1 st Law – Closed Systems

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

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

3. Conservation of Energy • Closed System • Isolated System

4. 1 3.0 p (bar) pVn=k 2a 0.1 0.2 V (m3) Example: Closed System • 4kg of a gas is in a piston–cylinder assembly. The gas undergoes a process for which: pV1.5 = constant Pi= 3 bar, Vi= 0.1 m3, Vf= 0.2 m3 u2 - u1= -4.6 kJ/kg, ΔKE = ΔPE = 0 • Determine Q in kJ

5. 1 3.0 p (bar) 2a pV1.5=k 0.1 0.2 V (m3) Solution: Closed System

6. Coolant:k = 150 W/m2∙KTf = 20°C 5 mm 5 mm 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 Tb = chip surface temperatureTf = coolant temperature = 20°C • Determine surface temperature of chip, in °C.

7. Coolant:k = 150 W/m2∙KTf = 20°C 5 mm 5 mm Solution: Steady State

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/s2. • Determine Q from the resistor to the air, in Btu, for a system consisting of • the air alone, • the air and the piston. patm =14.7 lb/in2mpiston = 100 lbApiston = 1 ft2 mair = 0.6 lbV2-V1 = 1.6 ft3∆uair = 18 Btu/lb a) b)

9. Solution 1: Different Boundaries patm =14.7 lb/in2mpiston = 100 lbApiston = 1 ft2 mair = 0.6 lbV2-V1 = 1.6 ft3∆uair = 18 Btu/lb a)

10. Solution 2: Different Boundaries patm =14.7 lb/in2mpiston = 100 lbApiston = 1 ft2 mair = 0.6 lbV2-V1 = 1.6 ft3∆uair = 18 Btu/lb b)