Thermo1 Chapter 04

Thermodynamics I

Spring 1432/1433H (2011/2012H)

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10:00am & Monday 8:00am - 9:00am

MEP 261 Class ZA

Dr. Walid A. AissaAssociate Professor, Mech. Engg. Dept.

Faculty of Engineering at Rabigh, KAU, KSA

Chapter #4October XX, 2011

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Office hours for Thermo 01 will be every

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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. Boles7th Edition, McGraw-Hill Companies,

ISBN-978-0-07-352932-5, 2008

Chapter 4

ENERGY ANALYSIS OF CLOSED SYSTEMSSYSTEMS

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.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 & -ve during compression

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

So, p = f (V) should be available.

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 andthe work done on the system.

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:

V = C , Hence dV = 0

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 process.

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

vv1 v2

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

But as p = CBut as p = C

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

Where, Wb is the total boundary work done

during the entire process as the piston moves.

Evaluation of Wb implies evaluation of v1 & v2 .

From Table A-5, psat corresponding to T =

160°C = 618.23 kPa. As psat > p (= 413.57kPa). 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, °°°°C v, m3/kg u, kJ/kg h, kJ/kg

150 0.47088 2564.4 2752.8

160 v u h

200 0.53434 2647.2 2860.9

Hence, vT= 160°°°°C, p = 0.4 MPa = 0.4836 m3/kg

From Table A-6, For p = 0.5 MPa

T, °°°°C v, m3/kg u, kJ/kg h, kJ/kg

Tsat =

151.83°°°°C

0.37483 2560.7 2748.1

151.83°°°°C

160 v u h

200 0.42503 2643.3 2855.8

Hence, vT= 160°°°°C, p = 0.5 MPa = 0.3833443 m3/kg

Hence by interpolation, For T = 160°CHence by interpolation, For T = 160°C

p, kPa v, m3/kg

0.4 0.4836

0.41357 v

0.5 0.3833443

Hence, v1= vT= 160°°°°C, p = 0.41357 MPa = 0.47 m3/kg

2) Evaluation of v2

Similarly,

For p =0.4 MPa

T, °C v, m3/kg u, kJ/kg h, kJ/kg

200 0.53434 2647.2 2860.9

204.4 v u h204.4 v u h

250 0.5952 2726.4 2964.5

Hence, vT= 204.4°°°°C, p = 0.4 MPa = 0.5397 m3/kg

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

T, °C v, m3/kg u, kJ/kg h, kJ/kg

200 0.42503 2643.3 2855.8

204.4 v u h

250 0.47443 2723.8 2961.0250 0.47443 2723.8 2961.0

Hence, vT= 204.4°°°°C, p = 0.5 MPa = 0.42938 m3/kg

Hence by interpolation, For T = 204.4°C

p, kPa v, m3/kg

0.4 0.5397

0.41357 v

0.5 0.429380.5 0.42938

Hence, v2= vT= 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, Wb= p *( V2 - V1 ) = 0.41357 MPa *

(2.377 m3 - 2.129 m3)= 0.41357×××× 106 Pa *

(2.377 - 2.129) m3 = 1.026 ×××× 105 J = 102.6 kJ.

c) For isothermal process T= Constant).

pV = mRT

Hence, pV = C

Hence, for isothermal process :

Hence, p = C V-1 (#-1)

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 m3Where, C = p1V 1= 100 kPa * 0.4 m3

= 40 kJ

Hence, From Eq. (#-2)

Wb= 40 kJ * ln (0.1 m3 /0.4 m3) = -55.45kJ

The negative sign indicates that the

work is done on the system.

d) For polytropic process ;

pV n = Constant).

As, pV n = CAs, pV n = C

Hence, p = C V-n (#-4)

Hence,

But, C = p2V2n = p1V1

n

Hence,

(4-9)

pV = mRT

But, From Equation of State

Hence,

p1V1 = mRT1& p2V2 = mRT2 (#-5)

By substituting by p2V2 & p1V1 from

Eq. (#-4) in Eq. (4-9) to get

(4-10)

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.

Fig. (EXAMPLE 4–4)

V1 = 0.05 m 3 , p1 = 200 kPa, k = 150

kN/m , V2 = 2*V1 , A = 0.25 m 2

Required?

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

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

Solution:

V2 = 2 *V1 = 2 * 0.05 m3 = 0.1 m3

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 isspring 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

i.e.

(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)

(4-12)

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

(4-14)

For a closed system undergoing a cycle,

the initial and final states are identical,

and thus,

For a cycle

∆∆∆∆E =0, thus

Q = W.

For a closed system undergoing a cycle,

the initial and final states are identical,

Hence, Eq. (4-11) can be re-written as

∆Esystem = E2 – E1 = 0 (#-6) &

E – E = 0 (#-7)i.e. E = EEin – Eout = 0 (#-7)i.e. Eout = Ein

Energy balance for a cycle can be

expressed in terms of heat and work

interactions as

Wnet, out = Qnet, in (4-16)i.e.

Energy balance (first-law) relation for

a closed system becomes

Q- W =∆E = ∆U+∆PE+∆KE (4-17)

where,

Qnet, in= Qin - Qout is the net heat input

Wnet, out= Wout - Win 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 Wb 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.

Qin

m = 25 g, p2= p1= 300 kPa

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

Qout= 3.7 kJ

From Eq. (4-17)

Qnet, in- Wnet, out=∆E= ∆U+∆PE+∆KE

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

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

Qnet, in- Wnet, out =∆E = ∆U+∆PE+∆KE 0 0

P = I2R = I V= 0.2 A* 120 V= 24 W

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

i.e. Qin= 7.2 kJ

(#-8)i.e., Qnet, in- Wnet, out =∆U

i.e. Qin= 7.2 kJ

Wnet, out = Wb

But, from Eq. (4-2)

(#-9)

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

re-written as:

Wb = p0 ( V2 – V1)

By substituting by Wb from Eq. (#-10) in

Eq. (#-9) to get:

(#-10)

Eq. (#-9) to get:

Wnet, out = Wb = p0 ( V2 – V1) (#-11)

By substituting from Eq. (#-11) in

Eq. (#-8) to get:

Qnet, in= Wnet, out +∆U=

= (p2 V2 – p1V1) + (U2 –U1) (#-12)

But, By definition; p V + U = H

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

Qnet, in= H2 –H1

Hence,

(#-13)

Qin= m(h2 –h1) (#-13′′′′)

From Table (A-5); Saturated water-

Pressure table

h1 = hg( p = 300 kPa) = 2724.9 kJ/kg

Qnet, in= Qin – Qout = 7.2 kJ- 3.7 kJ =3.5 kJ

Hence, 3.5 kJ = (25/1000) kg* (h2 -

2724.9 kJ/kg)

Hence, h2 = 2864.9 kJ/kg

It is clear that h2 (= 2864.9 kJ/kg) > hg

(p2 = 300 kPa)

Hence, state 2 is superheated stream.

Hence, from Table (A-6); superheated

streamstream150 2761.2

T2 2864.9

200 2865.9

Hence, T2 = 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 (c) the heat transfer for this process.

Solution:

Schematic and p-v diagram for

Example 4–6

m = 5 kg , p1 = 200 kPa , T1 = 25°C,

T2 = T1 = 25°C. V = ?, p2 = ?,

Qheat transfer with soroundings= ?,

From Table (A-5) for Saturated water-

(a) Evaluation of the volume of the tank, Vtank

From Table (A-5) for Saturated water-

Pressure table

Tsat (p =200 kPa ) =120.21 °CIt is clear that

T1 < [Tsat (p =200 kPa ) =120.21 °C].Hence, state 1 is compressed liquid.

It is clear that no data for compressed liquid at p =200 kPa.

Hence, v1 ≈ vf [T= 25°C]= 0.001003

m3/kg, u1 ≈ uf [T= 25°C]= 104.83 kJ/kg,

h1 ≈ hf [T= 25°C]= 104.83 kJ/kg.h1 ≈ hf [T= 25°C]= 104.83 kJ/kg.

Initial volume of the water (V1) is

V1 = m* v1 =5 kg * (0.001003 m3/kg) =

0.005 m3.

Total volume of the tank (Vtank )is twice

the initial volume of the water (V1)

Vtank = 2*V1 = 2* 0.005 m3= 0.01 m3

(b) Evaluation of final pressure, p22

v2 = Vtank / m = 0.01 m3 /5 kg = 0.002

m3/kg

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, vf = 0.001003 m3/kg , vg = At T2 = 25°C, vf = 0.001003 m3/kg , vg =

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 = psat (T = 25°C) = 3.1698 kPa

Temperature table

From Eq.(#-8)

i.e., Qnet, in- Wnet, out =∆U =m(u2 – u1 )0

(#-14)

From Table (A-4); Saturated water-

Temperature table

u2 = uf + (x* ufg) =104.83 +(2.30047 10 -5

* 2304.3) kJ/kg = 104.883kJ/kg

uf (T = 25°C) = 104.83 kJ/kg

Temperature table

ufg (T = 25°C) = 2304.3 kJ/kg

i.e., Qnet, in =m(u2 – u1 )

Hence, from Eq. (#-14)

Hence,

Qnet, in = 5 kg (104.883 –104.83) kJ/kgQnet, in = 5 kg (104.883 –104.83) kJ/kg

Qnet, in = 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; cv.

specific heat at

constant volume; cp.

specific heat at constant volume; cv.: 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; cp.: 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 cv.

Cv dT = du at constant volume

(4-20)

Formal definitions of cp.

Cp dT = dh at constant pressure

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 thatkJ/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

byp v = RT (3-10)

byp 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)

(#-15)

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.h = h(T) only (4-22)

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 = Cv (T) dT (4-23)v

dh = Cp (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 Cv and Cp as

functions of temperature.

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 Cp,avg, and Cv,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

temperature.

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

Cp dT= Cv dT + R dT

Cp = Cv + R

Hence,

(kJ/kg.K) (4–29)

In addition, Cp & Cv are related by ideal-

gas property called specific heat ratio k,

defined as;defined as;

k = Cp / Cv (4–31)

In molar basis;

(4–31)

where;

, &

(4–30)

To summarize

Three ways of calculating u

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

u1 = u (T=300 K) =214.07 kJ/kg

u2 = u (T=600 K) =434.787 kJ/kg

Hence, ∆u = u2 –u1 = 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 -2

c = 0.4802 -5 d = -1.966 -9

Ru = 8.31447 kJ/kmol.K

Cv (T) = Cp (T) - Ru

Cv (T) = (a- Ru) + bT+ cT2+dT3

But, from Eq. (4–30)

,

From Table (A–1); M (Air) =

28.97 kg/kmol

Cv (T) = [(a- Ru) + bT+ cT2+dT3]/M

Hence,

∆u = (1/M) {[(a- Ru) T+ b(T2/2) +

c(T3/3) +d(T4/4)]T=600K - [(a- Ru) T+

b(T2/2) + c(T3/3) +d(T4/4)]T=300K }

∆u = (1/ 28.97) {[(28.11 - 8.31447)

(600-300)]+ (0.1967 -2) [(6002-

3002) /2] + (0.4802 -5 [(6003 -

3003)/3] +(-1.966 -9 (6004-

3004)/4] } = 223 kJ/kg 300 )/4] } = 223 kJ/kg

Cv (T=600 K) = 0.764 kJ/kg.K

(c) Evaluation of ∆u using the average specific heat value (Table A–2b).

Cv (T=300 K) = 0.718 kJ/kg.K

Cv, average = [Cv(T=600 K) + Cv (T=300

K)]/2= (0.764+0.718)/2 kJ/kg.K= 0.741 kJ/kg.K

Hence, ∆u = Cv, average * [600 – 300] = Hence, ∆u = Cv, average * [600 – 300] = 0.741 kJ/kg.K * 300 K = 222.3 kJ/kg

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

of the helium gas.

Solution:

p , kPa

m= 0.5 kgm= 0.5 kg

T1= 30°C

p1= 3 kPa

m = 0.5 kg, T1= 30°C , p1= 3 kPa, P

=15 W , t = 30 min. (a) T2= ? and

(b) p2= ? . From Eq. (4-17)

0 00Q- W =∆E = ∆U+∆PE+∆KE

(4-17)

0

(#-18)

Q- W =∆E = ∆U+∆PE+∆KE

Wnet,out = Wout - Win

0

Hence,

(#-19)

By substituting from Eq. (#-19) in

Eq. (#-18) to get

Win = P *∆t

P *∆t =∆U=m (u – u ) (#-20)P *∆t =∆U=m (u2 – u1)

Hence,

∆t =30*60 = 1800 s

From Table (A-2) ,Cv= 3.1156 kJ/kg.K

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

T1= 30+273 K= 303 K

P *∆t =∆U=m (u – u )

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

3.1156 kJ/kg.K *(T2 - 303 K)

P *∆t =∆U=m (u2 – u1)

Hence,

T2 =320.32 K= 47.3 K

Treating Helium as an ideal gas. Hence,

pv = RT

As v2= v1

Hence, p2/T 2= p1/T 1

p2= p1* (T2/T1 )= 3 kPa * (320.32 K / 303 K)

Hence,

p2= 3 kPa * (320.32 K / 303 K)= 3.17 kPa

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

temperature of nitrogen..

Solution:

Schematic and P-V diagram for Example 4–9.

Qnet, in- Wb,out=∆E= ∆U+∆PE+∆KE

From Eq. (4-17) 0 0

Qnet, in= Qin –Qout=(IV.∆t)- Qout

i.e.i.e.

Qnet, in= [2*120 *(5*60)]/1000 kJ-

2.8 kJ = 69.2 kJ

Qnet, in=∆H

V1= 0.5 m 3 , p1= 400 kPa , T1= 27°C.

Treating Nitrogen as ideal gas. Hence,

pv = RT

From Table A-2, RN2 = 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

, Cp,N2 =1.039kJ/Kg.K

v1 = V1 /m

Hence, 0.2226 m 3 /Kg = 0.5 m 3 /m

m = 2.212 kg

Qnet, in=∆H= m*∆h= m*(h2-h1)Qnet, in=∆H= m*∆h= m*(h2-h1)

i.e., Qnet, in= m*Cp*(T2-T1)

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

*(T2-300 K)

T2= 330.1 K i.e., t2= 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.