Thermodynamics Solution Manual Pk Nag
Engineering Thermodynamics P. Nag Limited preview - 2008. Engineering Thermodynamics. Process reversible heat reversible process reversibly saturated liquid saturated vapour second law efficiency shown in Fig Solution specific heat specific volume steady flow surroundings temperature difference thermal thermodynamic thermometer throttling. Solutions to Basic and Applied Thermodynamics PK NAG Questions Solved with Solutions Solution manual is a complied answers from the Book Basic and.
P K Nag Exercise problems - Solved Thermodynamics Contents Chapter-1: Introduction Chapter-2: Temperature Chapter-3: Work and Heat Transfer Chapter-4: First Law of Thermodynamics Chapter-5: First Law Applied to Flow Process Chapter-6: Second Law of Thermodynamics Chapter-7: Entropy Chapter-8: Availability & Irreversibility Chapter-9: Properties of Pure Substances Chapter-10: Properties of Gases and Gas Mixture Chapter-11: Thermodynamic Relations Chapter-12: Vapour Power Cycles Chapter-13: Gas Power Cycles Chapter-14: Refrigeration Cycles Solved by Er. S K Mondal IES Officer (Railway), GATE topper, NTPC ET-2003 batch, 12 years teaching experienced, Author of Hydro Power Familiarization (NTPC Ltd). Introduction By: S K Mondal Chapter 1 1. Introduction Some Important Notes Microscopic thermodynamics or statistical thermodynamics Macroscopic thermodynamics or classical thermodynamics A quasi-static process is also called a reversible process Intensive and Extensive Properties Intensive property: Whose value is independent of the size or extent i.e. Mass of the system. E.g., pressure p and temperature T.
Extensive property: Whose value depends on the size or extent i.e. Mass of the system (upper case letters as the symbols). E.g., Volume, Mass (V, M).
If mass is increased, the value of extensive property also increases. E.g., volume V, internal energy U, enthalpy H, entropy S, etc. Specific property: It is a special case of an intensive property. It is the value of an extensive property per unit mass of system. (Lower case letters as symbols) e.g: specific volume, density ( v, ρ). Concept of Continuum The concept of continuum is a kind of idealization of the continuous description of matter where the properties of the matter are considered as continuous functions of space variables. Although any matter is composed of several molecules, the concept of continuum assumes a continuous distribution of mass within the matter or system with no empty space, instead of the actual conglomeration of separate molecules.
Describing a fluid flow quantitatively makes it necessary to assume that flow variables (pressure, velocity etc.) and fluid properties vary continuously from one point to another. Mathematical descriptions of flow on this basis have proved to be reliable and treatment of fluid medium as a continuum has firmly become established. For example density at a point is normally defined as 0 lim mρ ∀→ ⎛ ⎞= ⎜ ⎟∀⎝ ⎠ Here ∀ is the volume of the fluid element and m is the mass If ∀ is very large ρ is affected by the in-homogeneities in the fluid medium. Considering another extreme if ∀ is very small, random movement of atoms (or molecules) would change their number at different times. In the continuum approximation point density is defined at the smallest magnitude of ∀, before statistical fluctuations become significant. This is called continuum limit and is denoted by C∀. Lim C mρ ∀→ ∀ ⎛ ⎞= ⎜ ⎟∀⎝ ⎠ Page 3 of 265.
Introduction By: S K Mondal Chapter 1 One of the factors considered important in determining the validity of continuum model is molecular density. It is the distance between the molecules which is characterized by mean free path (λ).
It is calculated by finding statistical average distance the molecules travel between two successive collisions. If the mean free path is very small as compared with some characteristic length in the flow domain (i.e., the molecular density is very high) then the gas can be treated as a continuous medium. If the mean free path is large in comparison to some characteristic length, the gas cannot be considered continuous and it should be analyzed by the molecular theory. A dimensionless parameter known as Knudsen number, Kn = λ / L, where λ is the mean free path and L is the characteristic length. It describes the degree of departure from continuum.
Usually when Kn 0.01, the concept of continuum does not hold good. In this, Kn is always less than 0.01 and it is usual to say that the fluid is a continuum. Other factor which checks the validity of continuum is the elapsed time between collisions. The time should be small enough so that the random statistical description of molecular activity holds good.
In continuum approach, fluid properties such as density, viscosity, thermal conductivity, temperature, etc. Can be expressed as continuous functions of space and time.
The Scale of Pressure Absolute Pressure Gauge Pressure Vacuum Pressure Local atmospheric Pressure Absolute Pressure Absolute Zero (complete vacuum) At sea-level, the international standard atmosphere has been chosen as Patm = 101.325 kN/m2 Page 4 of 265. Introduction By: S K Mondal Chapter 1 Questions with Solution P. Nag Q1.1 A pump discharges a liquid into a drum at the rate of 0.032 m3/s. The drum, 1.50 m in diameter and 4.20 m in length, can hold 3000 kg of the liquid. Find the density of the liquid and the mass flow rate of the liquid handled by the pump. 12.934 kg/s) Solution: 2dVolume of drum = h 4 π × 2 3 3 3 3.50= m 4 7.422 m mass 3000 kg kgdensity 404.203m mVolume 7.422 mass flow rate Vloume flow rate density kg= 0.032 404.203 s kg12.9345 s π ×1 × 4.2 = = = = × × = = Q1.2 The acceleration of gravity is given as a function of elevation above sea level by 6g = 980.6 – 3.086 × 10 H− Where g is in cm/s2 and H is in cm.
If an aeroplane weighs 90,000 N at sea level, what is the gravity force upon it at 10,000 m elevation? What is the percentage difference from the sea-level weight? 89,716.4 N, 0.315%) Solution: 6g´ 980.6 3.086 10 10,000 100−= − × × × 2 2cm m977.514 9.77514s s= = sea 90,000W 90,000 N kgf 9.806 = = 9178.054 kgf= eteW 9178.054 9.77514 N= × 89716.765 N= ( ) 90,000 89716.765% less 100% 90,000 0.3147% less − = × = Q1.3 Prove that the weight of a body at an elevation H above sea-level is given by 2 0 2 mg dW g d H ⎛ ⎞= ⎜ ⎟+⎝ ⎠ Where d is the diameter of the earth. Solution: According to Newton’s law of gravity it we place a man of m at an height of H then Page 6 of 265. Introduction By: S K Mondal Chapter 1 Force of attraction = ( )2 GMm d H2 + (i) If we place it in a surface of earth then ( ) ( ) o2 o 2 GMmForce of attraction mg d 2 GMor g d 2 = = = d H 1.3 m ( ) ( ) ( ) ( ) ( ) 2 2 o 2 2 o GMmWeight W d H2 dmg 2 from equation. I d H2 dmg Pr oved. D 2H ∴ = + = + ⎛ ⎞= ⎜ ⎟+⎝ ⎠ Q1.4 The first artificial earth satellite is reported to have encircled the earth at a speed of 28,840 km/h and its maximum height above the earth’s surface was stated to be 916 km.
Taking the mean diameter of the earth to be 12,680 km, and assuming the orbit to be circular, evaluate the value of the gravitational acceleration at this height. The mass of the satellite is reported to have been 86 kg at sea-level. Estimate the gravitational force acting on the satellite at the operational altitude.
8.9 m/s2; 765 N) Solution: Their force of attraction = centrifugal force 2mvCentirfugal force r = 2 3 3 86 60 60 N 12680 10 916 10 2 760.65 N (Weight) ×⎛ ⎞× ⎜ ⎟×⎝ ⎠= ⎛ ⎞× + ×⎜ ⎟ ⎝ ⎠ = Q1.5 Convert the following readings of pressure to kPa, assuming that the barometer reads 760 mmHg: (a) 90 cmHg gauge (b) 40 cmHg vacuum (c) 1.2 m H2O gauge (d) 3.1 bar Solution: 760 mm Hg = 0.760 × 13600 × 9.81 Pa = 10139.16 Pa 101.4 kPa Page 7 of 265. Introduction By: S K Mondal Chapter 1 (a) 90 cm Hg gauge = 0.90 × 13600 × 9.81 × 10-3 + 101.4 kPa = 221.4744 kPa (b) 40 cm Hg vacuum = (76 – 40) cm (absolute) = 0.36 × 43.600 × 9.81 kPa = 48.03 kPa (c) 1.2 m H2O gauge = 1.2 × 1000 × 9.81 × 10-3 + 101.4 kPa = 113.172 kPa (d) 3.1 bar = 3.1 × 100 kPa = 310 kPa Q1.6 A 30 m high vertical column of a fluid of density 1878 kg/m3 exists in a place where g = 9.65 m/s2. What is the pressure at the base of the column. 544 kPa) Solution: p = z ρg = 30 × 1878 × 9.65 Pa = 543.681 kPa Q1.7 Assume that the pressure p and the specific volume v of the atmosphere are related according to the equation 1.4 52.3 10 pv = ×, where p is in N/m2 abs and v is in m3/kg. The acceleration due to gravity is constant at 9.81 m/s2. What is the depth of atmosphere necessary to produce a pressure of l.0132 bar at the earth’s surface?
Consider the atmosphere as a fluid column. 64.8 km) Page 8 of 265. Introduction By: S K Mondal Chapter 1 Solution: dp dh g= ρ 1.4 3 1 n 1.4 n n 1or dp dh g v gdhor v dp pv 2.3 10 2300 2300 2300 1or v where n p p 1.4 g dh 2300or dp p 2300or g dh dp p = × × = = × = ⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞= ⎜ ⎟ ⎝ ⎠ ⎛ ⎞= ⎜ ⎟ ⎝ ⎠ h dh p p + dp Zero line p = h gρ h dh p = h gρ H -hO ( ) ( ) ( ) H 101320n n 0 0 n 1 n 2300 dpor dh g p 2300or h 101320 0 2420m 2.42 km g 1 n − = ⎡ ⎤= − = =⎣ ⎦− ∫ ∫ Q1.8 The pressure of steam flowing in a pipeline is measured with a mercury manometer, shown in Figure.
Some steam condenses into water. Estimate the steam pressure in kPa. Take the density of mercury as 3 313.6 10 kg/m×, density of water as 103 kg/m3, the barometer reading as 76.1 cmHg, and g as 9.806 m/s2. Solution: 2o Hg H O p 0.50 g 0.03 g p+ × ρ × = × ρ × + 3 3or p 0.761 13.6 10 9.806 0.5 13.6 10 9.806 0.03 1000 9.806 Pa.
167.875 kPa = × × × + × × × − × × = Q1.9 A vacuum gauge mounted on a condenser reads 0.66 mHg. What is the absolute pressure in the condenser in kPa when the atmospheric pressure is 101.3 kPa?
13.3 kPa) Solution: Absolute = atm. – vacuum = 3 3101.3 – 0.66 × 13.6 × 10 × 9.81 × 10 kPa− = 13.24 kPa Page 9 of 265. Temperature By: S K Mondal Chapter 2 2. Temperature Some Important Notes Comparison of Temperature scale Relation: C 0 100 0 − − = F 32 212 32 − − = K 273 373 273 − − = 0 80 0 ρ − − = − − x 10 30 10 Questions with Solution P. Nag Q2.1 The limiting value of the ratio of the pressure of gas at the steam point and at the triple point of water when the gas is kept at constant volume is found to be 1.36605.
What is the ideal gas temperature of the steam point? 100°C) Solution: t p 1.36605 p = ( )v t p273.16 p 273.16 1.36605 373.15 C ∴ θ = × = × = ° Q2.2 In a constant volume gas thermometer the following pairs of pressure readings were taken at the boiling point of water and the boiling point of sulphur, respectively: Water b.p. 50.0 100 200 300 Sulphur b.p. 96.4 193 387 582 The numbers are the gas pressures, mm Hg, each pair being taken with the same amount of gas in the thermometer, but the successive pairs being taken with different amounts of gas in the thermometer. Plot the ratio of Sb.p.:H2Ob.p. Against the reading at the water boiling point, and extrapolate the plot to zero pressure at the water boiling point. This o1 0 0 C o F 2 1 2 373K 3 0 c m o 0 C o3 2 F 2 7 3 K 1 0 c m C F K o 8 0 o 0 Boiling Point Test Temperature Freezing Point x Page 11 of 265.
Work and Heat Transfer By: S K Mondal Chapter 3 3. Work and Heat Transfer Some Important Notes -ive W +ive W -ive Q +ive Q Our aim is to give heat to the system and gain work output from it. So heat input → +ive (positive) Work output → +ive (positive) − = =∫ ∫ v ff i f i v i W pdV pdv d Q = du + dW = − +∫ ∫ f f f i i i dQ u u dW − − += ∫ v f i f f i v i Q u u pdV Questions with Solution P. Nag Q3.1 (a) A pump forces 1 m3/min of water horizontally from an open well to a closed tank where the pressure is 0.9 MPa.
Compute the work the pump must do upon the water in an hour just to force the water into the tank against the pressure. Sketch the system upon which the work is done before and after the process. 5400 kJ/h) (b)If the work done as above upon the water had been used solely to raise the same amount of water vertically against gravity without change of pressure, how many meters would the water have been elevated?
91.74 m) (c)If the work done in (a) upon the water had been used solely to accelerate the water from zero velocity without change of pressure or elevation, what velocity would the water have reached? If the work had been used to accelerate the water from an initial velocity of 10 m/s, what would the final velocity have been? 42.4 m/s; 43.6 m/s) Solution: (a) Flow rate 1m3/hr. Pressure of inlet water = 1 atm = 0.101325 MPa Pressure of outlet water = 0.9 MPa Page 15 of 265. Work and Heat Transfer By: S K Mondal Chapter 3 ( ) 33 Power pv 1 m0.9 0.101325 10 kPa s60 kJ13.31 s ∴ = Δ = − × × = (b) So that pressure will be 0.9 MPa ∴ ρ = × = × 6 h g 0.9 MPa 0.9 10or h = m 91.743 m 1000 9.81 ( ) ( ) 2 2 2 1 2 2 2 1 2 2 2 1 1(c) m V V pv where m v 2 1or V V p 2 por V V 2 − = Δ = ρ ρ − = Δ Δ − = ρ 2 22 1 por V V 2 Δ= + ρ ( ) 62 2 2 0.9 0.101325 V 41.2 m / s. × − × = + = Q3.2 The piston of an oil engine, of area 0.0045 m2, moves downwards 75 mm, drawing in 0.00028 m3 of fresh air from the atmosphere. The pressure in the cylinder is uniform during the process at 80 kPa, while the atmospheric pressure is 101.325 kPa, the difference being due to the flow resistance in the induction pipe and the inlet valve.
Estimate the displacement work done by the air finally in the cylinder. 27 J) Solution: Volume of piston stroke = 0.0045 × 0.075m3 = 0.0003375m3 ∴ ΔV = 0.0003375 m3 as pressure is constant = 80 kPa So work done = pΔV = 80 × 0.0003375 kJ = 0.027 kJ = 27 J Final volume = 3.375×10-4 Initial volume = 0 3 m Q3.3 An engine cylinder has a piston of area 0.12 m3 and contains gas at a pressure of 1.5 MPa. The gas expands according to a process which is represented by a straight line on a pressure-volume diagram. The final pressure is 0.15 MPa. Calculate the work done by the gas on the piston if the stroke is 0.30 m. 29.7 kJ) Solution: Initial pressure ( 1p ) = 1.5 MPa Final volume (V1) = 0.12m2 × 0.3m Page 16 of 265.
Work and Heat Transfer By: S K Mondal Chapter 3 = 0.036 m3 Final pressure ( 2p ) = 0.15 MPa As initial pressure too high so the volume is neglected. Work done = Area of pV diagram ( ) ( ) 1 2 3 1 p p V 2 1 1.5 0.15 0.036 10 kJ 2 29.7 kJ = + × = + × × = Vneg. 0.36 m3 p 0.15 MPa 1.5 M aP Q3.4 A mass of 1.5 kg of air is compressed in a quasi-static process from 0.1 MPa to 0.7 MPa for which pv = constant. The initial density of air is 1.16 kg/m3. Find the work done by the piston to compress the air. 251.62 kJ) Solution: For quasi-static process Work done pdV given pV C= =∫ 2 1 v 1 1 1 1 2 2 v 2 1 1 1 1 1 1 1 2 1 1 2 2 1 1 31 1 1 2 dVp V p V pV p V C V V p Vp V l n p V V p p Vp V ln p p V 0.10.1 1.2931 ln MJ given p 0.1 MPa 0.7 m 1.5251.63 kJ V m 1.16 p 0.7 MPa = ∴ = = = ⎛ ⎞ = ∴ =⎜ ⎟ ⎝ ⎠ ⎛ ⎞ = ∴ =⎜ ⎟ ⎝ ⎠ = × × = = = = ρ = ∫ Q3.5 A mass of gas is compressed in a quasi-static process from 80 kPa, 0.1 m3 to 0.4 MPa, 0.03 m3. Assuming that the pressure and volume are related by pvn = constant, find the work done by the gas system.
–11.83 kJ) Solution: Given initial pressure ( )1p = 80kPa Initial volume ( )1V = 0.1 m3 Page 17 of 265. Work and Heat Transfer By: S K Mondal Chapter 3 Q3.7 A single-cylinder, single-acting, 4 stroke engine of 0.15 m bore develops an indicated power of 4 kW when running at 216 rpm.
Calculate the area of the indicator diagram that would be obtained with an indicator having a spring constant of 25 × 106 N/m3. The length of the indicator diagram is 0.1 times the length of the stroke of the engine.
505 mm2) Solution: Given Diameter of piston (D) = 0.15 m I.P = 4 kW = 4 × 1000 W Speed (N) = 216 rpm Spring constant (k) = 25 × 106 N/m Length of indicator diagram ( )dl = 0.1 × Stoke (L) Let Area of indicator diagram = ( ) da ∴ Mean effective pressure ( mp ) = d d a k l × ∴ m d d d d p LANand I.P. As 4 stroke engine 120 a k L A Nor I.P.
L 120 I.P l 120or a k L A N = × × × = × × × = × × × 2 2 d 2 6 2 4 2 2 Darea AI.P 0.1L 120 4 4k L D N and l 0.1L 4 0.1 120 4 1000 m 25 10 0.15 216 5.03 10 m 503mm − ⎡ ⎤π =× × × ⎢ ⎥= ⎢ ⎥× × π × × =⎢ ⎥⎣ ⎦ × × × × = × × π × × = × = Q3.8 A six-cylinder, 4-stroke gasoline engine is run at a speed of 2520 RPM. The area of the indicator card of one cylinder is 2.45 × 103 mm2 and its length is 58.5 mm.
The spring constant is 20 × 106 N/m3. The bore of the cylinders is 140 mm and the piston stroke is 150 mm. Determine the indicated power, assuming that each cylinder contributes an equal power. 243.57 kW) Solution: dm d ap k l = × 3 2 3 2 3 2 2.45 10 mm N mm N 120 10 Pa N / m 58.5 mm 1000m m m 837.607 kPa L 0.150 m × × ⎛ ⎞= × × ∴ × ⇒ = ⎜ ⎟× ⎝ ⎠ = = Page 19 of 265.
Work and Heat Transfer By: S K Mondal Chapter 3 2 2D 0.14A 4 4 N 2520 π π × = = = m 2 n 6 p LANI.P. N as four stroke 120 0.14 2520 6837.607 0.15 kW 4 120 243.696 kW = ∴ = × π × × = × × × = Q3.9 A closed cylinder of 0.25 m diameter is fitted with a light frictionless piston. The piston is retained in position by a catch in the cylinder wall and the volume on one side of the piston contains air at a pressure of 750 kN/m2. The volume on the other side of the piston is evacuated.
A helical spring is mounted coaxially with the cylinder in this evacuated space to give a force of 120 N on the piston in this position. The catch is released and the piston travels along the cylinder until it comes to rest after a stroke of 1.2 m. The piston is then held in its position of maximum travel by a ratchet mechanism. The spring force increases linearly with the piston displacement to a final value of 5 kN.
Thermodynamics Solution Manual Pk Nag Thermodynamics
Calculate the work done by the compressed air on the piston. 3.07 kJ) Solution: Work done against spring is work done by the compressed gas φ 0.25m 1.2m 120 5000Mean force 2 2560 N Travel 1.2m Work Done 2560 1.2 N.m 3.072 kJ + = = = ∴ = × = By Integration At a travel (x) force (Fx) = 120 + kx At 1.2 m then 5000 = 120 + k × 1.2 ∴ Fx = 120 + 4067 x Page 20 of 265. Work and Heat Transfer By: S K Mondal Chapter 3 1.2 x 0 1.2 0 1.22 0 2 W F dx 120 4067x dx x120x 4067 2 1.2120 1.2 4067 J 2 144 2928.24 J 3072.24J 3.072 kJ ∴ = = + ⎡ ⎤ = + ×⎢ ⎥ ⎣ ⎦ = × + × = + = = ∫ ∫ Q 3.l0 A steam turbine drives a ship’s propeller through an 8: 1 reduction gear. The average resisting torque imposed by the water on the propeller is 750 × 103 mN and the shaft power delivered by the turbine to the reduction gear is 15 MW. The turbine speed is 1450 rpm. Determine (a) the torque developed by the turbine, (b) the power delivered to the propeller shaft, and (c) the net rate of working of the reduction gear.
(a) T = 98.84 km N, (b) 14.235 MW, (c) 0.765 MW) Solution: Power of the propeller = Power on turbine shaft The net rate of working of the reduction gear = (15 – 14.235) MW = 0.7647 MW Q 3.11 A fluid, contained in a horizontal cylinder fitted with a frictionless leak proof piston, is continuously agitated by means of a stirrer passing through the cylinder cover. The cylinder diameter is 0.40 m. During the stirring process lasting 10 minutes, the piston slowly moves out a distance of 0.485 m against the atmosphere. The net work done by the fluid during the process is 2 kJ. The speed of the electric motor driving the stirrer is 840 rpm. Determine the torque in the shaft and the power output of the motor.
[pic] He was probably a hunter or a shepherd who was accustomed to spending months each year in the high country. [pic] Stone, wood, leather, and fibers[pic]/ 1st appearance of tools (bone, wood, stone) (Paleolithic “Old Stone Era”) What conclusions may be drawn from a study of the Iceman and the materials found with him about what he did and what he was most likely to have been? My study guide 101. He pulled an arrow out of. 11155 Words| 45 Pages strategic effects such as: A.
0.08 mN, 6.92 W) Page 21 of 265. Work and Heat Transfer By: S K Mondal Chapter 3 Solution: Change of volume = A L π = × π× = × = 2 2 3 3 d L 4 0.4 0.485 m 4 0.061 m As piston moves against constant atmospheric pressure then work = Δdone p V 0.485m φ = 0.4m M 101.325 0.061 kJ 6.1754 kJ = × = Net work done by the fluid = 2 kJ ∴ Net work done by the Motor = 4.1754 kJ There for power of the motor 34.1754 10 W 10 60 × = × 6.96 W PTorque on the shaft W = = 6.96 60 2 840 0.0791mN × = π × = Q3.12 At the beginning of the compression stroke of a two-cylinder internal combustion engine the air is at a pressure of 101.325 kPa. Compression reduces the volume to 1/5 of its original volume, and the law of compression is given by pv1.2 = constant. If the bore and stroke of each cylinder is 0.15 m and 0.25 m, respectively, determine the power absorbed in kW by compression strokes when the engine speed is such that each cylinder undergoes 500 compression strokes per minute. 17.95 kW) Page 22 of 265. Work and Heat Transfer By: S K Mondal Chapter 3 A B C bar p 0.2 0.4 0.8 V m1 3 50 pV = c1.3 1 1 2 2 5 5 Area under BC p V p V n 1 50 10 0.4 20.31 10 0.8 W 1.3 1 1.251MJ − = − × × − × × = − = 5 B B 3 B 3 C 1.3 5 1.3 B B C 1.3 1.3 C 5 Here p p 50 bar 50 10 Pa V 0.4m V 0.8m p V 50 10 0.4p V 0.8 20.31 10 Pa ⎡ ⎢ = = = ×⎢ ⎢ = ⎢ =⎢ ⎢ × ×⎢ = =⎢ ⎢ ⎢ = ×⎣ Total work = 2.251MJ Q3.14 A system of volume V contains a mass m of gas at pressure p and temperature T. The macroscopic properties of the system obey the following relationship: 2 ap + (V b) = mRT V ⎛ ⎞ −⎜ ⎟ ⎝ ⎠ Where a, b, and R are constants.
Obtain an expression for the displacement work done by the system during a constant-temperature expansion from volume V1 to volume V2. Calculate the work done by a system which contains 10 kg of this gas expanding from 1 m3 to 10 m3 at a temperature of 293 K. Use the values 4 2 315.7 10 Nm, 1.07 10 m a b −= × = ×, and R = 0.278 kJ/kg-K. 1742 kJ) Solution: As it is constant temp-expansion then ( ) ( ) ( )2 ap V b constant mRT k as T constant V ⎛ ⎞+ − = =⎜ ⎟ ⎝ ⎠ Page 24 of 265. Work and Heat Transfer By: S K Mondal Chapter 3 ( ) ( ) ( ) ( ) ( ) ( ) 1 1 2 22 2 1 2 2 1 2 2 2 1 2 2 1 2 1 2 1 2 1 12 1 1 2 a ap V b p V b k V V constant kaW p dV p V V b k a k adV or p V b V V b V a 1 1kln V b dv c V V V V b 1 1k ln a V b V V V ba 1 1p V b ln a V V b V ⎛ ⎞ ⎛ ⎞ ∴ + − = + − =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞= ∴ + =⎜ ⎟ −⎝ ⎠ ⎛ ⎞= − = −⎜ ⎟− −⎝ ⎠ ⎡ ⎤= − + − = +⎢ ⎥⎣ ⎦ ⎛ ⎞ ⎛ ⎞− = + −⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎛ ⎞ − = + − + −⎜ ⎟ −⎝ ⎠ ∫ ∫ ∫ 1V ⎡ ⎤⎛ ⎞ ⎢ ⎥⎜ ⎟ ⎝ ⎠⎣ ⎦ ( ) ( ) ( )2 ap V b constant mRT k as T constant V ⎛ ⎞+ − = =⎜ ⎟ ⎝ ⎠ Given m = 10 kg; T = 293 K; R = 0.278 kJ/kg.
Thermodynamics Solutions Manual Pdf
K ∴ Constant k = 10 × 293 × 0.278 kJ = 814.54 kJ a = 15.7 × 10 Nm4; b = 1.07 × 10-2m3 ⇒ V2 = 10m3, V1 = 1m3 ( ) ( ) 2 2 10 1.07 10 1 1W 814.54 ln a 1 1.07 10 10 1 1883.44 a 0.9 kJ 1883.44 157 0.9 kJ 1742.14kJ − − ⎛ ⎞− × ⎛ ⎞∴ = + −⎜ ⎟ ⎜ ⎟− × ⎝ ⎠⎝ ⎠ = − × = − × = Q3.15 If a gas of volume 6000 cm3 and at pressure of 100 kPa is compressed quasistatically according to pV2 = constant until the volume becomes 2000 cm3, determine the final pressure and the work transfer. 900 kPa, – 1.2 kJ) Solution: Initial volume ( 1v ) = 6000 cm3 = 0.006 m3 Initial pressure ( )1p = 100 kPa Final volume ( 2v ) = 2000 cm3 = 0.002 m3 If final pressure ( )2p ( ) ( ) × ∴ = = = 22 1 1 2 2 2 2 100 0.006p Vp 900 kPa V 0.002 Page 25 of 265.