CHAPTER5_SECTION5 - Chapter 5 The First Law of...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 5 The First Law of Thermodynamics Energy Balance for Charging and Discharging Processes 5-147 An evacuated bottle is surrounded by atmospheric air. A valve is opened, and air is allowed to fill the bottle. The amount of heat transfer through the wall of the bottle when thermal and mechanical equilibrium is established is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energies are negligible. 4 There are no work interactions involved. 5 The direction of heat transfer is to the air in the bottle (will be verified). Properties The gas constant of air is 0.287 kPa.m 3 /kg.K (Table A-1). Analysis We take the bottle as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u , respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance : m m m m m m m in out i out initial- = = = = system (since ) 2 Energy balance : E E E Q m h m u W E E ke pe in out in i i out initial- = + = 2245 = = 2245 2245 Net energy transfer by heat, work, and mass system Change in internal, kinetic, potential, etc. energies (since ) 2 2 Combining the two balances : ( 29 i in h u m Q- = 2 2 where kJ/kg 206.91 kJ/kg 290.16 K 290 kg 0.0096 ) K 290 )( K /kg m kPa 0.287 ( ) m 0.008 )( kPa 100 ( 2 17- A Table 2 3 3 2 2 2 = = = = = = = u h T T RT V P m i i Substituting, Q in = (0.0096 kg)(206.91 - 290.16) kJ/kg = - 0.8 kJ Q out = 0.8 kJ Discussion The negative sign for heat transfer indicates that the assumed direction is wrong. Therefore, we reverse the direction. 5-136 8 L Evacuated 100 kPa 17 C Chapter 5 The First Law of Thermodynamics 5-148 An insulated rigid tank is evacuated. A valve is opened, and air is allowed to fill the tank until mechanical equilibrium is established. The final temperature in the tank is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energies are negligible. 4 There are no work interactions involved. 5 The device is adiabatic and thus heat transfer is negligible. Properties The specific heat ratio air at room temperature is k = 1.4 (Table A-2)....
View Full Document

This homework help was uploaded on 04/18/2008 for the course EML 3007 taught by Professor Chung during the Spring '08 term at University of Florida.

Page1 / 22

CHAPTER5_SECTION5 - Chapter 5 The First Law of...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online