Transcript Document 7472728

First Law of Thermodynamics-The
Energy Equation (4)
Work transfer can also occur at the control surface when a
force associated with fluid normal stress acts over a distance.
W normalstress 
  V ndA
CS
W normalstress for particles on the wetted inside surface of the pipe is zero as V n is zero
W normalstress can be nonzero only where fluid enters and leaves the control volume
Work transfer can also occur at the control surface because
of tangential stress forces. Rotating shaft work is transferred
by tangential stresses in the shaft material. Combining all
the information

net  W net 
e


V
'

e

V
ndA

Q
V ndA


in

in
t CV
CS
CS
• Equation with total stored energy,

p V2
net  W shaftnet
e


V
'

(
u



gz
)

V
ndA

Q
CS  2
in
in
t CV
• For steady flow, and uniformly distributed properties,
p V2
p V2
p V2
(u    gz ) m  (u    gz ) m
CS (u    2  gz) V ndA  
 2
 2
flow
flow
out
in
•
For only one stream entering and leaving,
p V2
p V2
p V2
CS (u    2  gz) V ndA  (u    2  gz)out mout  (u    2  gz)in min
• When shaft work is involved, the flow is unsteady, at
least locally, e.g., the velocity and pressure at a fixed
location near the rotating blade of a fan is unsteady,
while upstream and downstream of the machine, flow is
steady.
Vout 2  Vin 2
p
p
m[u out  u in  ( )out  ( )in 
 g ( zout  zin )]  Q net  W shaftnet
in
in


2
One dimensional energy equation for steady-in-the meanflow valid for both incompressible and compressible flows.
Using enthalpy, following equation is obtained,
Vout 2  Vin 2
m[hout  hin 
 g ( zout  zin )]  Q net  W shaftnet
in
in
2
Example 1
• A pump delivers water at a steady rate of 300 gal/min
as shown in the figure. The change in water elevation
across the pump is zero. The rise in internal energy of
water associated with a temperature rise is 3000 ft.
lb/slug . Determine the power (hp) required by the
pump for an adiabatic process.
Example 2
Steam enters a turbine with a velocity of 30 m/s and
enthalpy h1 of 3348 kJ/kg. The steam leaves the turbine
as a mixture of vapor and liquid having a velocity of 60 m/s
and an enthalpy of 2550 kJ/kg. If the flow through the
turbine is adiabatic and changes in elevation are negligible,
determine the work output involved per unit mass of steam
through-flow.
Comparison of the Energy Equation with
the Bernoulli Equation
• When the one-dimensional energy equation for steadyin-the –mean flow is applied to a flow that is steady, the
equation becomes,
Vout 2  Vin 2
p
p
m[u out  u in  ( )out  ( )in 
 g ( zout  zin )]  Q net
in


2
Dividing by the mass flow rate, and rearranging,
pout Vout 2
pin Vin 2

 gzout 

 gzin  (u out  u in  qnet )

2

2
in
Q net
in
q

where net
Heat transfer rate per unit mass flow rate
in
m
Comparing with Bernoulli’s equation, it can be concluded
for incompressible frictionless flow,
u out  u in  qnet  0
in
Frictional loss
Useful or available energy
Loss of available energy
p V2

 gz
 2
u out  u in  qnet  loss
in
V2 2
p1 V12

 gz2  
 gz1 1 loss2

2
 2
p2
V2 2
Loss  K L
2
An important group of fluid mechanics problems involves onedimensional, incompressible, steady-in-the-mean flow with
friction and shaft work for pumps, blowers, fans and turbines.
For this kind of flow, expressing shaft work per unit mass, the
energy equation becomes,
Vout 2
pin Vin 2

 gzout 

 gzin  wshaft  (u out  u in  qnet )

2

2
netin
in
pout Vout 2
pin Vin 2
or ,

 gzout 

 gzin  loss

2

2
pout
Mechanical energy equation or extended Bernoulli’s
equation, involves energy per unit mass (ft. lb/slug=ft2/s2; or
N. m =m2/s2)
Example 3
• Axial flow fan- delivers 0.4 kW power
to the fan blades produce a 0.6 m axial
stream of air at 12 m/s.
V22
 wshaft  loss
2
netin

wshaft  loss
netin
wshaft
 0.752
netin
Energy equation in energy per unit weight involves heads
Vout 2
pin Vin 2

 zout 

 zin  hs  hl

2g

2g
pout
where
hs 
shaft
W net
in
mg

shaft
W net
in
Q
Energy Equation for Nonuniform Flows
If the velocity profile at any section where flow crosses the
control surface is not uniform, the energy equation is:
pout


 outVout 2
2
 gzout 
pin


inVin 2
2
 gzin  wshaft  loss
netin
where  =kinetic energy coefficient
For any velocity profile,  1, and=1 for uniform flow
The difference in loss calculated assuming uniform
velocity and actual velocity profiles in not large
compared to w shaft net in