ME 354 Tutorial, Week#9 Brayton Cycle With Intercooling, Reheat & Regeneration
ME 354 Tutorial, Week#9 Brayton Cycle With Intercooling, Reheat & Regeneration
ME 354 Tutorial, Week#9 Brayton Cycle With Intercooling, Reheat & Regeneration
Regenerator
10
Q in,1 Q in,2
5 Combustor
7 Reheat
Compressor Compressor Combustor 9
1 2 4 6 8
Wnet,out
2 Turbine 1 Turbine 2
Intercooler
3
1
Q out
To better visualize what is happening during the cycle we can draw a T-s process
diagram.
1
T
6 8
5
7 9
7s
9s
4
4s 10
2
2s
3 1
T (K) P (kPa)
1 300 100
2 300
2s 300
3 300 300
4 1000
4s 1000
5 1000
6 1400 1000
7 300
7s 300
8 1400 300
9 100
9s 100
10 100
2
Step 4: Calculations
Part a)
The thermal efficiency of the system can be expressed as the ratio of the sytems
net-work output to the systems heat input (we have heat inputs at the combustor
AND the reheat combustor) as shown in Eq1. Note: the heat input that occurs
from location 4 to 5 is an internal energy transfer, getting its energy from location
9 to 10 and thus does not qualify as a heat input into the system.
wnet ,out
th = (Eq1)
qin,1 + qin, 2
The wnet,out can be determined from the difference between the work output of the
turbines and the work input to the compressors. The work for each of these
devices can be determined from energy balances applied to the individual control
volumes to obtain Eq2. Note: We have applied the steady operating conditions
assumption and the assumption that ke & pe 0 in each of our energy
balances.
Since we are assuming that air is the working fluid and modeling it as an ideal
gas with constant specific heats at room temperature we can rewrite Eq2,
applying the ideal gas relation for enthalpy difference between two states [e.g. h2
- h1=cp(T2-T1)], as shown in Eq3.
The heat input can be determined from energy balances applied to the individual
control volumes (combustor and reheat combustor), similarly to what we did in
Eq2 & Eq3, to obtain Eq4.
From Eq3 & Eq4 we can see that we must determine the temperatures of
locations 1 through 9 before we can calculate the thermal efficiency.
Location 1
Location 2
We can use the isentropic efficiency of the compressor to determine the
temperature at Location 2 as shown in Eq5.
3
h2 s h1 c p (T2 s T1 ) T T1
c = = T2 = T1 + 2 s (Eq5)
h2 h1 c p (T2 T1 ) c
To find the temperature at Location 2 from Eq5 we must first determine the
temperature at Location 2 if the compression process were isentropic. We can
use an ideal gas relation for isentropic processes to find the temperature at state
2s as shown below.
k 1 0.4
T2 s P2 k 300[kPa] 1.4
= T2 s = 300[K ] = 410.62 K
T1 P1 100[kPa]
Substituting this back into Eq5 we can solve for the temperature at Location 2.
T2 s T1 410.62[ K ] 300[ K ]
T2 = T1 + = 300[ K ] + = 438.3 K
c 0.80
Location 3
T3=300 K (Given compressor 2 inlet conditions)
Location 4
Similarly to Location 2, we can use the isentropic efficiency of the compressor to
determine the temperature at Location 4, as shown in Eq6.
h4 s h3 c p (T4 s T3 ) T T3
c = = T4 = T3 + 4 s (Eq6)
h4 h3 c p (T4 T3 ) c
To find the temperature at Location 4 from Eq6 we must first determine the
temperature at Location 4 if the compression process was isentropic. We can
use the ideal gas relation for isentropic processes to find the temperature at state
4s as shown below.
k 1 0.4
T4 s P4 k 1000[kPa] 1.4
= T4 s = 300[K ] = 423.17 K
T3 P3 300[kPa]
Substituting this back into Eq6 we can solve for the temperature at Location 4.
T4 s T3 423.2[ K ] 300[ K ]
T4 = T3 + = 300[ K ] + = 453.96 K
c 0.80
4
Location 5
To find the temperature at Location 5 we can make use of the regenerators
effectiveness, as shown in Eq7.
Location 6
T6 = 1400 K (Given turbine 1 inlet conditions)
Location 7
We can use the isentropic efficiency of the turbine to determine the temperature
at Location 7, as shown in Eq8.
h6 h7 c p (T6 T7 )
t = = T7 = T6 t (T6 T7 s ) (Eq8)
h6 h7 s c p (T6 T7 s )
To find the temperature at Location 7 from Eq8 we must first determine the
temperature at Location 7 if the expansion process was isentropic. We can use
the ideal gas relation for isentropic processes to find the temperature at state 7s
as shown below.
k 1 0.4
T7 s P7 k 300[kPa] 1.4
= T7 s = 1400[K ] = 992.51 K
T6 P6 1000[kPa]
Substituting this back into Eq8 we can solve for the temperature at Location 7.
Location 8
T8 = 1400 K (Given turbine 2 inlet conditions)
Location 9
We can use the isentropic efficiency of the turbine to determine the temperature
at Location 9 as shown in Eq9.
h h c p (T8 T9 )
t = 8 9 = T9 = T8 t (T8 T9 s ) (Eq9)
h8 h9 s c p (T8 T9 s )
5
To find the temperature at Location 9 from Eq9 we must first determine the
temperature at Location 9 if the expansion process was isentropic. We can use
the ideal gas relation for isentropic processes to find the temperature at state 9s
as shown below.
k 1 0.4
T9 s P9 k 100[kPa] 1.4
= T9 s = 1400[K ] = 1023 K
T8 P8 300[kPa]
Substituting this back into Eq9 we can solve for the temperature at Location 9.
Location 5 contd
We can now go back to Eq7 and solve for the temperature at Location 5.
Our property table is now complete and we can solve for the thermal efficiency.
T (K) P (kPa)
1 300 100
2 438.3 300
2s 410.62 300
3 300 300
4 453.96 1000
4s 423.2 1000
5 969.43 1000
6 1400 1000
7 1074 300
7s 992.51 300
8 1400 300
9 1098.3 100
9s 1023 100
10 100
Substituting Eq3 and Eq4 into Eq1 we can solve for the thermal efficiency.
th =
[(T6 T7 ) + (T8 T9 )] [(T2 T1 ) + (T4 T3 )] = [(326) + (301.7)] [(138.3) + (153.96)]
(T6 T5 ) + (T8 T7 ) (430.57) + (326)
627.7 292.26
th = =0.443 or 44.3% Answer a)
756.57
6
Part b)
The net power developed with be the wnet,out (Eq3) multiplied by the mass flow
rate of air. We are given the mass flow rate of air in the problem statement as
being 5.807 kg/s.
kJ
W net ,out = m air wnet ,out = (5.807[kg / s])1.005 (627.7[ K ] 292.26[K ])
kg K
=1957.6 kW Answer b)
Part c)
The back work ratio is defined as the ratio of compressor work to the turbine
work.
wc ,in 292.26
bwr = = =0.466 or 46.6% Answer c)
wt ,out 627.7