IC engine and RAC Test 1

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

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

\(\omega = \frac{{0.622\; \times \;3.9396}}{{101.325\; - \;3.9396}}\)

Most high speed compression engines operate on Dual combustion cycle.

The intake charge in a diesel engine consists of:

• Air alone
• Air with lubricating oil
• Air with fuel
• Air with fuel and lubricating oil

In diesel engines, the intake charge is primarily air. This is because diesel engines rely on air for the combustion process, where fuel is injected into the highly compressed air to ignite. The absence of additional components like fuel or lubricating oil in the intake charge is crucial for the engine's efficient operation.

Thus, the correct composition of the intake charge is air alone, which is essential for optimal combustion and performance of diesel engines.

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

\( SHF = \frac{{10}}{{70 + 10}} = \frac{{10}}{{80}} = 0.125\)

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.

Cylinder block: It is the main supporting structure for the various components.
Inlet and exhaust valve: Regulate the incoming and outgoing charge to and form the cylinder.

Piston: It acts as moving boundary of the combustion system and forms the first link to transmit the gas force through connecting rod.

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%

The balance weights are provided for static and dynamic balancing of the rotating system.

Concept:

Compression ratio = \(r = \frac{{volume\;before\;compression\;}}{{volume\;after\;compression}}\)

Work done = Mean effective pressure × swept volume

swept volume = (v1 – v2)

Calculation:

Given Compression ratio (r) = 8, v₁ = 0.9 m³/kg, Net heat interaction for cycle = 1575 kJ/kg

For any cycle, the net heat interaction = Net work interaction

\(r = \frac{{{v_1}}}{{{v_2}}}\)

⇒  \({v_2} = \frac{{0.9}}{8} = 0.1125\frac{{{m^3}}}{{kg}}\)

Work done = Mean effective pressure × swept volume (v₁ – v₂)

∴ 1575 × 10³ = p_m × (0.9 – 0.1125)

p_m = 20 bar

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

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

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%

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)

In internal combustion engines, the gudgeon pin (UK, wrist pin or piston pin US) connects the piston to the connecting rod, and provides a bearing for the connecting rod to pivot upon as the piston moves.

Comparison of Engine Parameters

• Maximum Pressure: This parameter indicates the highest pressure reached during the engine cycle and is important for determining the strength and durability of the engine.
• Fuel Consumption: This parameter measures how efficiently the engine uses fuel to generate power. Lower fuel consumption is desirable as it indicates better fuel efficiency.
• Mean Effective Pressure: This parameter is a measure of the average pressure exerted on the piston during the entire engine cycle. It is a key indicator of the engine's performance and efficiency.
• Unit Power: Unit power refers to the power output of the engine per unit of displacement or volume. It is important for comparing the power output of engines with different cylinder dimensions.

 

Reason for Choosing Mean Effective Pressure as the Basis of Comparison

• Mean Effective Pressure is a comprehensive parameter that takes into account the overall performance and efficiency of the engine.
• It provides a more balanced comparison between engines of different cylinder dimensions, power, and speed.
• By considering the average pressure exerted on the piston throughout the engine cycle, Mean Effective Pressure offers a more accurate representation of the engine's performance compared to other parameters like maximum pressure or unit power.
• Therefore, choosing Mean Effective Pressure as the basis of comparison allows for a more informed evaluation of the engines and their overall efficiency.

Correct Statements about Volumetric Efficiency

• Statement 2: Volumetric efficiency is maximum at inlet mach number of 0.5.
• Statement 3: Inlet valve mach index is related to volumetric efficiency.

Explanation

• Statement 2 is correct because volumetric efficiency is indeed maximum at an inlet mach number of 0.5.
• Statement 3 is also correct as the inlet valve mach index is directly related to volumetric efficiency.

Therefore, the correct answer is option B: 2 and 3.

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

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

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 %

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

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 % 

Ans. (a) (A/F) for maximum power = 12 and (A/F) for Stoichiometric = 14.5

(A/F) for maximum fuel economy = 16.

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