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AGUS HARYANTO 01 March 2010. Objectives. Examine the moving boundary work or P.dV work. Identify the first law of thermodynamics for closed (fixed mass) systems. Develop the general energy balance for closed systems. Define the specific heat at constant volume and at constant pressure.

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Agus haryanto 01 march 2010


01 March 2010


  • Examine the moving boundary work or P.dV work.

  • Identify the first law of thermodynamics for closed (fixed mass) systems.

  • Develop the general energy balance for closed systems.

  • Define the specific heat at constant volume and at constant pressure.

  • Relate the specific heats to the changes in internal energy and enthalpy of ideal gases.

  • Changes in internal energy and enthalpy for incompressible substances.

  • Solve energy balance problems for closed (fixed mass) systems that involve heat and work interactions.

Moving boundary work

Wb = F.s

Wb = F.ds

Wb = PA.ds

Wb = P.dV

W b a under the curve
Wb = Aunder the curve

  • The area under the process curve on a P-V diagram is equal, in magnitude, to the work done during a quasi-equilibrium expansion or compression process of a closed system.

  • On the P-v diagram, it represents the Wb done per unit mass.)

W b depends on path
Wb depends on path

W b for systems undergo cycle
Wb for systems undergo cycle

Car engine
Car Engine

In a car engine, the boundary work done by the expanding hot gases is used to overcome friction between the piston and the cylinder, to push atmospheric air out of the way, and to rotate the crankshaft.

Ex 4 1 w b @ constant volume
EX. 4–1: Wb @ Constant Volume

A rigid tank contains air at 500 kPa and 150°C. As a result of heat transfer to the surroundings, the temperature and pressure inside the tank drop to 65°C and 400 kPa, respectively. Determine the boundary work done during this process.

Ex 4 1 w b @ constant volume1
EX. 4–1: Wb @ Constant Volume


P-V diagram of the process is shown.

The boundary work can be determined :

Ex 4 1 w b @ constant pressure
EX. 4–1: Wb @ Constant Pressure

A frictionless piston–cylinder device contains 10 lbm of steam at 60 psia and 320F. Heat is now transferred to the steam until the temperature reaches 400F. If the piston is not attached to a shaft and its mass is constant, determine the work done by the steam during this process.

Ex 4 1 w b @ constant pressure1
EX. 4–1: Wb @ Constant Pressure

AnalysisThe pressure of the steam remains constant during this process since both the atmospheric pressure and the weight of the piston remain constant. For a constant-pressure process (Mind that V = mv):


From the superheated vapor table (Table A–6E), the specific volumes are determined to be v1 = 7.4863 ft3/lbm at state 1 (60 psia, 320oF), v2 = 8.3548 ft3/lbm at state 2 (60 psia, 400oF). Substitutingthese values yields:

Example 4 3 isothermal compression
EXAMPLE 4–3: Isothermal Compression

A piston–cylinder device initially contains 0.4 m3 of air at 100 kPa and 80°C. The air is now compressed to 0.1 m3 in such a way that the temperature inside the cylinder remains constant. Determine the work done during this process.

Polytropic process
Polytropic Process

Pressure and volume are often related by PVn = C, where n and C are constants.

Polytropic process1
Polytropic Process

For ideal gas (PV = mRT), then:

Special cases (n = 1), then: (isothermal)

Example 4 4

  • A piston–cylinder device contains 0.05 m3 of a gas initially at 200 kPa. At this state, a linear spring that has a spring constant of 150 kN/m is touching the piston but exerting no force on it. Now heat is transferred to the gas, causing the piston to rise and to compress the spring until the volume inside the cylinder doubles. If the cross-sectional area of the piston is 0.25 m2, determine (a)the final pressure inside the cylinder, (b) the total work done by the gas, and (c) the fraction of this work done against the spring to compress it.


Final pressure = 200 + 120 = 320 kPa

Energy balance for closed systems

  • From Ch. 2:

  • In the rate form:

  • For constant rate:

Energy balance

  • Differential form:

  • Closed system undergo cycle:

  • Esystem = E2 - E1 = 0

  • Ein - Eout = 0

  • Ein = Eout.

  • No mass flow (closed system):

Energy balance first law relations summary
Energy balance (first-law) relations: Summary

where :

  • Q = Qnet,in

  • = Qin - Qout

  • (is the net heat input)

  • W = Wnet,out

  • = Wout - Win

    (is the net work output)

Example 4 5 heating gas @ p
EXAMPLE 4–5 Heating gas @ P

  • A piston–cylinder device contains 25 g of saturated water vapor that is maintained at a constant pressure of 300 kPa. A resistance heater within the cylinder is turned on and passes a current of 0.2 A for 5 min from a 120-V source. At the same time, a heat loss of 3.7 kJ occurs. (a) Show that for a closed system the boundary work Wband the change in internal energy U in the first-law relation can be combined into one term, H, for a constant pressure process. (b) Determine the final temperature of the steam.


(b) To find T2 we need other property. Work is only electrical work, which is determined from:

Apply the Equation resulted in (a)

Calculation cont d
Calculation (Cont’d)

Finally, from Table A-6:

Example 4 6 unrestrained expansion of water
EXAMPLE 4–6 Unrestrained Expansion of Water

  • A rigid tank is divided into two equal parts by a partition. Initially, one side of the tank contains 5 kg of water at 200 kPa and 25°C, and the other side is evacuated. The partition is then removed, and the water expands into the entire tank. The water is allowed to exchange heat with its surroundings until the temperature in the tank returns to the initial value of 25°C. Determine (a) the volume of the tank, (b) the final pressure, and (c) the heat transfer for this process.


(c) Under stated assumptions and observations, the energy balance on the system can be expressed as: