45 STEAM TABLE • From steam table, we can evaluate the thermodynamic properties like Specific Volume, Specific Internal Energy, Specific Enthalpy and Specific Entropy. Steam Table Saturated Water and Steam Temperature Based Pressure Based Superheated Steam
46 Saturation Temperature • At a given pressure, the temperature at which a pure substance changes phase is called saturation temperature (Ts). • For water – 100 °C is the saturation temperature at 1 atm pressure. Saturation Pressure • At a given temperature, the pressure at which a pure substance changes phase is called saturation pressure (Ps). • For water – 1 atm is the saturation pressure at Temperature 100 °C Saturated Liquid and Saturated Vapor Saturated Mixture f g Saturated Liquid Saturated Vapor Property A Af A g T (°C) Af = Property A at saturated liquid state Ag = Property A at saturated vapor state f g Property A Af A g T (°C) Saturated mixture Af < A < Ag X = 0 X = 1
47 Superheated Vapor • Superheated vapor table for P ≥ 80 bar: f g Property A Af A g T (°C) Superheated Vapor Degree of Superheat, DOS = T - Ts
48 Tutorial 1 A. Determine the phase, internal energy, volume, and entropy for steam at 14 bar with 2790 kJ/kg of enthalpy. Plot the t-v diagram. B. For a steam at 20 bar with 6.1 kJ/kgK entropy, calculate the: i. Specific volume ii. Specific enthalpy iii. Specific internal energy Show the point on p-v diagram. C. Find the dryness fraction, specific volume, and specific internal energy of steam at 105 bar and specific enthalpy 2100 kJ/kg. D. Steam at 100 bar has a specific volume of 0.02812 m3 /kg. Find the temperature, degree of superheat, specific enthalpy, and specific internal energy. E. Determine the phase and temperature of water at a state of 170 bar and 3028 kJ/kg enthalpy. F. Steam at 120 bar is at 500°C. Find the degree of superheat, specific volume, specific enthalpy, and specific internal energy. Show the point on p-v diagram. G. Steam at 160 bar has a specific enthalpy of 3139 kJ/kg. Find the temperature, degree of superheat, specific entropy, and specific internal energy. Show the point on p-v diagram.
49 INTERPOLATION Single interpolation Double interpolation Single Interpolation: Single interpolation is used to find the values in the table when one of the values is not tabulated. Example 1: Determine the saturation temperature at 77 bar Answer: Example 2: Determine the specific volume at 90 bar and 480˚c. Answer:
50 Double Interpolation: In some cases, a double interpolation is necessary, and it’s usually used in the Superheated Steam Table Double interpolation must be used when two of the properties (eg. temperature and pressure) are not tabulated in the Steam Tables. Example: Determine the specific enthalpy of superheated steam at 25 bar and 320oC. Answer: An extract from the Superheated Steam Tables:
51 Tutorial 2 A. Determine the specific enthalpy of dry saturated steam at 103 bar. B. Determine the specific enthalpy of steam at 15 bar and 275 oC. C. A superheated steam at 12.5 Mn/m2 is at 650oC. Determine its specific volume. D. Determine the degree of superheat and Entropy of steam at 10 bar and 380 °C.
52 IDEAL GAS • An ideal gas is defined as one in which all collisions between atoms or molecules are perfectly elastic and in which there are no intermolecular attractive forces. • Many familiar gases such as air, nitrogen, oxygen, hydrogen, helium, argon, carbon dioxide and others can be treated as ideal gas. IDEAL GAS LAW • The pressure (P), volume (V), and temperature (T) of an ideal gas are related by a simple formula called the ideal gas law. • The ideal gas equation of state predicts the p-v-t behavior of a gas. GAS CONSTANT • The gas constant R is different for each gas and is determined from :
53 Specific Heat • The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius. • In Thermodynamics, we interested in two kinds of specific heats: specific heat at constant volume (Cv) and specific heat at constant pressure (Cp). IDEAL GAS AT TWO DIFFERENT STATES • Ideal gas at two different states are related to each other by:
54 ANALYSIS OF PROCESS Tutorial 3 A. A quantity of a certain perfect gas is heated at a constant temperature from an initial state of 0.22 m3 and 325 kN/m2 to a final state of 170 kN/m2 . Calculate the final volume of the gas. B. A quantity of gas at 0.54 m3 and 345˚C undergoes a constant pressure process that causes the volume of the gas to decreases to 0.32 m3 . Calculate the temperature of the gas at the end of the process. C. 0.04 kg of a certain perfect gas occupies a volume of 0.0072 m3 at a pressure 6.76 bar and a temperature of 127˚C. Calculate the molecular weight of the gas. When the gas is allowed to expand until the pressure is 2.12 bar the final volume is 0.065 m3 . Calculate the final temperature. D. A certain perfect gas has specific heat as follows CP = 0.846 kJ/kg k and CV = 0.657 kJ/kg K. Find the gas constant and the molecular weight of the gas.
55 TOPIC 4: FIRST LAW OF THERMODYNAMICS CONCEPT OF THE FIRST LAW OF THERMODYNAMICS • The First Law is usually referred to as the Law of Conservation of Energy, i.e., Energy can neither be created nor destroyed, but rather transformed from one state to another. • In thermodynamics approach, the energy concept is synthesized from the prior assumption of heat and work. • When system undergoes a thermodynamics cycle, then THE NET HEAT SUPPLIED TO THE SYSTEM IS EQUAL TO THE NET WORK DONE TO THE SURROUNDINGS. • Q = W FORMS OF ENERGY: • In thermodynamics, there are 2 forms of energy group: 1. Macroscopic energy • Related to motion and influence some external effect: gravity, magnetism, surface tension. • Example: kinetic energy, potential energy. 2. Microscopic energy • Related to molecular structure of the system and the degree of the molecular activity. • Example: internal energy
56 INTERNAL ENERGY • Internal energy is the sum of all the energies a fluid possesses and stores within itself. If • u2 > u1 : internal energy increased (+ve) • u2 < u1 : internal energy decreased (-ve) ENERGY TRANSFER BY HEAT • Heat is the form of energy that is transferred between two systems (or a system and its surroundings) by virtue of a temperature difference. • Heat was considered to be a fluid that can spontaneously flow from a hot body to a cold body. • Heat is transferred by three mechanisms: o Conduction o Convection o Radiation • If heat energy flows into the system from the surroundings it is said to be positive. • If heat energy flows from the system to the surroundings it is said to be negative.
57 ENERGY TRANSFER BY WORK • Work transfer is defined as a product of the force and the distance moved in the direction of the force. • Mechanical forms of work: o Shaft work o Spring work o Work done on elastic solid bars. o Work associated with the stretching of a liquid film. o Work done to raise or to accelerate a body. • If work energy is transferred from the system to the surroundings, it is donated as positive. • If work energy is transferred from the surroundings to the system, it is donated as negative. ENERGY BALANCE • From the energy balance equation − = ∆ − = ∆ + ∆ + ∆ • Noting that a closed system does not involve kinetic and potential energy, the energy balance simplifies to − = ∆ • For a closed system undergoing a cycle, ∆ = Thus =
58 Energy balance For a closed system undergoing a cycle, TUTORIAL 1 A. The figure above shows a certain process, which undergoes a complete cycle of operations. Determine the value of the work output for a complete cycle, Wout. B. A system is allowed to do work amounting to 500 kJ whilst heat energy amounting to 800 kJ is transferred into it. Find the change of internal energy and state whether it is an increase or decrease. C. During a complete cycle operation, a system is subjected to the following: Heat transfer is 300 kJ supplied and 150 kJ rejected. Work done by the system is 200 kJ. Calculate the work transferred from the surrounding to the system. D. In a diesel engine, the amount of internal energy at the beginning of compression stroke is 55 kJ/kg. At the end of the stroke, the value of internal energy is 125 kJ/kg. The amount of work produced by the engine is 93 kJ/kg. Determine the amount of heat flows during the process and indicate either the heat was supplied or rejected by the engine. Qout = 3 kJ Qin = 10 kJ Wout = ? Win= 2 kJ SYSTEM
59 E. Determine the changes of internal energy for the system below: CLOSED SYSTEM/ NON-FLOW PROCESS • A close system (control mass) consists of a fixed amount of mass, and no mass can cross its boundary. • That is, no mass can enter or leave a close system. • But energy, in the form of heat or work can cross the boundary, and the volume of a close system does not have to be fixed. A closed system with a moving boundary
Constant Pressure (Isobaric), P1 = P2 Constant Volume (Isometric), V1 = V2 Constant Temper 11 1 = 22 2 1 1 = 2 2 11 1 = 22 2 1 1 = 2 2 1 1 1 P V = m R T & Work, W W = P(V2 – V1) or W = mR(T2 - T1) Work, W W = 0 Work, W = 11 ( or = 11 ( Heat, Q Q = mCp(T2 – T1) Heat, Q Q = mCv(T2 – T1) Heat, Q Q = W = 11 ( = 11 ( Internal Energy, U2 – U1 U2 -U1 = mCv(T2 -T1) Internal Energy, U2 – U1 U2 -U1 = mCv(T2 -T1) Internal Energy U2 -U1 = 0
60 rature (Isothermal), T1 = T2 Adiabatic (Isentropic), s1 = s2 Polytropic, PVn = constant 1 1 = 22 2 1 = 22 T T P P V V 2 1 2 1 1 1 2 1 = = − − 1 2 1 1 1 2 1 2 − − = = n n n V V P P T T & = ( 2 1 ) = 1 ( 2 1 ) ( 1 2 ) = 1 ( 1 2 ) Work, W = (1− 2) −1 or = 11−22 −1 Work, W = (1−2) −1 or = 11−22 −1 2 1 ) = 1 ( 2 1 ) 1 2 ) = 1 ( 1 2 ) Heat, Q Q = 0 Heat, Q = − − 1 y, U2 – U1 Internal Energy, U2 – U1 U2 -U1 = mCv(T2 -T1) Internal Energy, U2 – U1 U2 -U1 = mCv(T2 -T1)
61 TUTORIAL 2 A. A nitrogen (molecular weight 28) expands reversibly in a pneumatic cylinder at a constant pressure of 2.05 bar. The initial temperature was 25°C with a volume of 0.05m3 . The temperature rises to 500°C after the process. Assuming nitrogen to be a perfect gas (CP = 1.045 kJ/kgK), calculate: i. mass of nitrogen ii. work done by nitrogen. iii. heat flow during the expansion process B. 0.05 kg of air, initially at 130oC is heated at a constant pressure of 2 bar until the volume occupied is 0.0658 m3 . Calculate the heat supplied, work done and changes of internal energy of the gas. C. 3.4 kg of gas is heated at a constant volume of 0.92 m3 and temperature 17˚C until the temperature rise to 147˚C. If the gas is assumed to be a perfect gas, determine: i. the heat flow during the process. ii. the beginning pressure of the gas. iii. the final pressure of the gas. Given: CV = 0.72 kJ/kgK and R = 0.287 kJ/kgK D. Original state of a perfect gas are as follows: the mass of 1.7 kg, volume of 0.8 m3 and the temperature of 20˚C. The gas is heated at constant volume until the temperature reached 160˚C. Calculate the value of: i. initial pressure ii. final pressure iii. heat transferred. Given: CV = 718 J/kgK and R = 0.287 kJ/kgK E. By assuming the gas constant is 0.2968 kJ/kgK, find the mass of a gas at 300 kN/m2 and 45˚C with a volume of 0.046 m3 . The pressure is then increased to 1.27 MN/m2 at constant volume. Calculate: i. final temperature of the gas ii. heat transferred. iii. internal energy Take CP = 1.04 kJ/kgK F. 1 kg of nitrogen (molecular weight 28) is compressed reversibly and isothermally from 1.01 bar, 20oC to 4.2 bar. Calculate the work done and the heat flow during the process. Assume nitrogen to be a perfect gas. G. In an industrial process, 0.4kg of oxygen is compressed isothermally from 1.01bar and 220 C to 5.5bar. Determine the work done and heat transfer during the process. Assume that oxygen is perfect gas and take the molecular weight of oxygen to be M=32kg/k.mol.
62 H. In a thermally insulated reciprocating air compressor, air at 0.98 bar and 20˚C is compressed into one sixth of its original volume. Determine the pressure and temperature of the air after compression. If the compressor cylinder contains 0.05kg of air, calculate the required work input. For air, take γ = 1.4 and CV = 0.718 kJ/kgK I. Nitrogen (molecular weight 28) expands reversibly in a perfect thermally insulated cylinder from 2.5 bar, 200˚C to a volume of 0.09m3 . if the initial volume occupied was 0.03m3 , calculate the work done during the expansion. Assume nitrogen to be a perfect gas and take CV = 0.741 kJ/kgK J. A mass of 0.05kg of air at temperature of 40˚C and pressure of 1 bar is compressed polytropicly to 5 bar by following the law Pv1.25 = C. Determine: i. Final temperature ii. Final volume iii. Work transfer iv. Heat transfer v. Change of internal energy
63 OPEN SYSTEM/ FLOW PROCESS • An open system, or a control volume, involves mass flow. • Flow through these devices is best studied by selecting properly the region within the device as the control volume. • Both mass and energy can cross the boundary of a control volume. Open system in turbine • This process may be sub-divided into an unsteady flow process and steady flow process. • Criteria for steady flow process: o The mass of fluid flowing past any section in the system must be constant with respect to time. o The properties of the fluid at any particular section in the system must be constant with respect to time. o All transfer of work energy and heat which takes place must be done at a uniform rate.
64 APPLICATION OF OPEN SYSTEM Turbine: • A turbine is a device which uses a pressure drop to produce work energy which is used to drive an external load. Boiler: • In a boiler operating under steady conditions, water is pumped into the boiler along the feed line at the same rate as which steam leaves the boiler along the steam main, and heat energy is supplied from the furnace at a steady rate. Nozzle: • A nozzle utilises a pressure drop to produce an increase in the kinetic energy of the fluid.
65 Condenser: • In principle, a condenser is a boiler reverse. • In a boiler, heat energy is supplied to convert the liquid into vapour whereas in a condenser heat energy is removed to condense the vapour into a liquid. Throttle: • A throttling process is one in which the fluid is made to flow through a restriction, e.g., a partially opened valve or orifice, causing a considerable drop in the pressure of the fluid. Pump: • The action of a pump is the reverse of that of a turbine, i.e., it uses external work energy to produce a pressure rise.
66 THE STEADY FLOW ENERGY EQUATION
67 EQUATION OF CONTINUITY Tutorial 3 A. The velocity at the inlet section of a horizontal nozzle is 50 m/s and the specific enthalpy is 2980 kJ/kg. At the outlet section, the specific enthalpy is 2745 kJ/kg. If the heat loss from the nozzle is neglected, calculate: i. The increasing velocity of the working fluid at the outlet section of the nozzle. ii. The mass flow rate of the working fluid if the inlet cross-sectional area and the specific volume are 0.0932 m2 and 0.15 m3 /kg. iv. The outlet cross-sectional area of the nozzle if the specific volume is 0.46 m3 /kg. B. Steam enters a turbine with a velocity of 20 m/s and specific enthalpy 3990 kJ/kg. The steam leaves the turbine with a velocity of 35 m/s and specific enthalpy 1030 kJ/kg. The heat loss to the surroundings as the steam passes through the turbine is 20 kJ/kg. The steam flow rate is 42400 kg/h. Calculate the power generated by turbine in kW. C. Water flowed through a boiler with the velocity of 6 m/s, pressure of 0.98 bar and specific volume of 0.1 m3 /kg. Superheated steam produced has velocity of 50 m/s and pressure of 7 bar with specific volume of 0.2 m3 /kg. The specific internal energy in the outlet section is 100 kJ/kg greater than the inlet. If the heat transfer to the boiler is 40 kW, calculate the mass flow rate produced in kg/h.
68 D. In steady flow system, a fluid flows at the rate of 120 kg/min. It enters at pressure of 6 bar, velocity of 200 m/s, internal energy of 2250 kJ/kg and specific volume of 0.56 m3/kg. The fluid leaves the system at the pressure of 1.5 bar, velocity of 160 m/s, internal energy of 1500 kJ/kg and specific volume of 1.36 m3/kg. During its passage through the system, the fluid has a heat loss of 25 kJ/kg to the surroundings. Determine the power system in kilowatts, starting whether it is generated or provided to the system. Neglect the changes of any potential energy. E. Fluid with a specific enthalpy of 3200 kJ/kg enters a horizontal nozzle with negligible velocity at the rate of 15 kg/s. At the outlet, the specific enthalpy and specific volume of the fluid are 2050 kJ/kg and 2.05 m3 /kg respectively. Assuming an adiabatic flow process, determine the required outlet area of the nozzle.
69 TOPIC 5: SECOND LAW OF THERMODYNAMICS SECOND LAW OF THERMODYNAMICS • The first law of thermodynamics is concerned with the quantity of energy and transformation of energy from one form to another, ∑Q = ∑W. • A second Law of Thermodynamics is about: o For any process, a natural flow will always be from higher energy value to lower energy value. Heat transfer occurs spontaneously from a high temperature body to a low temperature body. Flow of heat is not possible from a cold body to a hot body without external work. Some external inputs or energy must be expanded to reverse the process. o For every real-world process, there will always be losses. Although the net heat supplied in a cycle is equal to the net work done, the gross heat supplied must be greater than the work done; some heat must always be rejected by the system, Wnet = QH - QL
70 HEAT ENGINE • Heat engine is a special device used in converting heat to work. • Characteristic of heat engines: o It receives heat from a high-temperature source. o It converts part of this energy (heat to work). o It rejects the remaining waste heat to a low-temperature sink. • Steam power plant is the best example for heat engine. • The main devices in steam power plant; o Pump o Boiler o Turbine o Condenser Schematic of heat engine. Part of the heat received by heat engine is converted to work, whereas the rest is rejected to a sink. Schematic of a steam power plant o Qin = amount of heat supplied to steam in boiler from a high-temperature source (furnace) o Qout = amount of heat rejected from steam in condenser to a low temperature sink (the atmosphere, a river, etc.) o Wout = amount of work delivered by steam as it expands in turbine o Win = amount of work required to compress water to boiler pressure
71 Thermal efficiency of heat engine. ℎ = ℎ = , It can be also expressed as ℎ = − Since , = − Tutorial 1 A. 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. B. A steam power plant produces 50 MW of net work while burning fuel to produce 150 MW of heat energy at the high temperature. Determine the cycle thermal efficiency and the heat reject by the cycle to the surroundings.
72 REVERSED HEAT ENGINE • Reversed heat engine is a system that operates in thermodynamics cycle that transfer heat from low temperature to high temperature media. • Example: Refrigerator/air-conditioner and heat pump. • The index of performance of a heat pumps or refrigerators are expressed in terms of the coefficient of performance. Reverse Heat Engine: Refrigerator/Air-conditioner • The objective of a refrigerator is to maintain the refrigerated space at low temperature by removing heat from it. = = , It can be also expressed as = − Since , = − QL = magnitude of heat removed from the refrigerated space at temperature TL QH = magnitude of heat rejected to the warm environment at temperature TH Wnet,in = net work input to the refrigerator
73 Reverse Heat Engine: Heat Pump • The objective of heat pump is to maintain a heated space at a high temperature by absorbing heat from a low-temperature source. • The work supplied to a heat pump is used to extract energy from the cold outdoors and carry it into the warm indoors. = = , It can be also expressed as = − Since , = − Tutorial 2 A. The food compartment of a refrigerator 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 (i) The coefficient of performance of the refrigerator (ii) The rate of heat rejection to the room that houses the refrigerator. B. An air conditioner removes heat steadily from a house at a rate of 750 kJ/min while drawing electric power at a rate of 6 kW. Determine (i) The COP of this air conditioner (ii) The rate of heat transfer to the outside air. C. 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 (i) The power consumed by the heat pump (ii) The rate at which heat is absorbed from the cold outdoor air. QL = magnitude of heat absorbed from cold outdoor air at temperature TL QH = magnitude of heat supplied into the warmer space at temperature TH Wnet, in = net work input to the heat pump.
74 CARNOT HEAT ENGINE CARNOT REVERSED HEAT ENGINE • Heat engine that operates on the reversible Carnot cycle is called Carnot heat engine. • A refrigerator or a heat pump that operates on the reversed Carnot cycle is called a Carnot refrigerator or a Carnot heat pump. Tutorial 3 A. 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 (i) The thermal efficiency of this Carnot engine and (ii) The amount of heat rejected to the sink per cycle. B. A Carnot refrigerator operates in a room in which the temperature is 22°C and consumes 2 kW of power when operating. If the food compartment of the refrigerator is to be maintained at 3°C, determine the rate of heat removal from the food compartment. C. An air-conditioning system operating on the reversed Carnot cycle is required to transfer heat from a house at a rate of 750 kJ/min to maintain its temperature at 24°C. If the outdoor air temperature is 35°C, determine the power required to operate this air-conditioning system.
75 CARNOT & RANKINE CYCLE • Carnot cycle is the most efficient cycle operating between two specified temperature limits. Process The Fluid is.. Heat/work 4 - 1 Heated reversibly and isothermally in boiler qin = h1 – h4 1 - 2 Expanded isentropically in a turbine wout = h1 – h2 2 - 3 Condensed reversibly and isothermally in a condenser qout = h2 – h3 3 - 4 Compressed isentropically by a compressor to the initial state win = h4 – h3 Carnot Efficiency Work Ratio Specific Steam Consumption = = − = = − = 3600 = 3600 − Tutorial 4 A. A steam power plant operates between a boiler pressure of 30 bar and a condenser pressure of 0.04 bar. For the Carnot Cycle, calculate: i.Cycle efficiency ii.Work ratio iii.Specific steam consumption Show the Carnot Cycle on T-s diagram. B. A steam power plant operates between a boiler pressure of 42 bar and a condenser pressure of 0.035 bar. Calculate for these limits the cycle efficiency and the work ratio for a Carnot cycle using wet steam. Show the Carnot Cycle on T-s diagram.
76 RANKINE CYCLE • Many of the impracticalities associated with the Carnot cycle can be eliminated by superheating the steam in the boiler and condensing it completely in the condenser. The cycle that results is the Rankine cycle. • Rankine Cycle is the ideal cycle for vapor power plants. Process The Fluid is.. Heat/work 4 - 1 Constant pressure heat addition in a boiler qin = h1 – h4 1 - 2 Isentropic expansion in a turbine wout = h1 – h2 2 - 3 Constant pressure heat rejection in a condenser qout = h2 – h3 3 - 4 Isentropic compression in a pump win = h4 – h3 @ win = Vf(P1 – P2) where Vf = 0.001 m3 /kg Rankine Efficiency Work Ratio Specific Steam Consumption = = − = = − = 3600 = 3600 −
77 Tutorial 5 A. A steam power plant operates between a boiler pressure of 50 bar and a condenser pressure of 0.065 bar. For the Rankine Cycle, calculate: i.Cycle efficiency ii.Work ratio iii.Specific steam consumption Show the Rankine Cycle on T-s diagram. B. A steam power plant operates between a boiler pressure of 42 bar and a condenser pressure of 0.035 bar. Calculate for these limits the cycle efficiency, the work ratio, and the specific steam consumption for a Rankine cycle with dry saturated steam at entry to the turbine. Show the Rankine Cycle on T-s diagram.
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