SF9 TD QB solution PDF

Branch: Mechanical Engineering Year / Sem: II/I (R22)Subject: THERMODYNAMICS Sub code: 22PC0ME04 QUESTION BANK Part –A: 2 Marks Questions: S.No...

Branch: Mechanical Engineering Year / Sem: II/I (R22)Subject: THERMODYNAMICS Sub code: 22PC0ME04 QUESTION BANK Part –A: 2 Marks Questions: S.No Short Question UNIT- I 1 What is meant by thermodynamic equilibrium? A system is in thermodynamic equilibrium if the temperature and pressure at all points are same ; It is only under thermodynamic equilibrium conditions that the properties of a system can be fixed. Thus for attaining a state of thermodynamic equilibrium the following three types of equilibrium states must be achieved : 1. Thermal equilibrium. The temperature of the system does not change with time and has same value at all points of the system. 2. Mechanical equilibrium. There are no unbalanced forces within the system or between the surroundings. The pressure in the system is same at all points and does not change with respect to time. 3. Chemical equilibrium. No chemical reaction takes place in the system and the chemical composition which is same throughout the system does not vary with time. 2 Why Constant Volume thermometer is preferred over a constant pressure gas Thermometer? The constant volume gas thermometer is preferred over a constant pressure gas thermometer in certain applications due to the following reason: The constant volume gas thermometer is considered more accurate compared to a constant pressure gas thermometer. This is because the volume of the gas is kept constant, eliminating the potential effects of variations in the external pressure. Any pressure changes observed can be solely attributed to temperature changes, allowing for more precise temperature measurements. 3 Distinguish between macroscopic and microscopic point of view s.no macroscopic microscopic 1 n this approach a certain quantity of The approach considers that the system is made up of matter is a very large number of discrete particles known as considered without taking into account molecules. These molecules have the events different velocities and energies. The values of these occurring at molecular level. In other energies are constantly changing with time. This words this approach to thermodynamics which is concerned approach to thermodynamics is directly with the structure of the matter is known as concerned with statistical thermodynamics. gross or overall behaviour. This is known as classical thermodynamics. 2 The analysis of macroscopic system The behaviour of the system is found by using requires statistical methods as the number of molecules is very simple mathematical formulae. large. So advanced statistical and mathematical  methods are needed to explain the changes in the system. 4 Discuss Quasi Static process. In thermodynamics, a quasi-static process is referred to as a slow process. It is a process that happens at an infinitesimally slow rate. A quasi-static process has all of its states in equilibrium. A quasi-static process is one in which the system is in thermodynamic equilibrium with its surroundings at all times. 5 What do you mean by a system ? System. A system is a finite quantity of matter or a prescribed region of space. A system, surrounded and influenced by its environment, is described by its boundaries, structure and purpose and is expressed in its functioning. 6 What are the properties of control volume? The surface of the control volume is referred as a control surface and is a closed surface. The surface is defined with relative to a coordinate system that may be fixed, moving or rotating. Mass, heat and work can cross the control surface and mass and properties can change with time within the control volume. 7 Extensive and intensive properties are differentiated on what basis ? 1. Intensive properties. These properties do not depend on the mass of the system. Examples : Temperature and pressure. 2. Extensive properties. These properties depend on the mass of the system. Example : Volume. Extensive properties are often divided by mass associated with them to obtain the intensive properties. For example, if the volume of a system of mass m is V, then the specific volume of matter within the system is V/m = v which is an intensive property. 8 Write down causes of irreversibility ? Four of the most common causes of irreversibility are friction, unrestrained expansion of a fluid, heat transfer through a finite temperature difference, and mixing of two different substances. These factors are present in real,irreversible processes and prevent these processes from being reversible 9 Derive work output for Piston displacement work. If the pressure of the fluid is greater than that of the surroundings, there will be an unbalanced force on the face of the piston.  Displacement Work.  Consider a piston cylinder arrangement as given in the Figure 2.4....  Force acting on the piston =Pressure ´Area.  =pA.  \ Work done =Force ´distance.  =pA ´dx.  =pdV. 10 Write Zeroth law of thermodynamics. Zeroth law of thermodynamics’ states that if two systems are each equal in temperature to a third, they are equal in temperature to each other.  UNIT-II 1 Write down SFEE. 2 Derive SFEE for Boiler. A boiler transfers heat to the incoming water and generates the steam 3 Discuss First law of thermodynamics The First Law of Thermodynamics states that energy cannot be created or destroyed; it can only be converted from one form to another. The First Law is used to categorise 'the performance of cyclic conversion systems like fossil-fired, steam power cycles or geothermal cycles. 4 What is Entropy? entropy, the measure of a system’s thermal energy per unit temperature that is unavailable for doing useful work. Because work is obtained from ordered molecular motion, the amount of entropy is also a measure of the molecular disorder, or randomness, of a system 5 Draw the layout of Heat Engine. 6 Derive COP of Refrigerator. the ratio of heat absorbed by the refrigerant while passing through the evaporator to the work input required to compress the refrigerant in the compressor ; in short it is the ratio between heat extracted and work done 7 Proove Equivalence of Kelvin-Planck and Clausius Statements Consider a higher temperature reservoir T1 and low temperature reservoir T2. Fig. 5.2 shows a heat pump which requires no work and transfers an amount of Q2 from a low temperature to a higher temperature reservoir (in violation of the Clausius statement). Let an amount of heat Q1 (greater than Q2) be transferred from high temperature reservoir to heat engine which devolops a net work, W = Q1 – Q2 and rejects Q2 to the low temperature reservoir. Since there is no heat interaction with the low temperature, it can be eliminated. The combined system 8 What is principle of Entropy increase? The principle of increase of entropy states that the entropy of the Universe also increases. It never decreases but remains constant only in a reversible process. For an isolated system: Δ S > 0 , for irreversible processes. Δ S = 0 , for reversible processes 9 Write about PMM-II. Perpetual motion machine of the second kind (PMM2): A fictitious machine that produces net-work in a complete cycle by exchanging heat with only one reservoir is called the PMM2. It violates the Kelvin-Plank statement 10 What is the concept of Clausius Inequality ? The Clausius inequality is a consequence of applying the second law of thermodynamics at each infinitesimal stage of heat transfer. The Clausius statement states that it is impossible to construct a device whose sole effect is the transfer of heat from a cool reservoir to a hot reservoir. UNIT-III 1 Draw PVT surface with expansion. 2 What is dryness fraction ? Dryness fraction is defined as the ratio of mass of dry steam (vapour) to combined mass of dry steam (vapour) & mass of liquid in mixture. It is denoted by x. 3 Write any two gas laws. Boyle’s law gives the relationship between the pressure of a gas and the volume of the gas at a constant  temperature. Basically, the volume of a gas is inversely proportional to the pressure of a gas at a constant temperature. P ∝ 1/V Charle’s law states that at constant pressure, the volume of a gas is directly proportional to the temperature (in Kelvin) in a closed system. Basically, this law describes the relationship between the temperature and volume of the gas. V∝T 4 What is throttling process ? The process of throttling involves utilizing a throttle valve to change a high-pressure fluid into a low- pressure fluid. Throughout the throttling process, enthalpy is constant 5 Describe briefly about “Mollier diagram”. The Mollier diagram is a graph used in thermodynamics to visualize the relationships between temperature, pressure, specific volume, enthalpy, and entropy of a substance. It's also known as the enthalpy-entropy chart. 6 State Vander Waals equation The equation is basically a modified version of the Ideal Gas Law which states that gases consist of point masses that undergo perfectly elastic collisions. However, this law fails to explain the behaviour of real gases. Therefore, the Van der Waals equation was devised, and it helps us define the physical state of a real gas. 7 Determine the enthalpy and entropy of steam and the pressure is 2MPa and the specific volume is 0.09m3/kg From the steam table, we find that the enthalpy of steam at 2 MPa and 0.09 m³ / kg is approximately 2,800 kJ/kg. From the steam table, we find that the entropy of steam at 2 MPa and 0.09 m³ / kg is approximately 7.5 kJ/(kg·K). 9 Write down formula for enthalpy of super heated steam. 10 Write Daltons Pressure Law. According to Dalton's law of partial pressures, the total pressure by a mixture of gases is equal to the sum of the partial pressures of each of the constituent gases. The partial pressure is defined as the pressure each gas would exert if it alone occupied the volume of the mixture at the same temperature.  Ptotal=PA+PB+... UNIT- IV 1 What is mass fraction? the mass fraction (also known as the weight fraction) of a component in a mixture is the ratio of the mass of that component to the total mass of the mixture. It is often expressed as a percentage 2 How to calculate mole fraction ? The mole fraction can be calculated by dividing the number of moles of one component of a solution by the total number of moles of all the components of a solution. It is noted that the sum of the mole fraction of all the components in the solution should be equal to 1. 3 Volumetric Analysis explain with one example. The mass of the nitrogen can be calculated from the volume it occupies under known conditions of temperature and pressure, and therefore the proportion of nitrogen in the sample can be determined. A volumetric method is also applied in the analysis of nitrates, which can be converted into nitric oxide, NO, a gas. 4 How to calculate Gas constant for Gas Mixture? The ratio of the molar gas constant(R) to the molar mass(M) of the gas mixture is called The specific gas constant. Denoted by Rspecific and mathematically expressed as – 5 What is wet bulb temperature? The wet-bulb temperature (WBT) is the temperature read by a thermometer covered in water-soaked (water at ambient temperature) cloth (a wet-bulb thermometer) over which air is passed. At 100% relative humidity, the wet-bulb temperature is equal to the air temperature (dry-bulb temperature); 6 Explain humidity ration. The humidity ratio or specific humidity is the weight of water vapour per unit weight of dry air (pound per pound or kg/kg). 7 Draw Psychrometric chart. 8 What is specific humidity at dry bulb temperature of 35 C and wet bulb temperature of 22 C. From the Psychrometric chart at 350C DBT and 220C WBT the Specific Humidity is 14 mm of Merucury. 9 Explain adiabatic saturation. Adiabatic saturation: When unsaturated air flows over a long sheet of water in an insulated chamber, the water evaporates, and the specific humidity of the air increases. Both the air and water are cooled as evaporation takes place. UNIT-V 1 Write down the processes in diesel cycle. Process 1-2 : isentropic compression Process 2-3: Constant pressure heat addition Process 3-4: Isentropic Expansion Process 4-1: Constant volume heat rejection 2 Draw the PV TS diagram for otto cycle. 3 Calculate the efficiency of otto cycle when the compression ratio is 6. 4 What is mean effective pressure ? Mean effective pressure, often known as MEP, is defined as the average pressure needed to act on the piston as it moves one displacement in order to produce work W. This pressure is required in order to calculate MEP. Pmep = Wnet/Vs 5 Draw the PV TS for Rankine cycle. Process 1-2: Isentropic pumpimg 2-3 : Constant pressure Heat addition in boiler 3-4 : Isentropic Expansion in turbine 4-1: Constant pressure heat rejection in condenser. 6 What is cut off ratio for diesel engine ? The cut-off ratio is defined as the ratio of the volume at the end of constant-pressure energy addition process to the volume at the beginning of the energy addition process. The compression ratios normally in the Diesel engines vary between 14 and 17.  7 Derive compression ratio for otto cycle. 8 Write down the processes in VCRS cycle. 9 Explain about carnot cycle. A Carnot heat-engine cycle described is a totally reversible cycle. That is, all the processes that compose it can be reversed, in which case it becomes the Carnot heat pump and refrigeration cycle. This time, the cycle remains exactly the same except that the directions of any heat and work interactions are reversed. 10 Why carnot cycle is not practically possible ? In real engines, the heat transfers at a sudden change in temperature whereas in a Carnot engine, the temperature remains constant. In our day to day lives, reversible processes can't be carried out and there is no such engine with 100 % efficiency. Thus, the Carnot cycle is practically not possible. Long Questions UNIT-1 1. Explain the state of equilibrium. Also discuss thermal, chemical and mechanical equilibrium with suitable examples. Answer:  Thermodynamic Equilibrium: A thermodynamic system is said to exist in a state of thermodynamic equilibrium when no change in any macroscopic property is registered if the system is isolated from its surroundings. An isolated system always reaches in the course of time a state of thermodynamic equilibrium and can never depart from it spontaneously. Therefore, there can be no spontaneous change in any macroscopic property if the system exists in an equilibrium state. A thermodynamic system will be in a state of thermodynamic equilibrium if the system is the state of Mechanical equilibrium, Chemical equilibrium and Thermal equilibrium. Mechanical equilibrium: The criteria for Mechanical equilibrium are the equality of pressures. Chemical equilibrium: The criteria for Chemical equilibrium are the equality of chemical potentials. Thermal equilibrium: The criterion for Thermal equilibrium is the equality of temperatures. 2. Explain the different types of systems with neat sketches and suitable examples Answer: Types of thermodynamic system On the basis of mass and energy transfer the thermodynamic system is divided into three types. 1. Closed system (Control Mass system) 2. Open system (Control Volume system) 3. Isolated system Closed system: A system in which the transfer of energy but not mass can takes place across the boundary is called closed system. The mass inside the closed system remains constant. For example: Boiling of water in a closed vessel. Since the water is boiled in closed vessel so the mass of water cannot escapes out of the boundary of the system but heat energy continuously entering and leaving the boundary of the vessel. It is an example of closed system.  Open system: A system in which the transfer of both mass and energy takes place is called an open system. This system is also known as control volume. For example: Boiling of water in an open vessel is an example of open system because the water and heat energy both enters and leaves the boundary of the vessel. Isolated system: A system in which the transfer of mass and energy cannot takes place is called an isolated system. For example: Tea present in a thermos flask. In this the heat and the mass of the tea cannot cross the boundary of the thermos flask. Hence the thermos flak is an isolated system. 3. Explain Zeroth law of Thermodynamics Answer: Zeroth Law of Thermodynamics: The Thermodynamics Zeroth Law states that if two systems are at the same time in thermal equilibrium with a third system, they are in equilibrium with each other. If an object with a higher temperature comes in contact with an object of lower temperature, it will transfer heat to the lower temperature object. If objects ‘A’ and ‘C’ are in thermal equilibrium with ‘B’, then object ‘A’ is in thermal equilibrium with object ‘C’. Practically this means all three objects are at the same temperature and it forms the basis for comparison of temperatures. If a=b; b=c then a=c Principles of Thermometry: Thermometry is the science and practice of temperature measurement. Any measurable change in a thermometric probe (e.g. the dilatation of a liquid in a capillary tube, variation of electrical resistance of a conductor, of refractive index of a  transparent material, and so on) can be used to mark temperature levels, that should later be calibrated against an internationally agreed unit if the measure is to be related to other thermodynamic variables. 4. Define the temperature. Name the different temperature scales in common use and establish a relation between Celsius and Fahrenheit scale. Answer: Temperature Scales All temperature scales are based on some easily reproducible states such as the freezing and boiling point of water, which are also called the ice-point and the steam-point respectively. A mixture of ice and water that is in equilibrium with air saturated with water vapour at 1atm pressure is said to be at the ice-point, and a mixture of liquid water and water vapour (with no air) in equilibrium at 1atm is said to be at the steam-point. Celsius and Fahrenheit scales are based on these two points (although the value assigned to these two values is different) and are referred as two-point scales. In thermodynamics, it is very desirable to have a temperature scale that is independent of the properties of the substance or substances. Such a temperature scale is called a thermodynamic temperature scale.(Kelvin in SI) 5. Write short notes on following : 1 Equality of temperature 2 Law of perfect gases 3 Process and cycle 4 Point Function & Path Function Answer: 1. Temperature is also measured by using several other temperature dependant properties. Two bodies (eg. Two copper blocks) in contact attain thermal equilibrium when the heat transfer between them stops. The equality of temperature is the only requirement for thermal equilibrium. 2. Perfect Gas Laws Boyle’s Law : Boyle’s Law Pressure is inversely proportional to volume: p∞ 1/v Robert Boyle noticed that when the volume of a container holding an amount of gas is increased, pressure decreases. Charles’ Law: Charles’ Law Volume is directly proportional to temperature: V = cT, where c > 0 is constant. Scientist Jacque Charles noticed that if air in a balloon is heated, the balloon expands. For an ideal gas, this relationship between V and T should be linear 3. Process & Cycle Process: When the system undergoes change from one thermodynamic state to final state due change in properties like temperature, pressure, volume etc, the system is said to have undergone thermodynamic process. Various types of thermodynamic processes are: isothermal process, adiabatic process, isochoric process, isobaric process and reversible process. Cycle: Thermodynamic cycle refers to any closed system that undergoes various changes due to temperature, pressure, and volume, however, its final and initial state are equal. This cycle is important as it allows for the continuous process of a moving piston seen in heat engines and the expansion/compression of the working fluid in refrigerators, for example. Without the cyclical process, a car wouldn't be able to continuously move when fuel is added, or a refrigerator would not be able to stay cold. Visually, any thermodynamic cycle will appear as a closed loop on a pressure volume diagram. Examples: Otto cycle, Diesel Cycle, Brayton Cycle etc. 4. Point and Path functions: Point function does not depend on the history (or path) of the system. It only depends on the state of the system. Examples of point functions are: temperature, pressure, density, mass, volume, enthalpy, entropy, internal energy etc. Path function depends on history of the system (or path by which system arrived at a given state). Examples for path functions are work and heat. Path functions are not properties of the system, while point functions are properties of the system. Change in point function can be obtained by from the initial and final values of the function, whereas path has to defined in order to evaluate path functions.  6. What do you mean by the term ‘ Property’? Prove that Heat and Work is not a point function. Solution: A thermodynamic system is said to exist in a state of thermodynamic equilibrium when no change in any macroscopic property is registered if the system is isolated from its surroundings. From the figure we can observe that W1-2 is different for black path and red path. Area under the curve gives the workdone in PV diagram. So we can say that as path changed work done also changes. Similarly on TS diagram curve changes for Heat transfer in the process. So Heat and Work both are path functions not the point functions. 7. A fluid at a pressure of 3 bar and with specific volume of 0.18 m3 /kg contained in a cylinder behind a piston expands reversibly to a pressure of 0.6 bar according to a law , P= c/V2 where c is a constant. Calculate the work done by the fluid on the piston. Solution: Givendata: Intial Pressure P1 = 3 Bar= 300 Kpa Initial Specific volume, v1 = 0.18m3/kg Final Pressure P2 = 0.6 Bar= 60 Kpa 8. A new scale N of temperature is divided in such a way that the freezing point of ice is 1000 N and the boiling point is 4000 N. What is the temperature reading on this new scale when the temperature is 1500C? At what temperature both the Celsius and the new scale reading would be the same? Solution : Given Data: ON New scale Boiling Point, TNB = 400 N Freezing Point, TFB = 100 N On Centigrade Scale we know that Boiling Point, TB = 100 C Freezing Point, TB = 0 C From temperature scales we know that At 150 C, on the new scale shows T= 450 N, At 150C, on new scale it shows, 450 N. ii) From temperature scales we know that T=50. 9. A reciprocating air compressor takes in 2 m3 /min at 0.11 MPa, 200C which is delivers at 1.5 MPa, 111 0C to an aftercooler where the air is cooled at constant pressure to 25 0C. The power absorbed by the compressor is 4.15 kW. Determine the heat transfer in compressor and the cooler. Solution: Givendata: Considering Process 1-2 in Compressor Initial Volume V1 = 2m3/min = 1/30 m3/sec Initial Pressure P1 = 0.11MPa = 0.11 x 106 N/m2 Initial Temperature T1 = 20°C = 293K Final Pressure, P2 = 1.5 x 106 N/m2 Final Temperature, T2 = 111°C = 384k Power supplied, W = 4.15KW Considering Process 2-3 in cooler, Final Temperature, T3 = –25°C Specific Heat, CP = 1.005 KJ/kgk Find Heat Transfer in Compressor, Q1-2 = ? Heat Transfer in Coller, Q2-3 = ? From SFEE Q – WS = mf [(h2 – h1) + 1/2( V22 – V12) + g(Z2 – Z1)] There is no data about velocity and elevation so ignoring KE and PE Q1-2 – W1–2 = m[cp(T2 – T1)]...(i) Now P1V1 = mRT1 m = (0.11 × 106 x 1/30)/(287 × 293) m= 0.0436Kg/sec; R= 8314/29 = 287 For Air From equation (i) Q1-2 – 4.15 × 103 = 0.0436[1.005 × 103 (111 – 20)] Q1-2 = 8.137KJ/sec For process 2 – 3; W2-3 = 0  Q2-3 – W2-3 = m[Cp(T2 – T1)] Q2-3 – 0 = 0.0436[1.005 x 103 (– 111 + 25)] Q2-3 = – 3.768KJ/sec 10. Pressure of the steam inside a boiler, as measured by pressure gauge, is 2 N/mm 2. The barometric pressure of the atmosphere is 765 mm of mercury. Find the absolute pressure of steam in N/m2 , kPa, bar and N/mm2. Solution: Given Data: Pressure in side boiler, Pg = 2 N/mm2 = 2 x 106 N/ m2 = 2 x 103 Kpa = 20 Bar Barometric pressure, h = 765mm of Hg for Patm Patm = ρgh = 13.6 × 103 × 9.81 × 765/1000 = 102063.16 N/m2 = 102063.16 Pa = 102.063 KPa = 1.02 Bar = 0.102 N / mm2 Absolute Pressure Pabs = Patm + Pg Pabs = 2 x 106 + 102063.16 N/m2 = 2102063.6 N/m2 Pabs = 2102.063 Kpa Pabs = 21.02 Bar  Pabs = 2.102 N/mm2 UNIT-II 1. A)Explain First law of thermodynamics. B) Explain and derive Steady Flow Energy Equation. Solution : A) The First law of thermodynamics states that energy is neither created nor destroyed. Thus the total energy of the universe is a constant. However, energy can certainly be transferred from one form to another form. The 1st law of thermodynamics can be mathematically stated as follows: §dQ = §dW B) The following assumptions are made in the system analysis: (i) The mass flow through the system remains constant. (ii) Fluid is uniform in composition. (iii) The only interaction between the system and surroundings are work and heat. (iv) The state of fluid at any point remains constant with time. 2. Derive the work done for following process: 1 Isochoric process 2 Isobaric process 3 Isothermal process 4 Adiabatic process 5 Polytrophic process Solution: 4. a) Explain Second Law of Thermodynamics. Prove that violation of Kelvin Plank statement leads to violation of Clausius statement. b) Prove that the violation of Clausius statement leads to violation of Kelvin Plank statement. Solution:  A) The second law of thermodynamics is a general principle which places constraints upon the direction of heat transfer and the attainable efficiencies of heat engines. Kelvin-Planck Statement: It is impossible to devise a cyclically operating device, which produces no other effect than the extraction of heat from a single thermal reservoir and delivers an equivalent amount of work. Heat engine with single thermal reservoir is not possible. Clausius statement It is not possible for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow. Energy will not flow spontaneously from a low temperature object to a higher temperature object. This precludes a perfect refrigerator. The statements about refrigerators apply to air conditioners and heat pumps, which embody the same principles. This is the "second form" or Clausius statement of the second law. Or It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lower-temperature body to higher-temperature body. Equivalence of Kelvin-Planck and Clausius Statements To prove that violation of the Kelvin-Planck Statement leads to a violation of the Clausius Statement, let us assume that Kelvin-Planck statement is incorrect. Consider another device 2 operating as a cycle, which absorbs energy QL as heat from a low temperature thermal reservoir at TL and rejects energy QH (QH=QL+W). Such a device does not violate Clausius statement. If the two devices are now combined, the combined device (enclosed by the dotted boundary) transfers heat QL from the low temperature reservoir at TL to a high temperature reservoir at TH without receiving any aid from an external agent, which is the violation of the Clausius statement. Likewise let us assume that the Clausius statement is incorrect. So we have a device 1, cyclically working transferring heat Q from a low temperature reservoir at TL to a high temperature thermal reservoir at TH. Consider another device 2, which absorbs heat Q1 from a high temperature reservoir at TH does work W and rejects energy Q as heat tot the low temperature reservoir at TL as shown If the two devices are combined (shown in figure by a dotted enclosure), then the combined device receives energy (Q1-Q) as heat from a thermal reservoir and delivers equivalent work (W=Q1-Q) in violation of the Kelvin-Planck statement. Therefore violation of Clausius statement leads to the violation of the Kelvin-Planck statement. Hence, these two statements are equivalent. 5. A cyclic heat engine operates between a source temperature of 800 0C and A Sink temperature of 300C.What is the least rate of heat rejection per KW net output of the engine? Solution: 6. In a steady flow process, a substance flows at the rate of 300 kg/min. It enters at a pressure of 6 bar ,a velocity of 300 m/s internal energy2000kj/kg and specific volume 0.4 m3 kj/kg. It leaves the system at a pressure of 0.1 MPa , a velocity of 150m/s, the internal energy1600 kj/kg and specific volume 1.2 m3. The inlet is 10 m above the outlet. During its passage through the system the substance has a work transfer of 3 MW to the surroundings. Determine the heat transfer in kj /s. Stating whether it is from or to the system. Solution: Given data: Mass flow rate , m = 300 Kg/Min = 5 Kg/Sec Entry conditions,  Pressure, P1= 6 Bar = 600 K Pa Velocity, C1 = 300 m/s Internal Energy u1 = 2000 Kj/Kg Specific Volume, v1 = 0.4 m3/ Kg Inlet Height, Z1 = 10 m Exit Conditions, Pressure, P2= 0.1M Pa = 100 K Pa Velocity, C2 = 150 m/s Internal Energy u2 = 1600 Kj/Kg Specific Volume, v2 = 1.2 m3/ Kg Work Transfer, W = 3 MW = 3000 KW Formula (2000+(600*0.4)+(3002/2)+(10*9.81))+Q= (1600+(100*1.2)+(1502/2))+3000 Q = (1600+(100*1.2)+(1502/2))+3000 - (2000+(600*0.4)+(3002/2)+(10*9.81)) Q= 4731 – 2285.09 Q= 2445.09 KW 7. Explain the difference between heat pump and refrigerator, also find the C.O.P. Solution : 8. A reversible heat engine operates between two reservoirs at temperature of 600 oC and 40oC. The engine derives a reversible refrigerator which operates between reservoirs at temperature of 40o C and -20o C. The heat transfer to the engine is 2MJ and the net work output of the combined engine and refrigerator plant is 360kJ. Find the heat transfer to the refrigerant and the net heat transfer to the reservoir at 40 o C. Also find these values if the efficiencies of heat engine and C.O.P. of refrigerator are each 40% of the maximum possible values. Solution:  9. What is Clausius inequality? Explain with its importance. Solution: 10  UNIT-III 1. A mass of 1.5kg and volume of 0.14m3 of certain gas at 40 bar is expended isentropically such that temperature falls to 500 K. Determine 1. Initial temperature of gas 2. Work done during the process 3. Pressure at end of expansion. Take R=0.287 kJ/kgK , and Cv=0.718 kJ/kgK Solution:  2. One Kg steam at a pressure of 4 bar and a dryness fraction of 0.963 is compressed isentropically until it is dry saturated. Heat is then supplied at constant pressure until the initial volume is attained and the steam is finally restored to its initial state by constant volume cooling. Evaluate the work and heat transfer in each step and verify that the net work done is equal to the difference between the heat supplied and heat rejected over the cycle Solution: Givendata: Mass of steam, m = 1 Kg 3. A cyclic steam power plant is to be designed for a steam temperature at turbine inlet of 3600C and an exhaust pressure of 0.08 bar. After isentropic expansion of steam in the turbine, the moisture content at the turbine exhaust is not to exceed 15%. Determine the greatest allowable steam pressure at he turbine inlet, and calculate the Rankine efficiency. Solution:  Explain following: (i) Avogadro’s Law (ii) Bleeding process (iii) Enthalpy 4. Explain following: (i) Avogadro’s Law (ii) Bleeding process (iii) Enthalpy Solution: Avogadro’s Law: Avogadro‟s law states. “Equal volumes of different perfect gases at the same temperature and pressure, contain equal number of molecules”. Bleeding process: Bleed is amount of steam output from turbine through pipe and exit from final stage of turbine. This bleed enters to feed water heater (low and high) and deaerator to increase unit efficiency. The bleed steam is actually the steam which has already performed work on turbine blades. Due to bleeding less heat is supplied to feed water in the boiler and it increases the mean temperature of heat addition which increases the efficiency Enthalapy: Enthalpy is the measurement of energy in a thermodynamic system. The quantity of enthalpy equals to the total content of heat of a system, equivalent to the system's internal energy plus the product of volume and pressure. H=u+Pv 5. Define Cp& Cv. Derive following expression: Cp-Cv =R Solution: specific heat capacity at constant pressure (Cp): It is defined as the amount of heat energy required raising or lowering the temperature of unit mass of the substance through one degree when the pressure kept constant. It is denoted by Cp.  specific heat capacity at constant volume(Cv): It is defined as the amount of heat energy required to raising or lowering the temperature of unit mass of the substance through one degree when volume kept constant. Consider a gas heated at constant pressure So, heat supplied, Q = mCp (T2 – T1) Work done, W = p(V2 – V1)= m R (T2 – T1) Change in internal energy, dU = mCv (T2 – T1) According to the first law of thermodynamics, Q = W + dU So, mCp (T2 – T1) = mR (T2 – T1)+ mCv(T2 – T1) Cp = r + Cv Cp – cv = R 6. Write short notes on “Mollier diagram”. Why do isobars on the Mollier diagram diverge from one another? Solution: Enthalpy–Entropy Chart An enthalpy–entropy chart, also known as the H–S chart or Mollier diagram, plots the total heat against entropy, describing the enthalpy of a thermodynamic system A vertical line in the h–s chart represents an isentropic process. The divergence of isobars is particularly significant for substances that undergo phase changes, such as water changing from a liquid to a vapor (or vice versa). During a phase change, the enthalpy of the substance changes significantly while the entropy remains relatively constant. This leads to a steeper slope of the isobars in the region of the phase change, causing them to diverge. 7. A steam pressure of holding capacity 4 m3 contains a mixture of saturated water and saturated steam at 2500C. The mass of the liquid present is 1 ton. Determine (i) Quality; (ii) Specific Volume; (iii) Specific Enthalpy; (iv) Specific Entropy and (v) Specific Internal Energy of steam. Solution: Volume, V = 4 m3 Temperature, T = 250°C Mass, m = 1000 kg From the Steam tables corresponding to 250°C vf = v1 = 0.001251 m3/kg vg = vs = 0.050037 m3/kg p = 39.776 bar Total volume occupied by the liquid, V1 = m1 × v1  = 100 × 0.001251 = 0.1251 m3. Total volume of the vessel, V = Volume of liquid + Volume of steam = V1 + VS 4= 0.1251 + Vs Vs= 3.875 m3 Mass of steam, ms = VS / vs = 3.875 / 0.050037 = 77.5 kg. Mass of mixture of liquid and steam, m = m1 + ms = 1000 + 77.5= 1077.5 kg Total specific volume of the mixture, v = 4/ 1077.5 = 0.00371 m3/kg We know that, v = vf + x vfg 0.00371 = 0.001251 + x (0.04885) x = 0.06 From Steam table corresponding to 250 °C, hf = 1085.8 KJ / kg hfg = 1714.6 KJ / kg sf = 2.794 KJ / kg K sfg = 3.277 KJ / kg K. Enthalpy of mixture, h = hf + x hfg = 1085.8 + 0.06 × 1714.6 = 1188.67 KJ / kg Entropy of mixture, s = sf + x sfg = 2.794 + 0.06 × 3.277 = 2.99 kJ / kg K. Internal energy, u = h –p v = 1188.67 –39.776×102 × 0.00418 = 1172 KJ / kg. 8. A rigid close tank of volume 3m3 contains 5 kg of wet steam at a pressure of 200 kPa. The tank is heated until the steam becomes dry saturated. Determine final pressure and heat transfer to the tank. Solution:  9. What is pure substance? Draw the phase equilibrium diagram for a pure substance on T- V plot with relevant constant property lines. Solution: Pure Substance A substance that has a fixed chemical composition throughout is called a pure substance such as water, air, and nitrogen. A pure substance does not have to be of a single element or compound. A mixture of two or more phases of a pure substance is still a pure substance as long as the chemical composition of all phases is the same. We will consider a phase change of 1 kg of liquid water contained within a piston-cycinder assembly as shown in Figure. As the water is heated at constant pressure, the temperature increases with a slight increase in specific volume until the system reaches point (f). saturated liquid state corresponding to 1.014 bar. The saturation temperature for water at 1.014 bar is 100oC. The liquid states along the line segment 1-f are called subcooled or compressed liquid states. When the system is at the saturated liquid state, 10. Draw PVT Surfaces and explain salient points on the curves. Solution:  The location of a point on the p-v-T surface gives the values of pressure, specific volume, and temperature at equilibrium. The regions on the p-v-T surface labeled solid, liquid, and vapor are single-phase regions. The state of a single phase is determined by any two of the properties: pressure, temperature, and specific volume. The two-phase regions where two phases exist in equilibrium separate the single-phase regions. The two-phase regions are: liquid-vapor, solid-liquid, and solid-vapor. Temperature and pressure are dependent within the two-phase regions. Once the temperature is specified, the pressure is determined and vice versa. The states within the two-phase regions can be fixed by specific volume and either temperature or pressure. UNIT IV 1. A sling psychrometer reads 390C dry bulb Temperature and 350C wet bulb Temperature. Find the humidity ratio, Relative humidity, dew point Temperature, specific volume, and enthalpy of air Solution:  2. 100m3 of air per min at 400C DBT and 15% relative humidity is passed through adiabatic humidifier. The air is coming out at 250C DBT and 200C WBT. Find: i) Dew Point Temperature ii) Relative Humidity iii) Water carried by the air per min. Solution:  3. A gas mixture contains 1 Kg of O2 and 3 Kg of N2. The pressure and temperature of the mixture are 1 bar and 270C. Determine: i) Mass fraction and mole fraction of each constituent ii) Average molecular weight of mixture iii) Partial Pressure of constituents iv) Specific gas constant v) Mixture volume vi) Mixture density. Solution: Given Data: Mass of O2 , mO2 = 1 Kg Mass of N2 , mN2 = 3 Kg Pressure P = 1 Bar Temperature, T= 27 C= 300K i) Mass and mole fractions Mass fraction of O2, mfO2 = mO2/m = 1/3=0.333 Mass fraction of N2, mfN2 = mN2/m = 2/3 = 0.666 Moles of O2, No2 = mo2/ MO2 = 1/32= 0.03125 Moles of N2, NN2 = mN2/ MN2 = 3/28= 0.107 Mole fraction of O2, YO2 = No2/N = 0.03125/0.13825 =0.226 Mole fraction of N2, YN2 = NN2/N = 0.107/0.13825 =0.774 ii) Average Molecular weight of mixture, M mix= (YO2*MO2) + (YN2*MN2) M mix= (0.226*32) + (0.774*28) M mix= 28.84 iii) Partial Pressure of constituents  Partial Pressure of O2, PO2 = YO2*P =0.226*1=22.6 Kpa Partial Pressure of N2, PN2 = YN2*P =0.764*1=76.4 Kpa iv) Specific gas constant R = Ru/M R = 8.314/28.84 R= 0.288 Kj/Kg K v) Mixture volume, V PV=mRT V=mRT/P = 4*0.288*300/100 =3.456 m3 Mixture Volume V= 3.456 m3 vi) Mixture density. = m/V= 4/3.456 = 1.157 kg/m3 4. Prove that the violation of Clausius statement leads to violation of Kelvin Plank statement. Solution: Consider a cyclically working device 1, which absorbs energy Q1 as heat from a thermal reservoir at TH. Equivalent amount of work W(W=Q1) is performed. Consider another device 2 operating as a cycle, which absorbs energy QL as heat from a low temperature thermal reservoir at TL and rejects energy QH (QH=QL+W). Such a device does not violate Clausius statement. If the two devices are now combined, the combined device (enclosed by the dotted boundary) transfers heat QL from the low temperature reservoir at TL to a high temperature reservoir at TH with out receiving any aid from an external agent, which is the violation of the Clausius statement. 5. A reversible adiabatic process begins at P1= 10 bar, T1 = 3000C and ends with P2= 2 bar. Find the specific volume and the work done per kg of fluid if (i) the fluid is air and (ii) the fluid is steam. Solution:  6. A mixture of hydrogen (H2) and oxygen (O2) is to be made so that the ratio of H 2 to O2 is 2:1 by volume respectively. Calculate i) the mass of O2 required ii) volume of the container Solution:  7. An air-water vapour mixture at 0.1 MPa, 300C, 80% relative humidity has a volume of 50 m3. Calculate Specific humidity, Dew point, WBT, mass of dry air and mass of water vapour. Solution: 9. Explain the following: i) Heating and dehumidification ii) Cooling and dehumidification. Solution: Cooling and dehumidification. The process in which the air is cooled sensibly and at the same time the moisture is removed from it is called as cooling and dehumidification process. Cooling and dehumidification process is obtained when the air at the given dry bulb and dew point (DP) temperature is cooled below the dew point temperature. Heating and Dehumidification The process in which the air is heated and at the same time moisture is removed from it is called as heating and dehumidification process. This process is obtained by passing the air over certain chemicals like alumina and molecular sieves. These elements have inherent properties due to which they keep on releasing the heat and also have the tendency to absorb the moisture. 10. What are gravimetric and volumetric analyses? Gravimetric analysis is a method in analytical chemistry to determine the quantity of an analyte based on the mass of a solid. Example: Measuring the solids suspended in the water sample – Once a known volume of water is filtered, the collected solids are weighed.  The principle behind the gravimetric analysis is that the mass of an ion in a pure compound and can be determined. Later, used to find the mass percent of the same ion in a known quantity of an impure compound. Volumetric analysis is a process used to determine the amount of the desired constituent by its volume. Hence it is a quantitative determination. Here, the volume of the constituent is measured via a titration (titrimetric analysis). UNIT-V 1. With the help of p-v and T-s diagram, show that for the same maximum pressure and temperature of the cycle and the same heat rejection, ηDiesel > ηDual > ηOtto Solution: For the same maximum pressure and temperature:  1-6-4-5: Otto cycle 1-7-4-5: Diesel cycle 1-2-3-45 Dual cycle Q1 is represented by: Area Under 6-4 for Otto Cycle Area Under 7-4 for Diesel Cycle Area Under 2-3-4 Dual Cycle As we can observe Q2 we can conclude that ηDiesel > ηDual > ηOtto 2. In an air standard otto cycle, the compression ratio is 7 and the compression begins at 1 bar and 313K. the heat added is 2510 kJ/kg. Find the (1). Maximum temp and pressure of the cycle (2) Work done per kg of air (3) Cycle efficiency and mean effective pressure. Take for air Cv=0.718kJ/kgK and R=287 J/kgK Solution:  3. (a) Derive an expression for efficiency of Ericsson cycle. (b) An Ericsson regenerative engine works between the temperature limit of 250C and 2300C. if the ratio of expansion is 2. Determine Work done per kg of air and efficiency of the cycle. Solution: Ericsson Cycle , which consists of two isothermal and two constant pressure processes. It is made thermodynamically reversible by the action of a regenerator. The p-v and T-s diagrams of the Ericsson cycle are shown in the figure. This cycle is used these days in the manufacture of closed-cycle type gas turbines. Following are the four Processes of an Ericsson cycle:  1-2 Process (Isothermal expansion or heat addition)  2-3 Process (Constant pressure or isobaric heat rejection)  3-4 Process (Isothermal compression)  4-1 Process (Constant pressure or isobaric heat absorption) efficiency, 4. Two engines are to operate on otto and diesel cycle with the following data: Maximum temperature=1500K; Exhaust temperature=700K; Ambient conditions= 1 bar and 300K Compare the compression ratios and maximum pressures and efficiencies of two engines. Solution: 5. An air engine, working on stirling cycle, has lower limit of temperature of 400 0C. The maximum and minimum pressure limits are 12 bar and 2 bar. If the expansion ratio of the cycle is 3 then find the ideal efficiency Solution:  6. Derive an expression for pressure ratio, temperature ratio and efficiency for otto cycle Solution  7. Derive an expression for pressure ratio, temperature ratio and efficiency for Diesel cycle SOLUTION:  8. Derive an expression for Efficiency in following cycles 1. Stirling Cycle 2. Air Standard Cycle 3. Bryaton Cycle Solution : 1. Stirling Cycle: Following are the four cycles of an Ideal Stirling Cycle: 1. Isothermal compression (Process 1-2) 2. Constant volume cooling (Process 2-3) 3. Isothermal expansion (Process 3-4) 4. Constant volume heating (Process 4-1) Process 1-2 Process 2-3 Process 3-4 Process 4-1 2. Air standard cycle: Otto cycle 3. Brayton cycle  9. Derive an expression of efficiency of Atkinston cycle Solution: The cycle comprises the following processes: adiabatic compression 1 → 2, isochoric heat addition 2 → 3, adiabatic expansion 3 → 4, and isobaric heat removal 4→ 1.  10. Derive an expression of efficiency of Rankine cycle Solution: