Energy Analysis of Closed Systems

Chapter 4

Energy Analysis of Closed Systems

4.1Boundary Work

Consider the gas enclosed in a piston-cylinder device below. The initial pressure is P, the total volume is V and the cross-sectional area of the piston is A. When the piston moves a distance ds in a quasi-equilibrium manner, the differential work done during this process is

The total boundary work done during the entire process is obtained by adding all the differential works from the initial to the final state

………… ( kJ )

The integral can only be evaluated when P is stated in terms of V , i.e. P = f (V).

A gas does work as it forces

the piston to move by ds

gas

Consider a quasi-equilibrium expansion process as shown below. On the P-V diagram, the differential area dA is equal to PdV, which is . The total area under the process curve 1–2 is obtained by adding the differential areas:

Area = A =

That is, “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”.

P 1

process path

dA = PdV

V1 dV V2

For different paths followed by the gas as it expands from state 1 to state 2, the area underneath the curve (and the work done) will be different. It is expected since work is a path function – it depends on path and end states.

P is the pressure at the inner surface of the piston. It is only equal to the gas pressure in the cylinder if the process is quasi-equilibrium.

During an expansion process, Wb is transferred from the system.
During a compression process, Wbis transferred to the system

4.1.1Boundary Work for Constant Volume Process

For a constant volume process, Wb = 0.

Example.

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

For a rigid tank, its volume is constant. Thus dV = 0, for which = 0.

500 kPa 400 kPa P, kPa

150 oC 65 oC

. cooling 500 1

1 V = constant 2

400 2

V

* Note : Wb = area under process curve

For constant volume, area = 0 !

4.1.2Boundary Work for a Constant-Pressure Process

For a constant-pressure process, = mP ….… (kJ)

where P = P1 = P2.

Example

A frictionless piston-cylinder device contains 5 kg steam at 400 kPa and 200 oC. Heat is transferred to the steam until the temperature reaches 250 oC. Determine the work done by the steam during this process.

P, kPa

1 2

5 kg heating 400

400 kPa

200 oC P = C 250 oC

1 2 v

. v1 =0.53434 v2 = 0.59520

Although not explicitly stated, this is a constant-pressure process since the weight of the piston and the atmospheric pressure are constant.

State 1 is at (400 kPa, 200 oC) superheated v1 = 0.53434 m3/kg

State 2 is at (400 kPa, 250 oC) superheated v2 = 0.59520 m3/kg

Thus,

=121.7 kJ

The +vesign indicates work is done by the system (steam) on the surroundings.

4.1.3Boundary Work for Isothermal Compression of an Ideal Gas

Example

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

P, kPa 2

0.4 m3 T = 80 oC

100 kPa compression PV = C

80 oC 0.1 m3 1

air T = C 80 oC 100

1 2 V

. 0.1 0.4 m3

Assume air is an ideal gas and the process is quasi-equilibrium.

= C

mRT is constant since m, R and T are constants.

where C = mRT1 = mRT2 = P1V1 = P2V2

can use here !

= – 55.45 kJ

Note : Since P1V1 = P2V2

The –ve sign indicates that work is done on the system by the surroundings, which is always true for compression processes.

4.1.3Boundary Work for a Polytropic Process

In practice, expansion and compression processes of gases often obey the equation : PVn = C, where n and C are constants. Such a process is called polytropic process. For such a process between state 1 and state 2,

Thus, =

since C = then,

= general equation !

------

For an ideal gas, PV = mRT , then the equation can be written as:

Wb= = ……………… ( kJ )

for .

------

For a special case, when n = 1, it follows that PV = C or P = for which

= …………. general, n = 1

where C = P1V1 = P2V2 . This is also the equation for an ideal gas undergoing an isothermal process !

Example

A piston-cylinder device contains 0.05 m3 of a gas initially at 200 kPa. At this state, a linear spring (k = 150 kN/m) is touching the piston but exerts no force on it. Heat is transferred to the gas causing the piston to rise and to compress the spring until the volume inside the cylinder doubles. 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.

P, kPa

. x

. 2

. P2

. Heating x=0 II

. 1

200 kPa V2 = 2V1 200

0.05 m3 = 0.1 m3 I

1 2 0.05 0.10 V, m3

(a)

The final volume (state 2) is, V2 = 0.2 V1 = 2(0.05) = 0.10 m3

From the undisturbed position, the spring is compressed a distance of,

= = 0.2 m

At state 2, the force exerted by the spring is, F = k = 150 (0.2) = 30 kN

At state 1, force balance on the piston (Fspring = 0), gives

At state 2, force balance on the piston gives,

200 + 120 = 320 kPa

(b) The work done is the area under the process curve (a trapezoid) from state 1 to state 2,

= 13 kJ

The work done is positive because integral of PdV is positive (P and dV are positive).

(c)

The rectangular area (region I) is work done against the piston and the atmosphere, and the

work done against the spring is represented by the triangular area (region II). Thus,

= 3 KJ

Note: A similar result could also be obtained from :

= = 3 kJ

4.2Energy Balance for Closed Systems

Energy balance for any system undergoing any kind of process is

……………………….... ( kJ )

Net energy transfer = Change in internal, kinetic,

by heat, work, mass potential, etc energies

or in rate form, it is written as

………………………. ( kW )

Rate of et energy transfer = Rate of change in internal, kinetic,

by heat, work, mass potential, etc energies

The energy balance on a per unit mass basis is,

………. ( kJ/kg )

and in differential form as,

For a closed system undergoing a cycle, the initial and final states are identical, thus

Since a closed system has no mass flow across its boundaries, the only energy interactions possible are work and heat transfer only. Therefore, for a cycle:

For a closed system, the general energy balance equation is :

where,

is net heat input

is net work output

Other forms of first-law relation for closed systems are :

per unit mass ( kJ/kg )
per unit mass; differential form ( kJ/kg )

Note : The 1st law cannot be proven mathematically !

Example

A piston-cylinder device contains 25 g 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. Determine the final temperature of the steam.

P, kPa

. Heating 1 2

. 300

300 kPa P = C

25 g

. heater

1 2

Let = work done on the system other than boundary work.

For the process, 0 0

= =

where enthalpy is defined as H = U + PV

Conclusion:

For constant-pressure process, we can use :

The electrical work done by the surroundings is (the heater is inside the system),

= 120 (0.2) 300 = 7200 J = 7.2 kJ

At state 1 (300 kPa, sat. vapor) = 2724.9 kJ/kg

= – 3.7 kJ (heat loss)

= – = – 7.2 kJ

– 3.7 – (– 7.2) = 0.025 h2 = 2864.9 kJ/kg

At state 2 (300 kPa, h = 2864.9) superheated steam T2 = 200oC

Example

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 oC, 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 the surroundings until the temperature in the tank returns to the initial value of 25 oC. Determine (a) the volume of the tank, (b) the final pressure, and (c) the heat transfer for this process.

P, kPa

Evacuated partition 1

. 200

. partition

200 kPa removed 5 kg

25 oC 25 oC

5 kg 2

1 2 v

(a)

State 1 (200 kPa, 25 oC) is compressed liquid = 0.001003 0.001 m3/kg

At state 1, volume occupied by water is : = 5 (0.001) = 0.005 m3

Total volume of tank = 2 V1 = 2 (0.005) = 0.01 m3

(b) At state 2, = 0.002 m3/kg

At 25 oC, vf= 0.00103 m3/kg, vg = 43.340 m3/kg.

Since vfv2vg state 2 is mixture. Thus P2 = Psat, 25C = 3.1698 kPa

At state 2, = 2.3 x 10-5

= 104.83 + (2.3 x 10-5)(2304.3) = 104.88 kJ/kg

The work term is zero, Wb and other forms of work are not present.

= 5 (104.88 – 104.83) = 0.25 kJ

The +ve sign indicates heat is transferred to the water during the process.

4.3Specific Heats

cv = specific heat at constant volume is “the energy required to raise the temperature of a unit

mass of a substance by one degree as the volume is maintained constant”.

cp = specific heat at constant pressure is “the energy required to raise the temperature of a unit

mass of a substance by one degree as the pressure is maintained constant”

cpcv because for the expansion process, work must be supplied to the system.

………… (kJ/kg.K or kJ/kg.oC)
….……… (kJ/kg.K or kJ/kg.oC)

cp and cvare properties, since they are expressed in terms of other properties.

4.3.1Internal Energy, Enthalpy, and Specific Heats of Ideal Gases

It has been demonstrated experimentally that for an ideal gas, u = u(T).

For an ideal gas, Pv = RT

Definition of enthalpy, h = u + Pv

Thus, h = u + RT

Since R is constant and u = u(T) h = h(T)

Since u and h depend only on temperature for an ideal gas, cp and cv also depend on temperature only. Thus, for ideal gases :

and

The change in internal energy or enthalpy for an ideal gas during a process from state 1 to state 2 is obtained by integrating these equations, to give :

…….. ( kJ/kg )

……… ( kJ/kg )

Although cvand cpare temperature dependent, using average specific heats simplifies the calculations :

For an ideal gas, h = u + RT can be differentiated to give

dh = du + RdT

Using the specific heats,

Dividing by dT,

gives cp – cv = R

Note : cp and cv can be used for any process and any system !

Example

Air at 300 K and 200 kPa is heated at constant pressure to 600 K. Determine the change in internal energy per unit mass using the average specific heat value (Table A-2b).

Average temperature = (300 + 900)/2 = 450 K

At this temperature from table, cv = 0.733 kJ/kg.K

Thus,

= 0.733 (600 – 300) = 220 kJ/kg

Example

An insulatedrigid tank initially contains 0.7 kg of helium at 27 oC and 350 kPa. A paddle wheel with a power rating of 0.015 kW is operated within the tank for 30 min. Determine (a) the final temperature and (b) the final pressure of the helium gas. Take cv as 3.1156 kJ/kg.oC.

P, kPa T2

0.7 kg Wsh 0.7 kg P2

27 oC

350 kPa paddle 27 oC

1 2 350

(a)

The amount of paddle-wheel work done on the system is

= 0.015 (30 x 60) = 27 kJ

The system is insulated Q = 0

0 – (– 27) = 0.7 (3.1156) (T2 – 27) T2 = 39.4 oC

(b) At state 1, and at state 2,

Thus, = 364.5 kPa

Note : In an ideal gas equation, P and T are absolute quantities !

Example

A piston-cylinder device initially contains 0.5 m3 of nitrogen at 400 kPa and 27 oC. An electric heater within the device is turned on an allowed to pass a current of 2 A for 5 min from a 120 V source. Nitrogen expands at constant pressure, and a heat loss of 2800 J occurs during the process. Determine the final temperature of nitrogen. Take R = 0.297 kJ/kg.K and cp = 1.039 kJ/kg.K, for nitrogen gas.

Electrical work done on nitrogen, We = VI = 120 (2) (5 x 60) = 72 000 J = 72 kJ

At state 1, = 2.245 kg

Qin = – 2800 J = – 2.8 kJ

For constant-pressure process, =

– 2.8 – (– 72) = 2.245 (1.039) (T2 – 27 )

T2 = 56.7oC

Example

A piston-cylinder device initially contains air at 150 kPa and 27 oC. At this state, the piston is resting on a pair of stops, and the enclosed volume is 400 L. The mass of the piston is such that a pressure of 350 kPa is required to move it. The air heated until its volume doubles. Determine (a) the final temperature, (b) the work done by the air, and (c) the total heat transferred to the air. For air, R = 0.287 kJ/kg.K and cv = 0.823 kJ/kg.K.

. Heating Heating

150 kPa V = C 350 kPa P = C 350 kPa

400 L 400 L 800 L

27 oC

1 2 3

(a) = 1400 K

(b) From 1 to 2, Wb = 0. From 2 to 3, W23 is boundary work for a constant-pressure process.

Thus total boundary work done by the system is

= 350 kPa (0.8 m3 – 0.4 m3) = 140 kJ … +ve sign indicates work done by system

(c)

= 0.697 kg

Process 1–2 ………. = 0 for constant volume process

Process 2–3

Adding, gives or

= 771 kJ ……….. +ve sign indicates heat is transferred to system.