Gas Turbine
Exam Duration: 45 Mins Total Questions : 15
Thermal efficiency of a standard Otto cycle for compression ratio 6 is
- (a)
60%
- (b)
68%
- (c)
71%
- (d)
65%
\(\eta =1-\frac { 1 }{ { \left( r \right) }^{ \gamma -1 } } \)
\(=1-\frac { 1 }{ \left( 6 \right) ^{ 1.6-1 } } =1-\frac { 1 }{ { \left( 6 \right) }^{ 0.6 } } \)
= 0.65 or 65%
If engine produces 10 kW brake power while working with a brake thermal efficiency of 30%. If the calorific value of the fuel used is 40000 kJ/kg. Then, what is the fuel consumption?
- (a)
1.5 kg/h
- (b)
3.0 kg/h
- (c)
0.3 kg/h
- (d)
1.0 kg/h
Given that
BP = lOkW
\(\eta \)th = 30%
CV = 40000 kJ/kg
mf =?.
We know that,
\(\eta _{ th }=\frac { BP }{ { m }_{ f }.CV } \)
\(0.3=\frac { 10 }{ { m }_{ f }+40000 } \)
\({ m }_{ f }=\frac { 10 }{ 0.3\times 40000 } kg/s\)
\({ m }_{ f }=\frac { 10 }{ 0.3\times 40000 } \times 3600\\ \)
mf = 3 kg/h
An Otto cycle operates with volumes of 40 cm3 and 400 cm3 at Top Dead Centre (TOG) and Bottom Dead Centre (BDC) respectively. If the power output is 100 kW, what is heat input in kJ/s? Assume \(\gamma\) = 1.4.
- (a)
166
- (b)
145
- (c)
110
- (d)
9.3
Given that
V = 40 cm3 V + V = 400 cm3 \(\gamma\) = 1.4
Compression ratio (r) =\(\frac { { v }_{ s }+{ v }_{ c } }{ { v }_{ c } } =\frac { 400 }{ 40 } =10\)
Efficiency (Otto cycle) -\(\eta =1-\frac { 1 }{ { r }^{ \gamma -1 } } =1-\frac { 1 }{ 100 } =0.6\)
\(\eta =\frac { Out\quad put }{ Heat\quad input\quad } \)
or Heat input = \(\frac { Out\quad put }{ \eta \quad } =\frac { 100 }{ 0.6 } =166kJ/s\\ \)
An open cycle constant pressure gas turbine uses a fuel of calorific value 40000 kJ/kg with air fuel ratio of 80 : 1 and develops a net output of 80 kJ/kg of air. The thermal efficiency of the cycle is
- (a)
61%
- (b)
16%
- (c)
18%
- (d)
None of these
Given that
CV = 40000 kJ/kg, output = 80 x ma kJ
\(\frac { { m }_{ a } }{ { m }_{ f } } =80\)
We know that
\(\eta =\frac { Out\quad put }{ input\quad } =\frac { 80\times { m }_{ a } }{ { m }_{ f }\times CV } =0.16\)
= 16%
In a Morse test for a 2-cylinder, 2-stroke, spark ignition engine, the brake power was 9 kW whereas the brake powers of individual cylinders with spark cut-off were 4.25 kW and 3.75 kW respectively. The mechanical efficiency of the engine is
- (a)
90%
- (b)
80%
- (c)
45.5%
- (d)
52.5%
We know that
IP = BP + FP
If n cylinders are used in Morse test then
(lP)n = (BP)n + FP
(lP)n-1 = (BP)n-1+ FP
(IP)1st = (BP)n cylinder - (BP)n-1
= 9-4.25=4.75kW
(1P)2nd= 9 - 3.75 = 5.25 kW
Total IP = 4.75 + 5.25 = 10 kW
We know that,
\(\eta =\frac { BP }{ IP } =\frac { 9 }{ 10 } \times 100=90\)%
In a Brayton cycle, the value of optimum pressure ratio for maximum net work done between temperatures T1 and T3' where T3 is the maximum temperature and T1 is the minimum temperature is
- (a)
\({ r }_{ p }=\left( \frac { { T }_{ 3 } }{ { T }_{ 1 } } \right) ^{ \frac { \gamma }{ \gamma -1 } }\)
- (b)
\({ r }_{ p }=\left( \frac { { T }_{ 3 } }{ { T }_{ 1 } } \right) ^{ \frac { \gamma -1 }{ 2\gamma } }\)
- (c)
\({ r }_{ p }=\left( \frac { { T }_{ 3 } }{ { T }_{ 1 } } \right) ^{ \frac { \gamma }{ 2\left( \gamma -1 \right) } }\)
- (d)
\({ r }_{ p }=\left( \frac { { T }_{ 3 } }{ { T }_{ 1 } } \right) ^{ \frac { 2\left( \gamma -1 \right) }{ \gamma } }\)
The thermal efficiency of a gas turbine cycle with regeneration in terms of T3 (maximum temperature), T1 (maximum temperature), rp (pressure ratio) and k\(\left( =\frac { { C }_{ P } }{ { C }_{ V } } \right) \) IS given by
- (a)
\(1-\frac { { T }_{ 1 } }{ { T }_{ 3 } } { r }_{ p }=\left( \frac { K }{ K-1 } \right) \)
- (b)
\(1-\frac { { T }_{ 3 } }{ { T }_{ 1 } } { r }_{ p }=\left( \frac { K }{ K-1 } \right) \)
- (c)
\(1-\frac { { T }_{ 3 } }{ { T }_{ 1 } } { r }_{ p }=\left( \frac { K-1 }{ K } \right) \)
- (d)
\(1-\frac { { T }_{ 1 } }{ { T }_{ 3 } } { r }_{ p }=\left( \frac { K-1 }{ K } \right) \)
In an ideal air-standard gas turbine cycle the minimum and maximum temperatures are respectively 300 K and 1200 K. Then, optimal pressure ratio of cycle for maximum work output is (for air \(\gamma\) = 1.4)
- (a)
2.1
- (b)
1.3
- (c)
1.2
- (d)
3.5
Optimal pressure ratio
\(=\left( \frac { { T }_{ max } }{ { T }_{ min } } \right) \frac { \gamma -1 }{ 2\gamma } \)
\(=\left( \frac { 1200 }{ 300 } \right) ^{ \frac { 1.4-1 }{ 2\times 1.4 } }\)
= 1.217
In a gas turbine, hot gases enters at 30 bar, 1000 K and exit at 1 bar. Specific heat of gases are Cp = 0.98 kJ/kg-K and Cv = 0.7538 kJ/kg-K. The work developed by turbine is (assume turbine efficiency = 0.90)
- (a)
454.5 kJ/kg
- (b)
500 kJ/kg
- (c)
329.29 kJ/kg
- (d)
907.1 kJ/kg
For process (3-4 '),
\(\frac { { T }'_{ 4 } }{ { T }_{ 3 } } ={ \left( \frac { { P }_{ 4 }' }{ { P }_{ 3 } } \right) }^{ \frac { \gamma -1 }{ \gamma } }\)
\(=\left( \frac { 1 }{ 30 } \right) ^{ \frac { 1.3-1 }{ 1.3 } }\)
T4' = 0.45 x 1000
= 450K
\(\gamma =\frac { { C }_{ P } }{ { C }_{ V } } =\frac { 0.98 }{ 0.538 } =1.3\)
\({ \eta }_{ t }=\frac { { T }_{ 3 }-{ T }_{ 4 } }{ { T }_{ 3 }-{ T' }_{ 4 } } \Rightarrow 0.90=\frac { 1000-{ T }_{ 4 } }{ 1000-450 } \)
T4 = 495 K
Work done by turbine
= Cp (T3 - T4)
= 0.98 x (1000 - 495)
= 454 .5 kJ /kg
Consider a 4-stroke petrol engine which is supplied with 15 kg of air/kg of octane fuel for following data:
Air standard efficiency = 52%
Relative efficiency = 69%
Mechanical efficiency = 84%
CV of fuel = 42000 kJ/kg
Brake power = 75 kW
Indicated thermal efficiency is
- (a)
38%
- (b)
32.7%
- (c)
35.8%
- (d)
41.9%
\({ \eta }_{ airstandard }=1-\frac { 1 }{ { \left( r \right) }^{ \gamma -1 } } \)
\(\Rightarrow \quad 0.52=1-\frac { 1 }{ { \left( r \right) }^{ \gamma -1 } } \)
\(\Rightarrow\)r = 6.26
We know,
\({ \eta }_{ Irelative }=\frac { { \eta }_{ indicated } }{ { \eta }_{ airstandard } } \)
\(\therefore \quad { \eta }_{ { i }_{ th } }\)= 0.52x0.69 = 35.8%
Consider a 4-stroke petrol engine which is supplied with 15 kg of air/kg of octane fuel for following data:
Air standard efficiency = 52%
Relative efficiency = 69%
Mechanical efficiency = 84%
CV of fuel = 42000 kJ/kg
Brake power = 75 kW
Brake specific fuel consumption is
- (a)
0.284 kg/w-h
- (b)
0.127 kg/w-h
- (c)
1.243 kg/w-h
- (d)
None of these
\({ \eta }_{ mechanica }=\frac { Brakepower }{ Indicatedpower } \)
\(\Rightarrow 0.84=\frac { 75 }{ IP } \)
\(\Rightarrow \quad IP=\frac { 75 }{ 0.84 } =89.28KW\)
\(\Rightarrow { \eta }_{ { i }_{ th } }=\frac { Heate\quad quivalent\quad of\quad indicated\quad power }{ Energy\quad supplied\quad by\quad fuel } \quad \)
\(=\frac { IP\times 60 }{ { m }_{ f }\times CV } \)
\(\Rightarrow { m }_{ f }=\frac { 89.28\times 60 }{ 0.358\times 42000 } =0.3554kg/min\)
\(bsfc=\frac { { m }_{ f } }{ bp } =\frac { 0.3554\times 60 }{ 75 } =0.2843kg/W-h\)
In a 4-stroke 4-cylinder petrol engine during Morse test, the spark plug of cylinders were short-circuited without change of speed and corresponding brake loads were 123, 118, 116 and 130 N. If net brake load is 170 N then mechanical efficiency is
- (a)
80%
- (b)
81%
- (c)
91%
- (d)
88.08%
\({ \eta }_{ mechanical }=\frac { Brakeload }{ Indicatedload } \)
Indicated load = (170 - 123) + (170 -118) + (170 - 116) + (170 - 130)
= 193N
\(={ \eta }_{ m }=\frac { 170 }{ 193 } =88.08\)%
In a 4-stroke 4-cylinder petrol engine when tested at full load with torque arm of 36 cm gave 170 N net brake load at rated speed of 3200 rpm. Then, brake power per cylinder is
- (a)
5.12 kW
- (b)
6.27 kW
- (c)
20.5 kW
- (d)
None of these
T = ω X r= 170 x 0.36
= 61.21 N-m
\(bp=\frac { 2\pi NT }{ 60\times 1000 } =\frac { 2\pi \times 3200\times 61.21 }{ 60\times 1000 } \)
bp = 20.5 kW
Brake power cylinder =\(\frac { 20.5 }{ 4 } \)= 5.125 kW
In a 4-stroke 6-cylinder CI engine, bsfc = 0.3 kg/kW-h, bp = 240 kW, N = 1300 rpm, density of fuel = 900 kg/m3 Calculate the volume of fuel injected per cylinder per cycle is
- (a)
0.314 x 10-6 m3
- (b)
0.287 x 10-6 m3
- (c)
1.21 x 10-7 m3
- (d)
3.9 x 10-7 m3
The fuel consumption /hour
= bsfc x brake power
= 0.3 x 240
= 72 kg
For 4 stroke engine,
Number for cycles/hour =\(\frac { N }{ 2 } \times 60\)
\(=\frac { 1300 }{ 2 } \times 60=39000\)
Quantity of fuel consumption/cycle and per cylinder is
\(=\frac { 72 }{ 6\times 39000 } \)= 0.000307 kg
Volume of fuel injected/cylinder/cycle is
\(=\frac { m }{ \rho } =\frac { 0.000307 }{ 900 } \)0.314x10-6 m3
Efficiency of a diesel cycle will approach to Otto cycle, when
- (a)
diesel engine will operate at high speed
- (b)
cut off period of diesel cycle is reduced to zero
- (c)
diesel fuel is balanced with petrol
- (d)
None of the above