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MEC132 PF thermofluids > Lecture 5 > Flashcards

Lecture 5 Flashcards

1
Q

SFEE

A

steady flow energy equation

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2
Q

enthalpy

A

internal energy + work done

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3
Q

why are we able to simplify the SFEE to

Q - W = m (h2 -h1) (all with respect to time

A

as enthalpy is often by 1000 times the largest terms so can ignore kinectic and potential terms especially for gases

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4
Q

simplify the SFEE to bernoullis equation

A

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

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5
Q

for closed systemm

A

Q -W =m(u2 -u1)

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6
Q

for steady flow thermal systems

A

SFEE with enthalpy terms

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7
Q

for steady flow fluid systems with work

A

SFEE densities and internal energy removed

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8
Q

steady flow fluid systems

A

bernoullis equation

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9
Q

what does a turbine do

A

extracts KE from flow and turns it into motion

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10
Q

draw a diagram of a turbine

A

take in high temperature and pressure fluid

out goes low temperature and pressure fluid

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11
Q

compressor takes

A

low temperature low pressure fluid and compresses it doing work on the fluid

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12
Q

in a powerstation what is the name of compresses that pressurise the water

A

feed pumps

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13
Q

work is needed to drive a compressor or tubine

A

work is needed to drive a compressor work is extracted from a turbine

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14
Q

SFEE simplification for turbines and compressors

A

adiabatic Q = 0

gases and steam kinetic and potential energy terms are negligible

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15
Q

SFEE eqation for turbine and compressor

A

-W = mass flow rate (h2 - h1)

or workrate = mass flow rate (h1-h2)

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16
Q

SFEE wind and water turbines

A

kinetic energy term and enthalpy term

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17
Q

Maximum work out

A

always from reversible process

18
Q

isentropic is

A

maximum work possible but never achieved in areal system

19
Q

for ideal gas undergoing isentropic adiabatic process

A

n = gamma = cp/cv
ideal work is
mass flow rate * cp * (T1-T2x)
cp as open system

20
Q

for steam undergoing isentropic adiabatic process

A

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

21
Q

draw diagram of turbine isentropic and real case including equations

A 
W = m (h1 - h2s)
W = eff *m(h1-h2s)
22
Q

for an ideal gas going through turbine how do calculate the change in temperature

A

T1 - T2 = eff * (T1-T2s)

W = mass flow rate cp(T1 -T2s)

23
Q

in a turbine in comparison to the isentropic temperature what should the final temperature be

A

exit temperature is higher

24
Q

For steam turbine efficiency

A

W = m (h1-h2)
h2 = h1 - eff * (h1 -h2s)
find T2 by finding temp with that enthalpy

25
Q

should W be positive for a turbine

A

yes as turbine extracts work

it is work done by the system

26
Q

should W be positive or negative for a compressor

A

negative as compressor doing work on the system

27
Q

for a compressor you have to put more

A

work in than the isentropic amount

28
Q

feedpump is a

A

compresses liquid water in a powerstation

water incompressible therefore volume cannot change

29
Q

for an isentropic feed pump

A

ds = 0 and q = 0
so dh =v dp
at constant volume
dh = v(p2 - p1) = -w

30
Q

graph diagram of a feedpump

A

see powerpoint

31
Q

nozzle takes

A

a low velocity flow at high pressure and temperature and it undergoes an expansion increasing KE

32
Q

nozzle takes

A

a low velocity flow at high pressure and temperature and it undergoes an expansion increasing KE
enthalpy of fluid decreases

33
Q

nozzle

A

undergoes an expansion

34
Q

diffuser

A

expansion

35
Q

diffuser takes

A

high velocity flow and slows it down
increases the enthalpy of the fluid
pressure will increases

36
Q

nozzle and diffuser simplification of SFEE

A

everything including work term goes apart from speed and enthalpy

37
Q

isentropic nozzle drawing

A

see powerpoint

38
Q

if nozzle is not 100% efficient what would happen to exit flow

A

it would be slower but hotter

39
Q

often assume which end of nozzle and diffuser

A

that velocity is zero wide end of diffuser or nozzle

40
Q

real devices difference

A

heat transfer through devices

irreversible pressure loss in fluid flow (mostly due to turbulent losses)

MEC132 PF thermofluids (7 decks)

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