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BULACAN STATE UNIVERSITYCOLLEGE OF ENGINEERINGES 653: Basic ThermodynamicsOLAYON, JESS RYAN D.ECE 4CENGR. HAIVELL JOY C. MATIASTerminologies: Thermodynamics It is a branch of physical sciences that treats various phenomena of energy and the related properties of matter, especially of the laws of tra...

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BULACAN STATE UNIVERSITY COLLEGE OF ENGINEERING

ES 653: Basic Thermodynamics

OLAYON, JESS RYAN D. ECE 4C

ENGR. HAIVELL JOY C. MATIAS

Terminologies: 1.

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Thermodynamics - It is a branch of physical sciences that treats various phenomena of energy and the related properties of matter, especially of the laws of transformation of heat into other forms of energy and vice versa. Surroundings - It is the mass or region outside the system. Boundary - It is the real or imaginary surface that separates the system from its surroundings. It can be either fixed or movable. Thermodynamic system - It refers to the quantity of matter or certain volume in space chosen for study. Closed system - It is a system in which there is no transfer of matter across the boundary. It consists of a fixed amount of mass, and no mass can cross its boundary. That is, no mass can enter or leave a closed system. Open system - It is a system in which there is a flow of matter through the boundary. It usually encloses the device that involves mass flow, such as: compressor, turbine, or nozzle. Isolated system - It is a system in which neither mass nor energy cross the boundaries and it is not influenced by the surroundings. State properties - It refers to the physical condition of the working substance such as temperature, pressure, density, specific volume, specific gravity or relative density. Transport properties - It refers to the measurement of diffusion within the working medium resulting from molecular activity, like; viscosities, thermal conductivities, etc. Intensive properties - These are properties which are size independent such as temperature, pressure, and density. Extensive properties - These are properties which depend on the size or extent of the system. Temperature - It is an indication or degree of hotness and coldness and therefore a measure of intensity of heat. Absolute temperature - It is the temperature measured from absolute zero. Absolute zero - It is the temperature at which the molecules stop moving. It is equivalent to 0 K or 0 °R. Temperature interval - It is the difference between two temperature readings from the same scale, and the change in temperature through which the body is heated. Pressure - It is the force exerted per unit area. Absolute pressure - It is the true pressure measured above a perfect vacuum. Gage pressure - It is the pressure measured from the level of atmospheric pressure by most pressure recording instrument like pressure gage and open-ended manometer. Atmospheric pressure - It is the pressure obtained from barometric reading. Density - It is the mass per unit volume, and is also known as mass density. Specific volume - It is the volume per unit mass. Specific gravity - It is the ratio of the density of a certain substance to the density of water.

23. Heat - It is a form of energy associated with the kinetic random motion of large number of molecules. 24. Sensible heat - It is the heat needed to change the temperature of the body without changing its phase. 25. Latent heat - It is the heat needed by the body to change its phase without changing its temperature. 26. Sublimation - It is the term used to describe the process of changing solid to gas without passing the liquid state. 27. Deposition - It is the reversed of sublimation. It is the process of changing gas to solid without passing the liquid state. 28. Entropy - It is the measure of randomness of the molecules of a substance. 29. Enthalpy - It is the heat energy transferred to a substance at a constant pressure process. 30. Internal energy - It is the energy stored within the body. It is the sum of the kinetic energies of all its constituent particles plus the sum of all the potential energies of interaction among these particles. 31. First law of thermodynamics - It states that energy can neither be created nor destroyed; rather it can only be transformed from one form to another. 32. Second law of thermodynamics - It states that heat cannot be transferred from cold body to a hot body without an input of work. It similarly states that heat cannot be converted 100% into work. The bottom line is that an engine must operate between a hot and a cold reservoir. Also indicated, is that energy has different levels of potential to do work, and that energy cannot naturally move from one realm of lower potential to a realm of higher potential. 33. Kelvin-Planck statement - It is impossible to construct a heat engine which operates in a cycle and receives a given amount of heat from a high temperature body and does an equal amount of work. 34. Third law of thermodynamics - The total entropy of pure substances approaches zero as the absolute thermodynamic temperature approaches zero. 35. Zeroth law of thermodynamics - It states that when any two bodies are in thermal equilibrium with the third body, then they are in thermal equilibrium with each other. 36. Perfect gas - It is a theoretically ideal gas which strictly follows Boyle’s Law and Charle’s Law of gasses. 37. Thermodynamic process - It is any change in that a system undergoes from one equilibrium state to another. 38. Path - It is the series of states through which a system passes during a process. 39. Reversible process - It is the process that can be reversed without leaving any trace on the surroundings. That is, both the system and the surroundings are returned to their initial states at the end of process. It is also known as qua si-equilibrium process. 40. Irreversible process - It is the process that proceeds spontaneously in one direction but the other. Once having taken place, the process cannot reverse itself and always results in an increase of molecular disorder. 41. Cyclic process - It is a process which gives the same states/conditions after the system undergoes a series of processes. 42. Isometric process - It is an internally reversible constant volume process of a working substance. It is also known as isochoric or isovolumic process. 43. Isobaric process - It is an internally reversible constant pressure process of a working substance. 44. Isothermal process - It is an internally reversible constant temperature process of a working substance.

45. Isentropic process - It is an internally reversible constant entropy process of a working substance. It is also known as reversible adiabatic process. 46. Adiabatic process - It is a reversible process in which there is now flow of heat between a system and its surroundings. 47. Carnot cycle - It is the most efficient hypothetical cycle which is composed of four reversible processes: two isothermal and two adiabatic processes. 48. Rankine cycle - It is a thermodynamic cycle derived from Carnot vapour power cycle for overcoming its limitations. It composed of the following cycles: two isobaric and two adiabatic processes. 49. Otto cycle - It is a constant volume combustion cycle introduced by Nicholas A. Otto. It composed of two isentropic and two isometric processes. 50. Diesel cycle - It is a constant pressure combustion cycle introduced by Rudolf Diesel. It composed of two isentropic, one isobaric and one isometric processes.

Problems: 1.

In an experiment to determine the specific heat of copper, a piece of copper weighing 50 g is first heated to 100 ºC in steam. It is then immersed into water at 27 ºC. The water in the calorimeter weighs 100 g and the inner aluminum cap weighs 50 g. If the final temperature is 30 ºC, what is the specific heat of copper, specific heat of aluminum is 0.22 Cal/g- ºC. Given: mc = 50 g ∆Tc = 100 – 30 = 70 ºC mw= 100 g ∆Tw= 30 – 27 = 3 ºC Cw= 1.0 Cal/g-ºC mal= 10 g ∆Tal= 30 – 27 = 3 ºC Cal= 0.22 Cal/g-ºC Required: Cc=? Solution: 𝑄𝑙𝑜𝑠𝑡 = 𝑄𝑔𝑎𝑖𝑛𝑒𝑑 𝑄𝑐𝑜𝑝𝑝𝑒𝑟 = 𝑄𝑤𝑎𝑡𝑒𝑟 + 𝑄𝑎𝑙𝑢𝑚𝑖𝑛𝑢𝑚 Answer:

𝑚𝑐 𝐶𝑐 ∆𝑇𝑐 = 𝑚𝑤 𝐶𝑤 ∆𝑇𝑤 + 𝑚𝑎𝑙 𝐶𝑎𝑙 ∆𝑇𝑎𝑙 50(𝐶𝑐 )(70℃) = 100(1.0)(3℃) + 10(0.22)(3℃) 𝐶𝑐 = 0.095

𝐶𝑎𝑙 𝑔∙℃

Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 2.

At STP, the density of chlorine is 3.22 kg/m3. What is the weight of this gas when it is contained in a flask of 100 cubic centimeters at 24 ℃ and 100 kPa. Given: ρchlorine= 3.22 kg/m3 T= 24℃ V = 100 cm3 = 0.0001 m3 P = 100 kPa Required: mchlorine=? Solution: At STP: 𝑃 = 𝜌𝑅𝑇 𝑃 𝐽 101.325 𝑅= = 0.1153 = 𝑔 ∙ 𝐾 𝜌𝑇 3.22(273) When contained in flask: 𝑃𝑉 100,000(0.0001) 𝑚= = 𝑅𝑇 0.1153(24 + 273) Answer: 𝑚 = 0.292 𝑔 Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects

3.

How much work is necessary to compress air in an insulated cylinder from 0.20 m3 to 0.01 m3? Use T1=20℃, P1=100kPa and k=1.4 . Given: V1= 0.20 m3 T1= 20℃ P1= 100 kPa V2= 0.01 m3 k=1.4 Required: W=? Solution: 𝑃1 𝑉1𝑘 = 𝑃2 𝑉2𝑘 100(0.20)1.4 = 6,628.91 𝑘𝑃𝑎 𝑃2 = (0.01)1.4 𝑃2 𝑉2 − 𝑃1 𝑉1 𝑊= 𝑘−1 6,628.91(0.01) − 100(0.2) 𝑊= 1.4 − 1 Answer: 𝑊 = 115.72 𝑘𝐽

Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects

4.

A heat engine is operated between temperature limits of 1370℃ and 260℃. Engine is supplied with 14,142 kJ/kWh. Find the Carnot cycle efficiency in percent. Given: T1= 1370℃ + 273 = 1643 K T2= 260℃ + 273 = 533 K Required: ec = ? Solution: 𝑇1 − 𝑇2 𝑒𝑐 = 𝑇1 1643 − 533 𝑒𝑐 = 1643 Answer: 𝑒𝑐 = 67.56% Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects

5.

A closed vessel contains air at a pressure of 160 kN/m2 gauge and temperature of 30℃. The air is heated at constant volume to 60℃ with the atmospheric pressure of 759 mmHg. What is the final gauge pressure? Given: Patm= 759 mmHg = 101.20 kPa P1(gauge)= 160 kPa T1= 30℃ + 273 = 303 K T2= 60℃ + 273 = 333K Required: P2(gauge)= ? Solution: 𝑃1 = 𝑃1(𝑔𝑎𝑢𝑔𝑒) + 𝑃𝑎𝑡𝑚 𝑃1 = 160 + 101.20 = 261.20 𝑘𝑃𝑎 𝑃1 𝑃2 = 𝑇1 𝑇2 𝑃1 𝑇2 𝑃2 = 𝑇1 (261.20)(333) 𝑃2 = = 287.06 𝑘𝑃𝑎 303 𝑃2(𝑔𝑎𝑢𝑔𝑒) = 𝑃2 − 𝑃𝑎𝑡𝑚 𝑃2(𝑔𝑎𝑢𝑔𝑒) = 287.06 − 101.20 Answer:

𝑃2(𝑔𝑎𝑢𝑔𝑒) = 185.86 𝑘𝑃𝑎

Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 6.

A volume of 450 cm3 of air is measured at a pressure of 740 mmHg absolute and a temperature of 20 ℃. What is the volume in cm3 at 750 mmHg and 0℃? Given: V1= 450 cm3 P1= 740 mmHg T1= 20℃ + 273 = 293 K P2= 750 mmHg T2= 0℃ + 273= 273 K Required: V 2= ? Solution: 𝑃1 𝑉1 𝑃2 𝑉2 = 𝑇1 𝑇2 𝑃1 𝑉1 𝑇2 740(450)(273) = 𝑉2 = 750(293) 𝑃2 𝑇1 Answer: 𝑉2 = 408.25 𝑐𝑚 3

Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects

7.

A steam condenser receives 10 kg per second of steam with an enthalpy of 2570 kJ/kg. Steam condenses into liquid and leaves with an enthalpy of 160 kJ/kg. Cooling water passes through the condenser with temperature increases from 13 ℃ to 24℃. Calculate the cooling water flow rate in kg/s. Given: ms= 10 kg/s H1= 160 kJ/kg H2= 2570 kJ/kg T1= 13℃ T2= 24℃ Required: mw = ? Solution: 𝐻𝑒𝑎𝑡 𝑙𝑜𝑠𝑠 𝑏𝑦 𝑡ℎ𝑒 𝑠𝑡𝑒𝑎𝑚 = 𝐻𝑒𝑎𝑡 𝑔𝑎𝑖𝑛 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑚𝑠 (𝐻2 − 𝐻1 ) = 𝑚𝑤 𝐶𝑝𝑤 ∆𝑇𝑤 𝑚𝑠 (𝐻2 − 𝐻1 ) 𝑚𝑤 = 𝐶𝑝𝑤 ∆𝑇𝑤 10(2570 − 160) 𝑚𝑤 = 4.187(24 − 13) Answer: 𝑚𝑤 = 523 𝑘𝑔/𝑠 Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects

8.

Ammonia weighing 22 kg is confined inside a cylinder equipped with a piston which has an initial pressure of 413 kPa at 38℃. If 2900 kJ of heat is added to ammonia until its pressure and temperature are 413 kPa and 100℃, respectively. What is the amount of work done by the fluid in kJ? (Note: Molecular weight of NH3= 17) Given: P1= 413 kPa T1= 38℃ + 273= 311 K m= 22 kg P2= 413 kPa T2= 100℃ + 273 = 373 K Q= 2900 kJ MWNH3=17 kg/mol Required: W=? Solution: 𝑊 = 𝑃(𝑉2 − 𝑉1 ) Solving for V1 and V2: 𝑃1 𝑉1 = 𝑚𝑅𝑇1 𝑅𝑜 𝑘𝐽 8.314 𝑅= = 0.489 = 𝑘𝑔 ∙ 𝐾 17 𝑀𝑊 𝑚𝑅𝑇1 22(0.489)(311) 𝑉1 = = = 8.101 𝑚 3 𝑃1 413 𝑚𝑅𝑇2 22(0.489)(373) 𝑉2 = = = 9.716 𝑚3 𝑃2 413 𝑊 = 413(9.716 − 8.101) Answer: 𝑊 = 667 𝑘𝐽 Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects

9.

Compute the gas constant of a mixture of 10 kg of oxygen and 20 kg of nitrogen. Given: m(O2) = 10 kg MW = 32 kg/mol m(N2) = 20 kg MW = 28 kg/mol Required: Rmix = ? Solution: ∑ 𝑚𝑖 𝑅𝑖 𝑅𝑚𝑖𝑥 = ∑ 𝑚𝑖 8.314 8.314 10 ( ) + 20 ( 28 ) 32 𝑅𝑚𝑖𝑥 = 10 + 20

Answer:

𝑅𝑚𝑖𝑥 = 0.2845

𝑘𝐽

𝑘𝑔 ∙ 𝐾

Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 10. The maximum thermal efficiency possible for a power cycle operating between 1200 ℉ and 225℉ is: Given: T1 = 1200℉ + 460 = 1660 R T2 = 225℉ + 460 = 685 R Required: ec=? Solution: ℃ 1660 − 685 𝑒𝑐 = 1660 Answer: 𝑒𝑐 = 58.73% Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 11. A 30 kg iron was put in a container with water. The water is at 10℃ and the iron has an initial temperature of 493 K., until the iron was in thermal equilibrium with the water. Find the change in entropy. Given: m= 30 kg T2 = 10℃ + 273= 283K T1= 493K Required: ΔS=? Solution: 𝑇2 ∆𝑆 = 𝑚𝐶𝑝 ln 𝑇1 283 ∆𝑆 = 30(0.4) ln 493 Answer: 𝑘𝐽 ∆𝑆 = −6.6 𝐾 Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 12. Twenty grams of oxygen gas (O2) are compressed at constant temperature of 30℃ to 5% of its original volume. Find the work done on the system. Given: m = 20 g T = 30℃ + 273 = 303K V2 = 0.05V1 Required: W=? Solution: 𝑉2

𝑊 = − ∫ 𝑃𝑑𝑉 𝑉1

𝑉2

𝑉1

𝑅𝑜 𝐶𝑎𝑙 1.98 = 0.0619 = 𝑔∙𝐾 32 𝑀𝑊 0.05𝑉1 𝑊 = −(20)(0.0619)(303) ln 𝑉1 𝑅=

Answer:

𝑊 = −𝑚𝑅𝑇 ln

𝑊 = 1124 𝐶𝑎𝑙

Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects

13. A device produces 37.5 joules per cycle. There is one power stroke per cycle. Calculate the power output if the device is run at 45 rpm. Given: 37.5 J/cycle 45 rpm Required: Po Solution: 37.5 𝐽 45 𝑐𝑦𝑐𝑙𝑒 1 𝑚𝑖𝑛 ) 𝑃= ) ( 60 𝑠 ( 𝑚𝑖𝑛 𝑐𝑦𝑐𝑙𝑒 Answer: 𝑃 = 28.125 𝑊 Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 14. Five moles of water vapor at 100℃ and 1 atmospheric pressure are compressed isobarically to form liquid at 100 ℃. The process is reversible and the ideal gas laws apply. Compute the work, in joules, done on the system. Note: R= 0.0821 Latm/mol-°R, υf= 0.001044 m3/kg, MH2O= 18.016 kg/mol. Given: n= 5 mol T= 100℃ + 273 = 373 K P= 1 atm= 101.325 kPa Required: W=? Solution: 𝑃𝑉 = 𝑛𝑅𝑇 5(0.0821)(373) = 153 𝐿 = 0.153 𝑚3 𝑉= 1 𝑉2 = 𝑛𝑀𝐻2𝑂 𝜐𝑓 = 5(18.016)(0.001044) = 0.094 𝑚3 𝑊 = −𝑃 (𝑉2 − 𝑉1 ) 𝑊 = −101,325(0.094 − 0.153) Answer: 𝑊 = 5.97 𝑘𝐽 ≅ 6 𝑘𝐽 Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 15. A gas at 65 kPa, 200°C is heated in a closed, rigid vessel till it reaches to 400°C. Determine the amount of heat required for 0.5 kg of this gas if internal energy at 200°C and 400°C are 26.6 kJ/kg and 37.8 kJ/kg respectively. Given: m = 0.5 kg u1 = 26.6 kJ/kg u2= 37.8 kJ/kg Required: Q=? Solution: As the vessel is rigid therefore work done shall be zero. 𝑊=0 From first law of thermodynamics: 𝑄 = 𝑈2 − 𝑈1 + 𝑊 = 𝑚(𝑢2 − 𝑢1 ) + 0 𝑄 = 0.5(37.8 − 26.6) Answer: 𝑄 = 5.6 𝑘𝐽 Source: Applied Thermodynamics by Onkar Singh 16. Find out the pressure difference shown by the manometer deflection of 30 cm of Mercury. Take local acceleration of gravity as 9.78 m/s2 and density of mercury at room temperature as 13,550 kg/m3. Given: ρ= 13,550 kg/m3 h=30 cm g= 9.78 m/s2 Required: P=? Solution: 𝑃 = 𝜌𝑔ℎ = 13,550(30 × 102 )(9.78) Answer: 𝑃 = 39,755.70 𝑃𝑎 Source: Applied Thermodynamics by Onkar Singh

17. In a cinema hall with a seating capacity of 500 persons the comfort conditions are created by circulating hot water through pipes in winter season. Hot water enters the pipe with enthalpy of 80 kcal/kg and leaves the pipe with enthalpy of 45 kcal/kg. The difference in elevation of inlet pipe and exit pipe is 10 m with exit pipe being higher than inlet pipe. Heat requirement per person is 50 kcal/hr. Estimate the quantity of water circulated per minute, neglecting changes in velocity. Given: h1= 80 kcal/kg h2=45 kcal/kg z= 10m Required: m= kg/min Solution: 𝐻𝑒𝑎𝑡 𝑙𝑜𝑠𝑡 = −500(50) = −25000 𝑘𝑐𝑎𝑙/ℎ𝑟 𝑄 + 𝑚1 (ℎ1 + 𝑔𝑧1 ) = 𝑚 2 (ℎ2 + 𝑔𝑧2 ) 𝑄 + 𝑚 (ℎ1 − ℎ2 ) = 𝑚(𝑔𝑧2 − 𝑔𝑧1 ) −25000 × 103 (4.18) + 𝑚(80 − 45) = 𝑚 (9.81)(10) 𝑚 = 714.75 𝑘𝑔/ℎ𝑟 Answer: 𝑚 = 11.91 𝑘𝑔/𝑚𝑖𝑛 Source: Applied Thermodynamics by Onkar Singh 18. A gas is enclosed in a cylinder with a weighted piston as the top boundary. The gas is heated and expands from a volume of 0.04 m3 to 0.10 m3 at a constant pressure of 200 kPa. Find the work done on the system. Given: V1= 0.04 m3 V2= 0.10 m3 P= 200 kPa Required: W=? Solution: 2

𝑊 = ∫ 𝑃𝑑𝑉 1

Answer:

𝑊 = 𝑃 (𝑉2 − 𝑉1 ) 𝑊 = 200(0.10 − 0.04) 𝑊 = 12 𝑘𝐽

Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 19. In the process where the product of pressure and volume is constant, a gas compression is carried out from an initial pressure of 200 kPa to a final pressure of 800 kPa. Considering that the initial specific volume is 0.10 m3/kg, determine the work done per kilogram of gas. Given: P1= 200 kPa P2= 800 kPa ν= 0.10 m3/kg Required: Work per kilogram=? Solution: 𝑃1 𝑉1 = 𝑃2 𝑉2 (200)(0.10) 𝑚3 = 0.025 = 0.025 𝑉2 = 800 𝑘𝑔 𝑉2 𝑊 = 𝑃1 𝑉1 ln 𝑉1 0.025 𝑊 = 200(0.10) ln 0.100 Answer: 𝑘𝐽 𝑊 = −27.7 𝑘𝑔 Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects

20. A steady state device has the following conditions of the working substance at the entrance: pressure equals 100 psia and density is 62.4 lbm/ft3. If 10,000 ft3/min of this fluid enters the system, determine the exit velocity if the exit area is 2 ft 2. Given: P= 100 psia ρ= 62.4 lbm/ft3 υ1= 10,000 ft3/min A= 2 ft2 Required: υ2=? Solution: 𝑚′1 = 𝑚2′ 𝜌1 𝐴1 𝑉1 = 𝜌2 𝐴2 𝑉2 𝜌1 𝜐1 = 𝜌2 𝐴2 𝑉2 62.4(10,000) = 62.4(2)𝑉2 Answer: 𝑓𝑡 𝑉2 = 5000 𝑚𝑖𝑛 Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 21. In a nozzle air at 627°C and twice atmospheric pressure enters with negligible velocity and leaves at a temperature of 27°C. Determine velocity of air at exit, assuming no heat loss and nozzle being horizontal. Take C P = 1.005 kJ/kg.K for air. Given: T1= 627℃ + 273= 900 K T2= 27℃ + 273 = 300 K Required: υ2=? Solution: 𝐶2 = √2(ℎ1 − ℎ2 )

𝐶2 = √2𝐶𝑝 (𝑇1 − 𝑇2 )

Answer:

𝐶2 = √2(1.005)(900 − 300) 𝐶2 = 1098.2

𝑚 𝑠

Source: Applied Thermodynamics by Onkar Singh 22. Nitrogen is isentropically expanded from 620℉ to 60℉ with volumetric ratio (V 2/V1) equal to 6.22. If the value of the gas constant (R) is 0.0787 Btu/lbm-R, compute the work done by the gas. Given: T1= 620℉ T2= 60℉ R= 0.0787 Btu/lbm-R Required: W=? Solution: 𝑅(𝑇1 − 𝑇2 ) 𝑊= 𝑘 0.0787(620 − 60) 𝑊= 0.40 Answer: 𝐵𝑡𝑢 𝑊 = 99.22 𝑙𝑏𝑚 Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 23. Helium is compressed isothermally from 14.7 psi and 68 ℉. The compression ratio is 4. Determine the change in entropy of the gas if the gas is 0.4961 Btu/lbm-R. Given: V1/V2= 4 R= 0.4961 Btu/lbm-R P= 14.7 psi Required: ∆S Solution: 𝑉2 ∆𝑆 = 𝑅𝑙𝑛 𝑉1

Answer:

1 ∆𝑆 = 0.4961𝑙𝑛 41

𝐵𝑡𝑢 ∆𝑆 = −0.689 𝑙𝑏𝑚 ∙ 𝑅

Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 24. A carnot engine operates between 800 R and 1000 R. What is its thermal efficiency? Given: TL= 800 R TH= 1000 R Required: e=? Solution: 𝑇𝐿 𝑒 =1− 𝑇𝐻 800 𝑒 =1− 1000 Answer: 𝑒 = 20% Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 25. In a steam power plant 5 kW of heat is supplied in boiler and turbine produces 25% of heat added while 75% of heat added is rejected in condenser. Feed water pump consumes 0.2% of this heat added for pumping condensate to boiler. Determine the capacity of generator which could be used with this plant. Given: Qadd= 5000 J/s Wt= 0.25(5000) = 1250 J/s Qrejected= 0.75(5000) = 3750 J/s Wp= -(0.002)(5000) = 10 J/s Required: Capacity of generator Solution: 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 = 𝑊𝑡 − 𝑊𝑝 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 = 1250 − 10 Answer: 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 = 1.24 𝑘𝑊 Source: Applied Thermodynamics by Onkar Singh 26. A rigid and insulated tank of 1 m3 volume is divided by partition into two equal volume chambers having air at 0.5 MPa, 27℃ and 1 MPa, 500 K. Determine final pressure and temperature if the partition ...