GATE ME 8. IC engine and RAC Chapter IC engine and RAC Test 1 Mock Test Series 2022-23

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IC engine and RAC Test 1

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Weekly Quiz Competition

Question 1
2 / -0.33

An ideal petrol engine produce 0.98 kJ of work after consuming 0.070 gm of fuel. The calorific value of fuel is 43757 kJ/kg. What is the required compression ratio if specific heat air is assumed constant?

A
2.62

B
4.54

C
6.81

D
4.33

Solution

Concept:
Heat supplied = mass of fuel burnt × Calorific value
Q_s = ṁ × C.V
Q_s = 0.070 × 10^-3 × 43754
∴ Q_s = 3.06299 kJ
Now,
\({\rm{Thermal\;efficiency\;}} = \frac{{{W_{net}}}}{{{Q_{supplied}}}} = 1 - \frac{1}{{{r^{\gamma - 1}}}}\)
Calculation:
\({\eta _T} = \frac{{0.98}}{{3.06299}}\)
\(\therefore {\eta _T} = 1 - \frac{1}{{{r^{0.4}}}}\)
∴ r = 2.622
Question 2
2 / -0.33

A sample of air at 35°C and standard atmospheric pressure is taken. Then sample has 70% relative humidity
If the saturation pressure is 5.628 kPa, then the amount of moisture in kg/kg of dry air is

A
0.0055

B
0.040

C
0.025

D
0.055

Solution

Concept:
Relative humidity, ϕ = 0.70, Saturation pressure, P_s = 5.628 kPa
Standard atm. Pressure P = 101.325 kPa
ϕ = P_v/P_s (Calculate P_v)
\({\rm{Specific\;humidity\;}}\left( \omega \right) = \frac{{0.622\;{P_v}}}{{P - {P_v}}}\)
Calculation:
\(\phi = 0.70 = \frac{{{P_v}}}{{5.628}}\)
P_v = 3.9396 kPa
\(\omega = \frac{{0.622\; \times \;3.9396}}{{101.325\; - \;3.9396}}\)
∴ ω = 0.025 kg/kg of d.a.
Question 3
2 / -0.33

If room sensible heat = 10 kW and latent heat = 70 kW, then room sensible heat factor is

A
0.125

B
0.20

C
0.15

D
0.75

Solution

Concept:
The sensible heat factor is the ratio of Sensible heat to the Room total heat.
Room total heat = Sensible heat (SH) + Latent heat (LH)
Sensible Heat Factor is given by,
\(SHF = \frac{{SH}}{{LH + SH}} \)
Calculation:
Given, SH = 10 kW, LH = 70 kW
\( SHF = \frac{{10}}{{70 + 10}} = \frac{{10}}{{80}} = 0.125\)
Question 4
2 / -0.33

The chemical formula for refrigerant R – 114 is

A
CCl₂F₂

B
C₂Cl₂F₄

C
CCl₃F

D
CHCl F₂

Solution

Concept:
Formula for naming of refrigerant,
R (m - 1) (n + 1)P,
When, m → number of carbon atoms
n → Number of hydrogen atoms
P → Number of fluorine atoms
Here,
m – 1 = 1 → m = 2
n + 1 = 1 → n = 2
P = 4
So, there are 2 C, zero H, 4F and remaining balance by Cl.
Therefore, chemical formula is C₂ Cl₂ F₄
Question 5
2 / -0.33

The compressor work in a refrigerating machine with throttle valve is 20 kJ/kg, by using expander the increase in net refrigerating effect is 3.5 kJ/kg. The percentage change in work is____

A
+21.12%

B
+17.5%

C
-14.9%

D
-17.5%

Solution

Concept:
Compressor work with expansion valve = 20 kJ/kg
When expander is placed in place of expansion valve.
∴ Net compression work = 20 – 3.5
∴ Net compression work = 16.5 kJ/kg
Now,
\({\rm{\% \;change\;in\;work\;}} = \frac{{16.5 - 20}}{{20}} \times 100\)
∴ Change in work = -17.5%
Question 6
2 / -0.33

The compression ratio of air standard Otto cycle is 8. The initial conditions of air at the beginning of the compression stroke are 300K and 0.2MPa. The amount of energy added during the combustion process is 25kJ/mol. Work done(kJ) per mole of air ____

A
14-14.2

B
144-14.21

C
147-14.26

D
149-14.22

Solution

Concept:
\(\eta = \frac{{Work\;done}}{{Heat\;input}}\)
\(\eta = 1 - {\left( {\frac{1}{r}} \right)^{\gamma - 1}}\)
Calculation:
\(\eta = 1 - {\left( {\frac{1}{r}} \right)^{\gamma - 1}}\)
\(\eta = 1 - {\left( {\frac{1}{8}} \right)^{0.4}}\)
∴ η = 56.47%
Now,
Work done = η × Q₁
Work done = 0.5647 × 25
∴ Work done = 14.12 kJ/mol
Question 7
2 / -0.33

The supply air temperature is 15˚C and apparatus dew point is 12˚C for the cooling coil with bypass factor of 0.15. What is temperature (˚C) at inlet of cooling coil?

A
32-32

B
324-321

C
327-326

D
329-322

Solution

Concept:
\({\rm{Bypass\;factor\;}} = \frac{{{T_s} - DPT}}{{{T_i} - DPT}}\)
T_s = temperature of supply air, T_i = temperature at inlet of cooling coil, DPT = Dew point temperature
Calculation:
\(0.15 = \frac{{15 - 12}}{{{T_i} - 12}}\)
∴ T_i – 12 = 20
∴ T_i = 32°C
Question 8
2 / -0.33

Moist air with DBT = 40°C and relative humidity of 60 % is compressed to 5.05 bar and DBT = 150° C.
Using the following data, find the relative humidity of compressed air ______%.
R_H2O = 462 J/kg k, P_s at 40° C = 7.384 kPa, P_s at 150° C = 4.76 bar, Atm. Pressure = 1.01 bar (round off to two decimals)

A
4.60-4.70

B
4.604-4.701

C
4.607-4.706

D
4.609-4.702

Solution

Concept:
Relative humidity, ϕ = P_v/P_s
\({\rm{Specific\;humidity\;}}\left( \omega \right) = \frac{{0.622\;{P_v}}}{{P - {P_v}}}\)
Calculation:
P_atm = 101 kPa
T_s = 40°C P_s = 7.384 kPa
ϕ₁ = 0.6 = P_v/P_s
P_v1 = 4.430 kPa
\(\omega = \frac{{0.622\; \times \;4.4304}}{{101 - 4.4304}}\)
∴ ω = 0.0285 kg vap./kg of dry air
In compression, ω remains constant
ω₁ = ω₂
\(0.0285 = \frac{{0.622\;{P_{v2}}}}{{P - {P_{v2}}}}\)
\(\frac{{0.622\;{P_{v2}}}}{{505 - {P_{v2}}}} = 0.0285\)
P_v2 = 22.1252 kPa
ϕ₂ = 22.1252/476
∴ ϕ₂= 0.04648 ~ 4.65%
Question 9
2 / -0.33

Dry bulb temperature of the air is 25°C and dew point temperature at 1 atm pressure is 20°C. The saturation pressure of water at 20°C and 25°C are 0.02462 bar and 0.03813 bar respectively. The specific heat of the air and water vapour are 1.005 and 1.88 kJ/kgK respectively and latent heat of vaporization of water at 0°C is 2500 kJ/kg. The enthalpy (in J/kg) of air is

A
64-65

B
644-651

C
647-656

D
649-652

Solution

Concept:
h = 1.005 t + ω (2500 + 1.88t)
Calculation:
Given:
P = 1 atm = 1.013 bar and P_v = 0.02462 (at DPT)
Now,
specific humidity \((\omega) = \frac{{0.622{P_v}}}{{P - {P_v}}} \)
\(\omega = \frac{{0.622 \times 0.02462}}{{1.013 - 0.02462}}\)
∴ ω = 0.01549 kg/kg of dry air
Now,
Enthalpy (h) = 1.005 t + ω (2500 + 1.88t)
h = 1.005 × 25 + 0.01549 (2500 + 1.88 × 25)
∴ h = 64.42 J/kg of dry air.
Question 10
2 / -0.33

An engine based on a standard Otto cycle is operating with air as an medium with specific heat of 1.4 has a cylinder diameter of 5 cm and stroke length of 7.5 cm. If the clearance volume is 88.15 cc and the heat supplied per kg of air per cycle is 900 kJ/kg, then work output (in kJ) per kg of air is?

A
-291.6

B
0

C
291.6

D
1800

Solution

Concept:
In a otto cycle efficiency is given as
\(η = 1 - {\left( {\frac{1}{r}} \right)^{\gamma - 1}}\)
Where r is the compression ratio and γ is the ratio of specific heat
Also compression ratio
\(r = \frac{{volume\;before\;compression}}{{volume\;after\;compression}} = \frac{{{V_s} + {V_c}}}{{{V_c}}}\)
V_s = Stroke volume or volume of cylinder = \(\frac{\pi }{4}\;{d^2}\;l\)
V_c = Clearance volume
Where d is the diameter of the cylinder and l is the stroke length
Also
\(η = \frac{{Work\;ouput}}{{Heat\;Supplied}}\)
Calculation:
Given:
γ = 1.4, V_c = 88.15 cc, d = 5 cm, l = 7.5 cm
Heat Supplied = 900 kJ/kg
\({V_s} = \frac{\pi }{4}\;{d^2}\;l = \frac{\pi }{4} \times {5^2} \times 7.5 = 147.262\;cc\)
\(r = \frac{{{V_s} + {V_c}}}{{{V_c}}} = \frac{{88.15 + 147.262}}{{88.15}} = 2.674\)
∴ r = 2.674
Now
\(η = 1 - {\left( {\frac{1}{{2.674}}} \right)^{1.4 - 1}} \)
∴ η = 0.325
Now,
Work Output = η × Heat input
Work Output = 0.325 × 900
∴ Work Output = 292.5 kJ/kg
Question 11
2 / -0.33

A car has 4 cylinder engine of 80 mm bore and 90 mm stroke, what is the clearance volume of each cylinder and the cubic capacity of engine if compression ratio is 8?

A
67.266 cc, 1900 c

B
64.63 cc, 2000 cc

C
64.63 cc, 1810 cc

D
66.43 cc, 1900 cc

Solution

Concept:
\({\rm{Cubic\;capacity\;of\;engine\;}} = \frac{\pi }{4}{d^2}L \times {\rm{No}}.{\rm{\;of\;cylinders}}\)
Where
\({V_s} = \frac{\pi }{4}{d^2}L = {\rm{\;displacement\;volume\;or\;swept\;volume}}\)
\({\rm{Compression\;ratio\;}}\left( r \right) = \frac{{{V_c} + {V_s}}}{{{V_c}}}\)
V_c = clearance vol.
Calculation:
\({V_s} = \frac{\pi }{4}{d^2}L\)
\({V_s} = \frac{\pi }{4}\left( {{{0.08}^2}} \right)\left( {0.09} \right)\)
V_s= 4.5238 × 10^-4 m³
V_s = 0.45238 × 10^-3 × 10⁶ cm³
∴ V_s = 452.38 cc
Now,
Total cubic capacity = 4 × 452.38
∴ Total cubic capacity = 1809.56 cc
Now,
\(r = 8 = \frac{{{V_c} + {V_s}}}{{{V_c}}} \Rightarrow {V_L} = \frac{{{V_s}}}{{r - 1}}\)
∴ V_L = 64.63 cc
Question 12
2 / -0.33

44 kg of C₃H₈ is burnt with 1160 kg of air (Mol. Wt. = 29) to produce 88 kg of CO₂ and 14 kg of CO
\({C_3}{H_8} + 5{O_2} = 3C{O_2} + 4{H_2}O\)
What is the percent excess air used?

A
67-71

B
674-711

C
677-716

D
679-712

Solution

Calculation:
\({\rm{Actual\;air}} - {\rm{fuel\;ratio\;}} = \frac{{Actual\;air\;\left( {kg} \right)}}{{fuel\;\left( {kg} \right)}}\)
\({\left( {AF} \right)_{actual}} = \frac{{1160}}{{44}} = 26.36\)
Now,
\({\rm{Stoichiometric\;air\;fuel\;ratio}} = \frac{{5\left[ {{O_2} + 3.76\;{N_2}} \right]}}{{{C_3}{H_8}}}\)
\({\rm{Stoichiometric\;air\;fuel\;ratio}} = \frac{{5\left[ {2 \times 16 + 3.76 \times 28} \right]}}{{3 \times 12 + 1 \times 8}}\)
∴ Stoichiometric air fuel ratio = 15.6
Now,
\(Percentage\;excess\;air = \frac{{Actual\;AF\;ratio}}{{stociometric\;AF\;ratio}} \times 100 - 100\)
\(Percentage\;excess\;air = \left( {\frac{{26.36 - 15.6}}{{15.6}}} \right) \times 100\)
∴ Percentage excess air = 68.9 %
Question 13
2 / -0.33

Air enters a duct at 15°C and 80% relative humidity and leaves at 30°C with a flow rate of 2.5 m³/s. If the air remains constant at 101.3 kPa, then the rate of heat transfer is
(P_sat at 15°C = 1.6 kPa and at 30°C = 4.3 kPa)

A
29.61 kW

B
37.33 kW

C
41.34 kW

D
46.28 kW

Solution

Concept:
Enthalpy of moist air,
h = 1.005 t_d + ω(2500 + 1.88 t_d) (t_d → dry bulb temperature in °C)
Heat transfer rate, Q = ṁ_a (h₂ – h₁)
Calculation:
Given:
T₁ = 15°C = 288 K, T₂ = 30°C = 303 K, ϕ₁ = 0.8, V̇ = 2.5 m³/s, P_t = 101.3 kPa,
\({P_{{s_1}}} = 1.6\;kPa,\;{P_{{s_2}}} = 4.3\;kPa\)
Now,
\({\phi _1} = 0.8 = \frac{{{P_{{v_1}}}}}{{{P_{{s_1}}}}}\)
\(0.8 = \frac{{{P_{{v_1}}}}}{{1.6}}\)
\(\therefore {P_{{v_1}}} = 1.28\;kPa\)
Now,
\({\rm{Specific\;humidity\;}}\left( {{\omega _1}} \right) = \frac{{0.622\;\;{P_{{v_1}}}}}{{{P_t} - {P_{{v_1}}}}}\)
\({\omega _1} = \frac{{0.622 \times 1.28}}{{101.3 - 1.28}}\)
∴ ω₁ = 7.96 × 10^-3 kg/kg of dry air
As no moisture is added or removed,
∴ ω₁ = ω₂ = 7.96 × 10^-3
Now,
Enthalpy of moist air of entry,
h₁ = (1.005 × 15) + 7.96 × 10^-3 × (2500 + 1.88 × 15)
h₁ = 35.199 kJ/kg
h₂ = (1.005 × 30) + 7.96 × 10^-3 × (2500 + 1.88 × 30)
∴ h₂ = 50.498 kJ/kg
Now,
Mass flow rate of dry air,
\({\dot m_a} = \frac{{{P_a}\dot V}}{{{R_a} \cdot {T_1}}} = \frac{{\left( {101.3 - 1.28} \right)5}}{{0.287 \times 288}}\)
ṁ_a = 3.025 kg/s
Now,
Heat transfer rate = ṁ_a (h₂ – h₁)
Heat transfer rate = 3.025 × (50.498 – 35.199)
Heat transfer rate = 46.279 kJ/s
∴ Heat transfer rate = 46.28 kW
Question 14
2 / -0.33

More test was conducted for a 2-stroke spark ignition engine having 2 – cylinders and following data is obtained. Brake power = 14 kW Brake Power of individual cylinder = 5.4 kW, 4.80 kW.
What is the mechanical efficiency of this engine?

A
78.65%

B
76%

C
74.25%

D
68.75%

Solution

Concept:
Let the cylinders be 1 and 2
Net brake power bp_1,2
Indicated power = Brake power + Frictional power
Ip = bp + fp
\({\rm{Mechanical\;efficiency}} = \frac{{b{p_{1,2}}}}{{i{p_{1,2}}}}\)
ip_1,2 = net indicated power = ip₁ + ip₂
Calculation:
Given:
bp_1,2 = 14 kW, bp₁ = 5.4 kW, bp₂ = 4.80 kW
ip_1-2 = bp_1-2 + fp
ip₁ = bp₁ + fp
ip₂ = ip_1-2 – ip₁
∴ ip₂= bp_1-2 + fp – bp₁ – fp
ip₂ = bp_1-2 – bp₁
ip₂ = 14 – 5.4
∴ ip₂ = 8.6 kW
Now,
ip₁ = bp_1-2 – bp₂
∴ ip₁= 14 – 4.8
∴ ip₁= 9.2 kW
Now,
ip_1-2 = 9 – 2 + 8.6
∴ ip_1-2 = 17.8 kW
Now,
η_m = bp_1-2/ip_1-2 = 14/17.8
∴ η_m = 78.65 %
Question 15
2 / -0.33

Heat rejection factor for an ideal refrigerator working on vapor compression refrigeration cycle is 1.55. The heat absorbed in evaporator is found to be 50 kW. Keeping the heat rejection from the condenser as constant, the COP is now increased by 20 %. What will be the work done required (kW) now?

A
46.32

B
43.62

C
26.34

D
24.36

Solution

Concept:
\({\rm{Heat\;rejection\;factor}} = \frac{{Heat\;rejected\;from\;condensor\;\left( {const} \right)}}{{Heat\;absorbed\;by\;evaporator}}\)
Work = Q_condensor - Q_evaporator
\({\rm{COP}} = \frac{{{Q_{absorbed}}}}{{work\;done}}\;\;\;\;\left( {{\rm{This\;is\;increased\;by\;}}20{\rm{\% }}} \right)\)
Calculation:
\(1.55 = \frac{{{Q_{condensor}}}}{{50}}\)
∴ Q_c = 77.5 kW
W = 77.5 – 50
∴ W = 27.5 kW
Now,
\(COP = \frac{{50}}{{27.5}} = 1.8181\)
(COP)_new = 1.2 × 1.8181
\(1.2 \times 1.8181 = \frac{{{Q_{condensor}} - W'}}{{W'}}\)
(2.1818 W’ + W’) = Q_c
\(W' = \frac{{{Q_c}}}{{3.1818}}\)
∴ W' = 24.357 kW

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Re-Attempt Weekly Quiz Competition

IC engine and RAC Test 1

Result

IC engine and RAC Test 1

Score

-

Rank

-

SHARING IS CARING

Question 1

2.62

4.54

6.81

4.33

Solution

Question 2

0.0055

0.040

0.025

0.055

Solution

Question 3

0.125

0.20

0.15

0.75

Solution

Question 4

CCl2F2

C2Cl2F4

CCl3F

CHCl F2

Solution

Question 5

+21.12%

+17.5%

-14.9%

-17.5%

Solution

Question 6

14-14.2

144-14.21

147-14.26

149-14.22

Solution

Question 7

32-32

324-321

327-326

329-322

Solution

Question 8

4.60-4.70

4.604-4.701

4.607-4.706

4.609-4.702

Solution

Question 9

64-65

644-651

647-656

649-652

Solution

Question 10

-291.6

0

291.6

1800

Solution

Question 11

67.266 cc, 1900 c

64.63 cc, 2000 cc

64.63 cc, 1810 cc

66.43 cc, 1900 cc

Solution

Question 12

67-71

674-711

677-716

679-712

Solution

CCl₂F₂

C₂Cl₂F₄

CCl₃F

CHCl F₂