TITLE.PM5

332 ENGINEERING THERMODYNAMICS

DHARM
M-therm\Th6-2.PM5

(i)The irreversibility ;

(ii)The effectiveness.

Take for air : cp = 1.005 kJ/kg K, cv = 0.718 kJ/kg K, R = 0.287 kJ/kg K.

Solution. Mass of air, m = 1 kg

Initial temperature, T 1 = T 0 = 290 K

Final temperature, T 2 = 400 K

Initial pressure, p 1 = p 0 = 1 bar

Final pressure, p 2 = 6 bar.

(i)The irreversibility, I :

We know that reversible work,

Wrev. = Change in internal energy – T 0 × Change in entropy

= (u 2 – u 1 ) – T 0 c

T

T

R p

p eep

log^2 log

1

2

1

F

HG

I

KJ

− F

HG

I

KJ

L

N

M

M

O

Q

P

P

= cv (T 2 – T 1 ) – T 0 c

T

T

R p

p eep

log^2 log

1

2

1

F

HG

I

KJ

− F

HG

I

KJ

L

N

M

M

O

Q

P

P

= 0.718 (400 – 290) – 290 1 005

400

290

0 287^6

1

. logeeF. log
HG

I

KJ

− F

HG

I

KJ

L

N

M

O

Q

P

= 78.98 – 290 (0.3232 – 0.5142)

= 134.37 kJ/kg (–) [–ve sign indicates that work is done on the air]

Actual work is given by

Wactual = mR T T

n

() 12

1

−

−

= RT T

n

() 12

1

−

−

as m = 1 kg.

The index n for the compression is given by :

T

T

p

p

n

2 n

1

2

1

1

=

F

HG

I

KJ

−

∴^400

290

6

1

1

=F

HG

I

KJ

n−

n or 1.379 = ()6nn−^1

loge 1.379 =

n

n

F −

HG

I

KJ

1

loge 6 or 0.3213 = 1.7917

n

n

F −

HG

I

KJ

1

∴

n

n

−

=

1 0 3213

17917

.

.

= 0.1793 or n = 1.218

∴ Wactual =

0 287 290 400

1 218 1

.( )

(. )

−

−

= – 144.8 kJ

(–ve sign means that the work is done on the air)

Now, irreversibility = Wrev – Wactual

= – 134.37 – (– 144.8) = 10.43 kJ. (Ans.)

(ii)The effectiveness, ∈∈∈∈∈ :

∈ = W

W

rev

actual

. = −
−

134.37

144.8

= 0.928 or 92.8%. (Ans.)