Lecture 5 Flashcards
SFEE
steady flow energy equation
enthalpy
internal energy + work done
why are we able to simplify the SFEE to
Q - W = m (h2 -h1) (all with respect to time
as enthalpy is often by 1000 times the largest terms so can ignore kinectic and potential terms especially for gases
simplify the SFEE to bernoullis equation
internal energy does not change and heat is not added
use density rather than specific volume (v = 1/density)
treat density as constant
no work input
for closed systemm
Q -W =m(u2 -u1)
for steady flow thermal systems
SFEE with enthalpy terms
for steady flow fluid systems with work
SFEE densities and internal energy removed
steady flow fluid systems
bernoullis equation
what does a turbine do
extracts KE from flow and turns it into motion
draw a diagram of a turbine
take in high temperature and pressure fluid
out goes low temperature and pressure fluid
compressor takes
low temperature low pressure fluid and compresses it doing work on the fluid
in a powerstation what is the name of compresses that pressurise the water
feed pumps
work is needed to drive a compressor or tubine
work is needed to drive a compressor work is extracted from a turbine
SFEE simplification for turbines and compressors
adiabatic Q = 0
gases and steam kinetic and potential energy terms are negligible
SFEE eqation for turbine and compressor
-W = mass flow rate (h2 - h1)
or workrate = mass flow rate (h1-h2)
SFEE wind and water turbines
kinetic energy term and enthalpy term
Maximum work out
always from reversible process
isentropic is
maximum work possible but never achieved in areal system
for ideal gas undergoing isentropic adiabatic process
n = gamma = cp/cv
ideal work is
mass flow rate * cp * (T1-T2x)
cp as open system
for steam undergoing isentropic adiabatic process
entropy at the start = entropy at the end
entropy at start know for p and t
can calculate final as entropy will be the same
draw diagram of turbine isentropic and real case including equations
W = m (h1 - h2s) W = eff *m(h1-h2s)
for an ideal gas going through turbine how do calculate the change in temperature
T1 - T2 = eff * (T1-T2s)
W = mass flow rate cp(T1 -T2s)
in a turbine in comparison to the isentropic temperature what should the final temperature be
exit temperature is higher
For steam turbine efficiency
W = m (h1-h2)
h2 = h1 - eff * (h1 -h2s)
find T2 by finding temp with that enthalpy
should W be positive for a turbine
yes as turbine extracts work
it is work done by the system
should W be positive or negative for a compressor
negative as compressor doing work on the system
for a compressor you have to put more
work in than the isentropic amount
feedpump is a
compresses liquid water in a powerstation
water incompressible therefore volume cannot change
for an isentropic feed pump
ds = 0 and q = 0
so dh =v dp
at constant volume
dh = v(p2 - p1) = -w
graph diagram of a feedpump
see powerpoint
nozzle takes
a low velocity flow at high pressure and temperature and it undergoes an expansion increasing KE
nozzle takes
a low velocity flow at high pressure and temperature and it undergoes an expansion increasing KE
enthalpy of fluid decreases
nozzle
undergoes an expansion
diffuser
expansion
diffuser takes
high velocity flow and slows it down
increases the enthalpy of the fluid
pressure will increases
nozzle and diffuser simplification of SFEE
everything including work term goes apart from speed and enthalpy
isentropic nozzle drawing
see powerpoint
if nozzle is not 100% efficient what would happen to exit flow
it would be slower but hotter
often assume which end of nozzle and diffuser
that velocity is zero wide end of diffuser or nozzle
real devices difference
heat transfer through devices
irreversible pressure loss in fluid flow (mostly due to turbulent losses)
