Chemical Engineering Thermodynamics-I Lecture 3 PDF

CHEMICAL ENGINEERING THERMODYNAMICS-I MS. TUBA SIRAJ Lecturer Chemical Engineering NED University of Engineering and Technology LECTURE # 3  Specific heat capacity We know from experience that it takes dif...

CHEMICAL ENGINEERING THERMODYNAMICS-I MS. TUBA SIRAJ Lecturer Chemical Engineering NED University of Engineering and Technology LECTURE # 3  Specific heat capacity We know from experience that it takes different amounts of energy to raise the temperature of identical masses of different substances by one degree. For example, we need about 4.5 kJ of energy to raise the temperature of 1 kg of iron from 20 to 308C, whereas it takes about 9 times this energy (41.8 kJ to be exact) to raise the temperature of 1 kg of liquid water by the same amount  Specific heat capacity In thermodynamics, we are interested in two kinds of specific heats: Specific heat at constant volume Cv Specific heat at constant pressure Cp.  Specific heat capacity The specific heat at constant volume Cv 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. The energy required to do the same as the pressure is maintained constant is the specific heat at constant pressure Cp.  Specific heats in terms of other thermodynamic properties. Consider a fixed mass in a stationary closed system undergoing a constant-volume process (and thus no expansion or compression work is involved). The conservation of energy principle ein - eout = Δesystem for this process can be expressed in the differential form: 𝛿𝑒𝑖𝑛 − 𝛿𝑒𝑜𝑢𝑡 = 𝑑𝑢 The left-hand side of this equation represents the net amount of energy transferred to the system. From the definition of Cv , this energy must be equal to Cv dT, where dT is the differential change in temperature. 𝛿𝑒𝑖𝑛 − 𝛿𝑒𝑜𝑢𝑡 𝐶𝑣 = 𝑑𝑇 Cv → The change in 𝛿𝑒𝑖𝑛 − 𝛿𝑒𝑜𝑢𝑡 = 𝐶𝑣 𝑑𝑇 internal energy with 𝑪𝒗 𝒅𝑻 = 𝒅𝒖 𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑣𝑜𝑙𝑢𝑚𝑒 temperature at 𝜕𝑢 constant volume 𝐶𝑣 = 𝜕𝑇 𝑣  Specific heats in terms of other thermodynamic properties. An expression for the specific heat at constant pressure Cp can be obtained by considering a constant-pressure expansion or compression process. It yields: 𝜕ℎ Cp → the change in 𝐶𝑝 = enthalpy with 𝜕𝑇 𝑝 temperature at constant pressure Observations: these equations are property relations and as such are independent of the type of processes. They are valid for any substance undergoing any process. cv is related to the changes in internal energy and cp to the changes in enthalpy. Cv can be defined as the change in the internal energy of a substance per unit change in temperature at constant volume. Cp can be defined as the change in the enthalpy of a substance per unit change in temperature at constant pressure.  Enthalpy Total heat content in a system P const. V, T ΔU δQ = dU + δW 21δ𝑄 = 2 𝑑𝑈 + 1 2 δ𝑊 1 𝑄1−2 = 𝑈2 − 𝑈1 + 𝑃 𝑉2 − 𝑉1 𝑄1−2 = 𝑈2 + 𝑃2𝑉2 − 𝑈1 − 𝑃1𝑉1 𝑄1−2 = 𝐻2 − 𝐻1 𝑸1−2 = ∆𝐇 H= U + PV  Temperature dependency Joule 1843, submerged two tanks connected with a pipe and a valve in a water bath. Initially, one tank contained air at a high pressure and the other tank was evacuated. When thermal equilibrium was attained, he opened the valve to let air pass from one tank to the other until the pressures equalized. Joule observed no change in the temperature of the water bath and assumed that no heat was transferred to or from the air. Since there was also no work done, he concluded that the internal energy of the air did not change even though the volume and the pressure changed. Therefore, he reasoned, the internal energy is a function of temperature only and not a function of pressure or specific volume  Temperature dependency Internal energy is the function of temperature u = u (T) h= u + PV Since, PV = RT ∴h= u + RT h = h(T) du = Cv(T)dT For ideal gases. dh = Cp(T)dT u = u(T) 2 h = h(T) ∆u = 𝑢2 − 𝑢1 = ∫ 𝐶𝑣 𝑇 𝑑𝑇 1 Cv = Cv (T) 2 Cp= Cp (T) ∆h = ℎ2 − ℎ1 = ∫ 𝐶𝑝 𝑇 𝑑𝑇 1  Temperature dependency At low pressures, all real gases approach ideal-gas behaviour, and therefore their specific heats depend on temperature only. The specific heats of real gases at low pressures are called ideal-gas specific heats, or zero-pressure specific heats, and are often denoted Cpo and Cvo The use of ideal-gas specific heat data is limited to low pressures, but these data can also be used at moderately high pressures with reasonable accuracy as long as the gas does not deviate from ideal-gas behaviour significantly. Temperature dependency u2-u1= Cv,avg(T2-T1) Δu =Cv ΔT h2-h1= Cp,avg(T2-T1) Δh =Cp ΔT These equations are valid for any ideal gas undergoing any process 12  Energy balance for closed system Closed system = Control mass =only energy can go in and out of the system. 1. The increase in the energy of a potato in an oven is equal to the amount of heat transferred to it. (Work done = 0) 2. In the absence of any work interactions, the energy change of a system is equal to the net heat transfer. (Work done = 0)  Energy balance for closed system 3. The work (electrical) done on an adiabatic system is equal to the increase in the energy of the system (Heat transfer= 0) 4. The work (shaft) done on an adiabatic system is equal to the increase in the energy of the system. (Heat transfer= 0)  Energy balance for closed system Energy change of a system, ΔEsystem Energy change = Energy at final state – Energy at initial state ΔEsystem = Efinal - Einitial = E2 – E1 The energy change of a system during a process is equal to the net work and heat transfer between the system and its surroundings. Energy can exist in numerous forms such as internal (sensible, latent, chemical, and nuclear), kinetic, potential, electric, and magnetic, and their sum constitutes the total energy E of a system ΔE = ΔU + ΔKE + ΔPE Where, ΔU = m(u2 – u1) ΔKE = 1m(v22 – v12) 2 ΔPE = mg(z1 – z2) For stationary systems, ΔKE = Δ PE = 0; thus ΔE = ΔU. 19  Energy balance for closed system Mechanism of Energy transfer, Ein and Eout 1. Heat transfer 2. Work transfer 3. Mass flow 0 0 Ein – Eout = (Qin – Qout) + (Win – Wout) +(Emass, in – Emass, out) = ΔEsystem For close system; Emass = 0 (Qin – Qout) + (Win – Wout)= ΔE Assuming Qin>Qout and Wout> Win (Qnet, in – Wnet, out)= ΔE General form (KJ) Q− W = E Per unit time (Kj/sec or watt) q − w =  e Per unit mass (Kj/Kg) q−w=d Differential form 20  Energy balance for closed system For Cyclic Process If the initial and the final states are exactly identical(same), then it is called cycle. 1. This process would not be isolated. 2. On a PV diagram, a cyclic process appears as a closed curve. The internal energy must be zero since it is a state variable. For a cycle ΔE = 0, thus Q = W. 21  Examples Example 2-10: Cooling of a Hot Fluid in a Tank (YUNUS A.CENGEL 8th Edition) A rigid tank contains a hot fluid that is cooled while being stirred by a paddle wheel. Initially, the internal energy of the fluid is 800 kJ. During the cooling process, the fluid loses 500 kJ of heat, and the paddle wheel does 100 kJ of work on the fluid. Determine the final internal energy of the fluid. Neglect the energy stored in the paddle wheel.  Examples EXAMPLE 4–2 Boundary Work for a Constant-Pressure Process (YUNUS A.CENGEL 8th Edition) A frictionless piston–cylinder device contains 10 lbm of steam at 60 psia and 320F. Heat is now transferred to the steam until the temperature reaches 400F. If the piston is not attached to a shaft and its mass is constant, determine the work done by the steam during this process.  Examples EXAMPLE 4–3 Isothermal Compression of an Ideal Gas (YUNUS A.CENGEL 8th Edition) A piston–cylinder device initially contains 0.4 m3 of air at 100 kPa and 80C. The air is now compressed to 0.1 m3 in such a way that the temperature inside the cylinder remains constant. Determine the work done during this process.  Examples EXAMPLE 2.6: isothermal work (J.M. SMITH 7th Edition) A horizontal piston cylinder arrangement is placed in a constant temperature bath. The piston slides in the cylinder with negligible friction, and an external force holds it in place against an initial gas pressure of 14 bar. The initial gas volume is 0.03 m3. The external force on the piston is reduce gradually, and the gas expands isothermally as its volume doubles. If the volume of the gas is related to its pressure so that the product PV is constant, what is the work done by the gas in moving the external force? How much work would be done if the external forces were suddenly reduced to half of its initial value instead of being gradually reduced? Example 2-1: Energy of a waterfall (J.M. SMITH 7th Edition) Water flows over a waterfall 100 m in height. Take 1 kg of the water as the system and assume that it does not exchange energy with its surroundings a) What is the potential energy of the water at the top of the falls with respect to the base of the fall? b) What is the kinetic energy of the water just before it strikes the bottom? c) After the 1kg of water enters the stream below the falls, what change has occurred in its state? EXAMPLE 2–11 Acceleration of Air by a Fan A fan that consumes 20 W of electric power when operating is claimed to discharge air from a ventilated room at a rate of 1.0 kg/s at a discharge velocity of 8 m/s (Fig. 2–50). Determine if this claim is reasonable.