ME235Lecture%205

# ME235Lecture 5 - Thermodynamics Vm235 Lecture 5 Chapter 5 – The First Law Key concepts for today Ideal gas Examples p We now wish to apply it to

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1 Thermodynamics Vm235 Lecture 5 Chapter 5 – The First Law Key concepts for today: ¾ Ideal gas. ¾ Examples. ¾ We now wish to apply it to an open system – 1 st Law analysis of a control volume – Chapter 6. A piston/cylinder contains 50 kg of water at 200 kPa with a volume of 0.1 m 3 . Stops in the cylinder are placed to restrict the enclosed volume to a maximum of 0.5 m 3 . The water is now heated until the piston reaches the stops. Find the necessary heat transfer. P Solution: C.V. H 2 O ; m = constant ; Energy Conservation (Eq.5.11): Constant pressure example: V 1 2 0.1 0.5 m(e 2 – e 1 ) = m(u 2 – u 1 ) = 1 Q 2 - 1 W 2 ; Process: P = constant 1 W 2 = P dV = P 1 (V 2 – V 1 ) or, (mu 2 + P 2 V 2 ) – (mu 1 + P 1 V 1 ) = m(h 2 –h 1 ) = 1 Q 2 State 1: v 1 = 0.1/50 = 0.002 m 3 /kg => 2-phase as v 1 < v g ; (Table B.1.2) x 1 = v 1 – v f v fg = 0.002 – 0.001061 0.88467 = 0.001061 ; h 1 = 504.68 + 0.001061 × 2201.96 = 507.02 kJ/kg 3 Heat transfer from the energy equation: 1 Q 2 = m(u 2 – u 1 ) + 1 W 2 = m(h 2 – h 1 ) = 50 kg × (526.92 – 507.02) kJ/kg = 995 kJ ; [Note: 1 W 2 = P 1 (V 2 – V 1 ) = 200 × (0.5 – 0.1) = 80 kJ ] State 2: v 2 = 0.5/50 = 0.01 m /kg also 2-phase as v 2 < v g , same P x 2 = v 2 – v f v fg = 0.01 – 0.001061 0.88467 = 0.01010 ; h 2 = 504.68 + 0.01010 × 2201.96 = 526.92 kJ/kg

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2 . : gases ideal of behavior the describing Equation only gases ideal for u(T) u RT Pv = = Internal Energy and Enthalpy of Ideal Gases . ) , P , T (i.e. region gas ideal in the tables the checking by yourself satisfy now, For later. until proof the postpone will we Hence, so. do entropy to need but this prove ally mathematic can We remember fact to important An or v ρ remember fact to important An . ) ( : : only gases ideal for T h h Thus RT u h RT Pv and Pv u h Because = + = = + = 10 100 3/kg) 5000 5200 5400 ) 0.1 1 Specific Volume (m 4400 4600 4800 u and h (kJ/kg Specific Volume 0.01 1 Pressure (MPa) 4000 4200 Internal Energy Specific Enthalpy
3 : ), , ( : @ @ ) ( ; 0 dv v u dT T u du so v T u u general In T const v const gas ideal for v u u general in v C + = = = = ± ±² ± ±³ ´ ´ Ideal Gases (internal energy & enthalpy) ; ), , ( : ) ( & : . ; : 0 0 0 0 dP h dT h dh P T h h general in Again T C C dT C du now because great is simplicity This gas ideal an for dT du C Thus v v v v ⎛ ∂ + ⎛ ∂ = = = = = . ; ) ( & : 0 0 0 0 @ @ ) ( ; 0 gas ideal an for T C C dT C dh or dT dh C Thus P T P P P P T const P const gas ideal for P h h general in P C = = = = = ± ± ´ ´ Chap 5, more examples: A rigid tank A of volume 0.6 m 3 contains 3 kg water at 120 o C and the rigid tank B is 0.4 m 3 with water at 600 kPa, 200 o C. They are connected to a piston cylinder initially empty with closed valves. The pressure in the cylinder should be 800 kPa to float the piston. Now the valves are slowly opened and heat is transferred so the water reaches a uniform state at 250 o C with the valves open. C Find the final volume , pressure , work done and heat transfer in the process.

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## This note was uploaded on 09/15/2010 for the course ME 235 taught by Professor Borgnakke during the Fall '07 term at University of Michigan.

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ME235Lecture 5 - Thermodynamics Vm235 Lecture 5 Chapter 5 – The First Law Key concepts for today Ideal gas Examples p We now wish to apply it to

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