Chapter 1

Introduction: Basic Principles

Solution

(a) From the gas law

3

311

1

550 104.53 kg/m

287 423

p

RTρ ×= = =

×

From the conservation of mass law

1

40180 m/s

4.53 0.04908

mc

Aρ= = =

×ɺ

Ans

where 2 20.04908 m4

A dπ = =

From the conservation of energy equation the stagnation temperature is

2 2

101 1

179.9423 439.1 K

2 2 1004.5p

cT T

C= + = + =

×Ans

(b) 2 147 273 420 K T = + =

As the flow is adiabatic, the energy of the air is conserved along the pipe, and 02 01h h=

1.1.(a) Air flows adiabatically through a long straight circular section duct, 0.25 m diameter at

a measured mass flow rate of 40 kg/s. At a particular section along the duct the values of static

temperature T = 150 0

C and static pressure p = 550 kPa. Determine the average velocity of the

airflow and its stagnation temperature.

(b) At another station further along the duct, measurements reveal that the static temperature

has dropped to 147 0

C as a consequence of wall friction. Determine the average velocity and the

static pressure of the airflow at this station.

Also determine the change in entropy per unit of mass flow between the two stations.

For air assume that R = 287J/(kgK) and 1.4γ =

( )2 2

2 22 12 1 2 1 2 3 26373

2 2p p p

c cC T C T c c C∴ + = + ∴ = + − =

2 162.4 m/s c∴ = ANS

From the continuity equation

( )1 1 2 2 and the gas law / we arrive at the

following useful expression:

m c A c A p RTρ ρ ρ= = =ɺ

(c) The change in entropy between the two stations i:

2 2 2 2

1 1 1 1

ln ln ln ln1p

T p T ps C R R

T p T p

γγ

∆ = − = − −

= 3.81 J/(kg K) Ans

N.B.In this type of flow (called a FANNO FLOW) the entropy will always increase.

2 1 2

1 2 1

2

180 4200.9635

185.5 423

530 kPa

p c T

p c T

p

×= = =×

∴ =

Solution

(1) From eqn. (1.35)

1

20 11 1.2755

2

pM

p

γγγ −− = + =

0 2.551 barp∴ = Ans.

and 20 01

1 1.072 279.9 K2 1.072

T TM T

T

γ −= + = ∴ = = Ans.

(2) From eqn. (1.38)

0 2

0

11 2 111

21

p

n

m C TM M

A p

γγγ γ

γ

+− −− = + −

ɺ

and we have 2 20.07069 m

4nA dπ= = and

1p

RC

γγ

= =−

1040 J/(kg K)

Substituting values (and part evaluating) we have, for the LHS of the above equation

( )51040 300 0.07069 2.551 10 0.03097m m× ÷ × × =ɺ ɺ

and for the RHS of the equation

( )1 12 1

2 31 1.4 0.61 1.072 1.0782

21 0.4

MM

γγγ γ

γ

+− − −=

− × + = −

27.16 kg/sm∴ =ɺ ANS.

1.2. Nitrogen gas at a stagnation temperature of 300 K and a static pressure of 2 bar flows

through a pipe duct of 0.3 m diameter. At a particular station along the duct length the Mach

number is 0.6. Assuming the flow is frictionless, determine

(1) the static temperature and stagnation pressure of the flow,

(2) the mass flow of gas if the duct diameter is 0.3 m.

For nitrogen gas take

( )297 J/ kgK and 1.4R γ= =.

Solution

From eqn. (1.12)

( )

2 21 101 02 1 1 2 22 2

2 22 1 2 1

h

2 where 3.5 287 1004.5 J/kgK1p p

h h c h c

Rc C T T c C

γγ

= ∴ + = +

∴ = − + = = × =−

( )2 22 2 1004.5 200 150 100 100450c∴ = × − + =

2 316.9 m/sc = Ans

From eqn.(1.28a)

2 2

1 1

150 50ln ln 1.0045ln 0.287 ln

200 150

0.289 0.315 0.0263 kJ/kg ANS

p

T ps C R

T p

s

∆ = − = −

∴∆ = − + =

1.3. Air flows adiabatically through a horizontal duct and at a section numbered

(1) the static pressure p1 = 150 kPa, the static temperature T1 = 200 0C and the

velocity c 1 = 100 m/s. At a station further downstream the static pressure p 2 = 50

kPa and the static temperature T 2 =150 0 C. Determine the velocity c 2 and the

change in entropy per unit mass of air.

For air take ( )1.4 and 287 J/ kgKRγ = =

4

g 1.5

gg 1.7