Reversed Carnot Cycle The Carnot heat-engine cycle is a totally reversible cycle. Therefore, all the processes that comprise it can be reversed, in which case it becomes the Carnot refrigeration cycle.
- 72. 72 THE CARNOT PRINCIPLES 1. The efficiency of an irreversible heat engine is always less than the efficiency of a reversible one operating between the same two reservoirs. 2. The efficiencies of all reversible heat engines operating between the same two reservoirs are the same.
- 73. 73 Proof of the First Carnot principle. Let reversible HE run as refrigerator (dashed line). If heat from refrigerator is supplied directly to irrev. HE, can eliminate HT reservoir. But what do we have then? A HE operating from a single reservoir and producing work amounting to Wirrev Wrev If Wirrev > Wrev, net work is produced and this violate the 2nd Law. Hence Wirrev < Wrev.