ENERGY ANALYSIS OF CLOSED SYSTEMS

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CHAPTER 4

• ENERGY ANALYSIS OF CLOSED SYSTEMS

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Introduction

• In chapter 2, we considered various forms of
  energy and energy transfer, and we developed a
  general relation for conservation of energy
  balance.
• In chapter 3, we learned how to determine
  thermodynamic properties of substances.
• In chapter 4, we apply energy balance relation to
  closed systems.

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Objectives

• Examine the boundary work or PdV work
• Develop the general energy balance applied to
  close systems.
• Define the specific heat at constant volume and
  constant pressure.
• Relate the specific heats to the calculation of
  the changes in internal energy and enthalpy of
  ideal gases.

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Common Types of Work Energy

• The net work done by the system may be in two
  forms
• Boundary work
• Shaft work and electrical work as other work,
  that is, work not associated with a moving
  boundary.

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Boundary Work

• Expansion and compression work is often called
  boundary work.
• Temperature of air rises when it is compressed.
• This is because energy is transferred to the air
  in the form of boundary work.
• The increase in the energy of the system be equal
  to the boundary work done on the system.

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• Boundary Work
• Boundary work occurs because the mass of the
  substance contained within the system boundary.

• Consider the gas enclosed in the piston-cylinder
  device shown.
• Initial pressure of the gas is P, total volume V,
  cross sectional area of piston is A.
• If the piston is allowed to move a distance ds,
  the differential work done during this process is

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• The boundary work is equal to the area under the
  process curve plotted on the pressure-volume
  diagram.

• Note from the figure
• - P is the absolute pressure and is always
  positive.
• Volume change dV is ve during expansion (V
  increase) -ve during compression (V decrease)
• When dV is positive, Wb is positive.
• - When dV is negative, Wb is negative.

Pressure-volume diagram
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Some Typical Processes
Constant volume If the volume is held constant,
dV 0, and the boundary work equation becomes
Constant pressure If the pressure is held
constant, the boundary work equation becomes
P-V diagram for V constant P-V diagram for P
constant
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• Constant temperature, ideal gas
• If the temperature of an ideal gas system is held
  constant, then the equation of state provides the
  pressure-volume relation

Then, the boundary work is
Note The above equation is the result of
applying the ideal gas assumption for the
equation of state.
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Example 1 Boundary work for a constant-volume
process

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

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Example 2 Boundary work for a constant-pressure
process

• A frictionless pistoncylinder device contains 5
  kg of steam at 400 kPa and 200oC. Heat is now
  transferred to the steam until the temperature
  reaches 250oC. If the piston is not attached to a
  shaft and its mass is constant, determine the
  work done by the steam during this process.

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Example 3 Isothermal compression of an ideal gas

• A pistoncylinder device initially contains 0.4
  m3 of air at 100 kPa and 80C. 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.

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Quiz 4

• A mass of 5 kg of saturated water vapor at 300
  kPa is heated at constant pressure until the
  temperature reaches 200?C. Calculate the work
  done by the steam during this process.

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ENERGY BALANCE FOR CLOSED SYSTEMS
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Refresh

• Energy Balance The net change (increase or
  decrease) in the total energy of the system
  during a process is equal to the difference
  between the total energy entering and the total
  energy leaving the system during that process.

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Energy change of a system
(Total energy entering system) (Total energy
leaving system) (Change in the total energy
of the system)
Ein - Eout ?Esystem
? U ? KE ? PE
where ? U m (u2 u1 ) ? KE ½ (m
)(V22 - V12) ? PE mg (z2 z1 )
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Energy balance for closed system

• Closed system (control mass) no mass can
  enter or leave a system

0
Heat
Work
Mass
Closed System
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Various forms of the first law relation for
closed system
General Qnet,in - Wnet,out ?Esystem
Stationary system Q net,in - W net,out
?U ? KE ?PE 0
Per unit mass q - w ?e
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Example 4 Electric heating of a gas at constant
pressure

• 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.
  Show that for a closed system the boundary work
  Wb and the change in internal energy ?U can be
  combined in one term, ?H for a constant pressure
  process.

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• For a closed system (involve boundary work at
  constant pressure)

Q - W ?U ? KE ?PE 0
Q - Wother - Wb ?U
Q - Wother P(V2 - V2) U2 U1
Since P is constant, PP1 P2
Q - Wother (U2 P2V2) - (U1 P1V1)
Q - Wother H2 - H1
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Example 5 Unrestrained Expansion of Water

• A rigid tank is divided into two equal parts by a
  partition. Initially one side of the tank is
  contains 5 kg of water at 200 kPa and 25?C, and
  the other side is evacuated. The partition is
  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
• The volume of the tank
• The final pressure
• The heat transfer for this process

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Quiz 5

• A 0.5-m3 rigid tank contains refrigerant-134a
  initially at 160 kPa and 40 percent quality. Heat
  is now transferred to the refrigerant until the
  pressure reaches 700 kPa. Determine
• the mass of the refrigerant in the tank
• the amount of heat transferred.

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Specific Heat

• Energy required to raise the temperature of a
  unit mass substance by one degree.
• Common unit kJ/kg.K or kJ/kg.oC
• Two kinds of specific heat
• Specific heat at constant volume, cv
• Specific heat at constant pressure, cp

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• cp is always greater than cv
• At constant pressure, the system is allowed to
  expand and the energy is required for the
  expansion work.

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• In thermodynamics, the specific heats are defined
  as

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Internal energy, enthalpy and specific heat of
IDEAL GASES

• Ideal gas Pv RT
• Using definition of enthalpy and equation of
  state of ideal gas
• u and h depend only on temperature for an ideal
  gas.
• cp and cv also depend on temperature only.

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Integrate these equations (from state1 to state
2)

• Changes in the internal energy and enthalpy is

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Ideal gas constant pressure specific heats for
some gases
Complex molecules
Figure 4-24
Monatomic gas
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• Specific heats of gases with complex molecules
  (molecules with two or more atoms) are higher and
  increase with temp.
• Variation of specific heats with temp is smooth
  and linear over small temp intervals.
• Therefore specific heat functions in Eqn 4-25
  4-26 can be replaced by the constant average
  specific heat values.
• Then the integrations in these eqns can be
  performed, yielding

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• Specific heat values for some common gases are
  listed as a function of T in Table A-2b.
• cp,avg and cv,avg are evaluated from this table
  at average temp (T1T2)/2

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• Another observation from Fig 4-24 is that ideal
  gas specific heats of monatomic gases such as
  argon, neon and helium remain constant over the
  entire temp range.
• Thus, ?u and ?h of monatomic gases can easily be
  evaluated from Eqn 4-27 and 4-28.

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• Three ways to determine internal energy and
  enthalpy for ideal gases
• 1. Using tabulated u and h data (Table A-17).
• 2. Using cp and cv as a function of temperature
  (Table A-2c)
• 3. Using average specific heat (Take the average
  of T1 and T2 Table A-2b)

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Specific heat relation of ideal gases

• Relationship between cp and cv obtained by
  differentiating relation h u RT

• Replace dh by cpdT and du by cvdT, then divide
  by dT

kJ/kg.K
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Example 5 Evaluation of the ?u of an ideal gas

• Air at 300 K and 200 kPa is heated at constant
  pressure to 600 K. Determine the change in
  internal energy of air per unit mass, using
• Data from the air table (Table A-17)
• The functional form of the specific heat
  (Table A-2c)
• The average specific heat value (Table A-2b)

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Problem 4-52

• A rigid tank contains 10 kg of air at 350 kPa and
  27?C. The air is now heated until its pressure
  doubles. Determine
• The volume of the tank and
• The amount of heat transfer

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