CHATER ENERGY INTERACTION (HEAT AND WORK) Energy can cross the boundary of a closed system in two ways: Heat and Work. WORK The work is done by a force as it acts upon a body moving in direction of force. The magnitude of force and the distance moved parallel to force is known as the magnitude of mechanical work. In thermodynamics the work transfer is considered to be occurring between the system and surroundings. Work is said to be done by system if the entire effect on the things external to the system can be equated to raising of weight (weights may not be actually raised but net effect external to the system would be raising of a weight). Let us consider a battery and the motor as shown in Figure as a system. The motor is driving a fan. When a fan is replaced by a pulley and weight, the weight is raised by pulley driven by motor. In thermodynamics it is said that system is doing work as the sole effect external to the system can be equated to raising of a weight. Motor Fan Motor + - Battery + - Battery W BOUNDARY Figure Suppose the fan driven by the motor battery arrangement becomes the part of the system itself as shown in Figure, no energy transfer takes place across the boundary of system and surrounding, therefore, no work is done by the system. Motor Fan + - Battery BOUNDARY Figure No work transfer takes place Sign convention Work done by system is arbitrarily taken to be positive, and when work is done on the system, it is taken to be negative.
W System W System Surroundings (a) W is positive Surroundings (b) W is negative Types of work () Closed system work (non flow work) () Open system work (flow work) Figure In isolated systems no energy is transferred. Therefore, no work is obtained. Closed system work Consider gas enclosed in a piston cylinder device as shown in Figure 4. The initial pressure of gas is, the total volume is, and the cross sectional area of piston is A. If a piston is allowed to move a distance ds in a quasi static manner, the differential work done during this process is dw Fds Ads d ds Figure 4 Let a quasi static process occurs from as shown in Figure 5 da = d d Figure 5 On - diagram, differential area da is equal to d which is the differential work. Work of process = Area under curve when projected on volume axis. W dw= d - Closed system work = Area under - curve projected on volume axis
Conditions for applying W d () The system must be closed system () The work should cross the boundary () rocess must be reversible process (quasi static process) ATH FUNCTION AND OINT FUNCTION Consider two different process (different paths) a and b with same end points as shown in Figure 6. a W -a- b Figure 6 =Area under a on volume axis W -b- =Area under b on volume axis Since the area under each curve represents the work for each process, and is dependent on the path of system, the amount of work done is not a function of end states of the process and is depends on the path of system going from state to state. For this reason work is called path function and dw is an inexact differential. dw=w W -W - Work is path function and thus inexact differential. For a given state there is a definite value of property. The change in thermodynamic property of a system in any process is independent of the path of process during change of state and depends upon the initial and final state of the system. For example change in volume in process -a- and -b- is given by d roperties are state function or point function and does not depend upon path traced or in other words we can say that properties are exact differential. To distinguish an inexact differential dw from an exact differential d or d, the differential line is being cut by a line at its top. dw d d dw / is called integrating factor. Inexact differential dw when multiplied by integrating factor / becomes a exact differential d. For a cyclic process, the initial and final states of the system are the same and hence change in property is always zero.
d 0, d 0, dt 0 WORK DONE IN ARIOUS ROCESSES Isochoric or Constant olume or Isomeric rocess W d Since change in volume is zero for constant volume process W 0 mrt T Figure 7 Therefore for a process - as shown in figure 7 T T Isobaric or Constant ressure or Isopiestic process = mrt Since pressure is constant T For a process - as shown in figure 8 T T W d W d W ( ) Figure 8 Constant Temperature or Isothermal process = mrt Since T is constant and mass does not change in closed system Constant Therefore for a process - as shown in figure = C W d C W d W C ln( ) Figure 9 W ln ln
W ln ln W mrtln Expansion Compression Figure 0 W mrtln Since W ositive Therefore, work is said to be done by system in expansion process. Figure W mrtln Since W Negative Therefore, work is said to be done on the system in compression process. Adiabatic process A process is said to be an adiabatic process if there is no heat transfer from the system or to the system. For reversible adiabatic process C Where = adiabatic index ( always greater than ) =.4 for air W Figure 0 olytropic process n C > n > Where n is called polytropic index W n General equation for all the processes is ( k Constant pressure process k C C )
8 When k = 0 = C Constant volume process k C or ( k ) / k C / k / k C When k = C Isothermal process k C When k = = C Since = mrt T C Adiabatic process k C When k = C K = n K = K = K = 0 olytropic K = K = n k C K = When k = n n C K = Figure Slope of Isothermal process The equation of isothermal process hyperbola. Differentiating the above equation d d 0 d d d d d Where d represents slope of isothermal process on - diagram. Slope of Adiabatic process C d d Therefore, Slope of adiabatic curve = (Slope of Isothermal curve) C represents the equation of a rectangular
Ideal gas equation for various processes ( = mrt) Constant volume T T Constant ressure T T Constant Temperature Adiabatic C or C mrt C T C For a adiabatic process - T T T T T γ = T olytropic process n n n T T n T n T OEN SYSTEM WORK Open system work is equal to area under - curve on pressure axis.
d da = d Figure W d () Isochoric process W ( ) W ( ) Figure () Constant pressure process W d Since d = 0 W 0 Figure ) Isothermal process =C Differentiating the above equation d d 0 d d d d Figure 4
For Isothermal process open system work is equal to closed system work because rectangular hyperbola when projected on pressure axis and volume axis will give same area. W mrtln C ln W = ln = ln = ln = ln (4) Adiabatic ( ) W (5) olytropic n( - ) W = n- HEAT Heat is defined as the form of energy transfer between two systems or a system and surrounding by virtue of to temperature difference. Heat is energy in transition (like work) and is recognized only as it crosses the boundary of system. Thus in thermodynamics the term heat simply means heat transfer. Q m c T Where Q is the amount of heat transfer to a body of mass m having specific heat c due to temperature difference T. Specific heat (kj/kg K) Amount of heat required to raise the temperature of unit mass of a substance through unit degree temperature difference. Constant pressure process When heat is added to the system as shown in Figure 5, the gas expands raising the piston and maintaining constant pressure inside the system. Q mc T mc (T T ) p p f i where c p is the specific heat at constant pressure. t i m t f HEAT Figure 5
Constant volume process Consider a process of heat transfer as shown in Figure 6. Stoppers do not allow the piston to move upwards (Expansion is restricted). Thus process can be called as constant volume heat addition. Q = m c T T where c v is the specific heat at constant volume. v f i t i m t f Figure 6 In constant pressure heating, more heat is required as compared to constant volume process for equal rise in temperature of a given mass of gas. This is attributed to the fact that constant pressure heating involves expansion (cooling effect). c p v cp always greater than cv because c p includes internal energy and external work (expansion) where as c v includes only internal energy. air air.005 kj/kgk c = 0.77 kj/kgk (R) air = c p - c v = 0.87 kj/kgk HEAT
QUESTION BANK Q A mass of gas is compressed in a quasi-static process from 80 Ka, 0.m to 0.4 Ma, 0.0 m. Assuming that the pressure and volume are related by pv n = constant, find the work done by the gas system. (a).76 kj (b) 8.7 kj (c) 4.69 kj (d) 6.5 kj Solution Let subscript denotes the initial condition and subscript denotes the final condition n n = Taking log e on both sides ln + n ln = ln + n ln n ln ln = ln n ln = ln 0. 400 n ln = ln 0.0 80 n =.67.4 Work done W = n 80 0. 400 0.0 W = =.764 kj.4 Q Determine the total work done by a gas system following an expansion process A- B-C as shown in Figure. (bar) 50 A B v.=c (a) 0.5 MJ (b).00 MJ (c).5 MJ (d).0 MJ 0. 0.4 C 0.8 (m )
Solution 5 6 Area under AB = 0.4 0. 50 0 J = 0 W = MJ.. B B C C C 0. bar Area under BC = n B B C C 5 5 500 0.4 0.0 0.8 Area under BC =.5MJ. Total work = Total area under the curve ABC = +.5 =.5 MJ Common data for questions and 4 A rigid tank of 0.568 m volume contains air at 6.895 bar and. 0 C. The tank is equipped with a relief valve that opens at a pressure of 8.68 bars and remains open until the pressure drops to 8.74 bars. Assume that the temperature of the air remains constant during discharge and air in the tank behaves as ideal gas. Assume gas constant = 94.6 Nm/kg K for air. If a fire causes the value to operate once as described. Q What is the temperature just before the value opens? Q4 (a) 8 K (b) 48 K (c) 68 K (d) 88 K What is the mass of air lost due to fire? (a) 0.08 kg (b) 0.8 kg (c) 0.8 kg (d) 0.8 kg Solution CASE BEFORE FIRE ressure, 6.895 bar 6.895 0 5 N/m olume, 0.566m Temperature, T. 7 94. K 5 6.895 0 0.566 mass, m 4.5 kg RT 94.6 94. CASE JUST BEFORE THE ALE OENS
ressure, 8.68 bar 8.680 5 N/m 0.566 m m m mrt 5 8.680 0.566 T 67.9 K mr 4.594.6 Solution 4 ressure, 8.74 bar 8.74 0 5 N/m olume, 0.566 m Temperature, T T 67.9 K p 5 8.74 0 0.566 mass, m 4. K RT 94.6 67.9 Hence, mass of air lost due to fire m m 4.5 4. 0.8 kg Q5 At the beginning of the compression stroke of a two-cylinder internal combustion engine the air is at a pressure of 0.5 ka. Compression reduces the volume to /5 of its original volume, and the law of compression is given by pv. = Constant. If the bore and stroke of each cylinder is 0.5 m and 0.5 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. Solution d 0.5 Initial olume = L 0.5 m 0.0044 m 4 4 Initial pressure = 0.5 ka.
Q6 Final volume = = 0.000884 m 5.. =. Or = 699.4 700 ka.. Work done/unit stroke/unit cylinder (W) =. 0.5 0.0044 700 0.000884 W..056 kj. ve work means work is done on the system W500 ower = kw 60 7.09 kw A closed system undergoes process as shown in Figure - Isobaric olytropic( n =.4) Isothermal process = = 4 bar = bar = m Find (a) (b) Net work Comment on the result. Solution (a) r ocess Isothermal process 4 4 m rocess olytropic process
n = C.4.4 = 4.4.4 =.485 m (b) Net work = W + W + W W = 400.486 94.4 kj work done by system 400.486 00 4 W 486 kj work done by system n.4 W = ln 554.5 kj work done on system W net = W + W + W 5.8 kj A clockwise cycle on diagram is a work producing cycle and anticlockwise cycle is a work consuming cycle. Net work is the area of closed region for a cycle. Q7 An ideal gas is heated at constant volume until its temperature is times the original temperature, then it is expanded isothermally till it reaches its original pressure. The gas is then cooled at constant pressure till it restored to its original state. Determine the net work done per kg of gas if initial temperature is 50 K. [IES 00:0 marks] Solution -Constant volume heating -Isothermal expansion -Constant pressure cooling Given that T 50 K rocess -
T T 50 =050 K For a constant volume process T T w dv 0 (dv = 0) rocess - ( ) w RT ln.87 050 ln(). kj/kg rocess - w dv v v R T T 0.87 50 050 00.9 kj / kg w w w w 0. kj/kg net 4
GATE QUESTIONS Q A kw, 40 litres water heater is switched on for 0 minutes. The specific heat c p for water is 4. kj/kgk. Assuming all the electrical energy has gone into heating the water, increase of the water temperature in degree centigrade is (a).7 (b) 4.0 (c) 4. (d) 5.5 [00: Mark] Solution (c) Heat Supplied ower time kw 060 sec 400 KJ Heat Supplied mcpt 400 40 4.T ( liter kg) T 4. 0 C Q A frictionless piston cylinder device contains a gas initially at 0.8 Ma and 0.05 m. It expands quasi-statically at constant temperature to a final volume of 0.0 m. The work output (in KJ) during this process will be (a) 8. (b) (c) 554.67 (d) 80 [009: marks] Solution (a) = 0.8 Ma (, ) = 0.05 m = 0.0 m For isothermal process ( Work output, W =, ) ln 0.0 W 0.8 0 kn/m 0.05 m ln = 8. kn.m = 8. kj 0.05 Q4 A compressor undergoes a reversible, steady flow process. The gas at inlet and outlet of the compressor is designated as state and state respectively. otential and kinetic energy changes are to be ignored. The following notations are used: v = specific volume and = pressure of the gas. The specific work required to be supplied to the compressor for this gas compression process is
(a) dv (b) vd (c) v ( ) (d) (v v ) [009: Mark] Solution (b) Open system work vdp Since work is done on compressor w c vdp Q5 Solution (d) Heat and work are (a) Intensive properties (b) Extensive properties (c) oint functions (d) ath functions [0: Mark] Q6 A cylinder contains 5 m of an ideal gas at a pressure of bar. This gas is compressed in a reversible isothermal process till its pressure increases to 5 bar. The work in KJ required for this process is (a) 804.7 (b) 95. (c) 98.7 (d) 0. [0: Mark] Solution (a) 5 m bar 00 ka 5 bar 500 ka 5 Wisothermal ln 00 5 ln 804.7 kj Q7 A reversible thermodynamic cycle containing only three processes and producing work is to be constructed. The constraints are: () There must be one isothermal process () There must be one isentropic process () The maximum and minimum cycle pressures and the clearance volume are fixed (4) olytropic process are not allowed Then the number of possible cycles are: (a) (b) (c) (d) 4
[005: Marks] Solution (c) max min - adiabatic - Isothermal - Constant volume/ressure process 4 Such work producing cycles are possible with clockwise direction. Q8 Match items from groups,,, 4 and 5 Group Group Group Group 4 Group 5 When Differential Function henomenon added to the system, is E Heat G ositive I Exact K ath M Transient F Work H Negative J Inexact L oint N Boundary Solution (d) (a) F G J K M F H I K N (b) E G I K M E G I K M (c) F H J L N E H I L M (d) E G J K N F H J K M [006: Marks] Q9 In a steady state steady flow process taking place in a device with a single inlet and a single outlet, the work done per unit mass flow rate is given by outlet w vdp, inlet Where v is the specific volume and p is the pressure, The expression for w given above
Solution (c) (a) Is valid only if the process is both reversible and adiabatic (b) Is valid only if the process is both reversible and isothermal (c) Is valid for any reversible process. (d) Is incorrect; it must be outlet w pdv inlet [008: Marks] The expression outlet w vdp is valid for any reversible process. It does not depend whether inlet the process is isothermal or adiabatic. w pdv work is for a reversible process for a closed system. Q0 A gas expands in a frictionless piston cylinder arrangement. The expansion process is very slow, and is resisted by an ambient pressure of 00 Ka. During the expansion process, the pressure of the system (gas) remains constant at 00 Ka. The change in volume of gas is 0.0 m. What is the maximum amount of work that could be utilized from the above process? (a) kj (b) kj (c) kj (d) 4kJ [008: Marks] Solution (b) A gas expands due to pressure difference. The net pressure difference responsible for expansion is 00 00 = 00 ka For a constant pressure process W W 00 0.0 = kj Linked Answer Questions and A football was inflated to a gauge pressure of bar when the ambient temperature was 5 0 C. When the game started next day, the air temperature at the stadium was 5 0 C. Assume that the volume of the football remains constant at 500 cm. Q The amount of heat lost by the air in the football and the gauge pressure of air in the football at the stadium respectively equal (a) 0.6 J,.94 bar
(b).8 J, 0.9 bar (c) 6. J,.94 bar (d) 4.7 J, 0.9 bar Solution (d) Given bar gauge ( ) absolute gauge atm.0.0 bar 0. ka T 0 5 C 88 K T 0 5 C 78 K 500 cm 500 0 6 m [006: Marks] absolute 0. 500 Mass of air in football 6.0450 kg RT 0.87 88 Heat loss to surrounding Q mcv T T 6.0450 0.780 Q 0.047 kj For a constant volume process T T 78.0.9 bar (absolute) 88 gauge absolute atm 0.9 bar Q Gauge pressure of air to which the ball must have been originally inflated so that it would be equal bar gauge at the stadium is (a). bar (b).94 bar (c).07 bar (d).00 bar [006: Marks] Solution (c)
bar gauge ( ) absolute atm.0 bar gauge absolute T T absolute.08 bar absolute ( gauge ) absolute atm.07 bar