Q.1 (a) Give classification of Cochran boiler and Lancashire boiler. (03 marks) Q.1 (b) Write a short note on CNG as a fuel. (04 marks) Q.1 (c) (i) The absolute pressure in a compressed air tank is 200 kPa. What is the gauge pressure in the tank if atmospheric pressure is 1.01 bar? (ii) The temperature of a system rises by 130°C during a heating process. Express this rise in temperature in kelvins. (iii) A 4-kW resistance heater in a water heater runs for 3 hours to raise the water temperature to the desired level. Determine the amount of electric energy used in both kWh and kJ. (07 marks)

Solution Q.1 (a)

Classification of Cochran Boiler

Cochran boiler is classified as follows:

• Orientation of Axis: Vertical boiler.
• Type of Tubes: Multi-tubular fire-tube boiler.
• Method of Firing: Internally fired.
• Circulation Method: Natural circulation.
• Mobility: Stationary boiler.
• Pressure Rating: Low-pressure boiler.

Classification of Lancashire Boiler

Lancashire boiler is classified as follows:

• Orientation of Axis: Horizontal boiler.
• Type of Tubes: Fire-tube (having two large flue tubes).
• Method of Firing: Internally fired.
• Circulation Method: Natural circulation.
• Mobility: Stationary boiler.
• Pressure Rating: Low-to-medium-pressure boiler.

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Solution Q.1 (b)

Short Note on CNG (Compressed Natural Gas) as a Fuel

Compressed Natural Gas (CNG) is an eco-friendly fuel option widely used in internal combustion engines.

• Composition: It is primary composed of Methane (CH4), typically around 80% to 90%, with trace amounts of other hydrocarbons.
• State and Storage: It is compressed to a high pressure of 20 to 25 MPa (200 to 250 bar) and stored in high-strength cylindrical steel or composite tanks.
• Environmental Advantages:
• Burns cleaner than conventional petrol or diesel.
• Significantly reduces carbon monoxide (CO), nitrogen oxides (NOx), carbon dioxide (CO2), and particulate emissions.
• It contains virtually no sulfur, hence producing no sulfur oxides.
• Safety:
• CNG has a high self-ignition temperature (around 540°C), making it less likely to catch fire compared to gasoline.
• It is lighter than air, meaning that in the event of a leak, it quickly dissipates upward into the atmosphere instead of pooling on the ground.
• Applications: Extensively utilized as an alternative fuel in light-duty and heavy-duty automobiles (buses, auto-rickshaws, and cars) and stationary engines.

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Solution Q.1 (c)

(i) Absolute and Gauge Pressure Calculation

Given data:

• Absolute pressure (P_abs) = 200 kPa
• Atmospheric pressure (P_atm) = 1.01 bar

Conversion: 1 bar = 100 kPa P_atm = 1.01 * 100 kPa = 101 kPa

Formula: Pgauge = Pabs - P_atm

Calculation: P_gauge = 200 kPa - 101 kPa = 99 kPa

Answer: The gauge pressure in the tank is 99 kPa.

(ii) Temperature Rise in Kelvins

Given:

• Temperature rise (Delta_T) = 130°C

Principle: A change of 1 degree Celsius is exactly equivalent to a change of 1 Kelvin on the absolute scale. DeltaT(K) = DeltaT(°C)

Answer: The rise in temperature is 130 K.

(iii) Electrical Energy Consumption

Given data:

• Power rating (P) = 4 kW
• Operating time (t) = 3 hours

Calculation in kWh: Energy (E) = Power Time E = 4 kW 3 h = 12 kWh

Calculation in kJ: We know that 1 kWh = 3600 kJ E = 12 * 3600 kJ = 43,200 kJ

Answer: The electrical energy used is 12 kWh (or 43,200 kJ).

Q.2 (a) State the function of any three mountings in boilers. (03 marks) Q.2 (b) Classify engineering materials. (04 marks) Q.2 (c) 0.4 kg of gas is expanded isentropically from 10 bar and 340°C to 1 bar. It is then heated at constant volume to 3 bar and 340°C and then finally it is compressed isothermally until the initial pressure of 10 bar is attained. Draw the p-V diagram for these processes and find the value of the adiabatic index γ. Take Cp = 1.005 kJ/kgK. (07 marks)

OR

Q.2 (c) A gas expands from 450 kPa and 130 litres to 150 kPa and 260 litres. The decrease in enthalpy during the process is 55 kJ. Taking Cv = 718 J/kgK, determine (i) change in internal energy, (ii) value of Cp, and (iii) value of R. (07 marks)

Solution Q.2 (a)

Functions of Three Boiler Mountings

1. Safety Valve:

• Function: It prevents excessive pressure buildup inside the boiler shell. It automatically opens and releases excess steam into the atmosphere when the steam pressure exceeds the safe maximum working limit.

1. Water Level Indicator:

• Function: It displays the actual level of water inside the boiler shell during operation. This helps the operator monitor and maintain the correct water level, preventing dangerous situations like overheating or dry heating.

1. Pressure Gauge:

• Function: It measures and displays the pressure of steam generated inside the boiler. It is usually a Bourdon tube type dial gauge calibrated to read pressure in bar or kPa.

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Solution Q.2 (b)

Classification of Engineering Materials

Engineering materials are broadly classified into four major categories:

1. Metals and Alloys:

• Ferrous Metals: Contain iron as the primary constituent. Examples: Carbon steels, alloy steels, and cast iron.
• Non-Ferrous Metals: Do not contain iron as the principal constituent. Examples: Aluminium, copper, zinc, lead, and brass/bronze alloys.

1. Ceramics:

• Inorganic, non-metallic materials characterized by high melting points and hardness. Examples: Alumina, silicon carbide, glass, and clay products.

1. Polymers:

• Long-chain organic materials consisting of repeating molecular units.
• Thermoplastics: Soften on heating and harden on cooling. Examples: PVC, Polyethylene.
• Thermosetting Plastics: Do not soften on heating once cured. Examples: Bakelite, Epoxy.

1. Composites:

• Materials formed by combining two or more distinct materials to achieve superior properties. Examples: Fiberglass, Carbon-fiber reinforced polymers (CFRP).

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Solution Q.2 (c)

Process Analysis and Adiabatic Index (γ) Determination

Given parameters:

• Mass of gas (m) = 0.4 kg
• State 1 (Initial): P1 = 10 bar, T1 = 340°C = 340 + 273.15 = 613.15 K
• Process 1-2: Isentropic expansion to P2 = 1 bar
• Process 2-3: Constant volume heating to P3 = 3 bar and T3 = 340°C = 613.15 K
• Process 3-1: Isothermal compression back to P1 = 10 bar (at constant temperature T1 = T3 = 613.15 K)

1. Representation on p-V Diagram

`` Pressure (p) ^ | (1) Isothermal (3-1) 10bar +...... | / \ Isentropic (1-2) | / \ 3 bar +... \ | (3) \ 1 bar +............* (2) | |<--Const V (2-3) +---------------------------> Volume (V) ``

2. Analysis of Process 2-3 (Constant Volume Heating)

Since Process 2-3 occurs at constant volume (V2 = V3):

Formula: P2 / T2 = P3 / T3

Substitute values: 1 / T2 = 3 / 613.15 T2 = 613.15 / 3 = 204.38 K

3. Analysis of Process 1-2 (Isentropic Expansion)

For an isentropic process between State 1 and State 2:

T2 / T1 = (P2 / P1) ^ ((γ - 1) / γ)

Substitute the known temperatures and pressures: 204.38 / 613.15 = (1 / 10) ^ ((γ - 1) / γ) 0.3333 = (0.1) ^ ((γ - 1) / γ)

Taking natural logarithms on both sides: ln(0.3333) = ((γ - 1) / γ) ln(0.1) -1.0986 = ((γ - 1) / γ) (-2.3026)

((γ - 1) / γ) = -1.0986 / -2.3026 = 0.4771 1 - (1 / γ) = 0.4771 1 / γ = 1 - 0.4771 = 0.5229 γ = 1 / 0.5229 ≈ 1.912

Answer: The value of the adiabatic index (γ) is 1.912.

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Solution Q.2 (c) OR

Given parameters:

• Initial state: P1 = 450 kPa, V1 = 130 litres = 0.130 m³
• Final state: P2 = 150 kPa, V2 = 260 litres = 0.260 m³
• Decrease in enthalpy (Delta_H) = -55 kJ
• Constant volume specific heat (Cv) = 718 J/kgK = 0.718 kJ/kgK

(i) Determine the Change in Internal Energy (Delta_U)

Enthalpy (H) is defined as: H = U + p*V

For a change between states: DeltaH = DeltaU + Delta(pV) = Delta_U + (P2V2 - P1*V1)

Calculate initial and final pV values: P1V1 = 450 kPa 0.130 m³ = 58.5 kJ P2V2 = 150 kPa * 0.260 m³ = 39.0 kJ

Now find DeltaU: -55 kJ = DeltaU + (39.0 kJ - 58.5 kJ) -55 = DeltaU - 19.5 DeltaU = -55 + 19.5 = -35.5 kJ

Thus, the internal energy decreases by 35.5 kJ.

(ii) Determine the Value of Cp

For an ideal gas, changes in enthalpy and internal energy are given by: DeltaH = m * Cp * DeltaT DeltaU = m * Cv * DeltaT

Taking the ratio of these equations: DeltaH / DeltaU = (m Cp DeltaT) / (m * Cv * DeltaT) = Cp / Cv

Substitute the values: -55 / -35.5 = Cp / 0.718 1.5493 = Cp / 0.718 Cp = 1.5493 * 0.718 = 1.1124 kJ/kgK = 1112.4 J/kgK

(iii) Determine the Value of R

Using the gas relation: R = Cp - Cv R = 1.1124 kJ/kgK - 0.718 kJ/kgK = 0.3944 kJ/kgK = 394.4 J/kgK

Answers:

• (i) Change in internal energy (Delta_U) = -35.5 kJ
• (ii) Specific heat at constant pressure (Cp) = 1112.4 J/kgK
• (iii) Gas constant (R) = 394.4 J/kgK

Q.3 (a) What is a rigid coupling? What are its types? (03 marks) Q.3 (b) Draw a schematic diagram of the vapor compression refrigeration system. Where is the condenser located in a split air-conditioner? (04 marks) Q.3 (c) A rigid tank contains 10 kg of water at 90°C. If 8 kg of the water is in the liquid form and the rest is in the vapor form, determine (a) the pressure in the tank and (b) the volume of the tank. (07 marks)

OR

Q.3 (a) What is a friction clutch? What are its types? (03 marks) Q.3 (b) What is the difference between the working principle of vapor compression and vapor absorption refrigeration system? (04 marks) Q.3 (c) Find the dryness fraction of steam supplied in a combined separating-and-throttling calorimeter from following data: initial pressure = 10 bar, final pressure = 1 bar, water separated = 1.5 kg, steam discharged from throttling calorimeter = 20 kg, temperature of steam after throttling = 120°C. Take specific heat of superheated steam as 2.1 kJ/kg K. (07 marks)

Solution Q.3 (a)

Rigid Coupling

A rigid coupling is a type of mechanical shaft coupling used to connect two shafts that are perfectly aligned in both lateral and angular directions. It provides a solid and permanent connection and is unable to tolerate any shaft misalignment.

Types of Rigid Couplings

• Muff or Sleeve Coupling
• Clamp or Split-muff Coupling
• Flange Coupling (Protected and Unprotected types)

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Solution Q.3 (b)

1. Schematic Diagram of Vapor Compression Refrigeration System (VCRS)

`` +-----------------------+ | CONDENSER | <--- Heat Rejected +-----------------------+ ^ | | | High Pressure Liquid High Pressure | v Hot Vapor | +-------+ +----------+ | EXP. | | COMPRES- | | VALVE | | SOR | +-------+ +----------+ | ^ | Low Pressure Liquid/Vapor | v +-----------------------+ | EVAPORATOR | <--- Heat Absorbed +-----------------------+ ``

2. Condenser Location in Split Air-Conditioner

In a split air-conditioner, the condenser is located in the outdoor unit (installed outside the room being cooled), alongside the compressor and the condenser cooling fan, to dump heat into the ambient outdoor air.

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Solution Q.3 (c)

Given parameters:

• Total mass of water (m) = 10 kg
• Temperature (T) = 90°C
• Mass of liquid phase (m_f) = 8 kg
• Mass of vapor phase (m_g) = 10 kg - 8 kg = 2 kg

(a) Pressure in the Tank

Since both liquid water and steam co-exist in thermodynamic equilibrium, the mixture is in a wet saturated state. The pressure inside the tank is the saturation pressure of water at T = 90°C.

From Steam Tables at T = 90°C:

• Saturation Pressure (P_sat) ≈ 70.18 kPa (or 0.7018 bar)

(b) Volume of the Tank

From Steam Tables at T = 90°C, read specific volumes:

• Specific volume of saturated liquid (v_f) ≈ 0.001036 m³/kg
• Specific volume of saturated vapor (v_g) ≈ 2.3593 m³/kg

Total Volume (V) is the sum of the volumes of liquid and vapor phases: V = Vf + Vg = (mf * vf) + (mg * vg)

Calculate: Vf = 8 kg * 0.001036 m³/kg = 0.008288 m³ Vg = 2 kg * 2.3593 m³/kg = 4.7186 m³

Total Volume: V = 0.008288 m³ + 4.7186 m³ ≈ 4.7269 m³

Answers:

• (a) Pressure in the tank = 70.18 kPa
• (b) Volume of the tank = 4.7269 m³

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Solution Q.3 (a) OR

Friction Clutch

A friction clutch is a mechanical device used to connect a driving shaft (engine) and a driven shaft (gearbox) smoothly, utilizing the frictional resistance between mating surfaces to transmit rotational power. It enables the engine to be engaged or disengaged from the transmission while running.

Types of Friction Clutches

• Single-plate clutch
• Multi-plate clutch
• Cone clutch
• Centrifugal clutch

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Solution Q.3 (b) OR

Difference Between VCRS and VARS

| Feature | Vapor Compression System (VCRS) | Vapor Absorption System (VARS) | | :--- | :--- | :--- | | Primary Energy Input | Mechanical Work (Electricity to run compressor) | Thermal Energy (Steam, gas burners, waste heat) | | Key Component | Compressor is used to raise pressure | Absorber, Pump, Generator replace the compressor | | Refrigerant Flow | Single refrigerant (e.g., R-134a, R-410A) | Refrigerant-Absorbent pair (e.g., NH3-H2O, LiBr-H2O) | | Moving Parts & Wear | High mechanical wear due to compressor pistons/rotors | Minimal wear; only a small solution pump operates | | COP Range | Higher COP (normally 3.0 to 5.0) | Lower COP (normally 0.6 to 1.2) |

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Solution Q.3 (c) OR

Given parameters:

• Initial pressure (P1) = 10 bar
• Final pressure after throttling (P2) = 1 bar
• Mass of water separated (M) = 1.5 kg
• Mass of dry steam discharged (m) = 20 kg
• Temperature after throttling (T_sup) = 120°C
• Specific heat of superheated steam (C_ps) = 2.1 kJ/kgK

Step 1: Find properties from Steam Tables

At P2 = 1 bar:

• Saturation temperature (T_sat2) = 99.63°C
• Enthalpy of dry saturated steam (h_g2) = 2675.4 kJ/kg

At P1 = 10 bar:

• Sensible heat of liquid (h_f1) = 762.6 kJ/kg
• Latent heat of vaporization (h_fg1) = 2013.6 kJ/kg

Step 2: Determine dryness fraction from Throttling Calorimeter (x2)

Since the throttling process (State 2 to State 3) is isenthalpic (h2 = h3):

h2 = h3 hf1 + x2 * hfg1 = hg2 + Cps * (Tsup - Tsat2)

Substitute the values: 762.6 + x2 2013.6 = 2675.4 + 2.1 (120 - 99.63) 762.6 + x2 2013.6 = 2675.4 + 2.1 20.37 762.6 + x2 2013.6 = 2675.4 + 42.78 = 2718.18 x2 2013.6 = 2718.18 - 762.6 = 1955.58 x2 = 1955.58 / 2013.6 ≈ 0.9712

Step 3: Determine dryness fraction from Separating Calorimeter (x1)

x1 = m / (M + m) x1 = 20 / (1.5 + 20) = 20 / 21.5 ≈ 0.9302

Step 4: Calculate overall dryness fraction (x)

x = x1 x2 x = 0.9302 0.9712 ≈ 0.9034

Answer: The dryness fraction of the steam supplied is 0.9034 (or 90.34%).

Q.4 (a) Find the incorrect statements from the following and correct them. -Rigid flange coupling is used when axes of shafts are parallel but not in alignment. -Pin type flexible coupling is used where angular misalignment is more. -Universal coupling requires proper alignment of shaft axes. -Oldham’s coupling is for small angular misalignment. (03 marks) Q.4 (b) Describe main parts of a centrifugal pump in short. (04 marks) Q.4 (c) A Diesel engine has a compression ratio of 15 and heat addition at constant pressure takes place 6% of the stroke. Find the air standard efficiency of the engine. (07 marks)

OR

Q.4 (a) Draw a labeled diagram of a simple band brake. (03 marks) Q.4 (b) Explain any one type of rotary pumps with figure. (04 marks) Q.4 (c) In an air standard Otto cycle, the upper and lower limits of absolute temperatures are T3 and T1 respectively. Show that for maximum work, the ratio of compression should have the value r = (T3 / T1) ^ ( 1 / (2 * (gamma - 1)) ). (07 marks)

Solution Q.4 (a)

All four statements provided in the question are incorrect. Their corrections are detailed below:

1. Statement: Rigid flange coupling is used when axes of shafts are parallel but not in alignment.

• Correction: Rigid flange coupling is used when the axes of the two shafts are perfectly in alignment. If axes are parallel but not in alignment, Oldham’s coupling is used.

1. Statement: Pin type flexible coupling is used where angular misalignment is more.

• Correction: Pin-type flexible coupling is used where very small/slight angular, lateral, or axial misalignment exists. Universal coupling is used where angular misalignment is large.

1. Statement: Universal coupling requires proper alignment of shaft axes.

• Correction: Universal coupling is specifically designed to connect shafts whose axes intersect at a large angle (it does not require proper axial alignment).

1. Statement: Oldham’s coupling is for small angular misalignment.

• Correction: Oldham’s coupling is designed for shafts having lateral (parallel) misalignment, not angular misalignment.

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Solution Q.4 (b)

Main Parts of a Centrifugal Pump

• Impeller: The rotating wheel of the pump equipped with backward curved blades or vanes. It is keyed to the shaft and directly increases the kinetic energy of the liquid.
• Casing: An airtight chamber surrounding the impeller. It is designed to convert the kinetic energy of the water leaving the impeller into pressure energy before discharging it.
• Suction Pipe with Foot Valve and Strainer: The pipe that connects the sump to the impeller inlet (eye). The foot valve acts as a one-way non-return valve, and the strainer prevents debris from entering the pump.
• Delivery Pipe: The pipe connected to the outlet of the pump casing to lift water to the required discharge height. It includes a delivery valve to regulate the discharge flow rate.

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Solution Q.4 (c)

Given parameters:

• Compression ratio (r) = V1 / V2 = 15
• Heat addition takes place at 6% of the stroke
• Assumed adiabatic index for air (γ) = 1.4

Step 1: Establish volume relationships

Let:

• V1 = Volume at start of compression
• V2 = Clearance volume
• Stroke volume (Vs) = V1 - V2

Since r = V1 / V2 = 15: V1 = 15 V2 Vs = 15 V2 - V2 = 14 * V2

Step 2: Determine Cut-off Ratio (rc)

Let state 3 represent the end of the constant pressure heat addition. Volume change during constant pressure heat addition = V3 - V2

We are given: V3 - V2 = 0.06 Vs V3 - V2 = 0.06 (14 V2) = 0.84 V2 V3 = V2 + 0.84 V2 = 1.84 V2

Cut-off ratio (rc) = V3 / V2 = 1.84

Step 3: Calculate Air Standard Efficiency of Diesel Cycle

Formula: η = 1 - [ (1 / r^(γ - 1)) ( (rc^γ - 1) / (γ (rc - 1)) ) ]

Calculate individual terms:

• r^(γ - 1) = 15^(1.4 - 1) = 15^0.4 ≈ 2.9542
• rc^γ = 1.84^1.4 ≈ 2.3482
• rc^γ - 1 = 2.3482 - 1 = 1.3482
• γ (rc - 1) = 1.4 (1.84 - 1) = 1.4 * 0.84 = 1.176

Substitute the values back: η = 1 - [ (1 / 2.9542) (1.3482 / 1.176) ] η = 1 - [ 0.3385 1.1464 ] η = 1 - 0.3880 = 0.6120 or 61.20%

Answer: The air standard efficiency of the engine is 61.20%.

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Solution Q.4 (a) OR

Labeled Diagram of a Simple Band Brake

A simple band brake consists of a flexible steel band lined with friction material wrapped around a rotating drum.

`` Applied Force (P) | v +---+ (Lever Point B) | | Fulcrum (O) | T1 (Tight Side Tension) *-----------+----------------------+ | (Lever Point A) | | | | T2 (Slack Side Tension) | +------------------+ | | ..---.. | v ." ". v / \ | DRUM | | (Radius R) | \ / ". ." '---' ``

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Solution Q.4 (b) OR

External Gear Pump

The external gear pump is a common type of rotary positive-displacement pump.

`` +-----------------------+ | CASING | | +---------+ | | | GEAR A | | Inlet | +---+ (Driver) | Outlet (Suction) -> | | | | | -> (Discharge) | +---+ (Driven) | | | GEAR B | | | +---------+ | +-----------------------+ ``

Working Principle

1. Suction: As the teeth of the driver gear (Gear A) and driven gear (Gear B) disengage on the suction inlet side, a volume expansion occurs, creating a localized low pressure (vacuum) that draws fluid into the pump.
2. Trapping and Transport: The fluid is trapped in the spaces between the gear teeth and the interior contour of the pump casing. It is then transported along the outer periphery of the housing towards the discharge side.
3. Discharge: When the teeth re-engage on the outlet side, the volume reduces, squeezing and forcing the fluid out through the discharge port under pressure.

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Solution Q.4 (c) OR

Derivation for Compression Ratio for Maximum Work in Otto Cycle

Let the temperatures of the air standard Otto cycle be:

• T1 = Minimum temperature of the cycle (at start of compression)
• T2 = Temperature at end of compression
• T3 = Maximum temperature of the cycle (at end of heat addition)
• T4 = Temperature at end of expansion

Let r = compression ratio.

For isentropic processes 1-2 and 3-4: T2 = T1 * r^(γ - 1) T3 / T4 = r^(γ - 1) => T4 = T3 / r^(γ - 1)

Let us define x = r^(γ - 1). Therefore: T2 = T1 * x T4 = T3 / x

Net work output (W) of the Otto cycle per unit mass is: W = Heat Supplied - Heat Rejected W = Cv (T3 - T2) - Cv (T4 - T1) W = Cv (T3 - T1 x - T3 / x + T1)

For maximum work, differentiate W with respect to variable x and equate to zero: dW / dx = 0 Cv [ 0 - T1 - T3 (-1 / x²) + 0 ] = 0 -T1 + T3 / x² = 0 T3 / x² = T1 x² = T3 / T1 x = (T3 / T1) ^ (1/2)

Since x = r^(γ - 1): r^(γ - 1) = (T3 / T1) ^ (1/2) r = (T3 / T1) ^ ( 1 / (2 * (γ - 1)) )

Hence proved.

Q.5 (a) Which mechanical drive will you suggest in each of the following different situations: -When there is less space available. -Where ‘slip’ is permitted. -When high velocity ratio is required. 03

(b) What is isothermal efficiency, clearance ratio, and volumetric efficiency of a compressor? 04

(c) Distinguish between petrol engine and diesel engine. Use the following points: cycle of operation, compression ratio, fuel ignition, governing, engine speed, thermal efficiency, engine weight. 07

OR

(a) State three disadvantages of using gear drives. 03

(b) Give classification of air compressors. 04

(c) The following data refer to a test on I.C. engine: Indicated power = 42 kW, frictional power = 7 kW, engine speed = 1800 rpm, specific fuel consumption per B.P. = 0.30 kg/kWh, calorific value of fuel used = 43000 kJ/kg. Calculate: (i) mechanical efficiency, (ii) brake thermal efficiency, and (iii) indicated thermal efficiency. (07 marks)

Solution Q.5 (a)

1. When there is less space available: Gear Drive (highly compact).
2. Where ‘slip’ is permitted: Belt Drive (Flat belt or V-belt).
3. When high velocity ratio is required: Gear Drive (using gear train/worm-gear combination).

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Solution Q.5 (b)

Definitions in Compressors

1. Isothermal Efficiency:

• It is the ratio of work required during a theoretical isothermal compression process to the actual work consumed by the compressor during real compression.
• Isothermal Efficiency = Isothermal Work / Actual Work

1. Clearance Ratio (C):

• It is the ratio of the clearance volume (Vc) of the cylinder to the swept or stroke volume (Vs).
• C = Vc / Vs

1. Volumetric Efficiency:

• It is the ratio of the actual volume of free air delivered per stroke (measured at ambient conditions) to the swept volume (Vs) of the cylinder.
• Volumetric Efficiency = 1 + C - C * (P2 / P1) ^ (1/n)

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Solution Q.5 (c)

Distinction Between Petrol Engine and Diesel Engine

| Parameter | Petrol Engine | Diesel Engine | | :--- | :--- | :--- | | Cycle of Operation | Works on Otto Cycle (Constant Volume heat addition) | Works on Diesel Cycle (Constant Pressure heat addition) | | Compression Ratio | Low compression ratio (6 to 10) | High compression ratio (15 to 22) | | Fuel Ignition | Spark plug initiates ignition via electrical spark | Self-ignition due to hot, highly compressed air via injector | | Governing | Quantity governing (regulates intake mixture charge) | Quality governing (regulates fuel quantity injected) | | Engine Speed | High-speed engines | Low-to-medium-speed engines | | Thermal Efficiency | Lower thermal efficiency (~25% to 30%) | Higher thermal efficiency (~35% to 45%) | | Engine Weight | Lighter in weight due to lower peak pressures | Heavier in weight to withstand high compression pressures |

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Solution Q.5 (a) OR

Three Disadvantages of Using Gear Drives

1. High Cost: Manufacturing gear teeth requires specialized, high-precision machining and is relatively expensive.
2. Lubrication Maintenance: Requires regular, consistent lubrication to reduce wear, heat generation, and noise.
3. Rigidity: Because they are rigid, they cannot absorb shock or vibration, and minor shaft misalignments can cause catastrophic tooth failure.

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Solution Q.5 (b) OR

Classification of Air Compressors

Air compressors are categorized on several bases:

1. Based on Working Principle:

• Positive Displacement:
• Reciprocating (Single/Double acting, Single/Multi-stage)
• Rotary (Screw, Lobe, Vane, Scroll)
• Dynamic / Turbo Compressors:
• Centrifugal (Radial flow)
• Axial Flow

1. Based on Number of Stages:

• Single-stage
• Multi-stage (2-stage, 3-stage, etc.)

1. Based on Cooling Method:

• Air-cooled
• Water-cooled

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Solution Q.5 (c) OR

Given parameters:

• Indicated Power (IP) = 42 kW
• Frictional Power (FP) = 7 kW
• Engine Speed = 1800 rpm
• Specific Fuel Consumption per Brake Power (sfc_BP) = 0.30 kg/kWh
• Calorific value of fuel (CV) = 43,000 kJ/kg

(i) Calculate Mechanical Efficiency (η_m)

Brake Power (BP) = IP - FP BP = 42 kW - 7 kW = 35 kW

η_m = BP / IP = 35 / 42 ≈ 0.8333 or 83.33%

(ii) Calculate Brake Thermal Efficiency (η_bth)

Mass of fuel consumed per hour (mfhr) = sfcBP * BP mf_hr = 0.30 kg/kWh * 35 kW = 10.5 kg/h

Mass of fuel consumed per second (m_f) = 10.5 / 3600 kg/s ≈ 0.002917 kg/s

Brake Thermal Efficiency Formula: ηbth = BP / (mf CV) η_bth = 35 / (0.002917 43000) η_bth = 35 / 125.431 ≈ 0.2790 or 27.90%

Alternative formula directly: ηbth = 3600 / (sfcBP CV) = 3600 / (0.30 43000) ≈ 27.91%

(iii) Calculate Indicated Thermal Efficiency (η_ith)

Indicated Thermal Efficiency Formula: ηith = IP / (mf CV) η_ith = 42 / (0.002917 43000) = 42 / 125.431 ≈ 0.3348 or 33.48%

Alternatively using Mechanical Efficiency relationship: ηith = ηbth / η_m = 0.2791 / 0.8333 ≈ 33.49%

Answers:

• (i) Mechanical Efficiency = 83.33%
• (ii) Brake Thermal Efficiency = 27.91%
• (iii) Indicated Thermal Efficiency = 33.49%

A square prism, base 4 cm side and height 6.5 cm has its axis inclined at 45 to the H.P. and has an edge of its base, on the H.P. and perpendicular to V.P. Draw the orthographic projections.

Analysis of the Problem

Solid : Square Prism Base Dimensions : 4 cm × 4 cm Prism Height (Axis) : 6.5 cm Condition 1 : One edge of the base is on the Horizontal Plane (H.P.) and is perpendicular to the Vertical Plane (V.P.). Condition 2 : The axis of the prism is inclined at 45° to the H.P.

Geometric Principles

Since the axis is inclined at 45° to the H.P., the base of the prism is inclined at 90° - 45° = 45° to the H.P. One edge of this base rests on the H.P. and is perpendicular to the V.P.

Step-by-Step Construction Procedure

Step 1: Initial Position (Simple Position)

Draw a reference line XY . Assume the prism is resting flat on the H.P. with its axis vertical (perpendicular to H.P.) and one base edge perpendicular to the V.P. Draw the Top View (Plan) : Draw a square abcd of side 4 cm below XY . Position it such that sides ad and bc are perpendicular to the XY line. Draw the Front View (Elevation) : Project the points upwards to the XY line. Since the base is on the H.P., the front view of the base lies on XY . Draw a rectangle of width 4 cm and height 6.5 cm . Label the base vertices as a1'b1'c1'd1' (on XY ) and the top face vertices as a'b'c'd' (at a height of 6.5 cm ). Let the axis of this initial front view be vertical.

Step 2: Tilted Position (Final Projections)

We need to tilt the prism such that the base edge a1'd1' remains on the H.P. (on the XY line) while the axis is inclined at 45° to the H.P. This means the base itself is inclined at 45° to the H.P. Locate a point a1'(d1') on the XY line to represent the resting edge. Draw a line representing the tilted base through a1'(d1') at an angle of 45° to XY . Mark a length of 4 cm along this line to locate the opposite base corner b1'(c1') . From points a1'(d1') and b1'(c1') , draw perpendicular lines (at 90° to the tilted base line) of length 6.5 cm to represent the longer lateral edges. These lines will make an angle of 45° with XY . Connect the top endpoints to complete the tilted rectangle a'b'c'd'-a1'b1'c1'd1' . This is the final Front View . Project the final Top View :

Draw vertical projectors downwards from all eight vertices of the tilted front view. Draw horizontal projectors from the initial top view square abcd . Locate the intersecting points (e.g., the intersection of the vertical projector from a' and the horizontal projector from a yields the final plan point a ). Connect the points to complete the final plan. Use solid lines for visible boundaries and dashed lines for hidden features (such as the base edge resting on the ground which is covered by the body of the prism).

A pentagonal prism, base of 4 cm side and axis of 7 cm is resting on one of the base edges on the H.P. The longer surface containing that edge is inclined at 45 to the H.P. and parallel to the V.P. Draw the projections.

Clarification of Ambiguity

The problem statement contains a minor geometric contradiction: "The longer surface containing that edge is inclined at 45 to the H.P. and parallel to the V.P." In orthographic projection, any plane/surface that is parallel to the V.P. must be perpendicular to the H.P. (i.e., making a 90° angle with the H.P.). It cannot be inclined at 45° to the H.P.

Assumption to proceed : We assume the question implies that the resting base edge is on the H.P. and is parallel to the V.P. , and the rectangular (longer) face containing this edge is inclined at 45° to the H.P. This means the axis of the prism is parallel to the V.P. and inclined to the H.P. at 45° .

Step-by-Step Construction Procedure

Step 1: Auxiliary End View (Side View) of Base

Draw reference line XY . Draw an auxiliary projection plane X1Y1 perpendicular to XY to draw the side view first. In the side view, the pentagon will show its true shape because the axis is parallel to the V.P. Draw a regular pentagon of side 4 cm resting on one of its edges on the ground line ( XY ). Since the face containing this edge is inclined at 45° , rotate this pentagon in the side view such that the bottom edge remains on the XY line and the adjacent longer face is inclined at 45° to the H.P.

Step 2: Front View (Elevation)

Project the vertices horizontally from the side view to the main front view region. The length of the prism along the axis is 7 cm . Since the axis is parallel to the V.P., draw parallel lines of length 7 cm at an angle of 45° (or corresponding projected height) to form the lateral edges of the prism. Complete the front view by drawing the two parallel pentagonal end-faces.

Step 3: Top View (Plan)

Project the vertices from the front view downwards. Project the widths from the side view using a 45° miter line from X1Y1 . Intersect these projection lines to find the final vertices of the pentagonal prism in the top view. Connect the vertices, using solid lines for visible outlines and dashed lines for hidden edges.

Draw orthographic projection (plan and elevation and side elevation) of a rectangular plane(parallel to VP) of a size 4 cm x 6 cm, kept in such a way that the 4 cm side is making an angle of 30 with the H.P(horizontal plane).

Analysis of the Plane

Type : Rectangular plane of 4 cm × 6 cm Orientation : Parallel to the V.P. This means its Front View (Elevation) will show the true shape and size of the rectangle. Condition : The 4 cm side is inclined at 30° to the H.P. (which translates to 30° to the reference XY line in the front view). Distance Assumption : Assume the plane is located 2 cm in front of the V.P. and its lowest point is 1.5 cm above the H.P. for clarity.

Step-by-Step Drawing Steps

1. Front View (Elevation)

Draw reference line XY . Draw a line a'b' of length 4 cm making an angle of 30° with the XY line. Point a' is the lower vertex. From a' and b' , draw perpendicular lines of length 6 cm upwards to locate points d' and c' respectively (such that angle angle d'a'b' = 90° and angle c'b'a' = 90° ). Join c'd' to complete the rectangle a'b'c'd' of size 4 cm × 6 cm . This represents the true shape of the rectangle.

1. Top View (Plan)

Since the plane is parallel to the V.P., its projection on the H.P. (Top View) will be a straight line parallel to the XY line. Draw vertical projectors downwards from all four vertices ( a', b', c', d' ) of the front view. Draw a horizontal line parallel to XY at the chosen distance of 2 cm below XY . The intersections of the vertical projectors with this line give the points a, d (which coincide) and b, c (which coincide). The plan is represented by the line segment ad - bc , parallel to XY .

1. Side View (Side Elevation)

Draw a profile plane line X1Y1 perpendicular to XY . Project horizontal lines from the front view vertices ( a', b', c', d' ) toward the profile plane. Project the top view line segment to the profile plane and transfer it vertically using a 45° miter line. Since the plane is perpendicular to the Profile Plane, the side elevation will be a straight vertical line of height equal to the vertical span of the tilted rectangle in the front view.

Draw projection (plan and elevation and side elevation) of a hexagonal plane of each side 4 cm, kept in such a way that the plane is parallel to H.P. and perpendicular to V.P. The plane is kept 3 cm above H.P. and one side of the plane is 5 cm away from the V.P.

Analysis of the Plane

Shape : Regular hexagon, side = 4 cm Orientation : Parallel to the H.P. (hence perpendicular to the V.P.). Its true shape will be visible in the Top View (Plan) . Position 1 : 3 cm above the H.P. Position 2 : One edge of the hexagon is 5 cm away from the V.P. (We assume this edge is parallel to the V.P.).

Step-by-Step Drawing Steps

1. Top View (Plan)

Draw the reference line XY . Draw a horizontal line parallel to XY at a distance of 5 cm below it. This represents the edge closest to the V.P. On this line, mark a segment ab = 4 cm . Construct a regular hexagon abcdef of side 4 cm starting from the edge ab , such that the rest of the hexagon lies further below the line (away from XY ). Each interior angle of a regular hexagon is 120° .

1. Front View (Elevation)

Since the plane is parallel to the H.P., its front view will be a straight line parallel to XY . Draw a horizontal line at a height of 3 cm above the XY line. Project vertical lines upwards from all six vertices ( a, b, c, d, e, f ) of the top view. The intersections with the horizontal line give the points a'(b') , f'(c') , and e'(d') . The resulting line segment represents the elevation.

1. Side View (Side Elevation)

Draw a vertical line X1Y1 as the profile reference line. Project the heights from the front view (which is a horizontal line at 3 cm ) to the side view. Project the widths from the top view to the side view using a 45° miter line. The side view will be a horizontal line segment of length equal to the total depth of the hexagon ( 4 × cos(30°) × 2 ≈ 6.93 cm ), located at a height of 3 cm above the XY line.

A circle of dia 5 cm is perpendicular to VP, but making an angle of 45* with HP. Draw the orthographic projections.

Analysis of the Geometry

Object : Circle of diameter 5 cm Orientation : Perpendicular to V.P., making an angle of 45° with H.P. Visual outcome : Since the circle is perpendicular to V.P., its Front View is a straight line of length 5 cm (the diameter) inclined at 45° to the XY line. Its Top View will be an ellipse because the circular plane is inclined to the H.P.

Step-by-Step Drawing Steps

1. Front View

Draw the reference line XY . Draw a line segment a'b' of length 5 cm inclined at 45° to the XY line. Point a' can lie on the XY line or a short distance above it. Divide the line segment a'b' into 8 equal parts (or 12 parts for greater precision). Let's label the intermediate division points as 1', 2', 3', 4', 5', 6', 7' .

1. Top View (Plan)

Draw an auxiliary circle of diameter 5 cm below XY to act as a generator tool. Divide this circle into 8 equal parts. Project vertical lines downwards from the division points on the tilted line a'b' in the front view. From the center line of the circle generator, measure the lateral offsets (widths) of each of the 8 division points. Transfer these offsets onto the corresponding vertical projector lines in the top view. Connect the plotted points with a smooth curve to draw the resulting ellipse . The major axis of this ellipse remains 5 cm (parallel to XY ), and the minor axis is compressed to 5 × cos(45°) ≈ 3.54 cm .

What is Orthographic Projection?

Definition

Orthographic projection is a technical drawing method used to represent a three-dimensional object on a two-dimensional plane.

Key Principles

Parallel Projectors : The projection lines (projectors) originating from the object are parallel to each other. Orthogonal to Plane : The projectors intersect the projection plane at a right angle ( 90° ). Multi-View System : Since a single view cannot fully define a 3D object, multiple views are projected on mutually perpendicular planes (Horizontal Plane for Plan, Vertical Plane for Elevation, and Profile Plane for Side Elevation).

Name at-least 6 drafting instruments.

The essential instruments used in manual technical drafting include:

Drawing Board : A flat, rigid wooden board providing a smooth surface to support the drawing sheet. T-Square : An instrument used to draw accurate horizontal lines and align other tools like set-squares. Set-Squares ( 30°-60° and 45° ) : Used to draw vertical, perpendicular, and precise angular lines. Compass : Used for drawing circles, arcs, and transferring curved dimensions. Dividers : Used to step off equal distances and transfer dimensions from a scale to the drawing. Protractor : A semi-circular tool used to measure and lay out angles on the sheet.

(Alternative: Mini-Drafter, French Curves, Drawing Pencils of varying hardness like 2H, HB, H)

Define First Angle Method in Orthographic Projections

Definition of First Angle Projection

In the First Angle Projection method, the object is assumed to be placed in the First Quadrant (above the Horizontal Plane and in front of the Vertical Plane).

Key Characteristics

Position of Object : The object is placed between the observer and the plane of projection. View Arrangement : The Front View (Elevation) is drawn above the XY reference line. The Top View (Plan) is drawn below the XY line. The Left-Hand Side View is drawn on the right side of the front view. Standardization : This is the standard projection method widely adopted in India (BIS), Europe, and ISO systems.

Define Third Angle Method in Orthographic Projections

Definition of Third Angle Projection

In the Third Angle Projection method, the object is assumed to be placed in the Third Quadrant (below the Horizontal Plane and behind the Vertical Plane).

Key Characteristics

Position of Plane : The projection plane is assumed to be transparent and is placed between the observer and the object. View Arrangement : The Front View (Elevation) is drawn below the XY reference line. The Top View (Plan) is drawn above the XY line. The Left-Hand Side View is drawn on the left side of the front view. Standardization : This method is primarily used in North America (ANSI/US standards) and Japan.

Bisect a 60* angle.

Step-by-Step Geometrical Construction

To bisect a given angle of 60° to create two equal 30° angles:

Draw the Angle : Draw two lines OA and OB meeting at a common vertex O such that the angle angle AOB = 60° using a protractor. Draw First Arc : With O as the center and any convenient radius on your compass, draw an arc that cuts the line OA at point P and the line OB at point Q . Draw Intersecting Arcs :

Set the compass radius to a value greater than half of the distance PQ . With P as the center, draw an arc in the interior region of the angle. Maintaining the exact same radius, place the compass needle on Q and draw another arc intersecting the first arc at point R .

Complete the Bisector : Draw a straight line from the vertex O through the intersection point R . Result : The line OR is the bisector. Angle angle AOR = angle BOR = 30° .

Divide a 75 mm line in 9 equal parts

Step-by-Step Geometrical Construction (Acute Angle Method)

To divide a line of length 75 mm into exactly 9 equal segments without using decimal calculations:

Draw the Main Line : Draw a horizontal line segment AB = 75 mm using a scale. Draw Auxiliary Line : From end A , draw an inclined line AC making an acute angle (around 20° to 30° ) with AB extending downwards. Mark Equal Divisions :

Adjust your compass to a small, convenient fixed radius. Starting at point A , mark 9 consecutive points along the line AC . Label these points 1, 2, 3, 4, 5, 6, 7, 8, 9 such that A1 = 1-2 = 2-3 = dots = 8-9 .

Join the Endpoints : Draw a straight line connecting point 9 to point B . Draw Parallel Lines :

Using set-squares or a parallel rolling ruler, draw lines parallel to the line 9-B starting from each of the intermediate points 8, 7, 6, 5, 4, 3, 2, 1 . These parallel lines will intersect the original line segment AB at points 8', 7', 6', 5', 4', 3', 2', 1' .

Result : The segment AB is now divided into 9 mathematically equal parts, each measuring exactly 8.33 mm .

Trisect a 65 mm horizontal line.

Step-by-Step Geometrical Construction (Division into 3 Parts)

To divide a horizontal line AB = 65 mm into 3 equal segments (trisection):

Draw the Line : Draw a horizontal line segment AB = 65 mm using a scale. Draw Auxiliary Line : From point A , draw an inclined line AC making an acute angle (approx. 30° ) downwards. Mark 3 Divisions : Using a compass with a fixed convenient radius, mark 3 consecutive points along AC starting from A . Label these points 1, 2, 3 . Join End Points : Draw a straight line from point 3 to point B . Project Parallel Lines : Draw lines parallel to 3-B passing through points 2 and 1 to intersect the original line AB at 2' and 1' respectively. Result : The points 1' and 2' divide AB into three equal parts: A-1' , 1'-2' , and 2'-B . Each part measures precisely 21.67 mm .

Construct a pentagon having side of 40 mm.

Method 1: Using Interior Angles (Protractor Method)

Draw Base : Draw a horizontal line segment AB = 40 mm . Calculate Angles : The interior angle of a regular pentagon is calculated as: θ = ((5-2) × 180° / 5) = 108° \n3. Plot Angles :

At point A , measure an angle of 108° and draw a line segment AE = 40 mm . At point B , measure an angle of 108° and draw a line segment BC = 40 mm .

Locate Top Vertex D :

Set your compass to a radius of 40 mm . With C as the center, draw an arc. With E as the center, draw another arc intersecting the first one at point D .

Complete Pentagon : Join CD and ED with straight lines to get the regular pentagon ABCDE .

Method 2: Geometrical Construction (Three-Circle Compass Method)

Draw a line segment AB = 40 mm . With A as the center and radius AB , draw a circle. With B as the center and radius AB , draw a second circle. These circles intersect at points P (top) and Q (bottom). Join PQ with a light vertical line. With Q as the center and radius AB , draw a third circle. This circle cuts the vertical axis at R and the first two circles at points S and T . Draw a line starting from S passing through R to intersect the second circle at point C . Draw a line starting from T passing through R to intersect the first circle at point E . With C and E as centers and radius 40 mm , draw intersecting arcs above to locate D . Join A-B-C-D-E-A to complete the regular pentagon.