Thermo1 Chapter 04

Thermodynamics I

Spring 1432/1433H (2011/2012H) Saturday, Wednesday 8:00am

-10:00am & Monday 8:00am - 9:00am MEP 261 Class ZA

Dr. Walid A. Aissa

Associate Professor, Mech. Engg. Dept. Faculty of Engineering at Rabigh, KAU, KSA

Chapter #4

Announcements:

Dr. Walid’s e-mail and Office Hours [email protected]

Office hours for Thermo 01 will be every

Sunday and Tuesday from 9:00 – 12:00 am

Dr. Walid’s office (Room 5-213)

in Dr. Walid’s office (Room 5-213). Text book:

Thermodynamics An Engineering Approach Yunus A. Cengel & Michael A. Boles

7th _{Edition, McGraw-Hill Companies,}

Chapter 4

ENERGY ANALYSIS OF CLOSED

SYSTEMS

Objectives of CH4: To

• Examine the moving boundary work or P dV work.

• Identify the first law of thermodynamics as simply a statement of the conservation of energy principle for closed (fixed mass) energy principle for closed (fixed mass) systems.

• Develop the general energy balance applied to closed systems.

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

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

*Describe incompressible substances and

determine the changes in their internal energy and enthalpy.

and enthalpy.

*Solve energy balance problems for closed

(fixed mass) systems that involve heat and work interactions for general pure substances, ideal gases, and incompressible substances

Chapter 4

ENERGY ANALYSIS OF CLOSED SYSTEMS

4–1 ■ MOVING BOUNDARY WORK

One form of mechanical work frequently

encountered in practice is associated with the

expansion or compression of a gas in a piston–cylinder device.

the expansion and compression work is often called moving boundary work, or simply boundary work.

work.

The boundary work is +ve during expansion &

The total boundary work done during the entire process as the piston moves is

The boundary work done during a process

depends on the path followed as well as the end states

Net work done during a cycle is the difference between the work done by the system and

a) Constant Volume (V = C) process

EXAMPLE 4–1 Boundary Work for a

Constant-Volume Process.

A rigid tank contains air at 500 kPa

and 150°C. As a result of heat transfer 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

Solution:

b) For isobaric process (p = C).

EXAMPLE 4–2 Boundary Work for a

Constant-pressure Process.

A frictionless piston–cylinder device

contains 4.53 kg of steam at 413.57 kPa °

contains 4.53 kg of steam at 413.57 kPa and 160°C. Heat is now transferred to the steam until the temperature reaches 204.4 °C. If the piston is not attached to a shaft and its mass is constant, determine the work done by the steam during this

Solution: p , kPa p₀ = 413.57 kPa. m = 4.53 kg p = 413.57 kPa. v, m3/kg

T₁ = 1 2 T 160 °°°°C T _{2 = 204.4 °°°°C} v v_{1 v2}

The total boundary work done during the entire process as the piston moves is

But as p = C But as p = C

Hence, W_b= p *( V₂ - V₁ ) = p *m* ( v₂ - v₁ ) kJ

Where, W_b is the total boundary work done

during the entire process as the piston moves. Evaluation of W_b implies evaluation of v₁ & v₂ .

From Table A-5, p_sat corresponding to T = 160°C = 618.23 kPa. As psat > p (= 413.57

kPa). Hence, state 1 is superheated steam. From Table A-6, For p = 0.4 MPa

1) Evaluation of v₁

From Table A-6, For p = 0.4 MPa

T,

°°°°

160 v u h

From Table A-6, For p = 0.5 MPa T,

°°°°

°°°°

°°°°

Hence, v_{T= 160°°°°C, p = 0.5 MPa} = 0.3833443 m3/kg Hence by interpolation, For T = 160°C Hence by interpolation, For T = 160°C

p, kPa v, m3/kg

0.4 0.4836

0.41357 v

2) Evaluation of v₂ Similarly, For p =0.4 MPa T,

°

Similarly, From Table A-6, For p = 0.5 MPa T,

°

Hence by interpolation, For T = 204.4°C p, kPa v, m3/kg 0.4 0.5397 0.41357 v 0.5 0.42938 0.5 0.42938 Hence, v₂= v_{T= 204.4°°°°C, p = 0.41357 MPa} = 0.52473 m3/kg

& V₂= m * v₂ = 4.53 kg * 0.52473 m3/kg= 2.377 m3 .

Hence, V₁= m * v₁ = 4.53 kg * 0.47 m3/kg = 2.129 m3 .

Hence, W_b= p *( V₂ - V₁ ) = 0.41357 MPa * (2.377 m3 - 2.129 m3_{)= 0.41357×××× 10}6 Pa *

c) For isothermal process T= Constant).

pV = mRT

Hence, pV = C

Hence, for isothermal process :

Hence,

Where, C = p₁V ₁= p₂V ₂

(#-2) (#-3)

EXAMPLE 4–3 Isothermal

Compression of an Ideal Gas.

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 The air is now compressed to 0.1 m in such a way that the temperature

inside the cylinder remains constant. Determine the work done during this process.

V ₁ = 0.4 m3 ,p₁ = 100 kPa and T = 80°C. V ₂ = 0.1 m3.

Solution:

Hence, From Eq. (#-3)

Where, C = p V = 100 kPa * 0.4 m3 Where, C = p₁V ₁= 100 kPa * 0.4 m3

= 40 kJ

Hence, From Eq. (#-2)

The negative sign indicates that the work is done on the system.

d) For polytropic process ;

pV n = Constant).

As, pV n = C As, pV n = C

Hence,

But, C = p₂V₂ n

=

(4-9)

pV = mRT

But, From Equation of State Hence,

By substituting by p₂V₂ & p₁V₁ from Eq. (#-4) in Eq. (4-9) to get

EXAMPLE 4–4 Expansion of a Gas

against a Spring.

A piston–cylinder device contains 0.05 m 3 of a gas initially at 200 kPa. At

this state, a linear spring that has a 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 m 2, determine (a) the final

pressure inside the cylinder, (b) the total work done by the gas, and (c) total work done by the gas, and (c)

the fraction of this work done against the spring to compress it.

V₁ = 0.05 m 3 , p₁ = 200 kPa, k = 150 kN/m , V₂ = 2*V₁ , A = 0.25 m 2

Required?

p₂ = ?, W_b = ?, Fraction of work done

against the spring to compress it? against the spring to compress it?

Solution:

Displacement of the spring; x

i.e.; i.e.;

Force applied by the linear spring at the final state; F is

F = k x = (150 kN/m)*(0.2 m)= 30 kN

Additional pressure applied by the spring on the gas at this state; p is spring on the gas at this state; p is

Without the spring, the pressure of the

gas would remain constant at

200 kPa while the piston is rising.

But under the effect of the spring, the

pressure rises linearly from 200 kPa to pressure rises linearly from 200 kPa to 200 + 120 = 320 kPa at the final state.

p₂ = 320 kPa

(b) One way for finding the work done

is to plot the process on a P-V diagram and find the area under the process

curve. From Fig. (Example 4–4) the area under the process curve (a trapezoid) is

Hence; W = 13 kJ

+ve sign indicates that the work is done by the system.

( c) Total work consists of two

portions:

I) Work represented by the

rectangular area (region I) is done against the piston and the

against the piston and the atmosphere, and

II) the work represented by the

triangular area (region II) is done against the spring.

Thus,

- Work done against the piston and the atmosphere (region I),

- Fraction of work done against the spring to compress it; (region II),

4–2 ■ ENERGY BALANCE FOR CLOSED SYSTEMS

Energy balance for any system undergoing any kind of process (as shown in Chap. 2) was

expressed as:

(4-11) or, in the rate form, as

(4-11)

For constant rates, the total quantities

during a time interval

∆t are related to

the quantities per unit time as

(4-13) (4-13)

The energy balance can be expressed

on a per unit mass basis as

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

and thus, For a cycle

∆

∆

∆

∆

For a closed system undergoing a cycle, the initial and final states are identical, Hence, Eq. (4-11) can be re-written as

∆

E – E = 0 i.e. E = E (#-7)

E_in – E_out = 0 i.e. E_out = E_in (#-7) Energy balance for a cycle can be

expressed in terms of heat and work interactions as

W_{net, out} = Q_{net, in}

(4-16) i.e.

Energy balance (first-law) relation for

a closed system becomes

Q- W =

∆

E =

∆

U+

∆

PE+

∆

KE

where,

Q

= Q

- Q

is the net heat input

W

= W

- W

is the net work

output

Various forms of the first-law

relation for closed systems

EXAMPLE 4–5 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

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 W

and the

change in internal energy

_∆

U

in the first-law relation can be

combined into one term,

∆

H, for a

combined into one term,

∆

H, for a

constant pressure process.

(b) Determine the final temperature

of the steam.

Solution:

Schematic and P-v diagram for Example 4–5.

m = 25 g, p₂= p₁= 300 kPa

I = 0.2 A , V = 120 V, t= 5 min

Q_out= 3.7 kJ

From Eq. (4-17)

Q

- W

=

∆

E=

∆

U+

∆

PE+

∆

KE

As system is stationary, hence ∆PE=0,∆KE=0

Hence, Eq. (4-17) can be re-written as

Q

- W

=

∆

E =

∆

U+

∆

PE+

∆

KE

P = I

R = I V=

0.2 A* 120 V= 24 W

Q_in= P * t = 24 W* (5*60) s= 7200 J

i.e. Q_in= 7.2 kJ

(#-8)

i.e., Q

- W

=

∆

U

i.e. Q_in= 7.2 kJ

W

= W

But, from Eq. (4-2)

But, p₂ =p₁= p₀ . Hence, Eq. (4-2) can be

re-written as:

W_b = p₀ ( V₂ – V₁)

By substituting by W_b from Eq. (#-10) in Eq. (#-9) to get:

(#-10)

Eq. (#-9) to get:

W

= W

= p

(

V

– V

)

By substituting from Eq. (#-11) in Eq. (#-8) to get:

Q

= W

+

∆

U=

=

(

p

V

– p

V

) +

(

U

–U

)

(#-12) But, By definition;

p V

+

U

= H

Hence, Eq. (#-12) can be re-written as:

Q

=

H

–H

Hence,

(#-13)

From Table (A-5); Saturated water-Pressure table

h

= h

( p = 300 kPa) = 2724.9

kJ/kg

It is clear that h₂ (= 2864.9 kJ/kg) > h_g (p₂ = 300 kPa)

Hence, state 2 is superheated stream.

Hence, from Table (A-6); superheated stream stream 150 2761.2 T₂ 2864.9 200 2865.9 Hence, T_{2 = 199.5224 °C}

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

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

(b) the final pressure, and

Solution:

Schematic and p-v diagram for

m = 5 kg , p₁ = 200 kPa , T₁ = 25°C,

T₂ = T₁ = 25°C. V = ?, p₂ = ?,

Q_{heat transfer with soroundings}= ?,

From Table (A-5) for Saturated water-(a) Evaluation of the volume of the tank, V_tank

From Table (A-5) for Saturated water-Pressure table

T_{sat (p =200 kPa ) =120.21 °C}

It is clear that T₁ < [T_{sat (p =200 kPa ) =120.21 °C].}

It is clear that no data for

compressed liquid at p =200 kPa.

Hence, v

≈

v

[T= 25°C]= 0.001003

m

/kg,

u

≈

u

[T=

]= 104.83 kJ/kg,

h

≈

h

[T=

]= 104.83 kJ/kg.

h

≈

h

[T=

]= 104.83 kJ/kg.

Initial volume of the water (V₁) is

V

= m* v

=5 kg * (0.001003 m

/kg)

=

0.005

.

Total volume of the tank (V_tank )is twice the initial volume of the water (V₁)

V

= 2*V

= 2*

0.005

= 0.01

m

v

= V

/ m

= 0.01

m

/5 kg

= 0.002

Specifying conditions at state 2:

At state 2, T₂ = 25°C, v₂ = 0.002 m3_/kg

From Table (A-4); Saturated water-Temperature table At T₂ = 25°C, v_f = 0.001003 m3_{/kg , v} g = At T₂ = 25°C, v_f = 0.001003 m3_{/kg , v} g = 43.34 m3_/kg

Hence, state 2 is saturated liquid-vapor mixture.

i.e., x = 2.30047 -5

From Table (A-4); Saturated water-Temperature table

p₂ = p_sat (T = 25°C) = 3.1698 kPa Temperature table

From Eq.(#-8)

i.e., Q

- W

=

∆

U =m(u

– u

)

(#-14) From Table (A-4); Saturated water-Temperature table u₂ = u_f + (x* u_fg) =104.83 +(2.30047 10 -5 * 2304.3) kJ/kg = 104.883kJ/kg u_f (T = 25°C) = 104.83 kJ/kg Temperature table u_fg (T = 25°C) = 2304.3 kJ/kg

i.e., Q

=m(u

– u

)

Hence, from Eq. (#-14)

Hence,

Q

=

5 kg (104.883 –104.83) kJ/kg

Q

=

5 kg (104.883 –104.83) kJ/kg

Q

=

5 kg (104.883 –104.83) kJ/kg

= 0.265 kJ

The +ve sign indicates that heat is transferred to the water

4–3 ■ SPECIFIC HEATS

The specific heat is defined as the energy required to raise the temperature of a unit

mass of a substance by one degree.

specific heat at constant volume ; c_v. specific heat at constant volume ; c_p.

specific heat at constant volume

; c_v.: the energy required to raise the temperature of the unit mass of a

substance by one degree as the volume is maintained constant

specific heat at constant pressure

; c_p.: the energy required to raise the temperature of the unit mass of a

substance by one degree as the pressure is maintained constant

(4-19)

Formal definitions of c_v.

(4-20)

Formal definitions of c_p.

Eqs. (4-20) & (4-20) are properties

relations and as such are independent of the type of processes.

A common unit for specific heats is kJ/kg ·°C or kJ/kg·K. Notice that

kJ/kg ·°C or kJ/kg·K. Notice that

these two units are identical since T(°C)

T(K), and 1°C change in temperature is equivalent to a change of 1 K.

4–4 ■ INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF IDEAL GASES

Ideal gas is a gas whose temperature,

pressure, and specific volume are related by

p v = RT

by

p v = RT

u = u(T) only

But by definition; enthalpy and internal energy of an ideal gas r related by:

(4-21)

h= u + p v

(3-10)

By substitution by p v from Eq. (3-10) in Eq. (#-15) to get:

h= u + RT

Since R is constant and u = u(T). Hence,

(#-16)

Since R is constant and u = u(T). Hence,

enthalpy of an ideal gas is also a

function of temperature only:

i.e.

As u = u(T) only; Eq. (4-21) & h = h(T) only; Eq. (4-22) . Hence, the partial

derivatives; (4-19) and (4-20) turn to ordinary derivatives; i.e.

du = C_vv (T) dT (4-23)

dh = C_p (T) dT (4-24)

The change in internal energy or enthalpy for an ideal gas during a process from

state 1 to state 2 is determined by integrating these equations:

(4-25) and

(4-26) (4-26)

To carry out these integrations, we need to have relations for C_v and C_p as

The variation of specific heats with T over Small temperature intervals (a few hundred

degrees or less)may be approximated as linear.

Therefore the specific heat functions in

Eqs. 4–25 and 4–26 can be replaced by the

constant average specific heat values. Then

constant average specific heat values. Then

the integrations in these equations can be performed, yielding:

(4-27)

(4-28) (4-28)

The specific heat values for some

common gases are listed as a function of temperature in Table A–2b. The average

specific heats C_p,avg, and C_v,avg are

evaluated from this table at the average evaluated from this table at the average temperature (T₁ + T₂)/2, as shown in Fig. 4–26.

FIGURE 4–26 For small T intervals, the

specific heats may be assumed to vary linearly with T.

If the final temperature T₂ is not known,

the specific heats may be evaluated at T₁

or at the anticipated average temperature. Then T₂ can be determined by using these specific heat values. The value of T₂ can specific heat values. The value of T₂ can be refined, if necessary, by evaluating the specific heats at the new average

Specific Heat Relations of Ideal Gases

Differentiating the relation (#-16)

h = u + RT

to get

dh = du + RdT (#-17)

(#-16)

dh = du + RdT (#-17)

By substituting by du & dh from Eqs. (4-23)&(4-24) in Eq. (#-17) to get

C_p = C_v + R Hence,

(kJ/kg.K) (4–29)

In addition, C_p & C_v are related by ideal-gas property called specific heat ratio k, defined as;

defined as;

In molar basis;

(4–31)

where;

, _&

To summarize

EXAMPLE 4–7 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 (a)

energy of air per unit mass, using (a)

data from the air table (Table A–17),

(b) the functional form of the specific

heat (Table A–2c), and (c) the

average specific heat value (Table

A–2b).

Solution:

(a) Evaluation

_∆

u using Table A–17

u

= u (T=300 K) =214.07 kJ/kg

u

= u (T=600 K) =434.787 kJ/kg

Hence,

_∆

u = u

–u

= 434.79 kJ/kg

-214.07 kJ/kg = 220.72 kJ/kg

(b) Evaluation of

_∆

u using functional

form of the specific heat (Table A–2c)

kJ/kmol.K

a = 28.11, b = 0.1967

c = 0.4802

d = -1.966

C

(T) = C

(T) -

R

C

(T) = (a-

R

) + bT+ cT

+dT

,

From Table (A–1); M (Air) = 28.97 kg/kmol

C

(T) = [(a-

R

) + bT+ cT

+dT

]/M

∆

u = (1/M) {

c(T3/3) +d(T4/4)]_T=600K

-

[(a-

R

) T+

∆

u = (1/

28.97

) {

} = 223 kJ/kg

} = 223 kJ/kg

C

(T=600 K) = 0.764

kJ/kg.K

(c) Evaluation of

_∆

u using the

average specific heat value (Table

A–2b).

C

(T=300 K) = 0.718

kJ/kg.K

C

= [C

(T=600 K) + C

(T=300

K)]/2= (0.764+0.718)/2

kJ/kg.K=

Hence,

_∆

u = C

* [600 – 300] =

Hence,

_∆

u = C

* [600 – 300] =

EXAMPLE 4–8 Heating of a Gas in a Tank by Stirring.

An insulated rigid tank initially

contains 0.5 kg of helium at 30°C and

3 kPa. A paddle wheel with a power

rating of 15 W is operated within the

rating of 15 W is operated within the

tank for 30 min. Determine (a) the final

temperature and (b) the final pressure

Solution: p , kPa m= 0.5 kg m= 0.5 kg T₁= 30°C p₁= 3 kPa

m

= 0.5 kg, T

= 30°C , p

= 3 kPa, P

=15 W , t = 30 min. (a) T

= ? and

(b) p

= ? .

Q- W =

∆

E =

∆

U+

∆

PE+

∆

KE

Q- W =

∆

E =

∆

U+

∆

PE+

∆

KE

W

= W

- W

Hence,

(#-19)

By substituting from Eq. (#-19) in Eq. (#-18) to get

W

= P *

∆

t

P *

∆

t =

∆

U=m (u – u )

P *

∆

t =

∆

U=m (u

– u

)

∆

t =30*60 = 1800 s

From Table (A-2) ,

_C

v

= 3.1156 kJ/kg.K

Hence, Eq. (#-20) can be re-written as:

T

= 30+273 K= 303 K

P *

∆

t =

∆

U=m (u – u )

(

15/1000) kW* 1800 s = 0.5 kg *

3.1156 kJ/kg.K *(T

- 303 K)

P *

∆

t =

∆

U=m (u

– u

)

T

=320.32 K= 47.3 K

Treating Helium as an ideal gas. Hence, pv = RT

As v

= v

EXAMPLE 4–9 Heating of a Gas by a Resistance Heater

A piston–cylinder device initially contains 0.5 m 3 of Nitrogen gas at 400 kPa and

27°C. An electric heater within the device is turned on and is allowed to pass a

is turned on and is 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

Solution:

Q

- W

=

∆

E=

∆

U+

∆

PE+

∆

KE

From Eq. (4-17) _{0 0}

Q

= Q

–Q

=(IV.

∆

t)- Q

i.e. i.e.

Q

=

[2*120 *(5*60)]/1000

kJ-2.8

kJ = 69.2 kJ

V₁= 0.5 m 3 , p₁= 400 kPa , T₁= 27°C. Treating Nitrogen as ideal gas. Hence,

pv = RT

From Table A-2, R_N2 = 0.2968 kJ/Kg.K

p₁v₁ = RT₁ Hence,

400 kPa *v₁ = 0.2968 kJ/Kg.K *(27+273 K) v₁ = 0.55725 m 3 /Kg

v₁ = V₁ /m Hence, 0.2226 m 3 /Kg = 0.5 m 3 /m m = 2.212 kg

Q

=

∆

H= m*

∆

h= m*(h

-h

)

Q

=

∆

H= m*

∆

h= m*(h

-h

)

_Q

= m*Cp*(T

-T

)

69.2 kJ = 2.212 kg * 1.039 kJ/Kg.K

*(T

-300 K)

T

= 330.1

K

_t

₌

_{57.1 °C.}

Homework

4–4C, 4–5, 4–6, 4–7, 4–8, 4–9, 4–12, 4–13, 4–14, 4–18, 4–21, 4–28, 4–56, 4–61.