GEOTHERMAL POWER PLANT
AND
STEAM GENERATOR
From:
Alcorcon and Prime Reviewers
Geothermal Power Plant Problems (Alcorcon Riviewer 1-11)
1. Mass flow rate of ground water in a geothermal power plant is 1,500,000 kg/hr and the quality after
throttling is 30%. Determine the brake power of turbine if the change of enthalpy of steam at inlet and
outlet is 100 KJ/kg.
A. 68.5 MW
C. 87.5 MW
B. 64.5 MW
D. 89.5 MW
Solution:
mS = x mg
= 0.3(1,500,000)
1 ℎ𝑟
1 𝑚𝑖𝑛
= 450,000 kg/hr(60 𝑚𝑖𝑛)( 60 𝑠
)
mS = 125 kg/s
WT = mS (h3 – h4)
= 125(700)
= 87500 KW
WT = 87.5 MW
2. Ground water of geothermal power plant has an enthalpy of 700 KJ/kg and at turbine inlet is 2,750
KJ/kg and enthalpy of hot water in flash tank is 500 KJ/kg. What is the mass of steam flow entering the
turbine if mass flow of ground water is 45 kg/sec?
SOLUTION:
h2 = hf + x(hg – hf)
700 = 500 + x(2750 – 500)
x = 0.0888
ms = x mg
ms = 0.0888(45)
ms = 4 kg/sec (answer)
3. The enthalpy entering the turbine of a geothermal power plant is 2750 kJ/kg and mass rate of 1
kg/sec. The turbine brake power is 1000 kW condenser outlet has enthalpy of 210 kJ/kg. If the
temperature rise of cooling water in condenser is 8°C, what is the mass of cooling water requirement?
Solution:
WT = ms(h3 – h4)
1000 kW = 1 kg/sec (2750 kJ/kg – h4)
h4 = 1750 kJ/kg
Where:
QR = Q W
ms(h4 – h5) = mWCp(t2 – t1)
1 kg/sec(1750 kJ/kg – 1750 kJ/kg) = mW(4.187 kJ/kg-K)( 8°C - 0°C)
mW = 45.97 kg/sec
4. In a 12 MW geothermal power plant, the mass flow of the steam entering the turbine is 26 kg/sec.
The quality after throttling is 25% and enthalpy of ground water is 750 kJ/kg. Determine the overall
efficiency of the plant.
Solution:
ms = xmg
26 = 0.25mg
mg = 104 kg/sec
ɳp = 12000 / [(104)(750)]
ɳp = 15.38%
5. A 16,000 Kw geothermal plant has a generator efficiency and turbine efficiency of 90% and 80%,
respectively. If the quality after throttling is 20% and each well discharges 200 000 kg/hr. Determine the
number of wells are required to produce if the change of enthalpy at entrance and exit of turbine is 500
KJ/Kg.
Solution:
WT = ms (h3-h4)
16,000
= 𝑚𝑠 (500)
0.9(0.8)
ms= 44.44kg/sec.
ms = 160,000kg/hr.
160,000 = 0.20 mg
Mg = 800,000 kg/hr.
No. of wells = 800,000/200,000
No. of wells = 4 wells
6. A geothermal power plant draws pressurized water from a well at 20 Mpa and 300 ⁰C. To produce a
steam waster mixture in the separator, where the unflashed water is removed, this water is throttled to
a pressure of 1.5 Mpa. The flashed steam which is dry and saturated passes through the steam collector
and enters the turbine ar 1.5 Mpa and espands to 1 atm. The Turbine efficiency is 85% a a rated power
plant output of 10 MW. Calculate overall plant efficiency
SOLUTION:
At 1.5 Mpa (table 2)
ℎ3 =2792.2 KJ/kg
𝑠3=6.448
At 1 atm (100 ⁰C)
𝑠𝑡 =1.3069
ℎ𝑓 =419.04
𝑠𝑓𝑔 =6.048
ℎ𝑓𝑔 =2257
𝑠3=𝑠4 =𝑠𝑓 +𝑥4 𝑠𝑓𝑔
6.448=1.3069+𝑥4 6.048
𝑥4 =0.8495
ℎ4 =ℎ𝑓 +xℎ𝑓𝑔
ℎ4 =419.04+0.8495(2257)
ℎ4 =2336.4 KJ/kg
𝑊𝑇 =𝑚𝑠 (ℎ3 − ℎ4 )ƞ𝑇
10000 = 𝑚𝑠 (2792.2-2336.4)(0.85)
𝑚𝑠 =25.81 kg/sec
At 20 Mpa and 300 ⁰C
ℎ1 =1333.3 KJ/kg
At 1.5 Mpa
ℎ𝑓 =844.89
ℎ𝑓𝑔 =1947.3
ℎ1 =ℎ2 = ℎ𝑓 + 𝑥2 ℎ𝑓𝑔
1333.3=844.89+𝑥2 1947.3
𝑥2 =.25
(25.81x3600)=0.25(𝑚𝑔 )
𝑚𝑔 =371,664 kg/hr
ƞ𝑜𝑣𝑒𝑟𝑎𝑙𝑙 =
10000
(
371,664
)(1333.3)
3600
ƞ𝒐𝒗𝒆𝒓𝒂𝒍𝒍 =7.26 %
7. A flashed steam geothermal power plant is located where underground hot water is available as
saturated liquid at 700 kPa. The well head pressure is 600 kPa. The flashed steam enters a turbine at 500
kPa and expands to 15 kPan when it is condensed. The flow rate from the well is 29.6 kg/s. Determine
the power produced in kW.
Solution:
h1 = hf at 0.70 Mpa
h1 = 697.22 kJ/kg
At 500 kPa:
hf = 640.23
hfg = 2108.5
h3 = hg at 0.50 Mpa
h1 = h2 = hf + x2 hfg
697.22 = 640.23 + x2 (2108.5)
X2 = 0.027
ms = xmg
ms = 0.027(29.6)
ms = 0.80 kg/s
From Mollier Diagram:
h4 = 2211 kJ/kg
Power produced = ms (h3-h4)
Power produced = 0.8(2748.7 – 2211)
Power produced = 430.16 kW
8. A flashed steam geothermal power plant is located underground hot water is available as saturated
liquid at 700 Kpa. The well head pressure is 600 Kp. The flashed steam enters a turbine at 500 Kpa and
expands to 15 Kpa, when it is condensed. The flow rate from the well is 29.6 kg/sec. Determine the
cooling water flow in kg/sec if water is available at 30C and a 100C rise is allowed through the
condenser.
SOLUTION
ℎ1 = ℎ𝑓 at 0.70 Mpa
𝑘𝐽
ℎ𝑓 = 697.22 𝑘𝑔
At 500 Kpa:
ℎ𝑓 = 640.23
ℎ𝑓𝑔 = 2108.5
ℎ3 = ℎ𝑔 at 0.50 Mpa
ℎ3 = 2748.7
𝑘𝐽
𝑘𝑔
ℎ1 = ℎ2 = ℎ𝑓 + 𝑥2 ℎ𝑓𝑔
697.22 = 640.23 + 𝑥2 (2108.5)
𝑥2 = 0.027
𝑚𝑠 = 𝑥𝑚𝑔
𝑚𝑠 = 0.027(29.6)
𝑘𝑔
𝑚𝑠 = 0.80 𝑠𝑒𝑐
ℎ5 = ℎ𝑓 at 15 kpa (0.015 Mpa)
ℎ5 = 225.94
𝑘𝐽
𝑘𝑔
𝑚𝑠 (ℎ4 − ℎ5 ) = 𝑚𝑤 𝐶𝑝 (∆𝑡)
0.8(2211-225.94) = 𝑚𝑤 (4.187)(10)
𝒌𝒈
𝒎𝒘 = 𝟑𝟕. 𝟗𝟑 𝒔𝒆𝒄 (answer)
9. In a certain geothermal area, studies show that 1,500,000 kg/hr of pressurized ground water is
available at 2500 psia and 60 ℉. The water will be throttled to 250 psia to produce wet steam and this
mixture will be passed through a water separator to remove the water droplets so that the saturated
steam at 250 psia is available at the entrance of steam turbine for the propose of power plant.other
power plant data are as follows.
Discharge pressure of Turbine = 25 in Hg vacuum
Turbine engine efficiency = 75%
Mechanical loss = 1.5% of shaft power
Generator efficiency = 97%
Assume atmospheric pressure to be 30 in Hg. Determine the maximum amount of power in kilowatts
that the plant can generate.
Solution:
At 250 psi (17.232 MPa) and 60 ℉ (326.667℃)
h1 = 1490.9 KJ/kg
At 250 psi (1.7232 MPa)
hf = 875 KJ/kg
hfg = 1921
h3 = hg at 1.7232 MPa
h3 = 2796 KJ/kg
h1 = h2 = hf + x2 hfg
1490.9 = 875 + x2 (1921)
X2 = 0.3206
m s = x2 m g
ms = o.3206(1,500,00)
ms = 480,915 kg/hr
P4 = (30 - 25) x
0.101325/29.97
P4 = 0.016932 Mpa
From Mollier diagram:
h4 = 2085 KJ/kg
Shaft Power = (480,915/3600) (2796 - 2085) (1 – 0.015) (0.75)
Shaft Power = 70,167 kW
Plant Output = 70,167 (0.062)
Plant Output = 68,062 kW
Plant Output = 68,062 MW
10. a geothermal power plant has an output of 16,000 kW, and combined efficiency of 80% the
pressurized ground water at 175 bar and 280°C leaves the wells to enter the flash chamber maintained
at 14 bar. The steam collected enters the turbine at 14 bar and exhaust at 1 atm. If one well discharges
180,000 kg/hr of hot water, how many wells are required?
a. 2
c. 3
b. 4
d. 5
sol:
h₃ = hg at 14 bar ( 1.4 Mpa )
h₃ = 2790 KJ/kg
S₃ = Sg at 1.4 Mpa
S3= 6. 4693
1 atm( 100°C )
Sc = 1. 3069
ht = 419.04
Sfg = 5.0480
hg = 2257
S3 = Sc = S4 + X2sfg
6. 4693 = 1. 3069 + X4( 6.0480 )
x4 = 0.8536
H4 = 419.04 + 0.8536( 2257 )
H4 = 2345.55 Kj/kg
18,000.80 = Mg( h3 – h4 )
20,000 = Mg( 2790 – 2345.55 )
= 45 kg/sec
At 1.4 Mpa
h1 =830.30
hg = 1959.7
at 175 bar( 17.5 Mpa ) and 280°C ( table 4 )
h1 = 1231 KJ/kg
h1 = h2 = hf + X2hfg
1231 = 830.30 + X2( 1959.7 )
X2 = 0.2046
Mg = X2Mg
( 45 x 3600 ) = 0.2046Mg
Mg = 791699.25 kg/hr
No. of wells = 791699.25/180,000
No. of wells = 4.3984
No. of wells 5 wells
11. A liquid dominated geothermal plant with a single flash separator receives water at 204˚C. The
separator pressure is 1.04 Mpa. A direct contact condenser operates at 0.034 Mpa. The turbine has a
polytropic efficiency of 0.75. For a cycle output of 50 MW, what is the mass flow rate of the well-water
in kg/s
At 204˚C:
ℎ𝑓 = 870.51 KJ/kg
At 1.04 Mpa:
ℎ𝑓 = 770.38
ℎ𝑔 = 2779.6
ℎ𝑓𝑔 = 2009.2
𝑠𝑔 = 6.5729
At 0.034 Mpa:
ℎ𝑓 = 301.40
ℎ𝑓𝑔 = 2328.8
𝑠𝑓 = 0.9793
𝑠𝑓𝑔 = 6.7463
SOLUTION:
ℎ3 = ℎ𝑔 at 1.04 Mpa
ℎ3 = 2779.6 KJ/kg
Solving for ℎ4 :
𝑠3 = 𝑠4 = 𝑠𝑓 + 𝑥𝑠𝑓𝑔
6.5729 = 0.9793 + 𝑥4 (6.7463)
𝑥4 = 0.829
ℎ4 = 301.4 + 0.829(2328.8)
ℎ4 = 2232.3 KJ/kg
𝑊𝑇 = 𝑚𝑠 (ℎ3 − ℎ4 )
50,000 = 𝑚𝑠 (2779.6-2232.3)0.75
𝑚𝑠 = 121.8 kg/sec
Solving for 𝑥2 : (ℎ1 = ℎ2 )
ℎ1 = ℎ2 = ℎ𝑓 + 𝑥ℎ𝑓𝑔
870.51 = 770.38 + 𝑥2 (2009.2)
𝑥2 = 0.049836
𝑚𝑠 = 𝑥𝑚𝑔
121.8 = 0.049836𝑚𝑔
𝒎𝒈 = 2,444 kg/sec
Geothermal Power Plant Problems (Prime Riviewer 1-17)
1. A liquid dominated geothermal plant with a single flash separator receives water at 204 oC. The
separator pressure is 1.04 MPa. A direct contact condenser operates at 0.034 MPa. The turbine has a
polytropic efficiency of 0.75. For a cycle output of 50 MW, what is the mass flow rate of well-water in
kg/s ?
Solution:
At 204 oC :
hf = 870.51 kJ/kg
At 1.04 MPa :
hf = 770.38 kJ/kg
hfg = 2009.2 kJ/kg
hg = 2779.6 kJ/kg
Sg = 6.5729 kJ/kg
At 0.034 MPa :
hf = 301.40 kJ/kg
hfg = 2328.8 kJ/kg
Sf = 0.9793 kJ/kg
Sfg = 6.7463 kJ/kg
h3 = hg @ 1.04 MPa
= 2779.6 kJ/kg
Solving for h4 :
S3 = S4
= (Sf + x Sfg)4
6.5729 = [0.9793 + x4(6.74673)]
X4 = 0.829
h = hf + xhfg
h4 = 301.4 + 0.829 (2328.8)
= 2232.3 kJ/kg
Solving for the mass flow rate to the turbine, mS :
WT = mS (h3 – h4) eT
50, 000 = mS (2779.6 – 2232.2) (0.75)
mS = 121.8 kg/s
Solving for the quality x2 (after throttling)
h2 = h1
h1 = (hf + xhfg)2
870.51 = 770.38 + x2 (2009.2)
X2 = 0.049836
Solving for the mass flow rate of the well-water:
mS = x2 mg
121.8 = 0.049836 mg
mg = 2,443 kg/s
2. A geothermal power plant draws a pressurized water from well at 20 Mpa and 300°C. To produce a
steam-water mixture in the separator, where the unflashed water is removed, this water is throttled to
a pressure of 1.5 Mpa. The flashed steam which is dry and saturated passes through the steam collector
and enters the turbine at 1.5 Mpa and expands to 1 atm. The tubine efficiency is 85% at a rated power
output of 10 MW. Calculate the over-all plant efficiency.
Required:
eplant = ?
Solution:
At 1.5 Mpa
h3 = 2792.2 kJ/kg
S3 = 6.4448 kJ/kg.K
At 1 atm, 100°C
hf = 419.04 kJ/kg
hfg = 2257 kJ/kg
Sf = 1.3069 kJ/kg
Sfg = 6.048 kJ/kg.K
S3 = S4
S3 = Sf + x4Sfg
6.4448 = 1.3069 + x4(6.0480)
h4 = hf + x4hfg
= 419.04 + (0.8495)(2257)
= 2336.40 kJ/kg
Wt = ms (h3 – h4) et
10,000 = ms (2792.2 – 2336.4) 0.85
ms = 25.81 kg/s
At 20 Mpa, 300°C:
h1 = 1333.30 kJ/kg
At 1.5 Mpa
hf = 844.89 kJ/kg
hfg = 1947.3 kJ/kg
h1 = h2 = hf + x2hfg
1333.3 = 844.9 + x2 (1947.3)
X2 = 0.25
ms = x2mg
25.81 = 0.25 mg
mg = 103.24 kg/s
then;
𝑒𝑝𝑙𝑎𝑛𝑡 =
𝑡𝑢𝑟𝑏𝑖𝑛𝑒 𝑜𝑢𝑡𝑝𝑢𝑡
𝑚𝑔 ℎ1
=
10,000
103.24 (1333.33)
thus;
𝒆𝒑𝒍𝒂𝒏𝒕 = 𝟎.𝟎𝟕𝟐𝟔 𝒐𝒓 𝟕.𝟐𝟔%
3. A flashed steam geothermal power plant is located when underground hot water is available as
saturated liquid at 700 kPa. the well head pressure is 600 kPa. The flashed steam enters a turbine at 500
kPa and expands to 15 kPa, when it is condensed. The flow rate from the well is 29.6 kg/s. Determind
the power produced in kW.
Solution:
h1 = hf at 0.70 Mpa
= 697.22 kj/kg-K
At 500 kPa:
hf = 640.23 kj/kg
hfg = 2108.5 kj/kg
h3 = hg @ 0.50 Mpa
h3 = 2748.7 kj/kg
h1 = h2 = hf + x2hfg
697.22 = 640.23 + x2(2108.5)
X2 = 0.027
ms = 0.027 mg
= 0.027(29.6)
= 0.80 kg/s
From mollier chart :
h4 = 2,211 kj/kg
then;
power = ms (h3-h4)
= 0.80(2748.7-2211)
Thus,
Power = 430.16 Kw
4. In a certain geothermal area, studies show that 1,500,000 kg/hr of pressurized ground water is
available at 2500 psia and 620 °F. The water will be throttled to 250 psia to produce wet steam and this
mixture will be passed through a water separator to remove the water droplets so that saturated steam
at 250 psia is available at the entrance of steam turbine for the proposed power plant. Other data are as
follows:
Discharge pressure of turbine = 25 in. hg vacuum
Turbine Engine efficiency
= 75%
Machine loss
= 1.5% of shaft power
Generator efficiency
= 97%
Assume atmospheric pressure to be 30 in. hg. Determine the maximum amount of power in kw that the
plant can generate.
Solution:
At 2500 psi (17.232 mpa) and 620 °F (326.667 °C)
h1 = 1490.0 kj/kg
At 250 psi (1.7232 mpa)
hf = 875 kj/kg
hfg = 1921 kj/kg
h3 = hg @ 1.7232 mpa
= 2796 kj/kg
h1 = hf + x2hfg
1490.9 = 875 + x2 (1921)
x2 = 0.3206
ms = x2mg
= 0.3206(1,500,000)
= 480,915 kg/hr
P4 = 30 – 25 = 5 in. Hg
= 0.16932 mpa
From Mollier Chart:
h4 = 2085 kj/kg
Then;
480,915
)(2796
3600
Shaft Power = (
– 2085)(1 – 0.015)(0.75)
= 70,167 KW
Thus;
Plant Output = 70,167 KW (0.97)
= 68,062 KW
Plat Output = 68.06 MW
5. A flashed steam geothermal plant has pressurized underground water available at 102 kg/𝑐𝑚2 and
240C. In order to produce steam-water mixture, the ground water is passed and throttled to 5.4
kg/𝑐𝑚2 in a steam separator. The dry steam produced in the separator is fed to double flow impulse and
reaction turbine with guaranteed engine efficiency of 85%. The turbine is directly coupled to a 3 phase,
60 Hz, 80% power factor, 13800-V air cooled generator. Exhaust is to be direct-contact spray type main
condenser designed to operate a vacuum of 647.5 mm of Hg. Generator efficiency is 95%, ground water
is 459 kg/s. What is the enthalpy of steam-water mixture in the steam separator?
SOLUTION
At 𝑃1 = 102
𝑘𝑔
𝑐𝑚2
(10 𝑀𝑝𝑎)𝑎𝑛𝑑 𝑡1 = 240°𝐶
Thus; from table,
𝒉𝟏 = 𝟏𝟎𝟑𝟖. 𝟏𝟎 𝒌𝑱/𝒌𝒈
6. Based on the data in problem No. 5, what is the enthalpy of steam entering the turbine?
Solution:
At P3 = P2 = 5.4 kg/cm2 (0.53 Mpa)
h3 = hg = 2751.3 kJ/kg
thus;
hentering = 2751.3 kJ/kg
7. A flashed steam geothermal plant has pressurized under ground water available at 102 kg/cm2 and
240ºC. In order to produce steam-water mixture, the ground water is passed and throttled to 5.4 kg/cm²
in a steam separator. The dry steam produced in the separator is fed to double flow impulse and
reaction turbine with guaranteed engine efficiency of 85%. The turbine is directly coupled to a 3 phase,
60 Hz, 80% power factor, 13800-V air cooled generator. Exhaust is to be direct-contact spray type main
condenser designed to operate a vacuum of 647.5 mmHg. Generator efficiency is 95%, ground water is
459 kg/s.
Base on data in problem, what is the enthalpy of the exhaust steam.
A. 2876.68 kJ/kg
B. 1874.67 kJ/kg
Solution:
At P4 = - 647.5 mmHg + 760 mmHg
= 112.5 mmHg = 0.015 Mpa
hf = 225.94 kJ/kg
hfg = 2373.1 kJ/kg
Sf = 0.7549 kJ/kg
Sfg = 7.2536 kJ/kg.K
At P3 = P2 = 0.53 Mpa
S1 = Sg = 6.8017 kJ/kg
S3 = S4 = Sf + xSfg
6.8017 = 0.7549 + x4(7.2536)
x4 = 0.8336
C. 2751.37 kJ/kg
D. 2204.15 kJ/kg
h4 = hf + x4hfg
= 225.94 + 0.8336 (2373.1)
Thus,
h4 = 2204.15 kJ/kg
8. Based on the data in problem No. 5, what is the mass flow rate of steam entering the turbine?
Solution:
At P2 = 5.4 kg/cm2 = 0.52 Mpa
hf = 649.78 kJ/kg
hfg = 2101.5 kJ/kg
h1 = h2 = hf + xhfg
x2 = 0.1848
then,
ms = x2mg
ms = 0.1848(459.33)
thus;
ms = 84.88 kg/s
9. Based on the date in problem No. 5, what is the mass flow rate of steam entering the turbine?
A. 84.88 kg/s
C. 94.88 kg/s
B. 74.88 kg/s
D. 104.88 kg/s
Solutions:
At P2 = 5.4 kg/cm2 = 0.53 Mpa
hf = 649.78 kJ/kg
hfg = 2101.5 kJ/kg
h1 = h2 = hf + hfg
1038.1 = 649.78 + x2 (2101.5)
x2 = 0.1848
then;
m s = x2 m g
= 0.1848 (459.33)
thus;
ms = 84.88 kg/s
10. Based on the data in problem no. 5, what is the apparent power developed by the generator in KVA?
Solution:
Apparent Power =
Power outpy
Power Factor
Apparent Power =
37501.98
0.80
thus;
Apparent Power =46,877.475 KVA
11. A liquid dominated geothermal plant with a single flash separator receives water at 204°C. The
separator pressure is 1.04 Mpa. A direct contract condenser operates at 0.034 Mpa. The turbine has a
polytrophic efficiency of 0.75. For a cycle output of 50MW, what is the mass flow rate of the well-water
in kg/s?
A. 2871
B. 2100
C. 186
D. 2444
Solutions:
At 204 °C
ht = 870.51 KJ/kg
at 1.04 Mpa:
ht = 770.38
hg = 2779.6
hfg = 2009.2
Sg = 6.5729
At 0.034 Mpa:
ht = 301.40
St = 0.9793
Hfg = 2328.8
Sfg = 6.7463
Solutions:
h3 = hg at 1.04 Mpa
h3 = 2779.6 KJ/kg
Solving for h4:
S3 = S4 = Sf + XSfg
6.5729 = 0.9793 + X4 (6.7463)
x4 = 0.829
h4 = 301.4 + 0.829(2328.8)
h4 = 2232.3 KJ/kg
Wt = ms (h3-h4)
50,000 = ms(2779.6-2232.3)0.75
ms=121.8 kg/sec
Solving for x2: (h1 =h2)
h1 = h2 = hf + xhfg
870.51 = 770.38 + x2(2009.2)
x2 = 0.049836
ms = xmg
121.8 = 0.049836mg
mg = 2,444 kh/sec
12. Based on the data in problem 5, how many production wells are needed to supply the plant with
pressurized water if each well has a capacity of 127,200 kg/hr?
Solution:
No. of wells =
=
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑟𝑜𝑢𝑛𝑑 𝑤𝑎𝑡𝑒𝑟
𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑤𝑒𝑙𝑙
459.44(3600)𝑘𝑔/ℎ𝑟
127,200 𝑘𝑔/ℎ𝑟
Thus;
No. of wells = 13 wells
13. Based on the data in problem no. 5, determine the thermal efficiency of the geothermal plant?
A. 9.88%
C. 8.69%
B. 9.57%
D. 7.86%
Solution:
EPLANT
=
=
Turbine Output
Mfh1
37501.98
459.33×1038.1
Thus;
Eplant
=
0.0788 or 7.86%
14. In a geothermal power plant, the mass flow rate of ground water is 400 kg/s and the quality after
throttling is 20%. If the turbine power is 80 MW, what is the change in enthalpy of steam at the inlet and
outlet of the turbine?
Solution:
Wt = ms(h3 – h4)
Wt = ms∆h
Solving for ms
ms = xmg
ms = 0.20(400)
ms = 80 kg/s
thus,
80,000 = 80 ∆h
∆h = 1000 kJ/kg (answer)
15. The mass flow rate of ground water in a 20 MW geothermal power plant is 150 kg/sec and its
enthalpy is 1000 kJ/kg. If the quality after throttling is 28% and the over-all plant efficiency is 20%, what
is the flow rate of steam entering the turbine?
Solution:
𝑚𝑠 = 𝑥𝑚𝑔
Solving for 𝑚𝑔 :
𝑒𝑝𝑙𝑎𝑛𝑡 =
𝑡𝑢𝑟𝑏𝑖𝑛𝑒 𝑜𝑢𝑡𝑝𝑢𝑡
𝑚𝑔 ℎ1
0.20 =
20000
𝑚𝑔 (1000)
𝑚𝑔 = 100 𝑘𝑔/ sec
𝑚𝑠 = 0.28(100)
Thus;
𝒎𝒔 = 𝟐𝟖 𝒌𝒈/𝒔𝒆𝒄
16. The turbine power of a geothermal power plant is 2 MW and the enthalpy at the inlet of the turbine
is 2500 kJ/kg. Steam flows at the rate of 2.5 kg/s. if the enthalpy at the outlet of the condenser is 300
kJ/kg and the temperature rise of the cooling water is 10oC, what is the volume flow rate of the cooling
water?
SOLUTION:
𝑉𝑤 =
𝑚𝑤
𝜌𝑤
Solving for h4 :
Wt = ms (h3 – h4)
2000 = 2.5 (2500 – h4)
h4 = 1700 kJ/kg
Solving for mw :
Qw = QR
mw ( 4.187 )( 10 ) = 2.5 ( 1700 – 300 )
mw = 83.59 kg/s
𝑉𝑤 =
Thus;
83.59
= 0.08359 𝑚3 ⁄𝑠
1000
Vw = 300.924 m3/hr
17. In the geothermal power plant, the enthalpies of the ground water and the turbine inlet are
1500kJ/kg and 3500kJ/kg respectively. If the enthalpy of the hot water in the flash tank is 700kJ/kg and
the mass flow rate of he steam is 15kg/s, what is the mass flow rate of ground water?
𝑚𝑠 = 𝑥𝑚𝑔
Solving for x:
ℎ2 = ℎ𝑓 + 𝑥ℎ𝑔
1500 = 700 + 𝑥(3500 − 700)
X = 0.2857𝑚𝑔
Then;
15 = 0.2857𝑚𝑔
𝒎𝒈 = 𝟓𝟐. 𝟓 𝒌𝒈/𝒔
Steam Generator Problems (Prime Reviewer 1-14)
1. A steam boiler on test generates 885,000lb of steam in a 4 hr period. The average steam pressure is
400 psia, the average steam temperature is 700 F, and the average temperature of the feed water
supplied to the boiler is 280 F. If the boiler efficiency for the period is 82. 5% and if the coal has a
heating values of 13850 Btu/lb as fired, find the average amount of coal burned in short tons per hour.
A. 9.84 short tons per hour
B. 10.75 short tons per hour
C. 12.05 short tons per hour
D. 11.45 short tons per hour
Solution:
At 400 psia and 700 F
h2= 1362. Btu/lb
At 280 F
h1= 249.1 Btu/lb
ebo 
ms (h2  h1 )
m f Qh
885,000
(1362.7  249.1)
4
0.825 
m f (13850)
lb 1shortton
(
)
hr 2000lbs
shortton
m f  10.78
hr
m f  21,653
2. A boiler operating at 11kg/cm² is required to generate a minimum of 50,000 kg/hr of saturated steam.
Feedwater enters the boiler at 80 C. The furnace is designed to fire coal at an average rate of 4,800
kg/hr and boiler efficiency is 85%. Compute the developed boiler horsepower.
SOLUTION:
At 11 kg/cm² (1.0971 MPa)
hS = hg =2781 kJ/kg
At 80 C
hf = 334.91 kJ/kg
Developed boiler Hp =
=
𝑚𝑠(ℎ𝑠−ℎ𝑓)
35,322
50,000(2781−334.91)
35,322
Thus;
Developed Boiler Hp = 3462.56 Hp
3. A water tube boiler evaporated 5.05 kg of water per second from a feed water temperature of 104.44
o
C to steam at 1241.1 Kpa and quality of 0.97 ; weight of coal fuel per second 0.57 kg ; higher value of
coal as fired 11,800 Btu/ lb . Determine the rate of heat absorption in Kj/kg the boiler horsepower , and
the efficiency of steam generating units .
A.
B.
C.
D.
11,558 Kj/ kg , 1178 Bo. Hp , 73.87 %
12,345 Kj/kg , 1234 Bo. Hp , 84.35 %
14,567 Kj/ Kg , 1045 Bo. Hp . 67.35 %
10,275 Kj/ kg , 1475 Bo. HP , 83.25 %
SOLUTION :
At 104.44 OC
H1 = 437.78 Kj/ kg ( interpolated)
At 1241.1 Kpa
Hf = 805.29 Kj/ kg ( interpolated)
Hfg = 1980.6 Kj/ kg ( interpolated )
H2 = 805.29 + .097 (1980 .6 )
= 2726.5 Kj/kg
Rate of Heat Absorption :
= ms ( h2 – h1 )
= 5.05 ( 2726.5 -437.28 )
= 11,558 Kj/s
Boiler Horsepower :
=
=
𝑚𝑠 ( 3600 )( 𝐻2−𝐻1)
35,322
5.05 ( 3600 ) ( 2726.5−437.78)
35,322
= 1,178 Bo. Hp
Boiler Efficency :
𝑚𝑠 ( 𝐻2 – 𝐻1 )
Ebo =
𝑚𝑓𝑄ℎ
=
11,558
0.57 ( 27,450 )
Thus;
Ebo = 0.7387 or 73.87 %
4. A waste heat recovery boiler produces 4.8 MPa (dry saturated) steam from 104 degrees Celsius feed
water. The boiler receives energy from 5 kg/s of 954 degrees Celsius dry air. After passing through the
waste heat boiler, the temperature of the air has been reduced to 343 degrees Celsius. How much
steam is produced in kg per second?
Note: At 4.8 MPa dry saturated, h=2796 kJ/kg
Solution:
𝐻𝑒𝑎𝑡 𝐿𝑜𝑠𝑠 = 𝐻𝑒𝑎𝑡 𝐺𝑎𝑖𝑛
𝑀𝑔. 𝐶𝑔 (𝑇1 − 𝑇2) = 𝑀𝑠(ℎ𝑠 − ℎ𝑓)
ℎ𝑠 = ℎ𝑔 @ 𝑇𝑎𝑏𝑙𝑒 2. 𝑆𝑎𝑡𝑢𝑟𝑎𝑡𝑖𝑜𝑛: 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒𝑠; 4.8 𝑀𝑝𝑎
ℎ𝑠 = 2796
𝑘𝐽
𝑘𝑔. °𝐾
ℎ𝑓 = ℎ𝑓 @ 𝑇𝑎𝑏𝑙𝑒 1. 𝑆𝑎𝑡𝑢𝑟𝑎𝑡𝑖𝑜𝑛: 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒𝑠; 104°𝐶
ℎ𝑓 = 435.92
𝑘𝐽
𝑘𝑔. °𝐾
therefore Mass of steam;
5𝑘𝑔 1𝑘𝐽
(1227 − 616)°𝐾
𝑠 . 𝑘𝑔. °𝐾 .
𝑀𝑠 =
(2796 − 435.92)𝑘𝐽
𝑘𝑔
𝐌𝐬 = 𝟏. 𝟐𝟗
𝐤𝐠
𝐬
5. A steam expands adiabatically in a turbine from 2000 kPa, 400 °C to 400 kPa, 250 °C. What is the
effectiveness of the process in percent assuming an atmospheric temperature of 15 °C. Neglect changes
in kinetic and potential energy.
Steam properties:
At 2000 kPa and 400 °C
At 400 kPa and 250 °C
h = 3247.6 kJ/kg
h = 2964.2 kJ/kg
S = 7.1271 kJ/kg.K
S = 7.3789 kJ/g.K
A. 79.62%
C. 82.45%
B. 84.52%
D. 74.57%
Solution:
Let; E = Effectiveness
Q
E = Q+Qs
Solving for Q :
Q = h1 – h2
= 3247.6 – 2964.2
= 283. 4 kJ/kg
Solving Qs :
Qs = T ( S2 – S1)
= ( 15 + 273 ) ( 7.3789 – 7.127 )
= 72.55 Kj/kg
Then ;
283.4
E = 283.4+72.55
= 0.7962
Thus ;
E = 79.62 %
6. A water tube boiler has a capacity of 1000 kg per hour of steam .The factor of evaporation is 1.3,
boiler rating is 200 % , boiler efficiency is 65%, heating surface area is 0.91 square meter per boiler
horsepower , and the heating value of fuel is 18,400 kCal per kg .The total coal available in the bunker is
50,000 kg .Determine the no. of hours to consume the available fuel.
Let, t = no.of hours to consume the available fuel.
t=
50,000
𝑚𝑓
solving for mass of fuel , mf
ℎ𝑠−ℎ𝑓
F.E = 2257
hs-hf = 2257 FE
= 2257(1.3)
E bo.=
𝑚𝑠(ℎ𝑠−ℎ𝑓)
𝑚𝑓(𝑄ℎ)
1000(2257)(1.3)
0.65 =(𝑚𝑓)(18,400)(4.187)
mf = 58.592
𝑘𝑔
ℎ𝑟
then,
t=
50,000
58.592
t = 853.36 hrs
7. Two boilers are operating steadily on 91,000 kg of coal contained in a bunker. One boiler is producing
1591 kg of steam per hour at 1.2 factor of evaporation and an efficiency of 65% and another boiler
produced 1364 kg of steam per hour at 1.15 factor of evaporation and an efficiency of 60%. How many
hrs will the coal in the bunker run the boilers if the heating value of coal is 7,590 kCal/kg?
A. 230.80 hrs
C. 350.35 hrs
B. 280.54 hrs
D. 300.54 hrs
Solution:
Let: t = total of number of hours the coal in the bunker run the boilers
t=
91,000
𝑚𝑡
Solving for the total fuel consumed, mt
For Boiler 1:
ebo.1 =
𝑚𝑠1(ℎ𝑠−ℎ𝑓)
𝑚𝑓1𝑄ℎ
ebo.1 =
𝑚𝑠1[𝐹𝐸1(2257)]
𝑚𝑓1𝑄ℎ
0.65 =
1591[1.2(2257)]
𝑚𝑓1[7590(4.187)]
mf1 = 208.605 kg/hr
For Boiler 2:
ebo.2 =
𝑚𝑠2[𝐹𝐸2(2257)]
𝑚𝑓2𝑄ℎ
0.60 =
1364[1.15(2257)]
𝑚𝑓2[7590(4.187)]
mf2 = 185.673 kg/hr
then;
mt = m1 + m 2
mt = 208.605 + 185.673
= 394.278 kg/hr
Thus;
t=
91,000 𝑘𝑔
𝑘𝑔
ℎ𝑟
394.278
t = 230.80 hrs.
8. Coal with a higher heating value of 6700 kCal/kg is consumed at the rate of 600 kg/hr in a steam
generator with a rated boiler horsepower of 200. The feed water temperature is 82 °C and steam
generated is at 10.2 kc/cm2 abs. saturated condition. The horsepower developed is 305. What is the
heating surface area?
SOLUTION
H.S = Rated Boiler Horsepower x 0.91
= 200 x 0.91
H.S = 182 m2
9. Find the rated boiler horsepower of a H.R.T. Boiler 60 in. in diameter 16 ft. long, and having seventy
and 3 in. O.D. tubes with 0.109 in. wall?
A. 95.09 Bo. Hp
B. 78.09 Bo. Hp
C. 89.09 Bo. Hp
D. 93.05 Bo. Hp
Solution:
Rated Bo. Hp =
𝐻𝑒𝑎𝑡𝑖𝑛𝑔𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑖𝑛 𝑓𝑡2
10
Solving for the heating surface Area, H.S. :
1
H.S. = 2 𝜋𝐷𝐿 + 𝜋𝑑𝑖𝐿𝑛 +
2
3
𝜋𝐷2
𝜋𝑑𝑜2𝑛
)− 4
4
(
Where:
D = 60 in. = 5 ft
do = 3 in. = 0.25 ft
L = 16 ft
t = 0.109 in. = 0.00908 ft
n = 70 pcs
di = 0.25 – 2(0.0098) = 0.2318 ft
H.S. =
1
2
𝜋(5)(16) + 𝜋(0.2318)(16)(70) +
2
3
(
𝜋(5)2
𝜋(0.25)2 (70)
)−
4
4
H.S. = 950.93 ft2
Then;
Rated Bo. Hp. =
𝟗𝟓𝟎.𝟗𝟑
𝟏𝟎
= 𝟗𝟓. 𝟎𝟗 𝑩𝒐. 𝑯𝒑
10. What is the heating surface area of a water tube boiler if the equivalent
rated boiler horsepower is 200?
A. 182
C. 450
B. 206
D. 198
Solution:
𝑅𝑎𝑡𝑒𝑑 𝐵𝑜. 𝐻𝑝. =
𝐻𝑆 𝑖𝑛 𝑚2
0.91
𝐻𝑆 = (200)(0.91)
thus,
𝑯𝑺 = 𝟏𝟖𝟐𝒎𝟐
11. The heating surface area of a fire tube boiler is 400 m2. What is the equivalent rated boiler
horsepower?
Solution:
Rated Bo. Hp =
HS in m2
1.1
=
400
1.1
Thus,
Rated Bo. Hp = 363.64
12. What would be the overall efficiency of a steam generator obtaining an evaporation equivalent to
10.95 kg of water from and at 100c per kg of coal containing 34,064 kj/kg heating value.
SOLUTION :
Ebo =
𝑀𝑠 ( 𝐻𝑠−𝐻𝑓)
𝑀𝑓𝑄ℎ
Where :
𝑀𝑠
𝑘𝑔 𝑜𝑓 𝑠𝑡𝑒𝑎𝑚
= 10.95
𝑀𝑓
𝑘𝑔 𝑐𝑜𝑎𝑙
Then,
Ebo =
10.95 (2627.1−419.04)
34,064
Thus,
Ebo= 0.73 or 73%
13. For a generation of dry and saturated steam at 1.0 MPa absolute, what is the percentage gain in
heat when the boiler feed-water is heated from 30ᵒC?
Solution:
At 1.0 MPa (dry and saturated steam)
hs = hg = 2778.1 kJ/kg
@ 30ᵒC (water)
hf1 = 125.79 kJ/kg
@ 90ᵒC (water)
hf2 = 376.92 kJ/kg
% Savings =
𝑄1−𝑄2
𝑄1
𝑋100
For Q1 and Q2
Q1 = hs - hf1
Q1 = 2778.1 – 125.79
Q1 = 2652.3 kJ/kg
Q2 = hs - hf2
Q2 = 2778.1-376.92
Q2 = 2401.18 kJ/kg
% Savings =
2652.3−2401.18
𝑋100
2652.3
% Savings = 9.47%
14. What is the percent rating of water tube boiler if the heating surface area is 400 m2 and the
developed boiler horsepower is 750?
A. 170.625%
B. 140.675%
C. 130.625%
D. 120.765%
Solution:
Percent Rating=
DevelopedBo.Hp
100
RatedBo.Hp
Percent Rating=
750
 100
400 / 0.91
Percent Rating= 170.625%