The First Law of Thermodynamics By: Yidnekachew Messele
It is the law that relates the various forms of energies for system of different types. It is simply the expression of the conservation of energy principle. Based on experimental observations, the first law of thermodynamics states that Energy can be neither created nor destroyed during a process; it can only change forms.
For the system shown above, the conservation of energy principle or the first law of thermodynamics is expressed as Total Energy Total Energy The Change in Total = Entering the System Leaving the System Energy of the System Ein Eout = Esystem The total energy of the system, E system, is given as E system = Internal Energy + Kinetic Energy + Potential Energy E system =U+KE+PE
If the system does not move with a velocity and has no change in elevation, it is called a stationary system, and the conservation of energy equation reduces to E E = U+ KE+ PE in E E = U in out out
Mechanisms of Energy Transfer, E in and E out Heat Transfer, (Q) Heat transfer to a system (heat gain) increases the energy of the molecules and thus the internal energy of the system and heat transfer from a system (heat loss) decreases. Work Transfer, (W) Work transfer to a system (i.e., work done on a system) increases the energy of the system, and work transfer from a system (i.e., work done by the system) decreases. Mass Flow, (m) When mass enters a system, the energy of the system increases because mass carries energy with it (in fact, mass is energy). Likewise, when some mass leaves the system, the energy contained within the system decreases.
The energy balance can be written more explicitly as E E = ( Q Q ) + ( W W ) + ( E E ) = E in out in out in out mass, in mass, out System { E E } Net energy transfer = { E } ( kj ) in out System Change in internal, kinetic, by heat, work and mass potential, etc..energies Or on a rate form, as { } { } E E = E ( kw ) in out Rate of net energy transfer by heat, work and mass System Rate change in internal, kinetic, potential, etc..energies
The first law and a closed system For the closed system where the mass never crosses the system boundary, then the energy balance is ( Qin -Q out ) + ( Win -W out ) = Esystem Closed system undergoing a cycle For a closed system undergoing a cycle, the initial and final states are identical, and thus W E = E E = system E in E 2 1 0 E = in = out E out 0 net, out = Qnet, in W net, out = Q net, in
If the total energy is a combination of internal energy, kinetic energy and potential energy E = U + KE + PE 2 2 mv ( 2 V1 ) Q12 W12 = ( U 2 U1) + + mg( Z2 Z1) 2 For negligible changes in kinetic and potential energy Q = U U + W ( ) 12 2 1 12
Internal energy and Enthalpy Internal energy The internal energy includes some complex forms of energy show up due to translation, rotation and vibration of molecules. It is designated by U and it is extensive property. Or per unit mass as, specific internal energy, U u = m If we take two phase as liquid and vapor at a given saturation pressure or temperature U U U = f + g mu = m fu f + mgug u = u + xu f fg
Enthalpy It is another extensive property which has a unit of energy and it is denoted by H. The enthalpy is a convenient grouping of the internal energy, pressure, and volume and is given by H = U + PV The enthalpy per unit mass is, ( ) h = u + Pv h = Q12 = U2 U1 + W12 W12 = Pdv = P( V2 V1 ) H m W 12 = PV2 PV1 = ( ) + ( ) Q = ( U PV ) + ( U PV ) Q U U PV PV 12 2 1 2 1 12 2 2 1 1 Q12 = H2 H1 h = hf xhfg
Specific Heat It defined as; the energy required to raise the temperature of a unit mass of a substance by one degree. It is an intensive property of a substance that will enable us to compare the energy storage capability of various substances. The unit is KJ or KJ. Kg In thermodynamics, we are interested in two kinds of specific heats: specific heat at constant volume and specific heat at constant pressure. KgK
The specific heat at constant volume can be viewed as the energy required to raise the temperature of the unit mass of a substance by one degree as the volume is maintained constant. Here the boundary work is zero because the volume is constant From first law Per unit mass δq = du δ q = du δ q = CvdT C CvdT v = du = dt du v
The specific heat at constant pressure C p can be viewed as the energy required to raise the temperature of the unit mass of a substance by one degree as the pressure is maintained constant. From first law Per unit mass δq = du + PdV = d ( U + PV) = dh δ q = dh δ q = p C dt C pdt = dh C p dh = dt p
Internal Energy, Enthalpy, and Specific Heats of Ideal Gases We defined an ideal gas as a gas whose temperature, pressure, and specific volume are related by Pv RT From the specific heat relation = u2 u1 = Cv dt Or taking average value of specific heat for narrow temperature difference ( ) dh C T dt du = C dt u2 u1 = C ( ) ave, v T2 T1 = p 2 1 p h2 h1 = C ( ) ave, p T2 T1 v h h C dt =
Relation between C P and C V for Ideal Gases dh h = u + RT dh = du + RdT Replacing by C dt and du by CvdT we have p C dt = C dt + RdT p v Cp = Cv + R At this point, we introduce another ideal-gas property called the specific heat ratio k, defined as K C v = C C p v R = K 1 C C p p = KCv v v C p = + R K KC = C + R C p = K R K 1
Internal Energy, Enthalpy, and Specific Heats of Solids and Liquids A substance whose specific volume (or density) is constant is called an incompressible substance. The specific volumes of solids and liquids essentially remain constant during a process. Therefore, liquids and solids can be approximated as incompressible substances. It can be mathematically shown that the constant-volume and constant-pressure specific heats are identical for incompressible substances. The specific heat can be expressed as C = C = C p v
Example
1. The initial pressure and volume of a pistoncylinder arrangement is 200kPa and 1m 3 respectively. 2000kJ of heat is transferred to the system and the final volume is 2m 3. Determine the change in the internal energy of the fluid.
2. A piston cylinder device initially contains 0.8 m3 of saturated water vapor at 250 kpa. At this state, the piston is resting on a set of stops, and the mass of the piston is such that a pressure of 300 kpa is required to move it. Heat is now slowly transferred to the steam until the volume doubles. Show the process on a P-v diagram with respect to saturation lines and determine (a) the final temperature, (b) the work done during this process, and (c) the total heat transfer.
3.A piston-cylinder contains steam initially at 1 Mpa, 450 o C and 2.5m 3. Steam ia allowed to cool at constant pressure until it first start condensing. Show the process on a T-v diagram with respect to saturatin lines and detrmine. a) The mass of the steam b) The final temperature c) The amount of heat tarnsfer.
4. A piston cylinder device initially contains steam at 200 kpa, 200 C, and 0.5 m 3. At this state, a linear spring is touching the piston but exerts no force on it. Heat is now slowly transferred to the steam, causing the pressure and the volume to rise to 500 kpa and 0.6 m 3, respectively. Show the process on a P-v diagram with respect to saturation lines and determine (a) the final temperature, (b) the work done by the steam, and (c) the total heat transferred.
The First Law and the Control Volume The conservation of mass and the conservation of energy principles for open systems or control volumes apply to systems having mass crossing the system boundary or control surface. In addition to the heat transfer and work crossing the system boundaries, mass carries energy with it as it crosses the system boundaries.
Hence the conservation of mass principle can be used to relate mass which entering and leaving a system. It can be expressed as The net mass transfer to or from a control volume during a process (a time interval t) is equal to the net change (increase or decrease) in the total mass within the control volume during that process (t). That is, Total mass entering Total mass leaving Net change in mass - = the CV during Δt the CV during Δt within the CV during Δt min mout = mcv ( kg) m m = dm / dt in out CV ( kg / s) m= ρv Thermodynamic processes involving control volumes can be considered in two groups: steady-flow processes and unsteadyflow processes.
Steady state process The flow through a control volume is at steady state if, the property of the substance at a given position within or at the boundaries of the control volume do not change with time. During a steady-flow process, the total amount of mass contained within a control volume does not change with time (m CV = constant). m m = dm / dt in out CV m in m = in = m out m out dm CV dt = m = 0 CV ρ V in V in in = ρ = V out out V out ρ in = ρout incompressible assumption V A = V A in in out out
Unsteady state process The properties within the control volume change with time but remain uniform at any instant of time. Typical example:- filling and empting processes where most of the cases average value of properties must be used. dm cv dt 0 dm cv = m i me dt dm dt cv = mi m e
Flow Work and The Energy of a Flowing Fluid Unlike closed systems, control volumes involve mass flow across their boundaries, and some work is required to push the mass into or out of the control volume. This work is known as the flow work, or flow energy.
Wflow F = FL PA = = PAL = PV ( kj ) wflow = Pv w = Pv and w = P v flow, in i i flow, exit e e W = ( Pv ) m and W = ( Pv ) m flow, in i i i flow, exit e e e W ( ) m ( ) = Pv Pv m flow e e e i i i W = W flow W cv
Development of energy balance The general representation of the first law of thermodynamics Q12 = W12 + E2 E1 The first law for open system will also have the same form, but W12 = Wflow + Wcv E = Internal Energy + Kinetic Energy + Potential Energy E=U+KE+PE 2 V e = u + ke + pe = u + + gz 2 The fluid entering or leaving a control volume possesses an additional form of energy, the flow energy Pv Then the total energy of a flowing fluid on a unit-mass basis (denoted by) becomes θ = Pv + e = Pv + ( u + ke + pe)
But the combination Pv + u has been previously defined as the enthalpy h. So the relation in the above equation reduces to For inlet For outlet General equation 2 V h ke pe h gz 2 θ = + + = + + ( kj / kg) e Pv u V gz Pv 2 2 i i+ i i= i+ + i+ i i e P v h V gz 2 2 e e+ e e= e+ + e e Pv h V gz 2 2 i i+ i i= i+ + i E E = de / dt in out system 0( steady) = 0 Q + W + mθ = Q + W + mθ in in out out in 2 2 V V Qin + Win + m( h + + gz) = Qout + Wout + m( h + + gz) 2 2 in out out
In such cases, it is common practice to assume heat to be transferred into the system (heat input) at a rate of Q, and work produced by the system (work output) at a rate of W, and then solve the problem. The first-law or energy balance relation in that case for a general steady-flow system becomes 2 2 V2 V 1 Q W= mh 2 h1+ + gz ( 2 z1) 2 2 2 V2 V1 q w= h2 h1+ + gz ( 2 z1) 2 When the fluid experiences negligible changes in its kinetic and potential energies (that is,ke = 0, pe = 0), the energy balance equation is reduced further to q w= h2 h1
Some Steady-Flow Engineering Devices
Nozzles and Diffusers Nozzles and diffusers are commonly utilized in jet engines, rockets, spacecraft, and even garden hoses. A nozzle is a device that increases the velocity of a fluid at the expense of pressure. A diffuser is a device that increases the pressure of a fluid by slowing it down. m in = m 1 2 out m = m = m E in = 2 2 V i V e Qnet + mi hi + + gzi = W net + me he + + gze inlet 2 exit 2 E out 2 2 V i V e mi hi + = me he + 2 2 V = 2( h h ) + V 2 e i e i
Turbines In steam, gas, or hydroelectric power plants, the device that drives the electric generator is the turbine. As the fluid passes through the turbine, work is done against the blades, which are attached to the shaft. As a result, the shaft rotates, and the turbine produces work. m in = 1 2 m out m = m = m E in = 2 2 V i V e Qnet + mi hi + + gzi = W net + me he + + gze inlet 2 exit 2 E out mh= mh+ W i i e e out W out = mh ( i he)
Compressors Compressors, as well as fans, are devices used to increase the pressure of a fluid. Work is supplied to these devices from an external source through a rotating shaft. 2 2 V i V e Qnet + mi hi + + gzi = W net + me he + + gze inlet 2 exit 2 m in = m out W net = mh ( e hi ) W net = mh ( i he)
Pumps The work required when pumping an incompressible liquid in an adiabatic steady-state, steady-flow process is given by 2 2 V2 V 1 Q W= m h2 h1+ + gz ( 2 z1) 2 The enthalpy difference can be written as ( ) ( ) ( ) h h = u u + Pv Pv 2 1 2 1 2 1 The pumping process for an incompressible liquid is essentially isothermal, and the internal energy change is approximately zero. Since v 2 = v 1 = v the work input to the pump becomes 2 2 V2 V 1 W= m vp ( 2 P1) + + gz ( 2 z1) 2 W = m v( P2 P1) ( ) W = m v P P in, pump 2 1
Throttling Valves Throttling valves are any kind of flow-restricting devices that cause a significant pressure drop in the fluid. The pressure drop in the fluid is often accompanied by a large drop in temperature, and for that reason throttling devices are commonly used in refrigeration and air-conditioning applications. m 2 2 V i V e Qnet + mi hi + + gzi = W net + me he + + gze inlet 2 exit 2 in = m out mh = mh i i e e h i = h e
Mixing Chambers The mixing of two fluids occurs frequently in engineering applications. The section where the mixing process takes place is called a mixing chamber. The ordinary shower is an example of a mixing chamber. m in = m m + m = m 1 2 3 m = m m out 2 3 1 E in = V V Q m h gz W m h gz 2 2 i e net + i i + + i = net + e e + + e inlet 2 exit 2 E out mh+ mh = mh 1 1 2 2 3 3 mh + m m h = mh 1 1 3 1 2 3 3 m ( h h ) = m ( h h ) 1 1 2 3 3 2 ( h3 h2) 1 3 m = m ( h h ) 1 2
Heat Exchangers Heat exchangers are normally well-insulated devices that allow energy exchange between hot and cold fluids without mixing the fluids. m in = m out E in = E out m = m = m 1 2 w m = m = m 3 4 R 2 2 V i V e Qnet + mi hi + + gzi = W net + me he + + gze inlet 2 exit 2 m h + m h = m h + m h 1 1 3 3 2 2 4 4 mw( h h ) = m ( h h ) 1 2 R 4 3
Example
1.Steam at 0.4 MPa, 300 o C, enters an adiabatic nozzle with a low velocity and leaves at 0.2MPa with a quality of 90%. Find the exit velocity, in m/s.
2. The power output of an adiabatic steam turbine is 5 MW, and the inlet and the exit conditions of the steam are as indicated in Figure below. a) Compare the magnitudes of h, ke, and pe. b) Determine the work done per unit mass of the steam flowing through the turbine. c) Calculate the mass flow rate of the steam.
3. Nitrogen gas is compressed in a steady-state, steady-flow, adiabatic process from 0,1 Mpa, 25 o C. During the compression process the temperature become 125 o C. If the mass flow rate is 0.2kg/s, determine the work done on the nitrogen, in kw. (use c p =1.039kJ/kg.K)
4. Saturated steam at 0.4 MPa is throttled to 0.1MPa, 100 o C. Determine the quality of the steam at 0.4MPa.
5. Steam at 0.2MPa, 300 o C, enters a mixing chamber and is mixed with cold water at 20 o C, 0.2MPa, to produce 20kg/s of saturated liquid at 0.2MPa. What are the required steam and cold water flow rates?
6. Two tanks (Tank A and Tank B) are separated by a partition. Initially Tank A contains 2-kg steam at 1 MPa and 300 C while Tank B contains 3-kg saturated liquid vapor mixture with a vapor mass fraction of 50 percent. Now the partition is removed and the two sides are allowed to mix until the mechanical and thermal equilibrium are established. If the pressure at the final state is 300 kpa, determine (a) the temperature and quality of the steam (if mixture) at the final state and (b) the amount of heat lost from the tanks.
7. Air enters an adiabatic nozzle steadily at 300 kpa, 200 C, and 30 m/s and leaves at 100 kpa and 180 m/s. The inlet area of the nozzle is 80 cm 2. Determine (a) the mass flow rate through the nozzle, (b) the exit temperature of the air, and (c) the exit area of the nozzle.
8. Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 10 MPa, 450 C, and 80 m/s, and the exit conditions are 10 kpa, 92 percent quality, and 50 m/s. The mass flow rate of the steam is 12 kg/s. Determine (a) the change in kinetic energy, (b) the power output, and (c) the turbine inlet area.
9. Refrigerant-134a is throttled from the saturated liquid state at 700 kpa to a pressure of 160 kpa. Determine the temperature drop during this process and the final specific volume of the refrigerant.
10. In steam power plants, open feedwater heaters are frequently utilized to heat the feedwater by mixing it with steam bled off the turbine at some intermediate stage. Consider an open feedwater heater that operates at a pressure of 1000 kpa. Feedwater at 50 C and 1000 kpa is to be heated with superheated steam at 200 C and 1000 kpa. In an ideal feedwater heater, the mixture leaves the heater as saturated liquid at the feedwater pressure. Determine the ratio of the mass flow rates of the feedwater and the superheated vapor for this case.
12. An adiabatic air compressor is to be powered by a direct-coupled adiabatic steam turbine that is also driving a generator. Steam enters the turbine at 12.5 MPa and 500 C at a rate of 25 kg/s and exits at 10 kpa and a quality of 0.92. Air enters the compressor at 98 kpa and 295 K at a rate of 10 kg/s and exits at 1 MPa and 620 K. Determine the net power delivered to the generator by the turbine.