mec 520 f15 lecture 7(1)

04/18/2023 MEC 520 – Energy Technology Thermodynamics 1

LECTURE 7MEC 520

OCTOBER 12, 2015

Rankine Cycle Analysis


• Review Steady-State, Steady-Flow Cycle Analysis of an Ideal Rankine Cycle

• Explain how a superheater changes the Rankine cycle model

• Explain how a reheat line changes the Rankine cycle model.

• Explain how regenerative heating changes the Rankine cycle model.

OBJECTIVES


VAPOR POWER SYSTEM MODEL


RANKINE CYCLE

Tools Conservation of mass Conservation of energy (First Law) Entropy balance (Second Law) Thermodynamic properties

21 hhm

W

Energy balances:For the turbine:

For the condenser:

32 hhm

Q

6 5

. .c w

Qh h

m

For the pump:

4

4 3 3 4 33int.

P

rev

WWh h vdP v p p

m m

For the boiler:

41 hhm

Q


WORK AROUND IDEAL RANKINE CYCLE

Process 2 → 3 : heat flows out through the condenserp2 (=p3) is usually givenT2 =T3 = Tsat(p2) look up sg, sl, hg, and hl s2=sg=s1, h2=hg, s3=sl, h3=hl

Process 1 → 2 : Isentropic expansion of the fluid through the turbine

T1 and p1 are usually given find h1 and s1 using the superheat tables

Process 3 → 4 : isentropic compression by pump

34334

,0

ppvhh

liquidibleincompressvdpdhds

Process 4 → 1 : heat flow into fluid in boiler

p4 =p1 41 hhmQboiler

in

PT

in

cycle

Q

WW

Q

W

41

32

41

3421 1hh

hh

hh

hhhh


NONIDEAL RANKINE CYCLE

For real processes, there is some heat transfer with the surroundings, and thus an increase in entropy. In the turbine and pump, this can be quantified using the isentropic efficiencies.For Turbine:

SST

realTT hh

hh

mW

mW

21

21

For Pump:

34

34

hh

hh

mW

mW S

realP

SPP

But, for a real process, the inefficiencies associated with combustion are more significant. These, however, are external to the power cycle, so are not analyzed as part of the Rankine Cycle analysis.


There are several thermodynamic enhancements that can be made to the basic Rankine cycle power plant to improve efficiency and increase power production per unit of fuel. Superheating: Heating of the fluid past a

saturated vapor and well into the superheated region by the steam generator

Reheating: Reheating fluid after it has passed through one turbine and then passing through a second turbine at lower pressure.

Regeneration Feedwater Heating : use steam bled off from the turbine to partially reheat the condensate from the condenser.

MODIFIED RANKINE CYCLE


As the temperature is raised, note that the sat. vapor location also shifts to the left, so that when the fluid expands through the turbine, a lower quality of the fluid is obtained.

SUPERHEAT AND REHEAT

Net work is decreased as area under the curve get narrower and there are physical effects on the turbine that involve more blade wear. It is more desirable to have the exit quality of the steam, x2, be above 90%

How can the cycle maintain high boiler Pressure and Temperature and still maintain a higher exit quality? Superheat the steam


HOW IS STEAM SUPERHEATED?


REHEATING

When Reheating is implemented, there is a multistage stage turbine (often identified as a High Pressure and Low Pressure turbine.)Before vapor starts to condense (reaches saturation), the vapor is sent back to the boiler and reheated

This affects the Ideal Rankine Cycle model because there are additional states to determine properties for, a second pass through the turbine, and a second heat exchange with the boiler. Therefore, identify additional properties states (1, 2, 3, and 4)


1ST LAW EQUATIONS AS APPLIED TO THE TURBINE AND BOILER

_ 1 20 ( )HP turbineW m h h

_ 3 40 ( )LP turbineW m h h

Turbine Reheat Equation

2 30 ( )reheatQ m h h Boiler Reheat Equation


REGENERATIVE FEEDWATER HEATING. (REGENERATION)

Steam is diverted to use in a mixing chamber to preheat the condensate prior to entering boiler. This is useful since water is typically cooled below saturated level out of the condenser.

Note: This rerouting will diminish the net work output. The reduction in Qboiler should be less than the reduction of Wcycle.


OPEN FEEDWATER ANALYSIS

Mass Balance: 321 mmm

#1

#2

#3Often will use the term “blend fraction”

1

2

m

my

consequently,

1

31m

my

Energy Balance: ii hm 0

6153221133220 hmhmhmhmhmhm

65261

15

1

32

1

2 10 hhyyhhm

mh

m

mh

m

m

Thus:

52

56

hh

hhy


REGENERATIVE VAPOR POWER CYCLE

211 hhm

WT

For the turbines:

For the condenser:

431 hhym

Qout

For the pump:

4567 1 hhyhhm

WP

For the boiler:71 hh

m

Qin

322 1 hhym

WT

• The change in enthalpy Dh of steam condensation in the feed water heaters raises the temperature of the feed water close to the saturation temperature of the extracted steam.

• The heat addition and heat rejected is significantly less, resulting in increased efficiency.


EXAMPLE

An Ideal Rankine cycle with the state properties given below includes one open feedwater heater operating at 100 psi. Saturated liquid exits the open feedwater heater at 100 psi. The mass flow rate of steam into the first turbine state is 1.4 x 106 lbm/hr. Determine

State 1 2 3 4 5 6 7T (°F) 1100p (psi) 1600 100 1 1 100 100 1600xh (Btu/lbm)s (Btu/lbm K)

(a) The net power developed, in Btu/hr.(b) The thermal efficiency.(c) The mass flow rate of cooling water,

if T = 20°F.


EXAMPLE

State 1 2 3 4 5 6 7

T (°F) 1100

p (psi) 1600 100 1 1 100 100 1600

x

h (Btu/lbm)

s (Btu/lbm K)

Start by finding the state properties

State 1: Using Table with T = 1100 F and p = 1600 psi, find h1 = 1547.7 and s1 = 1.6315

State 2: Using Table with s2=s1 and p = 100 psi,

find h2 = 1210.7

State 3: Using Table with s3=s1, and p = 1 psi,

find x3 = 81.2% and then, find h3 = 911.2

State 4: Using Table with sat. liquid and p = 1 psi, find v4 = vf = 0.01614, h4 = hf = 69.74, and s4 = sf = 0.1327


State 1 2 3 4 5 6 7

T (°F) 1100

p (psi) 1600 100 1 1 100 100 1600

x s.h. s.h. 0.812 0.00 liq. 0.00 liq.

h (Btu/lbm) 1547.7 1210.7 911.2 69.74 70.04 298.6 303.5

s (Btu/lbm K) 1.6315 1.6315 1.6315 0.1327 0.4744

Start by finding the state properties

State 5: Using h5 - h4 =v4 (p5 - p4)

find h5 = 70.04

State 6: Using Table with sat. liq. and p = 100 psi, find v6 = 0.01774 h6 = 298.6, and s6 = 0.4744

State 7: Using h7 - h6 =v6 (p7 - p6)

find h7 = 303.5

State 1 2 3 4 5 6 7

T (°F) 1100

p (psi) 1600 100 1 1 100 100 1600

x

h (Btu/lbm)

s (Btu/lbm K)


State 1 2 3 4 5 6 7

T (°F) 1100

p (psi) 1600 100 1 1 100 100 1600

x s.h. s.h. 0.812 0.00 liq. 0.00 liq.

h (Btu/lbm) 1547.7 1210.7 911.2 69.74 70.04 298.6 303.5

s (Btu/lbm K) 1.6315 1.6315 1.6315 0.1327 0.4744

Find heat in provided in boiler:

1 7 10 ( )inQ m h h

1 1 7( )inQ m h h 6(1.4 10 / )(1547.7 303.5) /m mlb hr Btu lb

91.74 10 /Btu hr

then


State 1 2 3 4 5 6 7

T (°F) 1100

p (psi) 1600 100 1 1 100 100 1600

x s.h. s.h. 0.812 0.00 liq. 0.00 liq.

h (Btu/lbm) 1547.7 1210.7 911.2 69.74 70.04 298.6 303.5

s (Btu/lbm K) 1.6315 1.6315 1.6315 0.1327 0.4744

Now find how the mass flow splits from the turbine

6 6(0.200)(1.4 10 / ) 0.28 10 /m mlb hr lb hr

6 5

2 5

298.6 70.040.200

1210.7 70.04

h hy

h h

therefore:

6 63 1(1 ) (1 0.200)(1.4 10 / ) 1.12 10 /m mm y m lb hr lb hr

2 1m y m

c


State 1 2 3 4 5 6 7

T (°F) 1100

p (psi) 1600 100 1 1 100 100 1600

x s.h. s.h. 0.812 0.00 liq. 0.00 liq.

h (Btu/lbm) 1547.7 1210.7 911.2 69.74 70.04 298.6 303.5

s (Btu/lbm K) 1.6315 1.6315 1.6315 0.1327 0.4744

Find work from turbine

2 1 2 3 1 30 ( ) ( )turbineW m h h m h h

6

6

(0.28 10 / )(1547.7 1210.7) /

(1.12 10 / )(1547.7 911.2) /

m m

m m

lb hr Btu lb

lb hr Btu lb

6 6 694.36 10 712.88 10 807.24 10 /Btu hr

2 1 2 3 1 3( ) ( )turbineW m h h m h h


State 1 2 3 4 5 6 7

T (°F) 1100

p (psi) 1600 100 1 1 100 100 1600

x s.h. s.h. 0.812 0.00 liq. 0.00 liq.

h (Btu/lbm) 1547.7 1210.7 911.2 69.74 70.04 298.6 303.5

s (Btu/lbm K) 1.6315 1.6315 1.6315 0.1327 0.4744

Find heat provided to pumps:

1 3 4 50 ( )pumpW m h h

6 6(1.12 10 / )(69.74 70.04) / 0.336 10 /m mlb hr Btu lb Btu hr

Pump 1:

Pump 2:

1 3 4 5( )pumpW m h h

2 1 6 7( )pumpW m h h 6 6(1.4 10 / )(298.6 303.5) / 6.860 10 /m mlb hr Btu lb Btu hr


State 1 2 3 4 5 6 7

T (°F) 1100

p (psi) 1600 100 1 1 100 100 1600

x s.h. s.h. 0.812 0.00 liq. 0.00 liq.

h (Btu/lbm) 1547.7 1210.7 911.2 69.74 70.04 298.6 303.5

s (Btu/lbm K) 1.6315 1.6315 1.6315 0.1327 0.4744

Therefore the Net Work

1 2net turbine pump pumpW W W W 6 6 6 6807.24 10 6.860 10 0.336 10 800.0 10 /Btu hr

Plant thermal efficiency:

6

6

800 10 /45.98%

1740 10 /net

in

W Btu hr

Q Btu hr


State 1 2 3 4 5 6 7

T (°F) 1100

p (psi) 1600 100 1 1 100 100 1600

x s.h. s.h. 0.812 0.00 liq. 0.00 liq.

h (Btu/lbm) 1547.7 1210.7 911.2 69.74 70.04 298.6 303.5

s (Btu/lbm K) 1.6315 1.6315 1.6315 0.1327 0.4744

for the condenser:

3 3 40 ( ) ( )cw cwIn cwOutm h h m h h

6 6(911.2 69.74) /(1.4 10 / ) 58.9 10 /

(1 / )( 20 )m

m m

Btu lblb hr lb hr

Btu lbm R R

2

3 4 3 43 3

0 ( )cwcwIn cwOut H cwIn cwOut

h h h hm m m

h h c T T


Solution:

We can see that Work of the turbine can be found as:Wturbine = (h1 – h2) + (h3 –h4)

and, Wnet = Wt – Wpump

where, Wpump = v5(p5 – p6)


What we know;

p1= 2500 psia, T1 = 1000°F

From Steam Tables we find h1 = 1457.5, s1= 1.5269

Steam expands icentropically ( so s2 = 1.5269

We’re told that steam expands to reheat pressure of 500 psia so next we find enthalpy (h2) of steam at 500 psia and s2 = 1.5269 from diagram or calculator. We find h2 = 1265.6.

Now steam exits the turbine and is sent back to the boiler at 500 psia to be reheated back to 1000°F

Thus we find hs at p3= p2 = 500 psia, T3 = 1000°F


At p3 = 500 psia, T3 = 1000°F we find h3= 1520.3 Btu/lband s3= s4 = 1.7371

Lastly, we’re told that the second turbine section discharge steam pressure is 1 psia so, for p4 = 1 psia and s3 = 1.7371 we find from mollier diagram or calculator that h4 = 970.5 Btu/lb

Now that we have h1, h2, h3, and h4 we can calculate turbine work as:

Wturbine = (h1 – h2) + (h3 –h4) = (1457.5 – 1265.6) + (1520.3 – 970.5)

Wturbine = 741.7 Btu/lb

The next step is to find pump work (Wp) so that we can calculate net work (Wn)


The next step is to find pump work (Wp) so that we can calculate net work (Wn). We saw that:

Wp = v5(p1 – p6)

We find v5 as saturated water at 1 psia from sat. steam tables as v5 = 0.01614 ft3/lb therefore,

Wp = v5(p1 – p6) = 0.01614 ft3/lb(2500lb/in2 – 1 lb/in2)()

Wp = 7.465 Btu/lb

We can now find net work (Wn) as Wt – Wn :

Wn = 741.7 Btu/lb - 7.465 Btu/lb = 734.24


The question asked for efficiency of the cycle which is”

ɳth =

We find qadded (qa) as (h1 – h6) + (h2 – h3)

We already have h1, h2 and h3 from above as:

h1 = 1457.5 Btu/lbh2 = 1265.6 Btu/lbh3 = 1520.3 Btu/lb

All we need now is h6 which is h5 plus Wp. At point 5 the steam has been condensed back to a saturated liquid at 1 psia.


We find h5 from saturated steam tables at 1 psia as:

h5 = 69.73 Btu/lb therefore;

h6 = 69.73 Btu/lb + 7.465 Btu/lb (from above Wp) = 77.19

so now we have;

h1 = 1457.5 Btu/lbh2 = 1265.6 Btu/lbh3 = 1520.3 Btu/lbh6 = 77.19 Btu/lb

We can now calculated heat added (qa)

qa = (h1 – h6) + (h2 – h3)qa = (1457.5 - 1265.6) + (1520.3 - 77.19)qa = 1635.01 Btu/lb


We now how everything we need to calculate cycle efficiency. We previously saw that cycle efficiency is:

ɳth =

Wn = 734.24 Btu/lb

qa = 1635.01 Btu/lb

ɳth =

mec 520 f15 lecture 7(1)

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