DIESEL
CHIMNEY
POWERPLANT
AND
0.44 = Pi / 0.05(44,000)
Pi = 968 kW
ASSIGNMENT:
1. Determine the indicated mean
effective pressure of an engine in psi
having a brake mean effective
pressure of 750 kPa and 80%
mechanical efficiency.
a. 136 psi*
b. 137 psi
c. 138 psi
d. 140 psi
4. A 750 kW diesel electric plant has a
brake thermal efficiency of 34%. If the
heat generated by fuel is 9,000,000
kJ/hr, what is the generator efficiency?
a. 85.33%
b. 65.88%
c. 75.55%
d. 88.23%*
Solution:
ntb = Pb / mf Qh
Solution:
0.34 = Pb / (9,000,000 / 3600)
nm = Pmb / Pmi
Pb = 850 kW
0.80 = 750 kPa / Pmi
Pmi = 937.5 kPa x (14.7 psi / 101.325
kPa)
Pmi = 136 psi
2. Determine the friction power of an
engine if the frictional torque
developed is 0.3 kilonewton-meter
running at 1200 rpm.
a. 40.6 kW
b. 37.7 kW
c. 36.5 kW
d. 50.3 Kw
nG = 750/850
nG = 88.23%
5. A 16-cylinder V-type diesel engine is
directly coupled to a 5000 kW AC
generator. If generator efficiency is
90%, calculate the brake horsepower
of the engine.
a. 7447 hp*
b. 6468 hp
c. 8542 hp
d. 7665 hp
Solution:
Solution:
Pf = 2πTn
nG = Gen Output / Pb
Pf = 2π(0.3)(1200/60)
0.9 = 5000 / Pb
Pb = 5555.55 kW (1 hp / 0.746 kW)
Pf = 37.70 kW
3. What is the power developed in the
cylinder if indicated thermal efficiency
is 44%, the engine uses 0.05 kg/s fuel
with heating value of 44,000 kJ/kg?
Pb = 7447.12 hp
a. 1000 kW*
b. 775
6. Determine the brake power of the
engine having a brake thermal
efficiency of 35% and uses 25 ͦAPI fuel
consumption of 40 kg/hr.
c. 36.5 kW
d. 588 kW
a. 165.84 kW
b. 173.52 kW*
c. 186.52 kW
d. 160.67 Kw
Solution:
Solution:
nti = Pi / mf Qh
Qh = 41,130 + 139.6 (API)
Qh = 41,130 + 139.6 (25)
w2 = 1.0947 kg/m3
Qh = 44,620 kJ/kg
nm = Pb / Pi
ntb = Pb / mf Qh
0.85 = 500 / Pi
0.35 = Pb / (40/3600) (44,620)
Pi = 588.23 kW
Pb = 173.52 kW
Pf = P i - Pb
7. Determine the specific gravity of fuel
oil having a heating value of 44,899.2
kJ/kg.
Pf = 588.23 - 500 = 88.23 kW
a. 0.90*
b. 0.80
c. 0.877
d. 0.893
(Pi2 / Pi1) = (w2 / w1)
(Pi2 / 588.23) = (1.0947 / 1.1765)
Pi2 = 547.336 kW
Pb2 = 547.336 - 88.23
Solution:
Qh = 41,130 + 139.6 (API)
Pb2 = 459.106 kW
API = 27
9. What is the displacement volume of
300mm x 4400mm, 4-stroke, 1200
rpm, 8-cylinder diesel engine?
SG = 141.5 / (131.5+API)
a. 0.243 m3/s
b.
c. 5.75 m3/s
d. 1.25 m3/s
44,899.2 = 41,130 + 139.6 (API)
2.262
m3/s
SG = 141.5 / (131.5+27)
SG = 0.8927
8. A 500-kW diesel engine operates at
101.3 kPa and 27 ͦC in Manila. If the
engine will operate in Baguio having
93 kPa and 23 ͦC, what new brake
power will develop if mechanical
efficiency is 85%?
a. 600 kW
b. 754 kW
c. 459 kW*
d. 971 Kw
Solution:
w1 = P/RT
w1 = 101.325 / (0.287) (27+273)
w1 = 1.1765 kg/m3
w2 = P/RT
w2 = 93 / (0.287) (23+273)
Solution:
VD = (π/4) D2 L N c
VD = (π/4) (0.3)2 (0.4) (1200/ 2x60) (8)
VD = 2.262 m3/s
10. What is the friction horsepower of
a 300-kW diesel engine having a
mechanical efficiency of 86%?
a. 86.5 hp
b. 87.5 hp
c. 90.5 hp
d. 65.5 hp*
Solution:
nm = Pb / Pi
0.86 = 300 / Pi
Pi = 348.84 kW
Pf = P i - Pb
Pf = 348.84 - 300
Solution:
Pf = 48.84 kW x (1 hp/ 0.746 kW)
mg = ma + mf - mash
Pf = 65.46 hp
A/F = ma / mf
11. Determine the output power of a
diesel power plant if the engine and
generator efficiency is 83% and 95%,
respectively. The engine uses 25 ͦAPI
fuel and has a fuel consumption of
0.08 kg/s.
16 = ma / mf
a. 2795 kW
b. 8642 kW
c. 9753 kW
d. 2815 Kw*
Solution:
Qg = m f Qh
Qg = 0.08 [41,130+139.6(25)]
Qg = 3569.6 kW
Generator Output = 3569.6 (0.83)
(0.95)
ma = 16 mf
mg = 16 mf + mf - 0.1mf
mg = 16.9 mf
mg = 16.9 (2500)
mg = 42,250 kg/hr
14. The gas density of chimney is 0.75
kg/m3 and air density of 1.15 kg/m3. If
the driving pressure is 0.25 kPa,
determine the height of chimney.
a. 54.6 m
b. 63.7 m*
c. 74.6 m
d. 68.5 m
Solution:
Generator Output = 2814.63 Kw
hw = H (da - dg)
12. Determine the piston speed of a
250 mm x 300 mm diesel engine
running at 1200 rpm.
0.25 = H (1.15 - 0.75)
a. 6 m/s
b. 12 m/s*
c. 18 m/s
d. 5 m/s
Solution:
Piston Speed = 2 L N
Piston Speed = 2 (0.30) (1200/60)
Piston Speed = 12 m/s
13. A boiler uses 2500 kg of coal per
hour and air required for the
combustion in 16 kg per kg coal. If ash
loss is 10%, determine the mass of the
gas entering the chimney.
a. 42,250 kg/hr*
b. 78,300 kg/hr
c. 85,452 kg/hr
d. 33,800 kg/hr
H = 63.71 m
15. The actual velocity of gas entering
in a chimney is 8 m/s. The gas
temperature is 25 ͦC and pressure of
98 kPa with a gas constant of 0.287
kJ/kg - K. Determine the chimney
diameter if mass of gas is 50,000
kg/hr.
a. 1.57 m
b. 1.81 m
c. 3.56 m
d. 1.39 m*
Solution:
PgVg = mgRgTg
98(Vg)
=
(25+273)
(50,000/3600)
Vg = 12.12 m3/sec
(0.287)
Vg = A x v
12.12 = (π/4) D2 (8)
D = 1.39 m
HYDRO-ELECTRIC & GAS TURBINE
POWERPLANT.
1. A hydro-electric power plant
consumes 52,650,000 KW-hrs. per
annum. Expected flow is 1665 m³/min
and overall efficiency is 65%. What is
the net head?
a. 30 m
b. 31 m
c. 32 m
d. 34 m*
SOLUTION:
𝐺𝑒𝑛.𝑂𝑢𝑡𝑝𝑢𝑡
Eff.net = 𝑊𝑎𝑡𝑒𝑟𝑃𝑜𝑤𝑒𝑟
0.65 =
52650000/8760
𝑊𝑎𝑡𝑒𝑟𝑃𝑜𝑤𝑒𝑟
Water Power = 9246.575 KW
Water Power = wQh
9246.575 = 9.81(1665/60)
h = 33.966 m / 34 m
2. In a hydro-electric power the tail
water level fixes at 480 m. The net
head is 27 m and head loss is 4% of
the gross head. What is the head
water elevation?
a. 500.34 m
b. 508.12 m*
c. 456.34 m
d. 567.34 m
SOLUTION:
SOLUTION:
hg = H.W. Elev – T.W. Elev.
hg = 195 – 160
hg = 35 m
h = hg + hL
h =35 – (0.05)(35)
h = 33.25 m
h = hg + hL
Water Power = wQh
27 = hg + 0.04hg
Water Power = 9.81(10)(33.25)
hg = H.W. Elev – T.W. Elev.
Water Power = 3261.825 KW
28.125 = H.W. Elev – 480
H.W. Elev = 508.125 m
3. The available flow of water is 25
m³/sec at 30 m elevation. If a hydroelectric plant is to be installed with
turbine efficiency of 85% and
generator efficiency of 90%, what
maximum power that the plant could
generate?
5. The flow of the river is 20 m³/sec
and produced a total brake of 6,000
KW. If it is porposed to install two
turbines each has 85% efficiency, what
is the available head?
a. 35 m
b. 37 m
c. 39 m
d. 36 m*
a. 4658.5 KW
b. 3478.5 KW
SOLUTION:
c. 5628.5 KW*
d. 4758.5 KW
Water Power = 6,000/0.85
SOLUTION:
Water Power = 7058.82 KW
Water Power = wQh
7058.82 = 9.81(20)h
Water Power = 9.81(25)(30)
h = 35.98 m / 36 m
Water Power = 7357.5 KW
6. Two turbines generates a total
brake power of 5000 KW. If one unit is
thrice the capacity of the other, find the
capacity of smaller unit.
Gen. Output = 7357.5 KW (0.85)(0.9)
Gen. Output = 5628.5 KW
4. For a proposed hydro-electric plant,
the tail water and head water elevation
is 160 m and 195 m, respectively. If
available flow is 10 m³/sec and head
loss of 5% of water available head,
what is the water power?
a. 1250 KW*
b. 3450 KW
c. 2456 KW
d. 5763 KW
a. 3261.8 KW*
b. 4254.6 KW
W2 = 3W 1
c. 5874.5 KW
d. 2456.5 KW
5000 = W 1 + 3 W 2
SOLUTION:
WT = W1 +W 2
W1 = 1250 KW
SOLUTION:
7. In a hydro-electric plant the brake
power is 1800 KW running at 450 rpm
and net head of 30 m. Determine the
specific speed of the turbine.
N=
a. 60.29 rpm
b. 65.29 rpm
c. 75.29 rpm
d. 71.29 rpm*
120𝑓
𝑃
300 =
120(60)
𝑃
P = 24 poles
SOLUTION:
Hp = 1800 x 1Hp/0.746 KW
Hp = 2414.87 Hp
Ns =
𝑁√𝐻𝑃
ℎ^5⁄4
Ns =
450√2414.87
(30𝑥3.281)^5⁄4
10. The penstock of hydro-electric
plant is 0.5x0.5 m with the velocity of
5.5 m/sec has a head of 20 m. What is
the output of the turbine if turbine
efficiency os 87%?
a. 845.32 KW
b. 789.34 KW
c. 654.56 KW
d. 234.56 KW*
Ns = 71.29 rpm
SOLUTION:
8. The specific speed of turbine is 75
rpm and running at 450 rpm. If the
head is 20 m and generator efficiency
is 90%, what is the maximum power
delivered by the generator.
Q=Axv
a. 450.5 KW
b. 650.5 KW
Water Power = wQh
c. 650.5 KW*
d. 780.5 KW
Water Power = 9.81(1.375)(20)
75 =
Q = 1.375 m³/sec
Water Power = 269.775 KW
SOLUTION:
Ns =
Q = (0.5 x 0.5)(5.5)
Turbine Output = 269.775(0.87)
𝑁√𝐻𝑃
ℎ^5⁄4
Turbine Output = 234.70 KW
450√𝐻𝑝
(20𝑥3.281)^5⁄4
Hp = 968.92
11. An air standard Brayton cycle has
a pressure ratio of 12. Find the thermal
efficiency of the engine.
Gen. Output = (968.92 x 0.746)(0.9)
a. 34.23%
b. 50.83%*
Gen. Output = 650.53 KW
c. 56.32%
d. 65.23%
9. For a generator running at 300 rpm
and 60 Hz, find the number of
generator poles.
SOLUTION:
a. 24 poles*
c. 18 poles
b. 8 poles
d. 20 poles
e=1e=1-
1
(𝑅𝑝)
𝑘−1
𝑘
1
(12)
1.4−1
1.4
e = 50.83%
12. An air standard Brayton cycle has
an air leaving the high-temperature
heat exchanger at 850℃ and leaving
the turbine at 310℃. What is the
thermal efficiency?
T2
= (8)^
298
1.4−1
1.4
T2 = 539.81°𝐾
T3
T4
P3
=( P4)^
1.4−1
1.4
T3 = 1100 + 273
a. 42.21%
b. 23.34%
T3 = 1373 °𝐾
c. 48.08%
d. 56.34%
1373
SOLUTION:
Wc = m(1)( 539.81 – 298)
T4 = 310 + 237
Wc/m = 241.81 KJ/kg
T4 = 583 °𝐾
= (rp) ^
WT = mCp(T3 – T4)
1.4−1
1123
= (rp) ^
583
1.4
WT = m(1)( 1373 – 757.95)
1.4−1
WT/m = 615.05 KJ/kg
1.4
WNET = 615.05 - 241.81
rp = 9.919
e=1-
1.4
Wc = mCp(T2 - T1)
T3 = 1123 °𝐾
T4
1.4−1
T4 = 757.95°𝐾
T3 = 850 + 273
T3
T4
= (8)^
WNET = 373.24 KJ/kg
1
1.4−1
(9.919)
1.4
e = 48.08%
13. An air standard Brayton cycle has
a pressure ratio of 8. The air properties
at the start of compression are 100
kPa and 25℃. The maximum
allowance temperature is 1100℃.
Determine the net work.
14. The air standard Brayton cycle has
a net power output of 100 kw. The
working substance is air, entering the
compressor at 30℃, leaving the high
temperature heat exchanger at 750℃
and leaving the turbine at 300℃.
Determine the mass flow rate of air.
a. 1698 kg/hr*
b. 1543 kg/hr
d. 2344 kg/hr
a. 373.24 KJ/kg*
b. 283.45 KJ/kg
c. 1543 kg/hr
c. 321.34 KJ/kg
d. 398.23 KJ/kg
SOLUTION:
SOLUTION:
P2
P1
T1 = 303°𝐾
=8
T1 = 25 + 273
P2
=( P1)^
T1
T3 = 750 + 273
T3 = 1023 °𝐾
T1 = 298°𝐾
T2
T1 = 30 + 273
1.4−1
1.4
T4 = 300 + 273
T4 = 573°𝐾
T2
T3
= T4
T1
T2
303
=
1023
573
T2 = 540.96°𝐾
Wc = mCp(T2 - T1)
Wc = m(1)( 540.96 – 303)
1. Mass flow rate of ground water in a
geothermal powerplant is 1,500,000
kg/hr and the quality after the throttling
is 30%. Determine the brake power of
turbine if the change in enthalpy of
steam at inlet and outlet is 700 KJ/kg.
A. 68.5 MW
B. 87.5 MW*
C. 64.5 MW
D. 89.5 MW
Wc = 237.96 m
SOLUTION:
WT = mCp(T3 – T4)
ms = x(mg) = 0.3 (1,500,000)
WT = m(1)( 1023 – 573)
ms = 450,000 kg/hr = 125 kg/sec
WT = 450 m
Wt = ms (h3-h4) = 125(700) = 87,500
KW
WNET = W T + W c
100 = 450 m + 237.96 m
Wt = 87.5 MW
m = 0.4716 kg/s (3600)
m = 1697.79 kg/hr
15. The compressor for an actual gas
turbine requires 300 KJ/kg of work to
quadruple the inlet pressure. The inlet
air temperature is 100℃. Determine
the compressor air exit temperature.
a. 234°𝐾
b. 542°𝐾
c. 653°𝐾
d. 673°𝑲*
SOLUTION:
T1 = 100 + 273
T1 = 373°𝐾
Wc = mCp(T2 - T1)
A. Ground water of geothermal
powerplant has an enthalpy of 700
KJ/kg and turbine inlet is 2750 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?
A. 3.27 kg/s
B. 2.27 kg/s
C. 4.27 kg/s
D. 9.27 kg/s
SOLUTION:
H2 = Hf + X(Hfg-hf)
700 = 500 + X(2750-500)
X = 0.0888
ms = x * mg = 0.0888 * 45 = 4 kg/sec
Wc/m = Cp(T2 - T1)
300 = 1(T2 – 373)
T2 = 673°𝑲
GEOTHERMAL
2. The enthalpy entering the turbine of a
geothermal powerplant is 2750 KJ/kg
and mass rate of 1 kg/sec. The turbine
brake power is 1000 KW condenser
outlet has an enthalpy of 210 KJ/kg. If
the temperature rise of cooling water in
condenser is 8◦C, what is mass of
cooling water requirement?
A. 44 kg/sec
C.46 kg/sec
B. 45 kg/sec
D. 47 kg/sec
SOLUTION:
SOLUTION:
WT = ms(h3-h4) ; (16,000/0.9*0.8) =
ms (500)
Ms = 44.44 kg/sec = 160,000 kg/hr
Wt = ms ( h3-h4); 1000 = 1(2750-h4);
h4 = 1750 KJ/kg
160,000 = 0.20(Mg)
Qr = Qw
Mg = 800,000 kg/hr
1(1750-210) = mw(4.187)(8)
No. of wells = 800,000/200,000
= 4 wells
Mw = 45.97 kg/sec
3. In a 12 MW geothermal powerplant,
the mass flow of steam entering the
turbine is 26 kg/s. The quality after
throttling is 25% and enthalpy of
ground water is 750 KJ/kg. Determine
the overall efficiency of the plant?
A. 7.4 %
B.9.6 %
C.5.4 %
D. 15.4 %*
SOLUTION:
Ms = x Mg
26 = 0.25 Mg
Mg = 104 kg/sec
5. A geothermal powerplant draws a
pressurized water from a 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
turbine efficiency is 85% at a rated
power output of 10 MW. Calculate
overall plant efficiency.
A. 7.29 %
B. 12.34 %
C. 9.34 %
D. 19.45 %
SOLUTION:
%overall = 12,000/(104*750) = 15.38%
@ 1.5 MPa h3 = 2792.2 KJ/kg, s3 =
6.4448
@ 1 atm (100 C) sf = 1.3069, sfg =
6.048, hf = 419.04, hfg = 2257
H4 = 2336.4 KJ/kg
X4 = 0.8495
4. 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
KG/kg.
A. 4 wells*
B. 5 wells
C. 6 wells
D. 8 wells
Wt = ms (h3-h4) %T
10,000 = ms (2792.2 – 2336.4)(0.85)
ms = 25.81 kg/sec
@ 20 MPa and 300 C h1= 1333.3
KJ/kg
@ 1 MPa hf = 844.89, hfg = 1947.3, x2
= 0.25
ms = x2 mg
(25.81 x 3600) = 0.25 (mg)
Mg = 371,664 kg/hr
%t = (10,000)/(371,644/3600)(1333.3)
= 7.26 %
6. A flashed steam geothermal
powerplant 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 Kpa, when it is
condensed. The flow rate from the well
is 29.6 kg/s. Determine the power
produced in KW.
A. 430.13 KW
B. 540.23 KW
C. 370.93 KW
D, 210.34 KW
SOLUTION:
H1 = 697.22 KJ/kg, h3 = 2748.7 KJ/kg,
hf = 640.23, hfg = 2108.5
697.22 = 640.23 + (X) (2108.5)
X2 = 0.027
ms = x mg
ms = 0.027 (29.6) = 0.80 kg/sec
Power produced = 0.80(2748.7 –
2211) = 430.16 KW
7. A 18,000 KW geothermal plant has a
generator efficiency of 90% and 80%
respectively. If the quality after
throttling is 20% and each well
discharges 400,000 kg/hr , determine
the number of wells are required to
produce if the change of enthalpy at
entrance and exit of turbine is 500
KG/kg.
A. 4 wells
wells D. 8 wells
B. 6 wells
C. 2
SOLUTION:
WT = ms(h3-h4) ; (18,000/0.9*0.8) =
ms (500)
Ms = 50 kg/sec = 180,000 kg/hr
180,000 = 0.20(Mg)
Mg = 900,000 kg/hr
No. of wells = 900,000/400,000 = 2.25
wells
8. In an ideal Rankine cycle, the steam
throttle condition is 4.10 MPa and
440◦C. If the turbine exhaust is 0.105
MPa, determine the thermal efficiency
of the cyle.
A. 20.34 %
B. 27.55 % C. 34.44 %
D. 43.12 %
SOLUTION: h1 = 3305.7, s1 = 6.8911,
x1 = 0.925, h2 = 2508.54, h3 = 423.24,
h4 = 427.412
Qa = h1-h4 = 3305.7-427.412 =
2878.29 KJ/kg
Wt = h1-h2 = 3305.7-2508.54 = 797.16
KJ/kg
Wp = h4-h3 = 427.412- 423.24 = 4.172
KJ/kg
Wnet = Wt-Wp = 797.16 -4.172 =
792.99 KJ/kg
%t = Wnet/ Qa = 792.99/2878.29 =
27.55 %
9. In a Rankine cycle, saturated liquid
water @ 1 bar is compressed
isentropically to 150 bar. First by
heating in a boiler, and then by
superheating at constant pressure of
150 bar, the water substance is
brought to 750 K. After adiabatic
reversible expansion in a turbine to 1
bar, it is then cooled in a condenser to
saturated liquid. What is the thermal
efficiency?
A. 23.45 %
B. 16.23 % C. 34.24 %
D. 18.23 %
SOLUTION: h1=3240.5, s1=6.2549,
x=0.8176, h2= 2263.6, h4 = 433, h3=
417.46
Wp = h4=h3 = 433 - 417.46 = 15.54
KJ/kg
Wt = h1-h2 = 3240.5 – 2263.6 = 976.9
KJ/kg
Efficiency = (976.9-15.54)/(3240.5433) = 34.24 %