Power Plant Engineering

3
 MT.
Example 5. The incident beam of sunlight has power density of 0.9 kW/m
2
 in the direction of the
beam. The angle of incidence 
θ
 is 60°. Calculate power collected by the surface having total flat area of
100 m
2
.
A = 100 m
2
I
bn
= 0.9 kW/m
2
Sun
Ray
N
θ
= 30°
S
 Fig. 2.48.
Solution. Equivalent solar power falling on the surface S
P
N
(Watts) = I
N
(W/m
2
) × A (m
2
)
 I
N
(kW/m
2
) = Ibn . cos 
θ
= (0.9 × 10
3
) × 0.5 W/m
2
= 4.5 × 10
3
W/m
2
P
N
(kW) = (0.45 × 10
3
W/m
2
) × 100 m
2
= 0.45 × 10
5
W = 0.045 mW.
Example 6. When a photovoltaic cell is exposed to solar insulation of 950 W/m
2
, the short circuit
current is 220 A/m
2
 both based on a unit area of the exposed junction. The open circuit voltage is 0.60
V and the temperature is 300 K. Calculate

---

112
POWER PLANT ENGINEERING
(a) reversed saturation current
(b) the voltage that maximizes the power
(c) the load current that maximizes the power
(d) the maximum power
(e) the maximum conversion efficiency
(f) the cell area for an output of 1 kW at the condition of maximum power.
Solution.
/
/
s
o
I
A
I
A
= exp 
ac
eV
kT






– 1 = 1.193 × 10
10
o
I
A
= 
10
220
1.193 10
×
= 1.8 × 10
–8
A/m
2
The voltage Vm
p
that maximizes the power is given by
exp 
mp
eV
kT






1
mp
eV
kT


+




= 1 + 
/
/
g
o
I
A
I
A
V
mp
= 0.52 V
mp
I
A
= 
/
1
/
mp
mp
eV
kT
eV
kT
−
s
o
I
I
A
A


−




= 210 A/m
2
max
A
ρ
= 
mp
I
A






V
mp
= 109 W/m
2
η
max
= 
max
/
/
in
P
A
P
A
= 
109
950
= 11.5%
A = 
out required
max
/
P
P
A
= 
1000
109
= 9.17 m
2
.
Example 7. Estimate the average daily global radiation on a horizontal surface at Ahmedabad
(22°00
′
 N 73°10
′
E) during the month of April. If the average sunshine hours per day are 10. Assume a
= 0.28 and b=0.48.
Solution. Let
I
sc
= Solar constant = 4870.8 kJ/m
2
.hr
H
o
= Daily global radiation on a m
2
horizontal surface at the location on a clear sky day in the
month H
o
is calculated from as
H
o
= 
24
π
I
sc
360
1
0.033 cos
(sin
. sin
cos
. cos . sin )
365
n
s
s
s


+
α
+
α




H
g
= Daily global radiation (monthly average) for a horizontal surface at the location is calcu-
lated by using value of H
o
, a, b as
H
g
 = H
o
a + b 






h
m
L
L
kJ/m
2
. day

---

NON-CONVENTIONAL ENERGY RESOURCES AND UTILISATION
113
The L
h
are hours per day (average of the month)
L
m
are maximum day hours in the month.
a, b are Angstrom’s constants.
Angle of declination 
δ
, from above Eqn.
δ
= 23.45 sin 
360
(284
)
365
n


+




For April 15, n is calculated as :
Jan.
Feb.
March
April
= n
31
+
28
+
31
+
15
= 105
= 23.45 sin 
360
(389)
365






= 23.45 sin 383.67 = 23.45 × 0.4 = 9.41°
From above Eqn. sunshine hour angle 
ω
s
ω
s
= cos
–1
(– tan 
φ
. tan 
δ
)
= cos
–1
(– tan 22° . tan 9.41°)
= cos
–1
(0.40 × 0.61) = cos
–1
(0.064)
= 86.33°
ω
s
is converted from degrees to radians. 180° = 
π
radians
86.33° = 
86.33
180
× 
π
= 1.507 rad
Maximum length L
m
= 
2
15
× 86.33° hours = 11.51 hours
L
h
= 10 hours (given). Now H
o
as
H
o
= 
24
π
I
sc
360
1
0.033 cos
. sin
. sin
365
+ ω
− ω


+
×
ω
φ
δ




∫
s
s
s
n
+ cos 
φ
. cos 
δ
. cos 
ω
s
)
Substituting above calculated values, we get
= 37210 [1 + 0.033 cos (360/365) × 105] × (1.507 sin 22°. sin 9.41 + cos 22° . cos 9.41 . sin 86.3)
= 37210 [1 + 0.033 × (– 0.23) × 1.507 × 0.375 × 0.16 + 0.93 × 0.987 × 0.998]
= 37210 [1.0076 × (0.09 + 0.916)]
H
0
= 37210(1.09) = 40559 kJ/m
2
day
H
g
= H
o
h
m
L
a
b
L




+










= 40559 [0.28 + 0.48 (10/11.51)]
= 40559 × 0.697 = 28270 kJ/m day
Average global radiation per day in April = 28270 kg/m
2
 day.

---

114
POWER PLANT ENGINEERING
Example 8. Calculate the day length on a horizontal surface at New Delhi (28°35
′
 N, 77°12
′
 E)
on December 1.
Solution. Day length = (2/15) cos
–1
(– tan 
Φ
. tan 
δ
)
Day length = 
2
15
cos
– 1
[– tan 28.58° tan (– 22.11°)]
= 10.30 h.
Example 9. Determine solar time corresponding to 1430 h (IST) at Bombay (190°07
′
 N, 72°51
′
E) on July 1. In India standard time is based on 82.50° E.
Solution. Solar time = 1430 h – 4(82.50 – 72.85) minutes + (– 4 minutes)
= 1430 h – 38.6 minutes – 4 minutes
= 1347 h.
Example 10. A 100 mW vapour-dominated system uses saturated steam from a well with a shut-
off pressure of 28 bar. Steam enters the turbine at 5.5 bar and condenses at 0.15 bar. The turbine
polytropic efficiency is 0.82 and the turbine-generator combined mechanical efficiency is 0.9. The cool-
ing tower exist is at 20°C. Calculate the necessary steam flow, the cooling water flow and the plant
efficiency and heat rate if reinjection occurs prior to cooling tower.
Solution.
7
1
S.T.
2
5
Cooling
Tower
Direct
Contact
Condenser
Steam-jet
Ejector
Condenser Pump
Alternate
Rejectors
Rejector Well
Ground
6
5
4
Fig. 2.49. Vapour-dominated Power Plant.
T
3
5
2
′
2
S
1
7
6
Fig. 2.50. T-s Diagram.

---

NON-CONVENTIONAL ENERGY RESOURCES AND UTILISATION
115
h
6
= h
g
at 28 bar = 2801.7 kJ/kg
h
l
at 5.5 bar = 2801.7 kJ/kg
t
1
= 176°C (20°C superheat)
s
1
= 6.897 kJ/kg-K
v
1
= 0.356 m
3
/kg
s
2s
at 0.15 bar = s
1
= 6.897 = 0.7327 + x
4s
(7.306)
x
2s
= 0.8437
 h
2s
= 218.7 + 0.8437 (2377.24)
= 2224.38 kJ/kg
Isentropic work = h
1
– h
2
,
 s
= 2801.7 – 2224.38 = 577.32 kJ/kg
Actual turbine work = 0.82 × 577.32 = 473.4 kJ/kg.
h
2
= h
1
– W
T
= 2801.7 – 473.4 = 2328.3 kJ/kg
Ignoring pump work
h
3
= h
4
= 218.7 kJ/kg
h
5
= h
f
at 20°C = 88.5 kJ/kg
Turbine steam flow = 
3
100 10
(473.4
0.9)
×
×
= 234.7 kg/s
= 0.845 × 10
6
kg/hr
Turbine volume flow = (0.845 × 10
6
× v
1
) /60
= (0.845 × 10
6
× 0.356)/60
= 5013.4 m
3
/min
Cooling water flow to condenser, m
5
m
5
(h
3
– h
5
) = m
2
(h
2
– h
3
)
m
5
= (2328.3 – 218.7) m
2
/(218.7 – 88.5)
= 2109.6 × 0.845 × 
6
10
130.2
= 13.7 × 10
6
kg/hr
Heat added = h
6
– h
4
= 2801.7 – 218.7 = 2583 kJ/kg
Plant efficiency = W
T
× 0.9 / (h
6
– h
4
) = (473.4 × 0.9)/2583 = 16.49%
Plant heat rate = 
3600
0.1649
= 21831.4 kJ/kWh.

---

116
POWER PLANT ENGINEERING
Example 11. A horizontal shaft, propeller type wind-turbine is located in area having following
wind characteristics :
Speed of wind 10 m/s at 1 atm and 15°C. Calculate the following:
1. Air density 
ρ
2. Total power density in wind stream, W/m
2
3. Maximum possible obtainable power density, W/m
2
4. Actual obtainable power density, W/m
2
5. Total power from a wind-turbine of 120 m dia.
6. Torque and axial thrust (Na) on the wind-turbine operating at 40 rpm and at maximum effi-
ciency of 42%.
Solution. 1. Air-density of wind
We use the equation 
ρ
= 
P
RT
...(1)
where
P = pressure of air, Pa
T = temperature of air, K
R = gas constant 287 J/kg.k 1 atm
Pressure 1 atm = 1.01325 × 10
5
Pa
Temperature 15°C = 15 + 273.15 = 378.15 K
Substituting in Eqn. (1)
Air density 
ρ
= 
P
RT
1.01325 × 
5
10
287
× 378.15 = 1.226 kg/m
3
.
2. For total power P
t
t
P
A
= 0.5
ρ
V
i
3
0.5 × 1.226 × 10
3
= 613 W/m
2
.