Steam Power Cycle Example Problem With Complete Solution

Example Problem with Complete Solution

5C-6 :

Analysis of a Steam Power Cycle

8 pts

A simple simple steam power plant operates on 8 kg/s of steam. Losses in the connecting pipes and through the various components are to be neglected. Calculate … a.) b.) c.) d.) e.) f.) g.) Data :

The power output of the turbine The power needed to operate the pump The velocity in the pump eit pipe The heat transfer rate necessary in the  boiler  The heat transfer rate reali!ed in the condenser  The mass flow rate of cooling water re"uired The thermal efficiency of the cycle

P $

!" 5"

#Pa %C

P! $! D!

8""" 5" "."5

#Pa %C m

P& '&

P+ $+

8""" 6""

#Pa %C

$cw,in $cw,ot

!" #Pa ".(! #g ap*#g total

!" 5"

%C %C

ead : Cycle problems of this type usually re"uire you to work your way around the cycle#  process by process until you have determined the values of all of the unknowns. This is a good approach here because the problem statement asks us to determine the values of unknowns in every process in the cycle. The only decision is where to to begin. $e can  begin with the turbine because that is the %st "uestion and also because we have enough information to answer part &a'. $e know $+ and P+# so we can determine /+. (tream ) is saturated miture with known P& and '&# so we can also determine /&. $ith $ith the usual assumtions about kinetic and potential energy# we can determine 0trb. *n fact# because we know the $ and P of streams % and + as well# we can analy!e the processes in this cycle in any convenient order. (o# we will let the "uestions posed in the problem determine the order in which we analy!e the processes. $e will apply the %st law to the  pump# the boiler and the condenser# in that order. ,se the (team Tables Tables in the 12S$

0ebboo# . 3ien : 4

ind :

P $ P! $! D!

8 !" 5" 8""" 5" "."5

#g*s #Pa %C #Pa %C m

P+ $+ P& '& $cw,in $cw,ot

8""" 6"" !" ".(! !" 5"

0trb 0pmp !

  

40 #0 m*s

7boil 7cond mcw t9

 40  40  #g*s 

Assmptions :

%+-

#Pa %C #Pa #g ap*#g total %C %C

Changes in kinetic and potential energy are negli"ible in all the  processes in the cycle The pump and turbine are adiabatic. All of the heat that leaves the working fluid in the condenser is transferred to the cooling water. o heat is lost to the surroundings.

;ations * Data * Sole : Part a.) 0egin by writing the %st Law for the turbine# assuming that changes in kinetic and  potential energy are negligible. This makes sense because we have no elevation# velocity or pipe diameter information to use.

1"n % *f we assume that the turbine is adiabatic# we can solve 1"n % for the shaft work of the turbine 2

1"n +

 ow# we must use the steam tables to determine /+ and /&. Let3s begin with stream . At a pressure of 8""" #Pa# the saturation temperature is 2

$sat

!(5." %C

0ecause $+ < $sat# we conclude that stream  is superheated steam and we must consult the (uperheated (team Tables. 4ortunately# there is an entry in the table for 8""" #Pa and6""oC.

/+

+6&!.& #=*#g

(tream ) is a saturated miture at !" #Pa# so we need to use the properties of saturated li"uid and saturated vapor at !" #Pa in the following e"uation to determine /& 2 At !" #Pa 2

/sat li;

!5.&! #=*#g

/sat ap

!6"8.( #=*#g

1"n 

/&  ow# we can plug /+ and /& back into 1"n + to answer part >a) 2

0trb

!&!".+ #=*#g (.???

40

Part b.) $rite the %st Law for the pump# assuming that changes in kinetic and potential energy are negligible. This makes sense because we have no elevation or velocity data and we are given only the outlet pipe diameter. Also# assume the pump is adiabatic# 7pmp @ ".

1"n )

1"n 5

 ow# we must determine / and /!. $e know the $ and P for both of these streams# so we should have no difficulty determining the / values.

$sat>P)

6"."58 %C

$  $sat# therefore we must consult the (ubcooled

$sat>P!)

!(5." %C

$ater Tables. $!  $sat# therefore we must consult the (ubcooled $ater Tables.

/

!"(.+5 #=*#g

/!

!6.!!#=*#g

 ow# we can plug / and /! back into 1"n 5 to answer part >b) 2

0pmp

-5&.(6" #0

Part c.) 6ere# we need to consider the relationship between velocity# specific volume and crosssectional area.

1"n 7 where 2

1"n 

A!

".""(6 ! m +

4rom the 12S$ 0ebboo#  2

B!

 ow# we can plug values into 1"n 7 to answer part >c) 2

!

"."""" + m *#g 86 &."(

m*s

Part d.) $rite the %st Law for the boiler# assuming that changes in kinetic and potential energy are negligible. This makes sense because we have no elevation# velocity or pipe diameter data. There is no shaft work in a boiler.

1"n 8

1"n 9

$e determined /! in part >b) and /+ in part >a)# so all we need to do is plug numbers into1"n 9. 7boil

!?.&"( 40

Part e.) $rite the %st Law for the condenser assuming that changes in kinetic and potential energy are negligible. This makes sense because we have no elevation# velocity or pipe diameter data. ,se the working fluid as the system so that :cond is the amount of heat transferred to the cooling water. There is no shaft work in a condenser.

1"n %;

1"n %%

$e determined / in part >b) and /& in part >a)# so all we need to do is plug numbers into1"n %%. 7cond

-?.688 40

Part f.) *n order to determine the mass flow rate of the cooling water# we must write the %st Law using the cooling water as our system. 4or this system# 7cw @ - 7cond because heat leaving the working fluid for the cycle enters the cooling water.

7cw

?.688 40

Assume that changes in kinetic and potential energy are negligible. This makes sense  because we have no elevation# velocity or pipe diameter data. There is no shaft work for the cooling water system. 1"n %+

$e cannot use the (team Tables to determine the enthalpy of the cooling water b ecause we do not know the pressure in either stream. The net best thing we can do is to use the specific heat of the cooling water to determine /cw using2

1"n %

*f we further assume that the specific heat of li"uid water is constant over the temperature range !"%C - 5"%C# than 1"n % simplifies to2 1"n %) $e can then combine 1"n %) with 1"n %+ to obtain 2 1"n %5

4inally# we can solve 1"n %5 for mcw 2

1"n %7

All we need to do is look up the average heat capacity of water  between !"%C and 5"%C.  *(T 2

CP,cw>5"%C) CP,cw>!"%C)

Part g.)

Let3s use 2

CP,cw

Then 2

mcw

#=*#g  #=*#g&.8&   &.8+

&.8

CP,cw

#=*#g&.8!?  

#=*#g 

&."5 #g*s

The thermal efficiency of this power cycle can b e determined directly from its definition.

1"n %

t9

Berify :  one of the assumptions made in this problem solution can be

".+5&?

verified. Answer s:

a.)

0trb

b.) c.) d.)

0pmp ! 7boil

(.?8 40

-55." #0 &. m*s !?.& 40 Download Solution PDF XLS TFT © B-Cubed, 2003, 2005, 2006. All rights reserved.

e.)

7cond

-?.?

40

f.) g.)

mcw t9

& ".+55

#g*s

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