ENERGY ANALYSIS: CLOSED SYSTEM

Transcription

1 ENERGY ANALYSIS: CLOSED SYSTEM A closed system can exchange energy with its surroundings through heat and work transer. In other words, work and heat are the orms that energy can be transerred across the system boundary. Sign convention: work done by a system is positive, and the work done on a system is negative. Heat transer to the system is positive and rom a system will be negative. System (-) (+) Q (+) (-) W Fig. : Sign convention or heat and work. Moving Boundary Work The expansion and compression work is oten called moving boundary work, or simply boundary work. We analyze the moving boundary work or a quasi-equilibrium process. Consider the gas enclosed in a piston-cylinder at initial and. I the piston is allowed to move a distance ds in a quasi-equilibrium manner, the dierential work is: δ W b Fds Ads d Fig. : the area under - diagram represents the boundary work. M. Bahrami MECH40 (F07) Energy Analysis: Closed System

2 The quasi-equilibrium expansion process is shown in Fig.. On this diagram, the dierential area under the process curve in diagram is equal to, which is the dierential work. Note: a gas can ollow several dierent paths rom state to, and each path will have a dierent area underneath it (work is path dependent). The net work or cycle work is shown in Fig. 3. In a cycle, the net change or any properties (point unctions or exact dierentials) is zero. However, the net work and heat transer depend on the cycle path. ΔU Δ ΔT Δ(any property) 0 or a cycle W net Fig. 3: network done during a cycle. olytropic rocess During expansion and compression processes o real gases, pressure and volume are oten related by n C, where n and C are constants. The moving work or a polytropic process can be ound: W polytopic d C n d n Since. For an ideal gas ( mrt) it becomes: W polytropic ( T ) mr T n, n ( kj ) The special case n is the isothermal expansion mrt 0 C, which can be ound rom: C, d d ln, n ( kj) W b isothermal Since or an ideal gas, mrt 0 at constant temperature T 0, or C/. M. Bahrami MECH40 (F07) Energy Analysis: Closed System

3 Example : olytropic work A gas in piston-cylinder assembly undergoes a polytropic expansion. The initial pressure is 3 bar, the initial volume is 0. m 3, and the inal volume is 0. m 3. Determine the work or the process, in kj, i a) n.5, b) n.0, and c) n0. Solution: Assumptions i. The gas is a closed system ii. The moving boundary is only work mode iii. The expansion is polytropic. a) n.5 W d We need that can be ound rom : n n (.06bar)( 0.m ) ( 3)( 0.) ( 3bar).06 bar 5 0 N / m kj W 7. 6 kj 3.5 bar 0 N. m b) n (isothermal), the pressure volume relationship is constant. The work is: W d ln 5 0 N / m kj W 3 bar 0 N. m ( 3bar)( 0. m ) ln 0.79 kj c) For n 0 (constant-pressure), the pressure-volume relation reduces to constant (isobaric process) and the integral become W ( - ). Substituting values and converting units as above, W30 kj. M. Bahrami MECH40 (F07) Energy Analysis: Closed System 3

4 Example : Mechanical work Calculate the work transer in the ollowing process: 3 Fig. 4: Schematic - diagram or Example. Solution: rocess - is an expansion ( > ) and the system is doing work (W >0), thus: W ( - ) + [0.5( + ) ] ( ) ( ) ( + ) / rocess - 3 is an isometric process (constant volume 3 ), so W 3 0 rocess 3 - is a compression ( 3 > ), work is done on the system, (W 3 < 0) W 3 - ( ) W cycle W net W + W 3 + W 3 ( ) ( ) / Note that in a cycle ΔU Δ ΔT Δ(any property) 0 First Law o Thermodynamics For a Closed System First law, or the conservation o energy principle, states that energy can be neither created nor destroyed; it can only change orms. The irst law cannot be proved mathematically, it is based on experimental observations, i.e., there are no process in the nature that violates the irst law. The irst law or a closed system or a ixed mass may be expressed as: net energy transer to (or rom) the system as heat and work net increase (or decrease) in the total energy o the system Q W ΔE (kj) M. Bahrami MECH40 (F07) Energy Analysis: Closed System 4

5 where Q net heat transer (ΣQ in ΣQ out ) W net work done in all orms (ΣW in ΣW out ) ΔE net change in total energy ( E E ) The change in total energy o a system during a process can be expressed as the sum o the changes in its internal, kinetic, and potential energies: ΔE ΔU + ΔKE + ΔE (kj) ΔU m ( u u ) ΔKE m ΔE mg ( ) ( z z ) Note: or stationary systems ΔE ΔKE 0, the irst law reduces to Q W ΔU The irst law can be written on a unit-mass basis: q w Δe (kj/kg) or in dierential orm: δq δw du (kj) or in the rate orm: δq δw du (kj/kg) For a cyclic process, the initial and inal states are identical, thus ΔE 0. The irst law becomes: Q W 0 (kj) Note: rom the irst law point o view, there is no dierence between heat transer and work, they are both energy interactions. But rom the second law point o view, heat and work are very dierent. Example 3: The Fist law Air is contained in a vertical piston-cylinder assembly itted with an electrical resistor. The atmospheric pressure is 00 ka and piston has a mass o 50 kg and a ace area o 0. m. Electric current passes through the resistor, and the volume o air slowly increases by m 3. The mass o the air is 0.3 kg and its speciic energy increases by 4. kj/kg. Assume the assembly (including the piston) is insulated and neglect the riction between the cylinder and piston, g 9.8 m/s. Determine the heat transer rom the resistor to air or a system consisting: a) the air alone, b) the air and the piston. M. Bahrami MECH40 (F07) Energy Analysis: Closed System 5

6 iston System boundary part a iston System boundary part b Air Air Fig. 5: Schematic or problem 3. Assumptions: Two closed systems are under consideration, as shown in schematic. The only heat transer is rom the resistor to the air. ΔKE ΔE 0 (or air) The internal energy is o the piston is not aected by the heat transer. a) Taking the air as the system, (ΔKE + ΔE + ΔU) air Q W Q W + ΔU air For this system work is done at the bottom o the piston. The work done by the system is (at constant pressure): W d ( ) The pressure acting on the air can be ound rom: A piston m piston g + atm A piston m A Thus, the work is piston piston g + atm ( 50kg)( 9.8m / s ) ( 0.m ) a N / m M. Bahrami MECH40 (F07) Energy Analysis: Closed System 6 ka + 00ka 04.9 ka 000a W (04.9 ka)(0.045m 3 ) 4.7 kj

7 With ΔU air m air Δu air, the heat transer is Q W + m air Δu air 4.7 kj + (0.3 kg)(4. kj/kg) 7.38 kj b) System consists o the air and the piston. The irst law becomes: (ΔKE + ΔE + ΔU) air + (ΔKE + ΔE + ΔU) piston Q W where (ΔKE ΔE) air 0 and (ΔKE ΔU) piston 0. Thus, it simpliies to: (ΔU) air + (ΔE) piston Q W For this system, work is done at the top o the piston and pressure is the atmospheric pressure. The work becomes W atm Δ (00 ka)(0.045m3) 4.5 kj The elevation change required to evaluate the potential energy change o the piston can be ound rom the volume change: Δz Δ / A piston m 3 / 0. m 0.45 m (ΔE) piston m piston g Δz (50 kg)(9.8 m/s )(0.45 m) 0.73 J 0. kj Q W + (ΔE) piston + m air Δu air Q 4.5 kj + 0. kj + (0.3 kg)(4. kj/kg) 7.38 kj Note that the heat transer is identical in both systems. Speciic Heats The speciic heat is deined as the energy required to raise the temperature o a unit mass o a substance by one degree. There are two kinds o speciic heats: Speciic heat at constant volume, c v : the energy required when the volume is maintained constant. Speciic heat at constant pressure, c p : the energy required when the pressure is maintained constant. The speciic heat at constant pressure c p is always higher than c v because at constant pressure the system is allowed to expand and energy or this expansion must also be supplied to the system. Let us consider a stationary closed system undergoing a constant-volume process, 0. Applying the irst law in the dierential orm: at constant volume (no work) and by using the deinition o c v, one can write: M. Bahrami MECH40 (F07) Energy Analysis: Closed System 7

8 c dt du or c v v u T v ( kj / kg. K) Similarly, an expression or the speciic heat at constant pressure C p can be ound. From the irst law, or a constant pressure process, Δ Δ. It yields: c p h T p ( kj / kg. K) Speciic heats (both c v and c p ) are properties; thereore, depend on the state and/or independent o the type o processes. Note: c v is related to the changes in internal energy u, and c p to the changes in enthalpy, h. It would be more appropriate to deine: c v is the change in speciic internal energy per unit change in temperature at constant volume. Similarly c p is the change in speciic enthalpy per unit change in temperature at constant pressure. Note: one important exception is two-phase mixtures; since the temperature remains constant while the internal energy and enthalpy o the mixture change. Speciic heats or ideal gases It has been shown mathematically and experimentally that the internal energy is a unction o temperature only, or ideal gases. u u(t) Using the deinition o enthalpy (h u + v) and the ideal gas equation o state (v RT), we have: h u + RT Since R is a constant and u is a unction o T only: h h(t) Thereore, at a given temperature, u, h, c v and c p o an ideal gas will have ixed values regardless o the speciic volume or pressure. For an ideal gas, we have: du c dh c v p ( T ) dt ( T )dt The changes in internal energy or enthalpy or an ideal gas during a process are determined by integrating: M. Bahrami MECH40 (F07) Energy Analysis: Closed System 8

9 Δu u Δh h u h c c v p ( T ) dt ( kj / kg) ( T ) dt ( kj / kg) As low pressures, all real gases approach ideal-gas behavior, and thereore their speciic heats depend on temperature only. The speciic heats o real gases at low pressures are called ideal-gas speciic heats (or zero-pressure speciic heats) and are oten denoted by c p0 and c v0. To carry out the above integrations, we need to know c v (T) and c p (T). These are available rom a variety o sources: Table A-a: or various materials at a ixed temperature o T 300 K Table A-b: various gases over a range o temperatures 50 T 000 K Table A-: various common gases in the orm o a third order polynomial For an ideal gas, we can write: RT h ( T ) u( T ) dh du R dt dt R c p c v In a mole basis:. where R u is the universal gas constant. The ratio o speciic heats is called the speciic heat ratio: The speciic heat ratio varies with temperature, but this variation is very mild. For monatomic gases, its value is essentially constant at.67. Many diatomic gases, including air, have a speciic heat ratio o about.4 at room temperature. Speciic heats or solids and liquids A substance whose speciic volume (or density) is constant is called incompressible substance. The speciic volumes o solids and liquids (which can be assumed as incompressible substances) essentially remain constant during a process. The constant volume assumption means that the volume-work (or boundary work) is negligible compared with other orms o energy. As a result, it can be shown that the constant-volume and constant-pressure speciic heats are identical or incompressible substances: M. Bahrami MECH40 (F07) Energy Analysis: Closed System 9

10 Speciic heats o incompressible substances are only a unction o temperature, The change o internal energy between state and can be obtained by integration: Δu u u C T kg ( ) dt ( kj / ) For small temperature intervals, a c at averaged temperature can be used and treated as a constant, yielding: ( ) Δ u cave T T The enthalpy change o incompressible substance can be determined rom the deinition o enthalpy (h u + v) h h (u u ) + v( ) Δh Δu + vδ (kj/kg) The term vδ is oten small and can be neglected, so: Δ Δ Δ Note: or constant-temperature processes such as in pumps Δ 0 : Δ Δ. For a process between states and ; it can be expressed as:. Example 4: Speciic heat and irst law Two tanks are connected by a valve. One tank contains kg o CO at 77 C and 0.7 bar. The other tank has 8 kg o the same gas at 7 C and. bar. The valve is opened and gases are allowed to mix while receiving energy by heat transer rom the surroundings. The inal equilibrium temperature is 4 C. Using ideal gas model, determine: a) the inal equilibrium pressure b) the heat transer or the process. Assumptions: The total amount o CO remains constant (closed system). Ideal gas with constant c v. The initial and inal states in the tanks are equilibrium. No work transer. M. Bahrami MECH40 (F07) Energy Analysis: Closed System 0

11 CO 8 kg, 7 C,. bar CO kg, 77 C, 0.7 bar alve The inal pressure can be ound rom ideal gas equation o state: m RT ( m + m ) t + + For tank and, we can write: m RT / and m RT /. Thus, the inal pressure, becomes: RT ( m + m ) RT ( m + m ) mrt ( 0kg)( 35K ) ( kg)( 350K ) ( 8kg)( 300K ) 0.7bar mrt + mt.bar b) The heat transer can be ound rom an energy balance: With W 0, ΔU Q W Q U U i T mt.05bar where initial internal energy is: Ui m u(t ) + m u(t ) The inal internal energy is: U (m + m ) u(t ) The energy balance becomes: Q m [u(t ) u(t )] + m [u(t ) u(t )] Since the speciic heat c v is constant Q m c v [T T ] + m c v [T T ] kj kj Q 5 kg. K kg. K The plus sign indicates that the heat transer is into the system. ( kg) ( 35K 350K ) + ( 8kg) ( 35K 300K ) 37. kj M. Bahrami MECH40 (F07) Energy Analysis: Closed System