MME 231: Lecture 06 he Laws of hermodynamics he Second Law of hermodynamics. A. K. M. B. Rashid Professor, Department of MME BUE, Dhaka oday s opics Relation between entropy transfer and heat Entropy change for reversible and irreversible processes he work function Common non-mechanical works 1
Relation between Entropy ransfer and Heat Sadi Carnot (1824) inventor of heat engines. Heat reservoir at high temperature, 1 Q 1 Heat Q 2 Engine Heat reservoir at low temperature, 2 W Schematic representation of the working principle of a heat engine Process : Superheated steam is passed from boiler (high temperature heat reservoir) to the cylinder using reversible isothermal process. Process : he steam performs work adiabatically by expanding against the piston. he temperature of the system (steam) is decreased. Process : he spent steam is exhausted isothermally to the atmosphere (low temperature heat reservoir). States and Processes of a Carnot cycle Process V: A fly wheel returns the piston adiabatically to its original position, thus completing the cycle and preparing for the next working stroke. 2
he Carnot Cycle P 0 V 3 Q 1 1 Q 3 2 0 1 V 3 2 (a) V (b) V he Carnot cycle in (a) P-V coordinate, and in (b) -V coordinate. he reversible paths used to complete the cycle 0-1-2-3-0 are: isothermal expansion, adiabatic expansion, isothermal compression V adiabatic compression. he overall efficiency of heat engine Work obtained Heat input - W Q - W = total work done on the system (heat engine) Q = heat absorbed by the system during stage = temperature of hot reservoir = temperature of cold reservoir (3.8) Using the first law, for the overall cyclic process: U = 0 = W + Q + Q (using W =0) W = - (Q + Q ) (3.9) 3
Pressure Substituting Eq.(3.9) into Eq.(3.8) will result Q Q Q Q Q - For the overall cyclic path ABA, Q/ = 0 0 A Volume A cyclic process ABA broken into a large number of Carnot cycles o dq rev / = 0 B hus, Q/ has the property of a state function. Clausius called this thermodynamic function as the entropy of the system, S. hus, the relation between heat absorbed and entropy transfer into a system becomes S = Q rev For infinitesimal change of process S = ds = dq rev dq rev = ds ; Q rev = ds 4
Entropy Changes for Different Processes Since entropy is a state function, for the process AB, S rev [AB] = S irr [AB] S rev,t + S rev,p = S irr,t + S irr,p S irr,t = S rev,t - S irr,p dq irr / < dq rev / For an isothermal process, is constant, and [ Q rev ] > [ Q irr ] (3.14) For all possible isothermal processes: Heat absorbed by a reversible process is the maximum. Example 3.2 A steel casting weighing 40 kg and at a temperature of 900 C is quenched in 2000 kg oil at 25 C. f there are no heat losses, what is the change in entropy of (a) the casting, (b) the oil, and (c) both considered together? C P (steel) = 0.5 kj/kg-k ; C P (oil) = 2.5 kj/kg-k Answer: o determine the final equilibrium temperature of oil and steel, we balance the energies of oil and steel casting at : (40 kg) (0.5 kj/kg-k) ( 1173 K) = 331.65 K + (2000 kg) (0.25 kj/kg-k) ( 298 K) = 0 5
(a) During heating the amount of heat absorbed by a system can be calculating using d Qrev = m C P d. hen (b) Change in entropy of the oil S = 2000 kg 0.25 kj kg-k (b) otal entropy change 331.65 K ln 298 K = +53.49 kj/k S = - 25.26 + 53.49 = 28.23 kj/k Note that although the entropy of the casting is decreased, total entropy change (for the oil and the steel casting) is increased. he Work Function For a finite process, the work done is defined as: W dw F(x). dx F = force dx = displacement dx P ext P dw = (P ext. A). (-dx) dw = P ext (-dv) Movement of piston 6
For reversible expansion of the gas, the value of dp would be so small that for all practical purposes, P ext = P hus, the work done on the system dw rev = - P dv dw irr = - P ext dv dw rev = -PdV W rev = - PdV Since for irreversible processes, P ext < P, work done by a reversible process is always the maximum. Some Common Non-Mechanical Works 1. Gravitational work, m f h where m f is the force acting on the system and h the height to which the system is lifted against gravity. 2. Electrical work, EdQ where E is the potential difference and dq the charge. 3. Surface work, gda where g is the surface tension and A the area. 4. Centrifugal work, m 2 r where m is the mass of the system, the angular velocity, and r the distance from axis of rotation. 7
Next Class Lecture 07 he Combined Statement Rashid/ Ch#3 8