1. Chap. 3 (Sec. 3-5 to End of Chapter) Mass Flow Rate =  A V (kg/s) Volume Flow Rate = A V (m 3 /s) V = velocity

2. Conservation of Mass (Continuity Principle) m in m out

3. Conservation of Mass (Continuity Principle) Steady state ? Change of Any quantity with time = 0  A V) in = (  A V) out Incompressible fluid ?

4. Flow Work W flow = PV w flow = Pv Total energy of moving fluid = e + Pv = u + p.e + k.e + Pv = h + p.e + k.e

5. First Law of Thermodynamics (ENERGY BALANCE) also known as Conservation of Energy Principle * CLOSED SYSTEMS * OPEN SYSTEMS (CONTROL VOLUME) STEADY FLOW UNSTEADY FLOW

6. E in – E out =  E system Net Energy transfer by heat, work, and mass Change in internal, KE, PE etc * CLOSED SYSTEMS (e.g: piston-cylinder etc.) Q – W =  E Q net, in – W net,out =  E system Moving boundary work, Shaft work, Paddle Work etc

7. Q – W =  E Q net, in – W net,out =  E system How the above equation simplifies for different situations ? for stationary systems for constant volume process for constant pressure process for many other situations given in your text book Moving boundary work, Shaft work, Paddle Work etc

8. Q – W =  E Q net, in – W net,out =  E system Moving boundary work, Shaft work, Paddle Work etc For example: Based on problem statement if you having the following information Closed System Stationary Adiabatic Constant Volume Process Paddle Work (w net, in ) Final Temperature ?

9. W net,in =  U system w net,in =  u system What will be my approach if (a) Substances like Steam, R134a if (b) ideal gases

10. if (a) Substances like Steam, R134a  U = m (u 2 -u 1 ) As long as you have any two properties, you can find rest How?  State (Saturated liquid, mixture, superheated)

11. if (b) ideal gases  U = m (u 2 -u 1 ) = m C v (T 2 – T 1 ) PV = mRT Pv = RT (PV/T) 1 = (PV/T) 2

12. STEADY – FLOW SYSTEMS Rate of Net Energy transfer in by heat, work, and mass Rate of Net Energy transfer out by heat, work, and mass = Mass balance ?

13. One inlet (1) and one exit (2) On a unit mass basis

14. In general, how above equation simplifies for (how to judge - based on function) Nozzles Diffusers Compressors, Pumps Turbines Mixtures, Heat Exchangers Throttling devices

15. m in – m out =  m system UNSTEADY – FLOW SYSTEMS Mass Balance  m in –  m out = (m 2 -m 1 ) system

16. E in – E out =  E system Energy Balance

17. Chap. 5 The Second Law of Thermodynamics (What is second law and how it helps) Statements (in words or schematic) Some concepts Heat Engine, Refrigerator, Heat Pump Reversible and Irreversible processes * The Carnot Cycle and it’s importance

18. 1 ST LAW Q net – W net =  E for cyclic devices  E = 0 Therefore W net = Q net W net = Q in - Q out

19. HIGH TEMPERATURE RESERVOIR AT T H  LOW TEMPERATURE SINK AT T L  W net, out HE QHQH QLQL

20. WARM ENVIRONMENT AT T H >T L  COLD REFRIGERATED SPACE AT TEMPERATURE T L  W net, IN REFRI. QHQH QLQL WARM ENVIRONMENT AT T H >T L (HOME)  COLD ENVIRONMENT AT TEMPERATURE T L  W net, IN HEAT PUMP QHQH QLQL

21. CARNOT PRINCIPLES 1. The efficiency of an irreversible heat engine is always less than of a reversible one operating between the same two reservoirs. 2. The efficiencies of all reversible heat engines operating between the same two reservoirs are the same (Independent of working fluid and it’s properties, the way cycle is executed, or the type of reversible engine)

22. FOR REVERSIBLE PROCESS ALONE