1. 1 st Law – Closed Systems GBET 120 Applied Thermodynamics

2. Conservation of mass Mass flow rate Closed System Open System Steady flow

3. Conservation of Energy Closed System Isolated System

4. Example: Closed System 4kg of a gas is in a piston–cylinder assembly. The gas undergoes a process for which: pV 1.5 = constant P i = 3 bar, V i = 0.1 m 3, V f = 0.2 m 3 u 2 - u 1 = -4.6 kJ/kg, ΔKE = ΔPE = 0 Determine Q in kJ p (bar) V (m 3 ) 1 2a 3.0 0.10.2 pV n =k

5. Solution: Closed System p (bar) V (m 3 ) 1 2a 3.0 0.10.2 pV 1.5 =k

6. Example: Steady State A silicon chip in a ceramic substrate at steady state. –Electrical power is 0.225 W. –δQ/dt to substrate is negligible. –The rate of energy transfer between chip and coolant is: Where: A = top surface area T b = chip surface temperature T f = coolant temperature = 20°C Determine surface temperature of chip, in °C. Coolant: k = 150 W/m 2 ∙K T f = 20°C 5 mm

7. Solution: Steady State Coolant: k = 150 W/m 2 ∙K T f = 20°C 5 mm

8. Example: Different Boundaries Air in a piston-cylinder assembly is heated with an electrical resistor. –The atmosphere exerts a pressure on the top of the piston. –The volume of the air slowly increases while its pressure remains constant. –The air and piston are at rest initially and finally. –The piston-cylinder material is a a good insulator –friction between the piston and cylinder can be ignored –The local acceleration of gravity is g =32.0 ft/s 2. Determine Q from the resistor to the air, in Btu, for a system consisting of a)the air alone, b)the air and the piston. p atm =14.7 lb/in 2 m piston = 100 lb A piston = 1 ft 2 m air = 0.6 lb V 2 -V 1 = 1.6 ft 3 ∆u air = 18 Btu/lb a)b)

9. Solution 1: Different Boundaries p atm =14.7 lb/in 2 m piston = 100 lb A piston = 1 ft 2 m air = 0.6 lb V 2 -V 1 = 1.6 ft 3 ∆u air = 18 Btu/lb a)

10. Solution 2: Different Boundaries p atm =14.7 lb/in 2 m piston = 100 lb A piston = 1 ft 2 m air = 0.6 lb V 2 -V 1 = 1.6 ft 3 ∆u air = 18 Btu/lb b)