Thermo 12

BULACAN STATE UNIVERSITY COLLEGE OF ENGINEERING

ES 653: Basic Thermodynamics

OLAYON, JESS RYAN D. ECE 4C

ENGR. HAIVELL JOY C. MATIAS

Terminologies: 1.

2. 3. 4. 5.

6.

7. 8.

9.

10. 11. 12. 13. 14. 15.

16. 17. 18.

19. 20. 21. 22.

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 quasi-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 𝑚𝑐 𝐶𝑐 ∆𝑇𝑐 = 𝑚𝑤 𝐶𝑤 ∆𝑇𝑤 + 𝑚𝑎𝑙 𝐶𝑎𝑙 ∆𝑇𝑎𝑙 50(𝐶𝑐 )(70℃) = 100(1.0)(3℃) + 10(0.22)(3℃) Answer: 𝐶𝑎𝑙 𝐶𝑐 = 0.095 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 101.325 𝐽 𝑅= = = 0.1153 𝜌𝑇 3.22(273) 𝑔∙𝐾 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 𝑃2 = = 6,628.91 𝑘𝑃𝑎 (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

4.

Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 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: e c= ? 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: V2= ? Solution: 𝑃1 𝑉1 𝑃2 𝑉2 = 𝑇1 𝑇2 𝑃1 𝑉1 𝑇2 740(450)(273) 𝑉2 = = 𝑃2 𝑇1 750(293) Answer: 𝑉2 = 408.25 𝑐𝑚3 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 NH 3= 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 8.314 𝑘𝐽 𝑅= = = 0.489 𝑀𝑊 17 𝑘𝑔 ∙ 𝐾 𝑚𝑅𝑇1 22(0.489)(311) 𝑉1 = = = 8.101 𝑚3 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 ( ) 32 28 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

𝑊 = −𝑚𝑅𝑇 ln

𝑉2 𝑉1

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

Answer: 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= ( )( ) 𝑐𝑦𝑐𝑙𝑒 𝑚𝑖𝑛 60 𝑠 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) 𝑉= = 153 𝐿 = 0.153 𝑚3 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 ) = 𝑚2 (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

𝑊 = 𝑃(𝑉2 − 𝑉1 ) 𝑊 = 200(0.10 − 0.04) Answer: 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 m 3/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 𝑉2 = = 0.025 = 0.025 800 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: 𝑚1′ = 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 ) 𝐶2 = √2(1.005)(900 − 300) Answer: 𝐶2 = 1098.2

𝑚 𝑠

Source: Applied Thermodynamics by Onkar Singh 22. Nitrogen is isentropically expanded from 620℉ to 60℉ with volumetric ratio (V2/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/lb m-R. Given: V1/V2= 4 R= 0.4961 Btu/lbm-R P= 14.7 psi Required: ∆S Solution: 𝑉2 ∆𝑆 = 𝑅𝑙𝑛 𝑉1

∆𝑆 = 0.4961𝑙𝑛

1 41

Answer: ∆𝑆 = −0.689

𝐵𝑡𝑢 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 is removed. Given: V= 1 m3 P1= 0.5 MPa T1= 27℃ + 273= 300 K P2= 1 MPa T2= 500 K Required: P f, T f Solution: 𝑃1 𝑉1 0.5 × 106 (0.5) 𝑛1 = = = 0.1002 𝑚𝑜𝑙 𝑅𝑇1 8314(300) 6 𝑃2 𝑉2 1 × 10 (0.5) 𝑛2 = = = 0.1203 𝑚𝑜𝑙 𝑅𝑇2 8314(500) Since the tank is insulated and rigid, we can assume ∆U = 0, W= 0, Q= 0, so writing ∆U ∆𝑈 = 𝑛1 𝐶𝑉 (𝑇𝑓 − 𝑇1 ) + 𝑛2 𝐶𝑉 (𝑇𝑓 − 𝑇2 ) = 0 𝑇𝑓 = 409.11 𝐾 (𝑛1 + 𝑛2 )𝑅𝑇𝑓 (0.1002 + 0.1203)(287)(409.11) 𝑃3 = = = 0.75 𝑀𝑃𝑎 𝑉 1 Answer: 𝑇𝑓 = 409.11 𝐾, 𝑃𝑓 = 0.75 𝑀𝑃𝑎 Source: Applied Thermodynamics by Onkar Singh

27. Ethyl alcohol vaporizes at 78℃, what is the change in entropy if 0.25 kg vaporizes at its boiling point considering that the latent heat of vaporization (Lv) is 1.0 x 105 J/kg. Given: m= 0. 25 kg Lv= 1.0 x 105 J/kg T= 78℃ + 273 = 351 K Required: ∆S =? Solution: 𝑄 = 𝑚𝐿𝑣 𝑄 = 0.25(1.0 × 105 ) 𝑄 = 2.5 × 104 𝐽 𝑄 2.5 × 104 ∆𝑆 = = 𝑇 351 Answer: ∆𝑆 = 71 𝐽/𝐾 Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 28. An ideal gas (at STP) occupies a volume of 22.4 L. While absorbing heat from the surroundings, the gas isobarically expands to 32.4 L. What is the change in the internal energy of the gas? Given: Vi= 22.4 L Vf= 32.4 L T= 273 K P= 101.325 kPa Required: ∆U=? Solution: 𝑊 = 𝑃∆𝑉 𝑊 = (101.325)(32.4) 𝑊 = 1.01 𝑘𝐽 𝑄 = 𝑊 + ∆𝑈 ∆𝑈 = 𝑄 − 𝑊 = 2.53 − 1.0 Answer: ∆𝑈 = 1.52 𝑘𝐽 Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 29. A small gasoline-powered engine leaf blower removes 800 J heat energy from a high temperature reservoir and exhausts 700 J to a low temperature reservoir. What is its engine’s thermal efficiency? Given: Qhot= 800 J Qcold= 700J Required: eth=? Solution: 𝑊𝑛𝑒𝑡 = 𝑄ℎ𝑜𝑡 − 𝑄𝑐𝑜𝑙𝑑 𝑊𝑛𝑒𝑡 = 800 − 700 = 100 𝐽 𝑊𝑛𝑒𝑡 100 𝑒𝑡ℎ = = 𝑄ℎ𝑜𝑡 800 Answer: 𝑒𝑡ℎ = 12.5% Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 30. The volumetric flow rate of standard air is 120 m3/s at dry bulb temperature of 18℃. Compute the standard air volume considering that standard air pressure is 101.325 kPa and standard air temperature at 21.11℃. Given: V1= 120 m3/s T1= 18℃ + 273= 291 K T2= 21.11℃ + 273 = 294.11 K Required: V2=? Solution: 𝑉1 𝑉2 = 𝑇1 𝑇2 𝑉1 𝑇2 𝑉2 = 𝑇1 120(2914.11) 𝑉2 = 291 Answer: 𝑉2 = 121.3 𝑚3 /𝑠 Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects

31. A gaseous mixture has a dew point temperature of 15℃. The total pressure is 143.27 kPa. Determine the amount of water vapor present in a 100 moles of the mixture if the saturation pressure at 15℃ is 1.7051 kPa. Given: P1= 143.27 kPa P2= 1.7051 kPa V1= 100 moles T= 15℃ Required: V2=? Solution: 𝑃2 𝑃1 = 𝑉2 𝑉1 𝑃2 𝑉1 𝑉2 = 𝑃1 1.7051(100) 𝑉2 = 143.27 Answer: 𝑉2 = 1.19 𝑚𝑜𝑙𝑒𝑠 Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 32. A sealed tank contains oxygen at 27℃ at a pressure of 2 atm. If the temperature increases to 100℃, what will be pressure inside the tank? Given: T1= 27℃ T2= 100℃ P1= 2 atm Required: P2=? Solution: 𝑃1 𝑃2 = 𝑇1 𝑇2 𝑃1 𝑇2 𝑃2 = 𝑇1 2(100 + 273) 𝑃2 = 27 + 273 Answer: 𝑃2 = 2.49 𝑎𝑡𝑚 Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 33. The pressure of the nitrogen has thermometer is 78 cm at 0℃. What is the temperature of a liquid in which the bulb of the thermometer is immersed when the pressure is seen to be 87.7 cm? Given: P1= 78 cm T1= 0℃ P2= 87.7 cm Required: T2=? Solution: 𝑃1 𝑃2 = 𝑇1 𝑇2 𝑇1 𝑃2 𝑇2 = 𝑃1 (0 + 273)(87.7) 𝑇2 = 78 Answer: 𝑇2 = 306.95 𝐾 = 34℃ Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 34. A 20 liter sample of gas exerts a pressure of 1atm at 25℃. It is allowed to expand into a 40 liter vessel that is held constant at 100℃, compute the final pressure? Given: P1= 1 atm T1= 25℃ V1= 20 L T2= 100℃ V2= 40L Required: P2=? Solution: 𝑃1 𝑉1 𝑃2 𝑉2 = 𝑇1 𝑇2

𝑇2 𝑃1 𝑉1 𝑉2 𝑇1 (100 + 273)(1)(20) 𝑃2 = 40(25 + 273) 𝑃2 =

Answer: 𝑃2 = 0.63 𝑎𝑡𝑚 Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 35. What is the horsepower required to isothermally compressed 800 ft3 of air per minute from 14.7 psia to 120 psia? Given: V= 800 ft3 P1= 14.7 psia P2= 120 psia Required: W in hp Solution: 𝑃1 𝑊 = 𝑃1 𝑉1 ln ( ) 𝑃2 (14.7)(144)(800) 14.7 𝑊= ln ( ) 33000 120 Answer: 𝑊 = −108 𝐻𝑝 Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 36. A building has to be maintained at 18℃ at all times. A heat pump is required for this when the temperature outside the building drops to -6℃, the building losses heat at the rate of 120,000 kJ/kg. What is the least power required to drive the heat pump? Given: Thigh= 18℃ +273 =291 K Tlow= -6℃ + 273= 267 K Qrejected= 120,000 kJ/kg Required: W=? Solution: 𝑇ℎ𝑖𝑔ℎ 𝐶𝑂𝑃 = 𝑇ℎ𝑖𝑔ℎ − 𝑇𝑙𝑜𝑤 291 𝐶𝑂𝑃 = = 12.125 291 − 267 𝑄𝑟𝑒𝑗𝑒𝑐𝑡𝑒𝑑 𝑊𝑛𝑒𝑡 = 𝐶𝑂𝑃 1 120000( ) 3600 𝑊𝑛𝑒𝑡 = 12.125 Answer: 𝑊𝑛𝑒𝑡 = 2.75 𝑘𝑊 Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 37. A heat engine is operated between temperature limits of 1370℃ and 260℃. Engine supplied with 14, 142 kJ per kwh. What is the carnot cycle efficiency? Given: Thigh= 1370℃ + 273 = 1643 K Tlow= 260℃ + 273 = 533 K Required: e=? Solution: 𝑇ℎ𝑖𝑔ℎ − 𝑇𝑙𝑜𝑤 𝑒= (100) 𝑇ℎ𝑖𝑔ℎ 1643 − 533 (100) 𝑒= 1643 Answer: 𝑒 = 67.56% Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 38. Two pounds of air initially at 60 psia and 600℉ expands insentropically until the temperature is 200℉. Compute the work done by the gas. Given: m= 2 lb T1= 600℉ T2= 200℉ P= 60 psia

Required: Solution:

W=? 𝑚𝑅(𝑇2 − 𝑇1 ) 1−𝑘 2(53.34)(200 − 600) 𝑊= (1 − 1.4)(778) 𝑊=

Answer: 𝑊 = 137 𝐵𝑡𝑢 Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 39. A carnot machine operates between a hot reservoir at 200℃ and a cold reservoir at 20℃. When operated as an engine, it receives 1000 kJ/kg, find the coefficient of performance, when operated as refrigerator. Given: Thigh= 200℃ Tlow= 20℃ Required: e=? Solution: 𝑇𝑙𝑜𝑤 𝑒= 𝑇ℎ𝑖𝑔ℎ − 𝑇𝑙𝑜𝑤 20 + 273 𝑒= (200 + 273) − (20 + 273) Answer: 𝑒 = 1.63% Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 40. A piston-cylinder system contains a gas which expands under a constant pressure of 1200 lbf/ft2. If the piston is displaced 12 in. during the process, and the piston diameter is 24 in., what is the work done in ft-lb by the gas on the piston? Given: P= 1200 lb/ft2 d= 24 in l= 12 in. Required: W=? Solution: 𝑊 = 𝑃𝑑𝑉 𝑙𝑏 𝜋(24)2 𝑖𝑛2 𝑓𝑡 3 𝑊 = 1200 [ ] (12𝑖𝑛) ( 3 3 ) 𝑓𝑡 4 12 𝑖𝑛 Answer: 𝑊 = 3768 𝑓𝑡 − 𝑙𝑏𝑓 Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 41. Helium (R=0.4968 Btu/lb-R) is compressed isothermally from 14.7 psia and 68℉. The compression ratio is 4. Calculate the work done by the gas in Btu/lbm. Given: R= 0.4968 Btu/lb-R T= 68℉ V1/V2= 4 Required: W/m in Btu/lbm Solution: 𝑉2 𝑊 = 𝑚𝑅𝑇 ln 𝑉1 𝑊 𝑉2 = 𝑅𝑇 ln 𝑚 𝑉1 𝑊 1 = (0.4968)(68 + 460) ln 𝑚 4 Answer: 𝑊 𝐵𝑡𝑢 = −364 𝑚 𝑙𝑏𝑚 Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 42. An ideal gas is 45 psig and 80℉ is heated in a closed container to 130℉. What is the final pressure? Given: P1= 45 psig T1= 80℉ T2= 130℉ Required: P2=?

Solution: 𝑃1 𝑃2 = 𝑇1 𝑇2 45 + 14.7 𝑃2 = 80 + 460 130 + 460 Answer: 𝑃2 = 65.23 𝑝𝑠𝑖𝑎 Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 43. Determine the amount of air in a room whose dimensions are 4 m x 5 m x 6 m at 100 kPa and 25℃. Given: V= 4(5)(6)= 120 m3 P= 100 kPa T= 25℃ Required: m=? Solution: 𝑃𝑉 𝑚= 𝑅𝑇 100(120) 𝑚= 0.287(25 + 273) Answer: 𝑚 = 140.3 𝑘𝑔 Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 44. An insulated rigid tank contains helium of 1.5 lb m at 80℉ and 50 psia. A paddle wheel with a power rating of 0.20 hp is operated within the tank for 30 mins. Determine the final temperature of the specific heat at constant volume of 0.753 Btu/lbm-℉. (Note: 1 hp = 2545 Btu/hr) Given: P= 0.20 hp t= 0.5 hr m= 1.5 lb Cv= 0.753 Btu/lbm-℉ T1= 80℉ Required: T2=? Solution: 𝑊 = 𝑃𝑡 𝐵𝑡𝑢 2545 ℎ𝑟 ) (0.5 ℎ𝑟) = 25.45 𝐵𝑡𝑢 𝑊 = 0.02 ℎ𝑝 ( ℎ𝑝 𝐸𝑖𝑛 − 𝐸𝑜𝑢𝑡 = ∆𝑈 𝑊 = ∆𝑈 𝑊 = 𝑚𝐶𝑣 (𝑇2 − 𝑇1 ) 25.45 = 1.5(0.753)(𝑇2 − 80) Answer: 𝑇2 = 102.5℉ Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 45. A 2.53 kJ of heat is absorbed by an ideal gas that occupies a volume of 22.4 liters at STP. If the gas expands isobarically to 32.4 liters, compute the change in the internal energy of the gas. Given: V1= 22.4 L V2= 32.4 L Q= 2.53 kJ Required: ∆U=? Solution: 𝑊 = 𝑃(𝑉2 − 𝑉1 ) 𝑊 = 101.325(32.4 × 10−3 − 22.4 × 10−3 ) 𝑊 = 1.01 𝑘𝐽 ∆𝑈 = 𝑄 − 𝑊 ∆𝑈 = 2.53 − 1.01 Answer: ∆𝑈 = 1.52 𝑘𝐽 Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects

46. An engineer is to design cyclic heat engine that operates between 150℃ and 27℃. If the engine is designed to do work of 100 J per 500 J of input work per cycle, compute the relative efficiency of the engine. Given: Thigh= 150℃ Tlow= 27℃ Wnet= 100 J Qhot= 500 J Required: erelative=? Solution: 𝑇ℎ𝑖𝑔ℎ − 𝑇𝑙𝑜𝑤 (150 + 273) − (27 + 273) 𝑒𝑚𝑎𝑥 = = 𝑇ℎ𝑖𝑔ℎ 150 + 273 𝑒𝑚𝑎𝑥 = 29.1 % The actual efficiency: eactual 𝑊𝑛𝑒𝑡 100 𝑒𝑎𝑐𝑡𝑢𝑎𝑙 = = = 20% 𝑄ℎ𝑜𝑡 500 The relative efficiency: erelative 𝑒𝑎𝑐𝑡𝑢𝑎𝑙 0.20 𝑒𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 = = 𝑒𝑚𝑎𝑥 0.291 Answer: 𝑒𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 = 68.73% Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 47. A cylinder is fitted with a piston that contains 0.10 mol of air at a temperature of 20℃. Find the work done if the piston is slowly pushed so that the air within the cylinder remains essentially in thermal equilibrium with the surroundings until the volume reaches half to its initial volume. Given: n= 0.10 mols T= 20℃ V2= 0.5V1 Required: W=? Solution: 𝑉2 𝑊 = 𝑃(𝑉2 − 𝑉1 ) = 𝑛𝑅̅𝑇 ln 𝑉1 0.5𝑉1 𝑊 = 0.1(8.314)(20 + 273) ln 𝑉1 Answer: 𝑊 = −169 𝐽𝑜𝑢𝑙𝑒𝑠 Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 48. How much is heat needed to double the pressure of one-half mole of helium gas confined in a rigid container if initially the gas is at standard pressure and temperature? Note: Helium C v= 3.10 kJ/kg-K, M= 4 kg/mol Given: P2= 2P1 n= 0.5 mol T= 273 K P1= 101.325 kPa Required: Q=? Solution: 𝑄 = 𝑚𝐶𝑣 ∆𝑇 At constant volume: 𝑃1 2𝑃1 = 𝑇1 𝑇2 𝑇2 = 2𝑇1 = 2(273) = 546 𝐾 The heat needed is: 𝑄 = 𝑛𝐶𝑣 ∆𝑇 𝑄 = (0.5)(12.4)(546 − 273) Answer: 𝑄 = 1692.6 𝐽 Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 49. When 0.05 m3 of air at STP (Standard Temperature and Pressure) is isothermally compressed to 0.010 m 3, compute the amount of heat flow from the gas. Given: V1= 0.05 m3 V2= 0.010 m3 T= 273 K P= 101.325 kPa Required: Q=? Solution: 𝑉2 𝑊 = 𝑃1 𝑉1 ln ( ) 𝑉1

0.010 𝑊 = (101.325)(0.05) ln ( ) 0.05 𝑊 = −8.154 𝑘𝐽 In the first law of thermodynamics: 𝑄 = ∆𝑈 + 𝑊 For isothermal, ∆U = 0 𝑄=𝑊 Answer: 𝑄 = −8.154 𝑘𝐽 Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects 50. How much heat was removed from a cubic meter of helium at STP when cooled at constant pressure to a final volume of 0.75 m3? Note: MWhelium= 4 kg/kmol, Cv= 3.10 kJ/kg-K. Given: V1= 1 m3 V2= 0.75 m3 T1= 273 K P= 101.325 kPa MWhelium= 4 kg Cv= 3.10 kJ/kg-K Required: Q? Solution: 𝑉1 𝑉2 = 𝑇1 𝑇2 273(0.75) 𝑇2 = = 204.75 𝐾 1 The mass of helium at STP: 𝑃𝑉 𝑚= 𝑅𝑇 101.325(1) 𝑚= 8.314 ( ) (273) 4 𝑚 = 0.179 𝑘𝑔 The change in internal energy: ∆𝑈 = 𝑚𝐶𝑣 ∆𝑇 ∆𝑈 = (0.179)(3.10)(204.75 − 273) ∆𝑈 = −37.87 𝑘𝐽 Using first law of thermodynamics: 𝑄 = ∆𝑈 + 𝑃(𝑉2 − 𝑉1 ) 𝑄 = −37.87 + 101.325(0.75 − 1) Answer: 𝑄 = −63.20 𝑘𝐽 Source: 2001 Solved Problems, Engineering Sciences and Applied Subjects