5.5.1 Nozzle and Diffuser
A nozzle is a device for increasing velocity or K.E. of a fluid at the expense of its pressure drop. A diffuser is a device for increasing the pressure of a fluid at the expense of its K.E.
Fig. 5.3 shows a convergent-divergent adiabatic nozzle using equation (5.9) between sections (1) and (2) we have
q + ( h+ v2 /2 + gz)in = ( h + v2 / 2 + gz)out + w
here q = 0, w = 0 Let us use subscript (1) for IN and (2) for OUT
h1 + v1 /2 + gz1 = h2 + v2 / 2 + gz2
h1 + v1 /2 + v2 /2
The SSSF energy equation for an adiabatic diffuser is
h2 – h1 = v1 – v2 /2
i.e., there is rise in enthalpy and pressure at the expense of K.E. (kinetic energy decreases).
Application of Nozzle
(i) Steam nozzles used at inlet to turbines.
(ii) Rocket nozzles used in rocket engine, missiles, space vehicles.
(iii) Injector nozzles used in combustion chamber.
Application of Diffuser
(i) Used at the inlet section of an aircraft engine.
(ii) Used in centrifugal compressors.
5.5.2 Throttling Process
It is a process of reducing pressure. Fig. 5.4 shows a partially opened valve. The passage of fluid flow is restricted.
Here q =0 , w =0 and v1 = v2 thus h1 = h2
Thus the throttling expansion process is an isenthalpic expansion process.
5.5.3 Steady Flow Through Heat Exchanger
In heat exchanger there are two fluids. One is a hot fluid (say A) and the other is a cold fluid (say B). Heat is transferred from hot fluid to the cold fluid. Thus heat lost by the hot fluid is equal to the heat gained by the cold fluid provided there is no loss of heat between the system and surroundings
Considering no heat and work transfer from the control surface and K.E. & P.E. change
negligible. We can write down
( mA hA)in + ( mB hB)in
= m( mA h A)out + ( m B h B)out
m( mA h A out) = m B( mA h AIN)
= – QA = QB
Condenser of a steam power plant is the example of the heat exchanger.
5.5.4 Turbines and Compressors
We get positive work as an output for turbine.
We give work as input to the compressor.
Fig. 5.6 shows a turbine which is insulated.
q = 0, v1 = v2 , z1 = z2 . thus
h1 = h2 + w
w = ( h1 – h2)
Example 5.1. A chilled water of 15 kg/s enters the system for air conditioning a tall building
with a velocity of 60 m/s at an height of 40m from the ground. The water leaves the system with
a velocity of 20 m/s at an height of 70 m. The enthalpies of water entering in and leaving out
are 30 KJ/kg and 50 KJ/kg respectively. The rate of workdone by a pump in the line is 40 KW.
Find out the rate at which heat is removed from the building.
Given m = 15 kg/ s
W= – 40KW
From eqn. (5.8) we have
Q + m ( h + v2 / 2 + gz) in = m ( h+ v2 / 2 + gz) out + W
Q = m( h2 –h1) + m ( v2 – v2 /2 + m g ( z2 – z1 ) + W
= 150 ( 50 – 30 ) + 15 ( 202 – 602 / 2) 1/1000 + 15 x 9.81 ( 70 -40 / 1000) – 40
= 240.41 kw = heat removed rate.
Example 5.2. A steam turbine receives steam at the rate of 0.42 kg/s. The inlet and outlet
conditions of steam are as follows.
Pressure 1.2 MPa 20 KPa
Temperature 1880C -
Enthalpy 2785 KJ/kg 2512 KJ/kg
Velocity 33.3 m/s 100 m/s
Elevation 3 m 0 m
If the heat lost to the surroundings is 0.29 KJ/s, find out the power output of the turbine under
steady flow conditions.
Solution. Given Q = – 0.29 KJ/s;m = 0.42 kg/s, h1 = 2785 KJ/ kg, V1 = 33.3 m/s, z1 =3m,
h2 = 2512 KJ/kg, V2 = 100 m/s, z2 = 0 m
-0.29 + 0.42 ( 2785 + 33.3 2 / 2000 + 9.81 x 3 / 1000)
= 0.42 ( 2512 + 1002 / 2000+9.81 x0/ 1000 ) + W
Example 5.3. An adiabatic horizontal nozzle receives steam at velocity of 60 m/s and enthalpy
of 3000 KJ/kg. It discharges it at an enthalpy 2762 KJ/kg, (a) find the velocity at exist from the
nozzle (b) if the inlet area is 0.1 m2 and the specific volume at inlet is 0.187 m /kg, find the mass
flow rate (c) if the specific volume at the nozzle exit is 0.498 m3 /kg, find the exit area of the nozzle.
Solution. For adiabatic (Q = 0), horizontal (z1 = z2) nozzle; we have
h2 – h2 = v1 – v2 / 2
2762 – 3000 = 602 – v2 / 2000
V2 = 692.5315 m/s .
m = A1 V1 / v1 =0.1 x 60 / 0.187 = 32.08 kg / s
Example 5.4. In an oil cooler, oil flows steadily through a bundle of metal tubes submerged in
a steady stream of cooling water. Under steady flow conditions, the oil enters at 90°C and leaves
at 30°C, while water enters at 25°C and leaves at 70°C. The enthalpy of oil at t°C is given by
h = 1.68 t + 10.5 X 10- 4 t2 KJ/kg.
What is the cooling water flow required for cooling 2.78 kg/s of oil ?
h = 1.68 t + 10.5 x 10-4 t2
h A in = 1.68 x 90 + 10.5 X 10- 4 x 902 = 159.7 KJ/kg
H A out = 1.68 x 30 + 10.5 x 10- 4 x 302 = 51.345 KJ/kg
From equation (5.12) we have
mA (h A in –h A out) = ma (h Bout-h Bin)
or 2.78 (159.7 – 51.345) = Mb x Cw (tBsout – Tb in)[A = hot liquid= oil, B =Water, i.e.; cold liquid]
Or 2.78 (159.7- 51.345) = m B X 4.1868 (70 – 25)
Or MB = 1.59 kg/ s. Ans.