Advanced Thermodynamics: Lecture 3 Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661
Moving Boundary Work: Closed systems Moving boundary work is the primary form of work involved in automobile engines. During their expansion, the combustion gases force the piston to move, which in turn forces the crankshaft to rotate. Also called PdV work. Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661
Moving Boundary Work: Closed systems We analyze the moving boundary work for a quasi- equilibrium process, Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition The initial pressure of the gas is P, the total volume is V, and the cross- sectional area of the piston is A. If the piston is allowed to move a distance ds in a quasi-equilibrium manner, the di ff erential work done during this process is δ W b = Fds = PAds = PdV Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661
Moving Boundary Work: Closed systems The total boundary work done during the entire process as the piston moves is obtained by adding all the di ff erential works from the initial state to the final state Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Z 2 Z 2 W b = Area under curve = A = dA = PdV 1 1 Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661
Work: Path Function Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661
Work: Cycle The net work done during a cycle is the di ff erence between the work done by the system and the work done on the system. Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661
Boundary Work for constant volume process A rigid tank contains air at 500 kPa and 150 o C. As a result of heat transfer to the surroundings, the temperature and pressure inside the tank drop to 65 o C and 400 kPa, respectively. Determine the boundary work done during this process. Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661
Boundary Work for constant pressure process A frictionless pistoncylinder device contains 10 lbm of steam at 60 psia and 320 o F. Heat is now transferred to the steam until the temperature reaches 400 o F. If the piston is not attached to a shaft and its mass is con- stant, determine the work done by the steam during this process. Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661
Polytropic process For a polytropic process the pressure and volume are related by the relation PV n = C Where n and C are constants. Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661
Work done during a Polytropic process Z 2 Z 2 CV − n dV W b = PdV = 1 1 = C V − n +1 � V − n +1 2 1 � n + 1 For and ideal gas PV = mRT , this equation becomes W b = mR ( T 2 � T 1 ) for n 6 = 1 1 � n For special case of n = 1, boundary work is Z 2 Z 2 ✓ V 2 ◆ CV − 1 dV = Cln W b = PdV = V 1 1 1 Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661
Isothermal compression of an Ideal gas A pistoncylinder device initially contains 0.4 m3 of air at 100 kPa and 80 o C. 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. Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661
Expansion of a Gas against a Spring A pistoncylinder device contains 0.05 m3 of a gas initially at 200 kPa. At this state, a linear spring that has a spring constant of 150 kN/m is touching the piston but exerting no force on it. Now heat is transferred to the gas, causing the piston to rise and to compress the spring until the volume inside the cylinder doubles. If the cross-sectional area of the piston is 0.25 m2, determine (a) the final pressure inside the cylinder, (b) the total work done by the gas, and (c) the fraction of this work done against the spring to compress it. Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661
Unrestrained Expansion of Water A rigid tank is divided into two equal parts by a partition. Initially, one side of the tank contains 5 kg of water at 200 kPa and 25 O C, and the other side is evacuated. The partition is then removed, and the water expands into the entire tank. The water is allowed to exchange heat with its surroundings until the temperature in the tank returns to the initial value of 25 O C. Determine (a) the volume of the tank, (b) the final pressure, and (c) the heat transfer for this process. Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661
Specific heats The specific heat is defined as the energy required to raise the temperature of a unit mass of a substance by one degree. It depends on how the process is executed. Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661
Specific heats The specific heat is defined as the energy required to raise the temperature of a unit mass of a substance by one degree. It depends on how the process is executed. Two kinds of specific heats: specific heat at constant volume c v and specific heat at constant pressure c p . c v 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 c p . Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661
Consider a fixed mass in a stationary closed system undergoing a constant–volume process. The conservation of energy principle for this process can be expressed in the di ff erential form as δ e in � δ e out = du The left–hand side of this equation represents the net amount of energy transferred to the system. From the definition of c v , this energy must be equal to c v dT , where dT is the di ff erential change in temperature. c v dT = du at constant volume ✓ ∂ u ◆ c v = ∂ T v c v is defined as the change in the internal energy of a substance per unit change in temperature at constant volume. Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661
Similarly for a constant pressure process, the energy balance can be written as δ e in � δ e out = dh From the definition of c p , this energy must be equal to c p dT , where dT is the di ff erential change in temperature. c p dT = dh at constant pressure ✓ ∂ h ◆ c p = ∂ T p c p is defined as the change in the internal energy of a substance per unit change in temperature at constant pressure. Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661
Joule’s experiment Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition u = u ( T ) For an ideal gas the internal energy is a function of temperature alone. Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661
Using the definition of enthalpy and the equation of state of an ideal gas, we have h = u + Pv Pv = RT h = u + RT ∴ h = h ( T ) Since u and h depend only on temperature for an ideal gas, the specific heats c v and c p also depend, at most, on temperature only. Then the di ff erential changes in the internal energy and enthalpy of an ideal gas can be expressed as du = c v ( T ) dT dh = c p ( T ) dT Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661
The change in internal energy or enthalpy for an ideal gas during a process from state 1 to state 2 is determined by Z 2 ∆ u = u 2 � u 1 = c v ( T ) dT 1 Z 2 ∆ h = h 2 � h 1 = c p ( T ) dT 1 Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661
The change in internal energy or enthalpy for an ideal gas during a process from state 1 to state 2 is determined by Z 2 ∆ u = u 2 � u 1 = c v ( T ) dT 1 Z 2 ∆ h = h 2 � h 1 = c p ( T ) dT 1 If variation of specific heats are smooth and approximately linear, the integrations can be replaced by u 2 � u 1 = c v , avg ( T 2 � T 1 ) h 2 � h 1 = c p , avg ( T 2 � T 1 ) Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661
Specific Heat Relations of Ideal Gases Di ff erentiating the relation, h = u + RT yields dh = du + RdT Replacing dh by c p dT and du by c v dT and dividing the resulting expression by dT , we obtain ✓ ◆ kJ c p = c v + R kg · K in Molar basis ✓ kJ ◆ c p = ¯ ¯ c v + R u kmol · K Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661
We introduce another ideal gas property called specific heat ratio k defined as k = c p c v For monatomic gases, its value is essentially constant at 1.667. Many diatomic gases, including air, have a specific heat ratio of about 1.4 at room temperature. Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661
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