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Thermodynamics

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Sidhant S / Bathinda

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  1. Chapter 3 THE FIRST LAW OF THERMODYNAMICS Energy interactions between a system and its surroundings across the boundary in the form of heat and work have been discussed separately in the previous chapter. So far, no attempt has been made to relate these interactions between themselves and with the energy content of the system. First law of thermodynamics, often called as law of conservation of energy, relating work, heat, and energy content of the system will be discussed in detail in this chapter. 3.1 First Law of Thermodynamics In its more general form, the first law may be stated as follows "When energy is either transferred or transformed, the final total energy present in all forms must precisely equal the original total energy' It is based on the experimental observations and can not be proved mathematically. All the observations made so far, confirm the correctness of this law. 3.2 First Law of Thermodynamics for a Closed System Undergoing a Process First law can be written for a closed system in an equation form as Energy entered Energy left Change in the energy into the system the system content of the system For a system of constant mass, energy can enter or leave the system only in two forms namely work and heat. Let a closed system of initial energy El receives Q units of net heat and gives out W units of work during a process. If E is energy content at the end of the process as given in Figure 3.1, applying first law we get a. Initial state Figure 3.1 Thermodynamics 1 [MIME3110] b. During the process 2 c, Final state First Law for a closed system 22
  2. Q W-(E2 El) Where the total energy content E Internal Energy Kinetic energy ...(3.1) Potential energy mgz The term internal energy usually denoted by the letter U is the energy due to such factors as electron spin and vibrations, molecular motion and chemical bond. Kinetic energy term is due to the system movement with a velocity C. For stationary systems this term will be zero. The term g is a constant of value 1 in SI unit. It will be dropped here after since SI unit is followed throughout the book. Potential energy term is due to the location of the system in the gravitational field. It remains constant for a stationary system. The unit of energy in SI is kJ. 3.3 The Thermodynamic Property Enthalpy Consider a stationary system of fixed mass undergoing a quasi-equilibrium constant pressure process Applying first law (212 where E w E E (I-J2 p(V2 E 1 UI) + m(C2 C 2) + mg(Z2 4) since it is a stationary system. also W Q 12 P2V2 1 PIVI PIVI) + (U2 The terms within brackets are all properties depending on the end states. This combination of properties may be regarded as a single property known as enthalpy. It is usually denoted by the letter H. ie H U+pV (or) h = u + pv Thermodynamics 1 [MIME3110] ...(3.3a) ...(3.3b) 23
  3. Where h is specific enthalpy in kJ/kg u is specific internal energy in kJ/kg and v is specific volume in m3/kg 3.4 Flow Energy Flow energy is defined as the energy required to move a mass into the a control volume against a pressure. Consider a mass of volume V entering into a control volume as given in the Figure 3.2 against a pressure p. control vollJtTIe Flow The Flow energy Therefore, Enthalpy mass volurneV Figure 3.2 Flow energv- Work done in moving the mass Force x distance PA x dx P x (Adx) Internal energy + Flow energy ...(3.4) 3.5 First Law of Thermodynamics for a Control Volume Mass simultaneously entering and leaving the system is a very common phenomenon in most of the engineering applications. Control volume concept is applied to these devices by assuming suitable control surfaces. To analyze these control volume problems, conservation of mass and energy concepts are to be simultaneously considered. Energy may cross the control surface not only in the form of heat and work but also by total energy associated with the mass crossing the boundaries. Hence apart from kinetic, potential and internal energies, flow energy should also be taken into account. Thermodynamics 1 [MIME3110] 24
  4. Conservation of mass Total mass entering the Total mass + leaving the control volume control volume Conservation of energy Net energy crossing the Total energy boundary in the form of heat and work associatedwiththe mass entering the control volume Control Volume m in Net change in the mass content of the control volume Total energy associatedwiththe mass leav ing the control volume m out Control Surface Net change in theenrgy contentof the control volume ...(3.5) Figure 3.3 First Law of Thermodynamics Applied to a control Volume As a rate equation, it becomes in +Zg h+—+Zg out ...(3.6) 3.6 The Steady-state Flow Process When a flow process is satisfying the following conditions, it is known as a steady flow process. 1. The mass and energy content of the control volume remains constant with time. 2. The state and energy of the fluid at inlet, at the exit and at every point within the control volume are time independent. 3. The rate of energy transfer in the form of work and heat across the control surface is constant with time. Thermodynamics 1 [MIME3110] 25
  5. Therefore for a steady flow process out also 0 in 2 2 out For problem of single inlet stream and single outlet stream +(Z2 Zl)g 2 This equation is commonly known as steady flow energy equation (SFEE). 3.7 Application of SFEE ...(3.7) ...(3.8) ...(3.9) ...(3.10) SFEE governs the working of a large number of components used in many engineering practices. In this section a brief analysis of such components working under steady flow conditions are given and the respective governing equations are obtained. 3.7.1. Turbines Turbines are devices used in hydraulic, steam and gas turbine power plants. As the fluid passes through the turbine, work is done on the blades of the turbine which are attached to a shaft. Due to the work given to the blades, the turbine shaft rotates producing work. Mass entering Control Surface Shaft work Mass leaving Figure 3.4 Schematic Representation of a Turbine Thermodynamics 1 [MIME3110] 26
  6. General Assumptions 1. Changes in kinetic energy of the fluid are negligible 2. Changes in potential energy of the fluid are negligible. 3.7.2 Compressors ...(3.11) Compressors (fans and blowers) are work consuming devices, where a low-pressure fluid is compressed by utilising mechanical work. Blades attached to the shaft of the turbine imparts kinetic energy to the fluid which is later converted into pressure energy. Mass leaving Shaft work Control Surface Mass entering Figure 3.5 Schematic Representation of a Compresso General Assumptions l. Changes in the kinetic energy of the fluid are negligible 2. Changes in the potential energy of the fluid are negligible Governing Equation Applying the above equations SFEE becomes Thermodynamics 1 [MIME3110] ...(3.12) 27
  7. 3.7.3 Pumps Similar to compressors pumps are also work consuming devices. But pumps handle incompressible fluids, whereas compressors deal with compressible fluids. out in Figure 3.6 Schematic diagram of a pump General Assumptions l. No heat energy is gained or lost by the fluids; 2. Changes in kinetic energy of the fluid are negligible. Governing Equation W] 'h) +(Z2 Zl)g] ...(3.13) As the fluid passes through a pump, enthalpy of the fluid increases, (internal energy of the fluid remains constant) due to the increase in pv (flow energy). Increase in potential energy of fluid is the most important change found in almost all pump applications. 3.7.4 Nozzles Nozzles are devices which increase the velocity of a fluid at the expense of pressure. A typical nozzle used for fluid flow at subsonic* speeds is shown in Figure 3.7. 1. 2. 3. General Assumptions In nozzles fluids flow at a speed which is high enough to neglect heat lost or gained as it crosses the entire length of the nozzle. Therefore, flow through nozzles can be regarded as adiabatic. That is = 0. There is no shaft or any other form of work transfer to the fluid or from the fluid; that is = 0. Changes in the potential energy of the fluid are negligible. Thermodynamics 1 [MIME3110] 28
  8. In Governing Equation c/ (112 IA) + 2 (Ill Control S urface Out 112) 2 3.7.5 Diffusers Diffusers are (reverse of nozzles) devices which increase the pressure of a fluid stream by reducing its kinetic energy. General Assumptions Similar to nozzles, the following assumptions hold good for diffusers. 1. 2. 3. Heat lost or gained as it crosses the entire length of the nozzle. Therefore, flow through nozzles can be regarded as adiabatic. That is Q = 0 There is no shaft or any other form of work transfer to the fluid or from the fluid; that is = 0. Changes in the potential energy of the fluid are negligible Thermodynamics 1 [MIME3110] 29
  9. Governing Equation (h2 3.7.6 Heat Exchangers (112 2 14) ...(3.14) 0 c 22 2 Devices in which heat is transferred from a hot fluid stream to a cold fluid stream are known as heat exchangers. Hot fluid out Cold fluid in General Assumptions cold fluid out Hot fluid in 1. Heat lost by the hot fluid is equal to the heat gained by the cold fluid. 2. No work transfer across the control volume. 3. Changes in kinetic and potential energies of both the streams are negligible. Governing Equation For both hot and cold streams As per the assumption, hot -0 cold The negative sign in the LHS is to represent that heat is going out of the system. lilh(hl 112) milc(h2 hi) Thermodynamics 1 [MIME3110] ...(3.15) 30
  10. 3.7.7 Throttling A throttling process occurs when a fluid flowing in a line suddenly encounters a restriction in the flow passage. It may be a plate with a small hole as shown in Figure 3.10 (a) a valve partially closed as shown in Figure 3.10 (b) a capillary tube which is normally found in a refrigerator as shown in Figure 3.10 (c) a porous plug as shown in Figure 3.10 (d) control surface surface Control surface Figure 3.10 Examples of tluottling processes Thermodynamics 1 [MIME3110] 31
  11. General assumptions 1. No heat energy is gained or lost by the fluid; ie., 2. There is typically some increase in velocity in a throttle, but both inlet and exit kinetic energies are usually small enough to be neglected. 3. There is no means for doing work; ie., o. 4. Changes in potential energy of the fluid is negligible. Governing Equation h 2 h 1 ...(3.16) Therefore, throttling is an isenthalpic process. 3.8 First Law for a Cyclic Process In a cyclic process the system is taken through a series of processes and finally returned to its original state. The end state of a cyclic process is identical with the state of the system at the beginning of the cycle. This is possible if the energy level at the beginning and end of the cyclic process are also the same. In other words, the net energy change in a cyclic process is zero. Path B Path A 2 Figure 3.11 First Law for a Cyclic Process Consider a system undergoing a cycle consisting of two processes A & B as shown in Figure 3.11 Net energy change AE + AE (QA Thermodynamics 1 [MIME3110] ..(3.17) ...(3.18) 32
  12. ie QA (or) QB WA W ...(3.19) ...(3.20) Hence for a cyclic process algebraic sum of heat tranfers is equal to the algebraic sum of work transfer. This was first proved by Joule, based on the experiments he conducted between 1843 and 1858, that were the first quantitative analysis of thermodynamic systems. 3.9 Energy is a property of a system Consider a system undergoing a process from statel to state2 along path A as shown in Figure 3.12. Let the system be taken back to the initial state 1 along two possible paths B and C. Process A, combined separately with process B and C forms two possible cycles. 1 Path C Path B Path A 2 Figure 3.12 Illustration to show that energy is property cycle IA2B1 AE cycle IA2C1 AE [WA + Wc] Thermodynamics 1 [MIME3110] ...(3.21) 33
  13. QA WA [QC WC] c ...(3.22) From Equation (3.21) and (3.22) it can be concluded that energy change in path B and path C are equal and hence energy is a point function depending only on the end states. It has been already shown that all the properties are point functions and hence energy is also a property of the system. 3.10 Specific Heat at Constant Volume and at Constant Pressure Specific heat at constant volume of a substance is the amount of heat added to rise the temperature of unit mass of the given substance by I degree at constant volume From first law for a stationary closed system undergoing a process dQ — pdV + dU or dq — pdv + du For a constant volume process or dQ dU or dq — du du - CdT ...(3.23) Similarly specific heat at constant pressure is the quantity of heat added to rise the temperature of unit mass of the given substance by 1 degree at constant pressure where dQ pdV + dU pdV + PV) dQ pdV + dH Vdp pdV dQ dH Vdp For a constant pressure process dp — 0 Hence dQ dH or dq — dh or dh - C dT Note (3.24) e For solids and liquids, constant volume and constant pressure processes are identical and hence, there will be only one specific heat. The difference in specific heats C — C = R— The ratio of sp. heat y =Cp/Cv Since h and u are properties of a system, dh = processes. Thermodynamics 1 [MIME3110] C dT and du-C dT, for all 34
  14. 3.11 Work Interaction in a Reversible Steady Flow Process In a steady flow process the work interaction per unit mass between an open system and the surroundings can be expressed in differential form as Also, Therefore, dq dw dw dq dw dh + CdC + gdZ dq — (dh + CdC +gdz) du + pdv (or) dh vdp dh — vdp — (dh + CdC + gdz) vdp — (CdC + gdz) g(Z2 2 For a stationary system 1 1 ...(3.26) 3.12 First law for an open system under unsteady flow conditions Many processes of engineering interest involve unsteady flow, where energy and mass content of the control volume increase or decrease. Example for such conditions are: 1) Filling closed tanks with a gas or liquid. 2) Discharge from closed vessels. 3) Fluid flow in reciprocating equipments during an individual cycle. To develop a mathematical model for the analysis of such systems the following assumptions are made. 1) The control volume remains constant relative to the coordinate frame. 2) The state of the mass within the control volume may change with time, but at any instant of time the state is uniform throughout the entire control volume. 3) The state of the mass crossing each of the areas of flow on the control surface is constant with time although the mass flow rates may be time varying. Thermodynamics 1 [MIME3110] 35
  15. Unlike in steady flow system, duration of observation At plays an important role in transient analysis. Let mass of the working fluid within the control volume before and after the observation be ml and m2 respectively. Applying mass balance we get, (1112 ml) cv Eml Emo Where Em is the mass entered the control volume during the interval At seconds. Emo is the mass left the control volume during the interval At seconds. By applying energy balance we get, —+Zg AE ...(3.27) in 2 2 out ...(3.28) Where E is the change in energy content of the control volume in At seconds. CV QC v is the heat energy entered into the control volume in At seconds. W is the work energy left the control volume in At seconds. h & ho are specific enthalpy of the inlet and outlet streams respectively. are the kinetic energy of the inlet and outlet streams respectively. Z g & Zog are the potential energy of inlet and outlet streams respectively. 3.13 Perpetual Motion Machine - I An engine which could provide work transfer continuously without heat transfer is known as perpetual motion machine of first kind. It is impossible to have such an engine as it violates first law of thermodynamics. Thermodynamics 1 [MIME3110] 36
  16. Exercises l. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. Define internal energy. Express mathematically first law of thermodynamic for the following. a. b. c. d. e. a closed system undergoing a process a stationary system of fixed mass undergoing a change of state a closed system undergoing a cycle. an open system. an open system with steady-state flow conditions. Define flow energy and enthalpy. For a stationary system of fixed mass undergoing a process such that its volume remains constant, - AU(T/F) dQ dh vdp for closed system undergoing a process (T/F). Define specific heat at (a) constant pressure (b) constant volume Determine the power of the cycle comprising four processes in which the heat transfers are : 50 kJ/kg, —20 kJ/kg, 71 J/kg and 12 kJ/kg having 100 cycles per minute. [48.3 kW] Write the steady flow energy equation and explain the terms involved in it. Show that energy is a property of the system. What are conditions for steady flow process? A piston-cylinder assembly contains Ikg or nitrogen at 100 kPa. The initial volume is 0.5 m . Heat is transferred to the substance in an amount necessary to cause a slow expansion at constant temperature. This process is terminated when the final volume is twice the initial volume. [34.710] 2 kg of air enclosed in a rigid container receives 0.2 kJ of paddle wheel work and 0.5 kJ of electrical energy per second. Heat loss from the system is 0.6 kJ/s. If the initial temperature is 250C what will be the temperature after 5 minutes? [45.90C] A well insulated, frictionless piston-cylinder assembly contains 0.5 kg of air initially at 750C and 300 kPa. An electric - resistance heating element inside the cylinder is energized and causes the air temperature to reach 1500C. The pressure of the air is maintained constant throughout the process. Determine the work for the process and the amount of electrical work. {Hint Qnet Thermodynamics 1 [MIME3110] W net = AU; W net=+W } electric 37
  17. 14. 15. 16. 17. 18. 19. [-26.9 kJ ; - 37.7] A cylinder contains 168 litres of a gas at a pressure of I bar and temperature of 470C. If this gas is compressed to one-twelfth of its volume, pressure is then 21 bar. Find a. index of compression b. change in internal energy c. heat rejected during compression Take C — 1.089 and cv = 0.837 both in kJ/kg [1.225 ; 41.81 kJ ; -14.0510] a. A mass of 10 kg is falling from a height of 100 m from the datum. What will be the velocity when it reaches a height of 20 m from the datum? Take the total heat loss from the mass when it falls from 100 m height to 20 m height is 5 kJ. [8.68 m/s] b. An insulated box containing carbon dioxide gas falls from a balloon 3.5 km above the earths surface. Determine the temperature rise of the carbon dioxide when box hits the ground. Take - 0.6556 kJ/kg [52.3700 A working substance flows at a rate of 5 kg/s into a steady flow system at 6 bar, 2000 kJ/kg of internal energy and 0.4 m [kg specific volume with a velocity of 300 m/s. It leaves at 10 bar, 1600 kJ/kg internal energy, 1.2 m [kg specific volume with a velocity of 150 m/s. The inlet is 10m above the outlet. The work transfer to the surroundings in 3 MW. Estimate the heat transfer and indicate the direction. [5630 kJ/s] An air compressor takes in air at 100 kPa, 400C and discharges it at 690 kPa, 2080C. The initial and final internal energy values for the air are 224 and 346 kJ/kg respectively. The cooling water around the cylinders removes 70 kJ/kg from the air. Neglecting changes in kinetic and potential energy, calculate the work. [100.216 kJ/kg] A perfect gas of c 1.1 kJ/kg flows through a turbine at a rate of 3 kg/s. The inlet and exit p velocity are 30 and 130 m/s respectively. The initial and final temperature are 6500C and 2500C respectively. Heat loss is 45 kJ/s. Find the power developed. [1251 kW] In a turbine 4500 kg/min of air expands polytropically from 425 kPa and 1360 K to 101 kPa. The exponent n it equal to 1.45 for the process. Find the work and heat. Thermodynamics 1 [MIME3110] [33939 kW ; -2927 kJ/s] 38
  18. 20. 21. 22. 23. 24. 25. Air expands through a nozzle from a pressure of 500 kPa to a final pressure of 100 kPa. The enthalpy decrease by 100 kJ/kg. The flow is adiabatic and the inlet velocity is very low. Calculate the exit velocity. [447.2 m/s] A closed system undergoes a cycle consisting of three process 1-2, 2-3 and 3-1. Given that Qi2 — 30 kJ, Q23 — 10 kJ, 5 kJ, 5 kJ and AE31 — 15 kJ, determine (231, wn, AE12 and AE23 [20 kJ ; 50 kJ ; 25 kJ ; -4010 ] The following cycle involves 3 kg of air : Polytropic compression from 1 to 2 where Pi 150 kPa, Tl — 360 K, P2 — 750 kPa and n — 1.1 ; constant-pressure cooling from 2 to 3; and constant - temperature heating from 3 to 1. Draw the PV diagram and find temperatures, pressures and volumes at each state and determine the net work and heat. [150 kPa ; 2.066 m3•, 360 K ; 750 kPa ; 0.478 m3•, 416.72 K ; 750 kPa ; 0.414 m3•, 360 K ; -35 kJ] A cycle, composed of three processes, is . Polytropic compression (n 1.5) from 137 kPa and 380C to state 2 ; constant pressure process from state 2 to state 3 ; constant volume process form state 3 and to state l. The heat rejected in process 3-1 is 1560 kJ/kg and the substance is air. Determine (a) (b) (c) (d) the pressures, temperatures and specific volumes around the cycle the heat transfer in process 1-2 the heat transfer in process 2-3 work done in each process and net work done in the cycle [137 kPa ; 0.6515 m3/kg ; 311.0 K ; 1095 kPa ; 0.1630 m3/kg ; 621.8 K ; 1095 kPa ; 0.6515 m3/kg ; 2487.0 K ; 44.44 kJ ; 1872.25 kJ ; -178kJ ; 534.9 kJ ; O ; 356.9 kJ] 0.15 m of air at a pressure of 900 kPa and 3000 C is expanded at constant pressure to 3 time its initial volume. It is then expanded polytropically following the law PV15 C and finally compressed back to initial state isothermally. Calculate (a) heat received (b) heat rejected (c) efficiency of the cycle [944.5kJ ; -224.906 kJ ; 0.291] 0.8 m3/kg and A piston and cylinder device contains I kg of air, Initially, v T — 298 K. The air is compressed in a slow frictionless process to a specific volume of 0.2 1.3 m [kg and a temperature of 580 K according to the equation PV m3/kg). If cv of air is 0.78 kJ/kg determine : Thermodynamics 1 [MIME3110] 0.75 ( p in bar, v in 39
  19. (a) (b) work and heat transfer (both in kJ) [ -137.85 kJ ; 82.11 10] 26. 27. 28. 29. 30. 31. 32. 33. The internal energy of a closed system is given by U 100 + 50 T + 0.04 T 2 in Joules, and the heat absorbed by Q = 4000 + 16 T in Joules, where T is in Kelvin. If the system changes from 500 K to 1000 K, what is the work done ? [47 kJ] One kg of air, volume 0.05 m , pressure 20 bar expands reversibly according to the law 1.3 pv C until the volume is doubled. It is then cooled at constant pressure to initial volume and further heat at constant volume so that it returns back to initial process. Calculate the network done by air. [21.98 kJ] Air at the rate of 14 kg/s expands from 3 bar, 1500C to Ibar reversibly and adiabatically. Find the exit temperature and power developed. Neglect the changes in kinetic and potential energy. [ 309 k ; 1.603 kW] Specific internal energy of a certain substance can be expressed as follows: u - 831.0 + 0.617 Where u is the specific internal energy in kJ/kg p is the pressure in k Pa v is the specific volume in m3/kg One kg of such substance expands from 850 kPa, 0.25 m3/kg to 600 kPa, 0.5 m3 [kg. Find the work done and heat transferred. [ 176.06 kJ ; 230 10] A cylinder of 8 cm internal diameter is fitted with a piston loaded by a coil spring of stiffness 140 N/cm of compression. The cylinder contains 0.0005 m3 of air at 150C and 3 bar. Find the amount of heat which must be supplied for the piston to a distance of 4 cm. Sketch the process on a p-V diagram. [ 0.417 10] Prove that for a polytropic process of index n. An air conditioning system for a computer room in a tower block draws in air on the roof at a height of 100 m with a velocity of 25 m/s. The air is at 280C. The air is discharged at a height of 10 m with a velocity of 2 m/s at 140C. The mass flow rate is 2 kg/s, and a heat transfer of — 40.73 kW cools the air before it is discharged. Calculate the rate of work for the air passing through the system. Take Cp for air as 1005 J/kgK. 10.23 kW] A diffuser reduces the velocity of an air stream from 300 m/s to 30 m/s. If the inlet pressure and temperature are 1.01 bar and 3150C, determine the outlet pressure. Find also the area required for the diffuser to pass a mass flow of 9 kg/s. Thermodynamics 1 [MIME3110] 40
  20. 34. 35. 36. 37. 38. 39. [4.586 bar, 0.17 A centrifugal air compressor operating at steady state has an air intake of 1.2 kg/min. Inlet and exit conditions are as follows: Properties Inlet Exit p (kPa) TOC u kJ/kg v m [kg 100 200 o 50 195.14 230.99 If the heat loss is negligible, find the power input. 0.784 0.464 [ 1.005 kW] A household gas cylinder initially evacuated is filled by 15 kg gas supply of enthalpy 625 kJ/kg. After filling, the gas in the cylinder has the following parameters : pressure enthalpy 10 bar ; 750 kJ/kg and specific volume = 0.0487 m3/kg. Evaluate the heat received by the cylinder from the surroundings. [1144.510] 0.56 m of air at 0.2 MPa is contained in a fully insulated rigid vessel. The vessel communicates through a valve with a pipe line carrying high pressure air at 300 K temperature. The valve is opened and the air is allowed to flow into the tank until the pressure of air in the tank is raised to IMPa. Determine the mass of air that enters the tank. Neglect kinetic energy of the incoming air. [3.72 kg] An insulated rigid tank contains 8 kg of air at 1.5 bar pressure and 310 K temperature. It is filled with air from a large reservoir at 15 bar and 335 K. If the air behaves as a perfect gas, make calculations for the amount of air added and its temperature. [47.6 kg ; 446.04K] A pressure vessel contains a gas at an initial pressure of 3.5 MN/m and at a temperature of 600C. It is connected through a valve to a vertical cylinder in which there is a piston. The valve is opened, gas enters the vertical cylinder, and work is done in lifting the piston. The valve is closed and the pressure and the temperature of the remaining gas in the cylinder are 1.7 MN/m and 250C, respectively. Determine the temperature of the gas in the vertical cylinder if the process is assumed to be adiabatic. Take y — 1.4. [267.6 K] A pressure vessel is connected, via a valve, to a gas main in which a gas is maintained at a constant pressure and temperature of 1.4 MN/m and 850C, respectively. The pressure vessel is initially evacuated. The valve is opened and a mass of 2.7 kg of gas passes into the pressure vessel. The valve is closed and the pressure and temperature of the gas in the pressure vessel are then 700 KN/m and 600C, respectively. Determine the heat transfer to or from the gas in the vessel. Determine the volume of the vessel and the volume of the gas before transfer. For the gas, take C = 0.88 kJ/kgK, cv — 0.67 . Neglect velocity of the gas in the main Thermodynamics 1 [MIME3110] [-248.2 kJ ; 0.27 m3 ; 0.145 ] 41

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