First Law of Thermodynamics Internal Energy (u), It is the energy associated with molecules. Or it is the heat energy stored in a gas. If a certain amount of heat is supplied to a gas, the result is that temperature of gas may increase or volume of gas may increase thereby doing some external work or both temperature and volume may increase, but it will be decided by the conditions under which the gas is heated. If during heating of the gas the temperature increases its internal energy will also increase. Internal energy is an extensive property, and it is generally expressed in KJ/Kg 𝑢 Specific internal energy, 𝑚 We do not know how to find the absolute quantity of internal energy in any substance; however, what is needed in engineering is the change of internal energy (ΔU). Law of Conservation of Energy Energy is neither created nor destroyed but only changes from one form to another; this is known as the law of the conservation of energy. The First Law of Thermodynamics can, therefore, be stated as follows When a system undergoes a thermodynamic cycle, then the net heat supplied to the system from the surroundings is equal to net work done by the system on its surroundings. Or For a closed system undergoing thermodynamic cycle, the net heat transfer ∑ 𝑄 is equal to net work transfer ∑ 𝑊 Hence ∑ 𝑄 = ∑ 𝑊 →valid for thermodynamic cycles. This is also expressed in the form ∮ 𝑑𝑄 = ∮ 𝑑𝑊 Where ∮ → Represents the sum for a complete cycle. The first law applies to reversible as well as irreversible transformations, for non-cyclic process, a more general formulation of first law of thermodynamics are required. A new concept which involves a term called internal energy fulfills this need. The First Law of Thermodynamics may also be stated as follows: Heat and work are mutually convertible but since energy can neither be created nor destroyed, the total energy associated with an energy conversion remains constant.\ No machine can produce energy without corresponding expenditure of energy, i.e., it is impossible to construct a perpetual motion machine of first kind Sample Adiabatic work In the sample given above, there are only two energy transfer quantities as the system performs a thermodynamic cycle. If the cycle involves many more heat and work quantities, the same result will be found. Expressed algebraically as (∑ 𝑊)𝑐𝑦𝑐𝑙𝑒 = 𝐽(∑ 𝑄)𝑐𝑦𝑐𝑙𝑒 Where J→ Joule's equivalent. This is also expressed in the form ∮ 𝑑𝑄 = 𝐽 ∮ 𝑑𝑊 Where J→ Proportionality constant and is known as Mechanical Equivalent of heat Cycle completed by a system with two energy interactions (adiabatic work transfer 𝑊1−2 followed by heat transfer 𝑄2−1 Consequences (results) of first law of thermodynamics Heat transfer is a path function From, (𝑑𝑄)1𝐴2+ (𝑑𝑄)2𝐵1 = (𝑑𝑊)1𝐴2 + (𝑑𝑊)2𝐵1 ………………….. (i) (𝑑𝑄 )1𝐴2+ (𝑑𝑄 )2𝐶1 = (𝑑𝑊 )1𝐴2 + (𝑑𝑊 )2𝐶1 …………………… (ii) Subtracting (ii) from (i) (𝑑𝑄)1𝐴2 - (𝑑𝑄)1𝐴2 + (𝑑𝑄)2𝐵1 - (𝑑𝑄)2𝐶1 = (𝑑𝑊)1𝐴2- (𝑑𝑊)1𝐴2 + (𝑑𝑊)2𝐵1 - (𝑑𝑊)2𝐶1 (𝑑𝑄)2𝐵1 - (𝑑𝑄)2𝐶1 = (𝑑𝑊)2𝐵1 - (𝑑𝑊)2𝐶1 But (𝑑𝑄)2𝐵1 ≠ (𝑑𝑄)2𝐶1 Hence (𝑑𝑄)2𝐵1 − (𝑑𝑄)2𝐶1 ≠ 0 Also (𝑑𝑊)2𝐵1 ≠ (𝑑𝑊)2𝐶1 Hence (𝑑𝑊)2𝐵1 - (𝑑𝑊)2𝐶1 ≠ 0 Therefore, though for the path B and C end points are same, but heat transfer is not the same and hence heat transfer is a path function. Important points with respect to heat transfer and work transfer i. Both heat and work transfers are path function ii. Both heat and work transfers are not properties iii. Both are inexact differentials iv. Both are boundary phenomenon Energy is a Property of a System Energy is a property of a system Consider a system which changes its state from state 1 to state 2 by following the path L, and returns from state 2 to state 1 by following the path M. So the system undergoes a thermodynamic cycle. (𝑑𝑄)1𝐿2+ (𝑑𝑄)2𝑀1 = (𝑑𝑊)1𝐿2 + (𝑑𝑊)2𝑀1 ………………….. (i) (𝑑𝑄)1𝐿2+ (𝑑𝑄)2𝑁1 = (𝑑𝑊)1𝐿2 + (𝑑𝑊)2𝑁1 …………………… (ii) Subtracting (ii) from (i) (𝑑𝑄)1𝐿2+ (𝑑𝑄)2𝑀1 - (𝑑𝑄)1𝐿2 - (𝑑𝑄)2𝑁1 = (𝑑𝑊)1𝐿2 + (𝑑𝑊)2𝑀1 - (𝑑𝑊)1𝐿2 - (𝑑𝑊)2𝑁1 (𝑑𝑄)2𝑀1 - (𝑑𝑄)2𝑁1 = (𝑑𝑊)2𝑀1 - (𝑑𝑊)2𝑁1 (𝒅𝑸 − 𝒅𝑾)𝟐𝑴𝟏 = (𝒅𝑸 − 𝒅𝑾)𝟐𝑵𝟏 The quantity (𝑑𝑄 − 𝑑𝑊) is same for path M and N, and hence it does not depend on path, it depends only on the end point. Therefore, (𝑑𝑄 − 𝑑𝑊) must be a property and this property is known as Energy. (𝑑𝑄 − 𝑑𝑊)2𝑀1 = (𝑑𝑄 − 𝑑𝑊)2𝑁1 = (𝑑𝑄 − 𝑑𝑊) = 𝑑𝐸 (𝑑𝑄 − 𝑑𝑊) = 𝑑𝐸 Therefore, dQ – dW = dE Thus dQ = dE + dW ……………(iii) This is the first law of thermodynamics for a closed system undergoing Reversible or irreversible process Because W=PdV → is valid for reversible process Different Forms of Stored Energy The symbol E refers to the total energy stored in a system. Basically there are two modes in which energy may be stored in a system: (a) Macroscopic energy mode (b) Microscopic energy mode Macroscopic energy mode The macroscopic energy mode includes the macroscopic kinetic energy and potential energy of a system. Let us consider a fluid element of mass m having the center of mass velocity V. The macroscopic kinetic energy 𝐸𝑘 of the fluid element by virtue of its motion is given by 𝑚𝑉 2 𝐸𝑘 = 2 If the elevation of the fluid element from an arbitrary datum is z, then the macroscopic potential energy 𝐸𝑝 by virtue of its position is given by 𝐸𝑝 = 𝑚𝑔𝑧 Microscopic energy mode The microscopic energy mode refers to the energy stored in the molecular and atomic structure of the system, which is called the molecular internal energy or simply internal energy. Internal energy is denoted by U. Other forms of energy which can also be possessed by a system are magnetic energy, electrical energy and surface (tension) energy. In the absence of these forms of energy, the total energy E of a system is given by E=𝐸𝑘 + 𝐸𝑝 + 𝑈 ……………. (iv) Where 𝐸𝑘 + 𝐸𝑝 → Macroscopic U→ Microscopic 𝐸𝑘 → Kinetic energy 𝐸𝑝 → Potential energy 𝑈 → Internal energy In the absence of motion and gravity 𝐸𝑘 → 0 𝐸𝑝 → 0 Therefore, from (iv) E=𝐸𝑘 + 𝐸𝑝 + 𝑈 E= 0+0+𝑈 E=U ……………………. (v) Or E=∆𝑈……………. (Vi) From (iii) dQ = dE + dW dQ = dU + dW Which is the same as; Q=dU+ W Considering the different forms of work transfer which may be present when only PdV work is present, the equations becomes; dQ =dE + PdV dQ = dU + PdV Or, in the integral form Q =∆E + ∫ 𝑃𝑑𝑉 Q =∆U + ∫ 𝑃𝑑𝑉 Specific Heat at Constant Volume The specific heat of a substance at Constant volume 𝐶𝑣 is defined as the rate of change of specific internal energy with respect to temperature when the volume is held constant. 𝑑𝑈 𝐶𝑣 = ( 𝑑𝑇 ) 𝑣 𝑇 For constant volume process, (∆𝑈)𝑣 = ∫𝑇 2 𝐶𝑣 𝑑𝑇 1 For a closed stationary system composed of a unit mass of a pure substance Q= ∆U + W dQ = dU + dW Or For a process in the absence of work other than w=PdV work dW= PdV dQ =dU + PdV When the volume is held constant (𝑄)𝑣 = (∆𝑈)𝑣 = 𝑇 2 ∫𝑇 𝑐𝑣 𝑑𝑇 1 Heat transferred at constant volume increases the internal energy of the system. If the specific heat of a substance is defined in terms of heat transfer, then, 𝑑𝑄 𝑐𝑣 = ( 𝑑𝑇 ) 𝑣 Where Q→ is not a property 𝑐𝑣 → Property of a substance Since U, T and V are properties, cv → is a property of the system The product m𝑐𝑣 = 𝐶𝑣 → heat capacity at constant Volume (J/K) Enthalpy One of the fundamental quantities which occur invariably in thermodynamics is the sum of internal energy (u) and pressure volume product (pv). This sum is called Enthalpy (h). The enthalpy of a substance, h, is defined as h = u + pv →It is an intensive property of a system (KJ/kg). The enthalpy of a fluid is the property of the fluid, since it consists of the sum of a property and the product of the two properties. Since enthalpy is a property like internal energy, pressure, specific volume and temperature, it can be introduced into any problem whether the process is a flow or a non-flow process The total enthalpy of mass, m, of a fluid can be, H = U + pV, where H = mh. For a perfect gas h = u + pv Heat transferred at constant pressure increases the enthalpy of a system Total Enthalpy H= mh 𝐻 h = 𝑚 (J/Kg) Specific Heat at Constant Pressure The specific heat at constant pressure 𝑐𝑝 is defined as the rate of change of enthalpy with respect to temperature when the pressure is held constant. 𝑑ℎ 𝑐𝑝 = (𝑑𝑇 ) 𝑝 Since h, T and p are properties, then, 𝑐𝑝 → is a property of the system 𝑐𝑣 , 𝑐𝑝 → should not be defined in terms of heat transfer of constant pressure Although (𝑑𝑄)𝑝 = 𝑑ℎ For a constant pressure process 𝑇 (∆ℎ)𝑝 = ∫𝑇 2 𝑐𝑝 . 𝑑𝑇 1 cp → is a property of the system. Just like cv , The heat capacity at constant pressure Cp is mcp = Cp → heat capacity at constant pressure (J/K) Energy of an Isolated System An isolated system is one in which there is no interaction of the system with the surroundings. For an isolated system, dQ = 0 & dW = 0 The first law gives dE = 0 E = constant Therefore, the energy of an isolated system is always constant. Perpetual Motion Machine of the First Kind-PMM1 The first law states the general principle of the law of conservation of energy, which states; Energy is neither created nor destroyed but only gets transformed from one form to another. There can be no machine which would continuously supply mechanical work without some other form of energy disappearing simultaneously. Such a fictitious machine is called a Perpetual Motion Machine of the First Kind (PMMI). The opposite of the above statement is also true. There can be no machine which would continuously consume work without some other form of energy appearing simultaneously. Limitations of the first law of thermodynamics The first law deals with the amounts of energy of various forms transferred between the system and its surroundings and with changes in the energy stored in the system. It treats work and heat interactions as equivalent forms of energy in transit and does not indicate the possibility of a spontaneous process proceeding in a certain direction. APPLICATION OF FIRST LAW OF THERMODYNAMICS TO NON-FLOW OR CLOSED SYSTEM 1. Reversible Constant Volume (or Isochoric) Process (v = constant) Reversible constant volume process In a constant volume process the working substance is contained in a rigid vessel, hence the boundaries of the system are immovable and no work can be done on or by the system. It will be assumed that constant volume implies zero work unless stated otherwise. Considering mass of the working substance unity and applying first law of thermodynamics to the process. 𝑄 = (𝑢2 − 𝑢1 ) + W The work done, W 2 W=∫1 𝑃𝑑𝑣 = 0 as dv=0 Q = (𝑢2 − 𝑢1 )= 𝑐𝑣 (𝑇2 − 𝑇1 ) Where 𝑐𝑣 → Specific heat at constant volume For mass, m, of working substance Q = 𝑈2 − 𝑈1 = m𝑐𝑣 (𝑇2 − 𝑇1 ) 2. Reversible Constant Pressure (or Isobaric) Process (p = constant). Reversible constant pressure process When the boundary of the system is inflexible as in a constant volume process, then the pressure rises when heat is supplied. Hence for a constant pressure process, the boundary must move against an external resistance as heat is supplied. Hence for a constant pressure process, the boundary must move against an external resistance as heat is supplied. Here, a cylinder behind a piston can be made to undergo a constant pressure process. Since the piston is pushed through a certain distance by the force exerted by the gas, then the work is done by the gas on its surroundings Considering unit mass of working substance and applying first law of thermodynamics to the process; 𝑄 = (𝑢2 − 𝑢1 ) + W Then, work done, W 2 W=∫1 𝑃𝑑𝑣 = 𝑃(𝑣2 − 𝑣1 ) Q=(𝑢2 − 𝑢1 ) + p(𝑣2 − 𝑣1 ) = 𝑢2 − 𝑢1 + p𝑣2 - p𝑣1 But h→ 𝑢 + 𝑝𝑣 Q=(𝑢2 + 𝑝𝑣2 ) - (𝑢1 + 𝑝𝑣2 ) = ℎ2 − ℎ1 Q= ℎ2 − ℎ1 =𝑐𝑝 (𝑇2 − 𝑇2 ) Where h → Enthalpy (specific) 𝑐𝑝 →Specific heat at constant pressure. For mass, m, of working substance Q=𝐻2 − 𝐻1 = m𝑐𝑝 (𝑇2 − 𝑇2 ) 3. Reversible Temperature (or Isothermal) Process (pv = constant, T = constant) A process at a constant temperature is called an isothermal process. When a working substance in a cylinder behind a piston expands from a high pressure to a low pressure there is a tendency for the temperature to fall. In an isothermal expansion heat must be added continuously in order to keep the temperature at the initial value. Similarly in an isothermal compression heat must be removed from the working substance continuously during the process. Reversible isothermal process Considering unit mass of working substance and applying first law to the process 𝑄 = (𝑢2 − 𝑢1 ) + W 𝑄 = 𝑐𝑣 (𝑇2 − 𝑇1 ) + W But 𝑇2 = 𝑇1 𝑄 = 𝑐𝑣 (𝑇1 − 𝑇1 ) + W 𝑄 = 0+W Then, work done, W 2 W=∫1 𝑝𝑑𝑣 𝐶 In this case, pv= constant or p= 𝑣 , c→ Constant 𝑉 𝑑𝑣 1 𝑣 W= ∫𝑉 2 𝐶 𝑣 𝑣 = 𝐶[log 𝑒 𝑣]𝑣21 = Clog 𝑒 𝑣2 1 For mass, m, of the working substance 𝑉 Q= 𝑝1 𝑣1 log 𝑒 𝑉2 1 𝑝2 𝑉 Q= 𝑝1 𝑣1 log 𝑒 𝑝 where 𝑉2 = 1 1 𝑝2 𝑝1 4. Reversible Adiabatic Process ( pvγ = constant) An adiabatic process is one in which no heat is transferred to or from the fluid during the process. Such a process can be reversible or irreversible. The reversible adiabatic non-flow process will be considered in this section. Considering unit mass of working substance and applying first law to the process Q = (𝑢2 − 𝑢1 ) + W But since for an adiabatic process Q→0 O = (𝑢2 − 𝑢1 ) + W W= (𝑢2 − 𝑢1 ) for any adiabatic process The above is true for an adiabatic process whether the process is reversible or not. For an adiabatic process to take place, perfect thermal insulation for the system must be available. Where, 𝑝𝑣 𝛾 = constant 5. Polytropic Reversible Process (𝑝𝑣 𝑛 = constant) It is found that many processes in practice approximate to a reversible law of form pv n = constant, where n is a constant. Both vapours and perfect gases obey this type of law closely in many non-flow processes. Such processes are internally reversible. 6. Free Expansion Consider two vessels 1 and 2 interconnected by a short pipe with a valve A, and perfectly thermally insulated. Initially let the vessel 1 be filled with a fluid at a certain pressure, and let 2 be completely evacuated. When the valve A is opened the fluid in 1 will expand rapidly to fill both vessels 1 and 2. The pressure finally will be lower than the initial pressure in vessel 1. This is known as free or un-resisted expansion. The process is highly irreversible; since the fluid is eddying continuously during the process. Now applying first law of thermodynamics (or nonflow energy equation) between the initial and final states. 𝑄 = (𝑢2 − 𝑢1 ) + W In this process, no work is done on or by the fluid, since the boundary of the system does not move. No heat flows to or from the fluid since the system is well lagged. The process is therefore, adiabatic but irreversible. 𝑢2 − 𝑢1 = 0 Since 𝑢2 = 𝑢1 In a free expansion, therefore, the internal energy initially equals the initial energy finally. For a perfect gas, U=𝑐𝑣 T ∴ For a free expansion of a perfect gas, 𝑐𝑣 𝑇1= 𝑐𝑣 𝑇2 where→ 𝑇1 = 𝑇2 That is, for a perfect gas undergoing a free expansion, the initial temperature is equal to the final temperature. ENGINEERING APPLICATION OF FIRST LAW TO STEADY FLOW PROCESS a) Control Volume For any system and in any process, the first law can be written as Q= ∆E + W………..… (i) Where E→ Represents all forms of energy stored in the system For Pure substance E= Ek + Ek + U ………. (ii) Ek → Kinetic energy Ep → Potential energy U→ Residual energy stored in the molecular structure of the substance From (i) and (ii) Q= ∆Ek + ∆Ek + ∆U + W………. (iii) This equation refers to system having a particular mass of substance, and is free to move from place to place. Note; This is mass transfer across the system boundary, the system is called an open system. Most of the engineering devices are open systems involving the flow of fluids through them. Flow process involving work and heat interactions Consider a steam turbine above in which steam enters at a high pressure, does work upon the turbine rotor, and then leaves the turbine at low pressure through the exhaust pipe. If a certain mass of steam is considered as the thermodynamic system, then the energy equation becomes Q= ∆Ek + ∆Ek + ∆U + W In order to analyze the expansion process in turbine the moving system is to be followed as it travels through the turbine, taking into account the work and heat interactions all the way through. Instead of concentrating attention upon a certain quantity of fluid, which constitutes a moving system in flow process, attention is focused upon the certain fixed region in space called a control volume through which the moving substance flows. To distinguish the two concepts o The system (closed) boundary usually changes shape, position and orientation relative to the observer, the control volume boundary remains fixed and unaltered. o Again, while matter usually crosses the control volume boundary • no such flow occurs across the system boundary. The broken line represents the surface of the control volume which is known as the control surface. This is the same as the system boundary of the open system. Since there is mass transfer across the control surface, a mass balance also has to be made. Sections l and 2 allow mass transfer to take place, and Q and Ware the heat and work interactions respectively. b) Steady Flow Energy Equation (S.F.E.E.) The rate at which the fluid flows through a machine or piece of apparatus is constant. This type of flow is called steady flow. Or A flow is said to be steady flow if properties do not vary with respect to time at any given section. Steady flow' means that the rates of flow of mass and energy across the control surface are constant. For steady flow, there is no accumulation of mass and energy in the control volume which mass entering is equal to mass leaving and energy entering is equal to energy leaving the control volume. Assumptions The following assumptions are made in the system analysis: The mass flow through the system remains constant. Fluid is uniform in composition. The only interaction between the system and surroundings are work and heat. The state of fluid at any point remains constant with time. In the analysis only potential, kinetic and flow energies are considered. Engineering applications of steady flow energy equation (S.F.E.E.) Water Turbine In a water turbine, water is supplied from a height. The potential energy of water is converted into kinetic energy when it enters into the turbine and part of it is converted into useful work, which is used to generate electricity. Considering center of turbine shaft as datum, the energy equation can be written as follows 𝐶1 2 𝐶2 2 (𝑈1 + 𝑃1 𝑉1 + 𝑍1 𝑔 + ) + 𝑄 = (𝑈2 + 𝑃2 𝑉2 + 𝑍2 𝑔 + )+𝑊 2 2 In case; Q=0 ∆𝑈 = 𝑈2 − 𝑈1 = 0 𝑉1 = 𝑉2 = 𝑉 𝑍2 = 0 Then, 𝐶1 2 𝐶2 2 (𝑃1 𝑉 + 𝑍1 𝑔 + ) = (𝑃2 𝑉 + 𝑍1 𝑔 + )+𝑊 2 2 W is positive because work is done by the system (or work comes out of the boundary). Steam or Gas Turbine In a steam or gas turbine, steam or gas is passed through the turbine and part of its energy is converted into work in the turbine. This output of the turbine runs a generator to produce electricity. The steam or gas leaves the turbine at lower pressure or temperature. Applying energy equation to the system. 𝑍1 = 𝑍2 (Where, ∆𝑍 = 0) 𝐶 2 𝐶 2 ℎ1 + 21 − 𝑄 = ℎ2 + 22 + 𝑊 The sign of Q is negative because heat is rejected (or comes out of the boundary). The sign of W is positive because work is done by the system (or work comes out of the boundary). Centrifugal Water Pump A centrifugal water pump draws water from a lower level and pumps to higher level as shown. Work is required to run the pump and this may be supplied from an external source such as an electric motor or a diesel engine. Here Q = 0 and ∆𝑈 = 0 as there is no change in temperature of water; 𝑉1 = 𝑉2 = 𝑉 Applying the energy equation to the system 𝐶 2 𝐶 2 𝑃1 𝑉1 + 𝑍1 𝑔 + 21 = 𝑃2 𝑉2 + 𝑍2 𝑔 + 22 − 𝑊 The sign of W is negative because work is done on the system (or work enters the boundary). Centrifugal Compressor A centrifugal compressor compresses air and supplies the same at moderate pressure and in large quantity. Applying energy equation to the system ∆Z = 0 (Generally taken) C 2 C 2 (h1 + 21 ) – Q = (h2 + 21 ) – W The Q is taken as negative as heat is lost from the system and W is taken as negative as work is supplied to the system. Or (h1 + C1 2 2 ) – Q = (h2 + C2 2 2 )–W Reciprocating Compressor The reciprocating compressor draws in air from atmosphere and supplies at a considerable higher pressure in small quantities (compared with centrifugal compressor). The reciprocating compressor can be considered as steady flow system provided the control volume include the receiver, which reduces the fluctuations of flow considerably. Applying energy equation to the system ∆𝑃𝐸 = 0 𝑎𝑛𝑑 ∆𝐾𝐸 = 0 Since these changes are negligible compared with other energies. ℎ1 − 𝑄 = ℎ2 − 𝑊 Boiler A boiler transfers heat to the incoming water and generates the steam. For this system, C 2 ∆Z = 0 And ∆ ( 22 ) = 0 W = 0 since neither any work is developed nor absorbed Applying energy equation to the system ℎ1 + 𝑄 = ℎ2 Condenser The condenser is used to condense the steam in case of steam power plant and condense the refrigerant vapour in the refrigeration system using water or air as cooling medium. For this system ∆𝑃𝐸 = 0 𝑎𝑛𝑑 ∆𝐾𝐸 = 0 (Since there values are very small compared with enthalpies) 𝑊 = 0 (Since neither any work is developed nor absorbed) Using energy equation to steam flow ℎ1 − 𝑄 = ℎ2 𝑤ℎ𝑒𝑟𝑒 Q→ Heat lost by 1 kg of steam passing through the condenser. Assuming there are no other heat interactions except the heat transfer between steam and water, then Q = Heat gained by water passing through the condenser Q = 𝑚𝑤 (ℎ𝑤2 − ℎ𝑤1 ) = 𝑚𝑤 𝑐𝑤 (𝑡𝑤2 − 𝑡𝑤1 ) But Q = ℎ1 − ℎ2 Then ℎ1 − ℎ2 = 𝑚𝑤 (ℎ𝑤2 − ℎ𝑤1 ) = 𝑚𝑤 𝑐𝑤 (𝑡𝑤2 − 𝑡𝑤1 ) 𝑤ℎ𝑒𝑟𝑒 𝑚𝑤 → Mass of cooling water passing through the condenser, 𝑐𝑤 → Specific heat of water Evaporator An evaporator is an equipment used in refrigeration plant to carry heat from the refrigerator to maintain the low temperature. Here the refrigerant liquid is passed through the evaporator and it comes out as vapour absorbing its latent heat from the surroundings of the evaporator. For this system ∆𝑃𝐸 = 0, ∆𝐾𝐸 = 0 𝑎𝑛𝑑 𝑊 = 0 (No work is absorbed or supplied) Applying the energy equation to the system ℎ1 + 𝑄 = ℎ2 Q is taken as + ve because heat flows from the surroundings to the system as the temperature in the system is lower than the surroundings Steam Nozzle In case of a nozzle as the enthalpy of the fluid decreases and pressure drops simultaneously the flow of fluid is accelerated. This is generally used to convert the part of the energy of steam into kinetic energy of steam supplied to the turbine. For this system, ∆𝑃𝐸 = 0, 𝑄 = 0 𝑎𝑛𝑑 𝑊 = 0 (No work is absorbed or supplied) Applying the energy equation to the system, h1 + C1 2 2 = h2 + C2 2 2 But C1 2 2 − C2 2 2 = h1 − h2 C1 2 − C2 2 = 2(h1 − h2 ) C2 2 = C1 2 + 2(h1 − h2 ) C2 = √C1 2 + 2(h1 − h2 ) Where velocity C is in m/s and enthalpy h in joules. If C1 << C2 , then C2 = √ 2(h1 − h2 ) C2 = √ 2∆ℎ