Midlands state university ; thermodynamics presentation
- 1. GROUP 5 BRIDGET SIWAWA DILLON MAFUKIDZE LIBERTY SHOKO CHIVAURA TATENDA BLESSING GODAMA MUTABISI
- 2. First law of thermodynamics • The first law of thermodynamics is a version of the law of conservation of energy, adapted for thermodynamic systems. The law of conservation of energy states that the total energy of an isolated system is constant; energy can be transformed from one form to another, but can be neither created nor destroyed.
- 3. • . The first law of thermodynamics states that the change in the internal energy ΔU of a closed system is equal to the amount of heat Q supplied to the system, minus the amount of work W done by the system on its surroundings.
- 4. • The first law is often formulated Δ U = Q − W where ∆U is the total change in internal energy of a system Q is the heat exchanged between a system and its surroundings W is the work done by or on the system
- 5. Second law of thermodynamics • The second law of thermodynamics states that the total entropy can never decrease over time for an isolated system, that is, a system in which neither energy nor matter can enter nor leave. The total entropy can remain constant in ideal cases where the system is in a steady state (equilibrium), or is undergoing a reversible process. In all spontaneous processes, the total entropy always increases and the process is irreversible. The increase in entropy accounts for the irreversibility of natural processes
- 6. • A more formal defination for entropy as heat moves around a system is given in the equation • dS=δQ/T • dS≥0 • Where dS is calculated by measuring how much heat has entered a closed system(δQ)divided by the common temperature(T)at the point where the heat transfer took place
- 7. Carnot cycle • A reversible heat engine absorbs heat QH from the high-temperature reservoir at TH und releases heat QL to the low-temperature reservoir at TL. The temperatures TH and TL remain unchanged. And we know from the 1st law of thermodynamics, work is done by the heat engine, W=QH+QL. Here QH>0 and QL<0.
- 8. The Carnot cycle consists of: • Isothermal expansion • Adiabatic expansion • Isothermal compression • Adiabatic compression
- 10. Isorthermal expansion • The 1st step in the Carnot Cycle is an isothermal expansion. • We slowly remove weights from the back of the piston. The pressure slowly decreases and the gas slowly expands to a larger volume. • During this process, the gas is going to cool infinitesimally. • But because the process moves very slowly and the system is in contact with the thermal reservoir at TH, the temperature of the gas in the cylinder remains at a constant value of TH throughout this process. • Heat is transferred from the hot reservoir to the system. This is QH. • This step is an isothermal expansion and the curve connecting states 1 and 2 is called an isotherm. • Notice the shape of the isotherm on the PV Diagram. This is the typical shape as long as no phase change occurs.
- 11. Adiabatic expansion • With the cylinder insulated, we pull more weight off the back of the piston. • The gas in the cylinder expands slowly. • The gas cools as it expands and no heat transfer occurs. • This is an adiabatic expansion and the curve connecting states 2 and 3 on the PV Diagram is called an adiabat. • Notice that the adiabat is steeper than the isotherm. • Does that make sense to you ? • Along the isotherm, heat is continuously added as the process moves from state 1 to state 2. This addition of heat causes the gas to expand little bit faster as the pressure drops. This shifts each point on the curve a little bit to the right. The result is that the isotherm is made less steep by the heat flowing in from the hot reservoir. • Along the adiabat, there is no flow of heat into the system, so the points on the curve are not shifted to the right and, voila, the adiabat is steeper than the isotherm! • Now, on with the cycle. What do you think the next step will be? • Remember that this is a cycle, so somehow the system must return to state 1. • The volume needs to decrease and the pressure needs to increase, so I hope you guessed that the next step will be a compression. • But what kind of compression?
- 12. Isorthermal Compression • In this step, we compress the gas inside the cylinder by slowly adding weight to the back of the piston. • As the gas is compressed, its temperatures rises an infinitesimal amount • But heat transfer to the cold reservoir keeps the temperature constant at T3 or TC. • So, the path from state 3 to state 4 is an isotherm. • Notice that the shape of this isotherm is similar to the shape of the isotherm at TH. • This is generally the case if TC is not too much less than TH. • But keep in mind that each isotherm has a slightly different shape. • So, what type of process do you think we need next ?
- 13. Adiabatic compression • I hope you realized that the last step in this cycle is an adiabatic compression. • To accomplish an adiabatic process, we must first remove the cold reservoir and then perfectly insulate the cylinder, just like we did in step 2-3. • We need to increase the pressure from P4 to P1, so we must add weight to the back of the piston. • As the pressure increases, the volume of the gas in the system decreases. • The adiabat is steeper than the path of the isothermal compression because the insulation prevents any heat loss and the energy that cannot escape just increases the rate at which the pressure increases. • That closes the loop on our PV Diagram and so it completes the reversible thermodynamic cycle called the Carnot Cycle.
- 14. Carnot cycle
- 15. • Step A-B, heat 𝑞2 is reversibly transferred from a heat reservoir at the temperature 𝑡2 to the working fluid, which isothermally and reversibly expands from state A to state B, performing work 𝑤1 equal to area ABba. • Step B-C, the working fluid undergoes a reversible adiabatic expansion from state B to state C, as a result of which its temperature decreases to 𝑡1, performing work 𝑤2, equal to area BCcb. • Step C-D, heat 𝑞1 is isothermally and reversibly transferred from the working fluid to a heat reservoir at the temperature 𝑡1. Work, 𝑤3, equal to area DCcd, is performed on the working fluid. • Step D-A, the working fluid is reversibly and adiabatically compressed during which its temperature increases from 𝑡1 to 𝑡2, and work 𝑤4, equal to area ADda, is performed on the working fluid. • During the cyclic process the work done by the substance is: 𝑤 = 𝑤1 + 𝑤2 − 𝑤3 − 𝑤4
- 16. and is equal to area ABCD. • The heat absorbed by the working fluid is 𝑞 = 𝑞2 − 𝑞1. • For the cyclic process, ∆𝑈 = 0 and from the first law: 𝑞 = 𝑤 Thus 𝑞2 − 𝑞1 = 𝑤 • and the efficiency, 𝜂 𝑐, of the Carnot engine is the work done on the surroundings divided by the heat input from the hot reservoir: 𝜂 𝑐 = 𝑤 𝑞2 = 𝑞2 − 𝑞1 𝑞2 = 1 − 𝑞1 𝑞2 • Consider a second working with a different substance between the same temperatures 𝑡1 and 𝑡2, and letting this engine be more efficient than the first one,
- 17. • Greater efficiency can be obtained in either two ways: 1. 𝑞2 is withdrawn from the heat reservoir at 𝑡2, and more work, 𝑤′, is obtained from it than was obtained from the first engine, i.e. 𝑤′˃ 𝑤. thus, the second engine rejects less heat, 𝑞1 ′ to the cold reservoir at 𝑡1 than does the first engine, i.e. 𝑞1 ′ ˂ 𝑞. 2. The same work is obtained by withdrawing less heat 𝑞2 ′ from the heat reservoir at 𝑡2, i.e. 𝑞2 ′ ˂ 𝑞2. Thus less heat, 𝑞′1 is rejected into the reservoir at 𝑡1, i.e. 𝑞1 ′ ˂ 𝑞1. • Now consider that the second engine is run in the forward direction and the first engine is run in the reverse direction. • From 1, for the second engine to run in the forward direction, 𝑤′ = 𝑞2 − 𝑞1 ′ , and for the first one to run in the reverse direction, −𝑤 = −𝑞2 + 𝑞1, and the sum of the two processes is: 𝑤′ − 𝑤 = (𝑞1 − 𝑞1 ′ ) i.e. an amount of work (𝑤′ – 𝑤) has been obtained from a quantity of heat (𝑞1 – 𝑞1 ′ ) without an other change occurring.
- 18. • This is contrary to human experience i.e. heat cannot be converted to work without leaving a change in any other body. (Kelvin’s statement: It is impossible, by means of a cyclic process, to take heat from a reservoir and convert it to work without, in the same operation, transferring heat to a cold reservoir). • From 2, for the second engine to run in the forward direction 𝑤 = 𝑞′2 − 𝑞′1, and for the first engine to run in the reverse direction −𝑤 = −𝑞2 + 𝑞1, and the sum of the two processes is 𝑞2 ′ − 𝑞2 = 𝑞1 ′ − 𝑞1 = 𝑞 i.e. the amount of heat 𝑞 at one temperature has been converted to heat at a higher temperature without any other change occurring. This corresponds to the spontaneous flow of heat up a temperature gradient which is more contrary to human experience. (Clausius’ statement: It is impossible to transfer heat from a cold body to a hot reservoir without, in the same process, converting a certain amount of work into heat).
- 19. Specific heat capacity • Is the heat needed to raise the temperature of a 1kg substance by 1K. It is heat capacity per unit mass of a substance • C = Q / ΔT • where C is specific heat, Q is energy (usually expressed in joules), and ΔT is the change in temperature (usually in degrees Celsius or in Kelvin). Alternatively, the equation may be written: • Q = CmΔT • Specific heat and heat capacity are related by mass: • C = m * S • Where C is heat capacity, m is mass of a material, and S is specific heat. Note that since specific heat is per unit mass, its value does not change, no matter the size of the sample. So, the specific heat of a gallon of water is the same as the specific heat of a drop of water.
- 20. Heat Conduction
- 21. Heat conduction • Is the flow of internal energy from the region of higher concentration to that of lower temperature by interaction of the adjacent particles (atoms, ions, molecules, electrons) in intervening space. • It is the rate at which heat is transferred. •
- 22. Heat conduction • Note: it's the rate (Φ) at which heat is transferred, not the amount (Q) of heat transferred. Q=-KA∆Tt/L • Q=specific heat. • K =Thermal conductivity • ∆T= change in temperature • T=time • L=length of material • A=cross sectional area.
- 23. Heat conduction Factors affecting the rate of heat transfer by conduction. • temperature difference • length • cross-sectional area • material
Editor's Notes
- #15: The working of the Carnot engine can be depicted by the following cycle (Carnot cycle):