TITLE.PM5

180 ENGINEERING THERMODYNAMICS

dharm

/M-therm/Th4-5.pm5

Capacity of electric motor :

Steady flow energy equation is given by

mu pv

C Zg

w 1111

2

+++ 2 1

F

HG

I

KJ

+ Q = mu pv

C Zg

w 2222

2

+++ 2 2

F

HG

I

KJ

+ W ...(i)

Considering the datum from suction 1, as shown

Z 1 = 0, Z 2 = 8.5 + 2.2 = 10.7 m

u 2 – u 1 = 0 ; Q = 0

Thus eqn. (i) reduces to

Wmpv pv Z Zg=−+−+w CC−

F

HG

I

KJ

L

N

M

M

O

Q

P

P

()() 11 2 2 1 2 1

2

2

2

2 ...(ii)

As water is incompressible fluid.

∴ v 2 = v 1 = v==^11 ρ 1000

The mass flow through inlet and exit pipe is given by

mdCw=× × ×=× × ×π ρ π dCρ

441

2

12

2

2 as ρ 1 = ρ 2 = ρ (for water)

∴ 50 = 4 π × (0.2)^2 × C 1 × 1000

∴ C (^12)
50 4
2 1000
= ×
π××()0.
= 1.59 m/s
and C (^22)
50 4
1 1000
= ×
π××()0.
= 6.37 m/s
Substituting the values in eqn. (ii), we get
W=××−××
F
HG
I
KJ+− × +
F −
HG
I
KJ
L
N
M
M
O
Q
P
P
50 1 10^1
1000
42 10^1
1000
0107 981^5937
2
55
22
.(.).1. 6.
= 50[– 320 – 104.96 – 19.02]
= 22199 J/s or 22.199 kJ/s ∼ 22.2 kW.
Hence capacity of electric motor = 22.2 kW. (Ans.)
Example 4.49. During flight, the air speed of a turbojet engine is 250 m/s. Ambient air
temperature is – 14°C. Gas temperature at outlet of nozzle is 610°C. Corresponding enthalpy
values for air and gas are respectively 250 and 900 kJ/kg. Fuel air ratio is 0.0180. Chemical
energy of fuel is 45 MJ/kg. Owing to incomplete combustion 6% of chemical energy is not released
in the reaction. Heat loss from the engine is 21 kJ/kg of air.
Calculate the velocity of the exhaust jet.
Solution. Refer Fig. 4.55.
Air speed of turbojet engine, Ca = 250 m/s
Ambient air temperature = – 14°C
Gas temperature at outlet of nozzle = 610°C
Enthalpy of air, ha = 250 kJ/kg
Enthalpy of gas, hg = 900 kJ/kg