BPSC AE Paper 5 (Mechanical) 25th March 2022 Official Paper (Previous Year Paper)

Explanation:

For a multistage compression process with perfect intercooling to the initial temperature at each stage, the work of compression is minimized when the pressure ratio across each stage is the same. This is known as optimal pressure staging.

Given the compression is from P₁ = 1 bar to P₄ = 16  bar and it's a four-stage compressor, we need to find the optimal pressure ratio r_p for each stage.

The overall pressure ratio rp is the final pressure divided by the initial pressure,$r_p = \frac{P_4}{P_1}$. Once we find the overall pressure ratio, the optimal pressure ratio for each stage is the fourth root of r_p (since there are four stages), and from there, we can calculate the pressures for each intercooling stage.

Calculating the overall pressure ratio: [$r_p = \frac{P_4}{P_1} = \frac{16~ \text{bar}}{1~\text{bar}} = 16$

Since it's a four-stage compressor, we take the fourth root for the optimal pressure ratio for each stage: $r_p = \frac{P_4}{P_1} = \frac{16~ \text{bar}}{1~ \text{bar}} = 16$

This means each stage must double its inlet pressure. Since we are specifically asked for the optimum pressure in the last intercooler (which would be the inlet pressure for the fourth stage), we need to calculate the pressures at each stage:

After the first stage: $r_p = \frac{P_4}{P_1} = \frac{16~ \text{bar}}{1~ \text{bar}} = 16$

After the second stage: $P_3 = P_2 \times r_p = 2~ \text{bar} \times 2 = 4~ \text{bar}$

After the third stage (and the pressure in the last intercooler): $P_4 = P_3 \times r_p = 4~ \text{bar} \times 2 = 8~ \text{bar}$

So, the optimum pressure in the last intercooler will be 8 bar.

Explanation:

• Pressure pulse transmission in a flowing mixture provides the basis for a theoretical formulation of two-phase critical flow and sonic velocity.
• Homogeneous and separated phases are considered, showing that phase pattern plays an important role in pulse transmission.
• Graphs are presented for critical flow and sonic velocity of steam-water mixtures, based on no phase change or heat transfer across the pulse.
• Pressure pulses move with sonic velocity but the object moves with half of the sonic velocity so the concentration of pressure pulse becomes larger ahead of the object.
• Mach number is less than 1, therefore pressure pulse velocity is higher than the source velocity. This will lead to a higher concentration of pressure pulse ahead of the source or object.

Explanation:

Thermal conductivity:

• It is property of material which allow it to conduct heat from higher temperature to lower temperature.
• SI unit of the thermal conductivity is Wm-1K-1.
• Higher the thermal conductivity more easily will be the heat conduction through the substance
• Thermal conductivity is a material property. It will not differ with the dimensions of a material, but it is dependent on the temperature
• Thermal conductivity depends on the chemical composition of the substance
• The thermal conductivity of the liquids is more than the gasses and the metals have the highest
• The thermal conductivity of the gases and liquids increases with the increase in temperature
• The thermal conductivity of the metal decreases with the increase in temperature
• Thermal conductivity is affected by phase change

Ks > Kl > Kg

increasing order of thermal conductivity

Air < Steam at 1 bar < Water at 100ºC < Wood

Important Points

• The thermal conductivity of gases is inversely proportional to molecular weight. Thus, thermal conductivity decreases with increasing molecular weight.
• The thermal conductivity of non-metallic liquids generally decreases with increasing temperature, the notable exceptions is being glycerin and water. Thermal conductivity of non-metallic liquids decreases with increasing molecular weight.

Explanation:

General heat conduction equation:

For a 3-dimensional body with variable properties the heat conduction equation is given by

$\left [ \frac{\partial }{\partial x}\left ( k\frac{\partial T}{\partial x} \right ) ;+;\frac{\partial }{\partial y}\left ( k\frac{\partial T}{\partial y} \right ) ;+;\frac{\partial }{\partial z}\left ( k\frac{\partial T}{\partial z} \right ) \right ];+;\dot{q}_g=\frac{\partial }{\partial t}(ρ cT)$

For constant properties 'k', 'ρ' and 'c' can be taken out-

$\frac{\partial^2 T}{\partial x^2} ;+; \frac{\partial^2T}{\partial y^2} ;+; \frac{\partial^2 T}{\partial z^2} ;+;\frac{\dot{q}_g}{k}=\frac{1}{\alpha}\frac{\partial T}{\partial t}$

where $\alpha =\frac{k}{\rho c}$

If the heat conduction is one-dimension with constant properties and without heat generation.

$\frac{\partial^2 T}{\partial x^2} =\frac{1}{\alpha}\frac{\partial T}{\partial t}$

$\therefore \frac{\partial T}{\partial t}={\alpha}\frac{\partial^2 T}{\partial x^2}$

$\therefore \frac{\partial T}{\partial t}\propto{}\frac{\partial^2 T}{\partial x^2}$

​Concept:

Additional Information

Scattering:

• The air molecules, water vapor, and dust particles scatter the solar radiation as it passes through the atmosphere.
• A part of this scattered radiation also reaches the atmosphere.

Explanation:

Following are the most commonly used temperature measuring devices.

• Glass thermometer and other thermometers
• Thermocouples
• Thermistors
• Resistance temperature detector
• Pyrometer etc.

Thermocouples

• A thermocouple is a device for measuring temperature consisting of a pair of wires of different metals or semiconductors joined at both ends. One junction is at the temperature to be measured, the second at a fixed temperature. The electromotive force generated depends upon the temperature difference.
• A thermocouple application is found in Kilns, Gas turbine exhaust, Diesel engines, Boiler, water heaters, ovens etc.
• For measuring temperature range of -200°C to 2000°C thermocouple is used and above 2000°C pyrometer will be used because a pyrometer is used without touching the object.

Pyrometer

• A pyrometer is a device for measuring high temperatures that use the radiation emitted by a hot body as a basis for measurement.

Thermometer

• A thermometer is an instrument for measuring temperature. It usually consists of a narrow glass tube containing a thin column of a liquid which rises and falls as the temperature rises and falls.

Concept:

Thermal resistance:

• Thermal resistance is defined as the ratio of the temperature difference between the two faces of a material to the rate of flow per unit area.

          ${\bf{i}}.{\bf{e}}.;{{\bf{R}}_{{\bf{thermal}}}} = \frac{{{\bf{\Delta T}}}}{{\bf{Q}}}$

• Its unit is K/W
• The concept of thermal resistance is used to solve composite layer problems.

​For a solid plate, thermal resistance is given by:

${{\bf{R}}_{{\bf{thermal}}}} = \frac{{\bf{L}}}{{{\bf{kA}}}}$

Since the slabs are in contact with each other side by side. Therefore, the thermal resistances will be in series.

The equivalent thermal resistance to the flow of heat is given by,

${R_{eq}} = {R_1} + {R_2}$

$\frac{{{L_{eq}}}}{{{k_{eq}}A}} = \frac{{{L_1}}}{{{k_1}A}} + \frac{{{L_2}}}{{{k_2}A}}$

Since ${L_1} = {L_2} = L$

$\frac{{2L}}{{{k_{eq}}}} = \frac{L}{{{k_1}}} + \frac{L}{{{k_2}}}$

$\frac{{{k_{eq}}}}{2} = \frac{{{k_1}{k_2}}}{{{k_1} + {k_2}}}$

${k_{eq}} = \frac{{2{k_1}{k_2}}}{{{k_1} + {k_2}}}$

Important Points

Thermal resistance for various shapes: 

Concept:

COP of a Carnot Refrigerator:

\({\bf{CO}}{{\bf{P}}{\bf{R}}} = \frac{{{{\bf{T}}{\bf{L}}}}}{{{{\bf{T}}{\bf{H}}} - {{\bf{T}}{\bf{L}}}}}\)

where TL = Lower temperature, TH = Higher temperature.

Calculation:

Given:

T_H = 303 K,T_L = 273 K

Now, COP is:

\({\bf{CO}}{{\bf{P}}{\bf{R}}} = \frac{{{{\bf{T}}{\bf{L}}}}}{{{{\bf{T}}{\bf{H}}} - {{\bf{T}}{\bf{L}}}}}\)

$COP = \frac{273}{303-273}=9.1$

The given specific enthalpies are

Exit of evaporator (h₁) = 232 kJ/kg, Inlet of condenser (h₂) = 283 kJ/kg,

Exit of condenser (h₃) = 116 kJ/kg

h₄ = 116 kJ/kg      --- From p-h curve we can see that h₃ = h₄.

Now,   

\(COP =\frac{{Refrigerationeffect}}{{Workdone}}\)

${\rm{COP}} = \frac{{{\rm{h}}1 - {\rm{h}}4}}{{{\rm{h}}2 - {\rm{h}}1}}$

Substitute the values, we get

${\rm{COP}} = \frac{{232 - 116}}{{283 - 232}}$

${\rm{COP}} = \frac{{116}}{{51}}$

∴ COP = 2.27

Explanation:

• Bell Coleman cycle is also known as Reversed Brayton cycle or Reversed Joule cycle
• The working fluid of the Bell Coleman refrigeration cycle is Air.
• This system of refrigeration is used for Air Craft refrigeration and it has lightweight.

where,

1. Process 1 - 2: isentropic compression
2. Process 2 - 3: constant pressure heat rejection
3. Process 3 - 4: isentropic expansion
4. Process 4 - 1: constant pressure heat absorption

Additional Information

Air Refrigeration System and Bell-Coleman Cycle or Reversed Brayton Cycle: 

• In the air refrigeration system, the air is taken into the compressor from the atmosphere and compressed.
• The hot compressed air is cooled in a heat exchanger up to the atmospheric temperature (in ideal conditions).
• The cooled air is then expanded in an expander. The temperature of the air coming out from the expander is below the atmospheric temperature due to isentropic expansion.
• The low-temperature air coming out from the expander enters into the evaporator and absorbs the heat. The cycle is repeated.

Concept:

Coefficient of Performance of Refrigerator

${\left( {C.O.P} \right)_R} = \frac{{{T_L}}}{{{T_H} - {T_L}}}$

Where TL is the lower temperature limit and TH is the high-temperature limit

COP is also given by 

$COP = \frac{{Desired;Effect}}{{{P_{in}}}};$

Calculation: 

Given:

Desired effect = 1 TR = 3.5 kW, 1 HP = 0.746 kW, 1 TR = \(\frac{3.5}{0.746}kW=4.69HP\)

$COP = \frac{{Desired;Effect}}{{{P_{in}}}};$

$P_{in} = \frac{{4.715}}{{{COP}}};$

For refrigeration the unit commonly used is the ton. This unit is derived from the number of Btu required to freeze 1 ton of water. Since the heat of fusion of ice is approximately 144 Btu/lb, then 288,000 Btu are required to freeze 2,000 lb of water. This is the standard definition of a commercial ton of refrigeration at a uniform rate during a 24 h period.

This capacity is also the same as 12,000 Btu/h or 200 Btu/min. The efficiency of a vapor compression system is given by the horsepower used per ton of refrigeration. If the refrig- eration is N tons, or 200 N Btu/min, and since 1 hp = 42.44 Btu/min, then

$\rm \frac{hp}{N}=\frac{200}{42.44 \ \gamma}=\frac{4.72}{\gamma}$

Equation represents the horsepower per ton of refrigeration required for the cycle.

Additional Information

A ton of refrigeration (TR), also called a refrigeration ton (RT), is a unit of power used, especially to describe the heat-extraction capacity of refrigeration and air conditioning equipment.

It is defined as the rate of heat transfer that results in the melting of 1 ton of pure ice at 0°C in 24 hours.

The unit conversions is:

• 1 TR = 210 kJ/min = 3.5 kW
• 1 kCal = 4.18 kJ
• 1 kJ = 0.239 k-cal
• 210 kJ = 50.19 k-cal
• 1 TR = 50.19 k-cal/min = 3011.52 k-cal/hr

Concept:

A psychrometric chart is represented as shown in the figure.

Dry Bulb Temperature: Actual temperature of gas or mixture of gases

Wet bulb temperature: Temperature obtained by an accurate thermometer having a wick moistened with distilled water

Dew point temperature: Temperature at which the liquid droplets just appear when the moist air is cooled continuously.

Relative humidity along saturation line is 100%.

Basic Processes in Conditioning of Air:

Sensible heating: Moisture content of air remains constant so specific humidity is constant, temperature increases as it flows over a heating coil

Sensible cooling: Moisture content of air remains constant so specific humidity is constant, but its temperature decreases as it flows over a cooling coil

Cooling and dehumidification: When moist air is cooled below its dew-point by bringing it in contact with a cold surface, some of the water vapour in the air condenses and leaves the air stream as a liquid, as a result, both the temperature and humidity ratio of air decreases.

Heating and Humidification: During winter it is essential to heat and humidify the room air for comfort. It is done by first sensibly heating the air and then adding water vapour to the air stream through steam nozzles, as a result, both the temperature and humidity ratio of air increases.

Cooling & humidification: Air temperature drops and its humidity increases. 

Heating and de-humidification: This process can be achieved by using a hygroscopic material, which absorbs the water vapour from the moisture. If this process is thermally isolated, then the enthalpy of air remains constant, as a result, the temperature of air increases.

Humidity ratio/Specific humidity (ω ) (kg/kg) of a given moist air sample is defined as the ratio of the mass of water vapour (mw) to the mass of dry air (ma) contained in the sample.

$w = \frac{{{m_w}}}{{{m_a}}} = 0.622\frac{{{P_v}}}{{P - {P_v }}}$

Explanation:

Kinematic viscosity:

Kinematic viscosity is defined as the ratio of dynamic viscosity and density of the fluid.

$kinematic~ viscosity= \frac {dynamic ~ viscocity}{density ~of ~fluid}$

$ν = \frac {μ}{ρ }$

• SI unit = m2 / s
• CGS unit = Stokes or cm2 / s
• 1 Stoke = 10-4 m2 / s

Additional Information

Dynamic Viscosity  ( µ )

• Viscosity can be defined as the measure of a fluid's resistance to deformation at a given flow.
• Caused by friction within a fluid.
• Result of intermolecular forces between the particles within a fluid.
• SI unit N-s/m2 or Pa-s.

Shear stress is given by,

$Shear;stress = ;\mu ;\frac{{du}}{{dy}}$

where,

µ = Dynamic Viscosity of fluid

$\frac{u}{y}$= Rate of shear deformation

Explanation:

For a Newtonian fluid,

shear stress, $\tau \propto \frac{{du}}{{dy}}$

or, $\tau = \mu \times \frac{{du}}{{dy}}$, where τ = shear stress,

$\frac{{du}}{{dy}}=\frac{{d\theta}}{{dt}}$ = rate of shear strains,

θ = displacement in fluid layer

A stream lined body is defined as the body whose surface coincides with the stream - lines, when the body is placed in a flow. In that case the separation of flow will take place only at trailing edge (or rearmost part of the body).

Behind a stream lined body, wake formation zone will be very small and consequently the pressure drag will be very small. Thus the total drag on the stream lined body will be due to friction (shear) only.

Concept:

Head loss due to friction is given by:

${\rm{h;}} = {\rm{;}}\frac{{{\rm{f}}L{{\rm{v}}^2}}}{{2{\rm{gd}}}}$

d = diameter of pipe, f = friction factor, L = length of pipe, and v = velocity of flow

Calculation:

Given:

D = 200 mm = 0.2 m, L = 500 m, Q = 0.32 m3/s, f = 0.0225, g = 9.8 m/s2

Head loss due to friction is

${\rm{h;}} = {\rm{;}}\frac{{{\rm{f}}L{{\rm{v}}^2}}}{{2{\rm{gd}}}}$

$Q = \frac{{\rm{\pi }}}{4}{D^2}V \Rightarrow V = \frac{{4Q}}{{\pi {D^2}}}$

$h=\frac{fL}{2gd} \times (\frac{4Q}{\pi D^2})^2$

$h=\frac{0.0225 \times 500}{2 \times 9.81 \times 0.2} \times (\frac{4 \times 0.32}{\pi \times 0.2^2})^2$

h = 297.76 m

Concept:

Kaplan turbine:

• In Kaplan turbine, water flows parallel to the axis of rotation of the shaft, hence it is called axial flow turbine.
• Head at the inlet of turbine is the sum of pressure energy and kinetic energy.
• During flow of water through runner, a part of pressure energy is converted into kinetic energy.
• It indicates that Kaplan turbine is a reaction turbine. When the vanes on the hub are adjustable, the turbine is known as Kaplan turbine.
• Kaplan turbine operates at low head, high discharge.
• It has a high specific speed.
• When vanes are not adjustable, turbine is called propeller turbine. Kaplan turbine components are shown in the figure below.

Explanation:

The maximum density of water is 1000 kg/m3 at 4°C temperature.

• The maximum water density occurs at 4°C, since two opposing effects are in equilibrium at this temperature.
• The water molecules in ice are in a crystal lattice that has plenty of empty space. When the ice melts into liquid water, the structure breaks, and the liquid density increases.
• The molecules move more quickly and get further apart at temperatures well above zero. Density decreases as the temperature increases. The water also contains several ice-like clusters at temperatures close to 0°C.

Explanation:

Condition of stable equilibrium for a floating body in terms of metacentric height (GM) as follows:

• Stable equilibrium : GM > 0 (M is above G)
• Neutral equilibrium : GM = 0 (M coinciding with G)
• Unstable equilibrium : GM < 0 (M is below G)
  

The conditions of equilibrium of a floating body are: (1) stable (2) unstable and (3) neutral.

1. Stable condition.

• The equilibrium of a floating body is stable if there is a righting moment to restore it to its original position whenever it is given a small angular displacement.
• In such equilibrium, the metacenter lies above the center of gravity i.e. metacentric height is positive.
• The external couple is balanced by the couple developed by the internal forces.

2. Unstable condition.

• The equilibrium of a floating body is said to be unstable if any small tilting of the body is accompanied by the development of forces or moments which tend to further increase the displacement of the body. In unstable equilibrium, the metacenter (M) lies below the center of gravity (G) i.e. the metacentric height is negative.
• External applied couple and couple developed due to internal forces acting in the same direction, thereby increasing the displacement and making the equilibrium unstable.

3. Neutral condition.

• The equilibrium of a body is said to be neutral when on the small tilting of the body, the body takes a new position. In this equilibrium, the metacenter and center of gravity coincide.
• As neither a righting moment nor an overturning moment acts on the body, the body remains at rest a new position.

Explanation:

• Manometers are devices in which columns of a suitable liquid are used to measure the difference in pressure between two points or between a certain point and the atmosphere.
• A manometer is needed for measuring large gauge pressures. It is basically the modified form of the piezometric tube. A common type of manometer is like a transparent "U-tube" as shown in Figure:

These are of two types:

1. Simple manometers
2. Differential manometers

A simple manometer consists of a glass tube having one of its ends connected to a point where pressure is to be measured and another end remains open to atmosphere.

Concept:

Pascal’s Law:

• It was given by Blaise pascal
• It is also known as principle of transmission of fluid pressure
• It states that the pressure change at any point in a confined incompressible fluid is transmitted throughout the fluid such that the same change occurs everywhere.
• Alternate definition:- The pressure applied to any part of the enclosed liquid will be transmitted equally in all directions through the liquid.

Applications:

• The underlying principle of hydraulic jack and hydraulic press.
• Force amplification in the braking system of most motor vehicles.
• Used in artesian wells, water towers and dams.

Calculation:

Given:

Diameter of plunger, d = 1.5 cm, Diameter of ram, D = 15 cm, weight exerted on the ram, F1 = 1000 kg

According to pascals law,

Pressure applied at plunger = Pressure exerted on the ram

$\frac{{{F_1}}}{{{A_1}}} = \frac{{{F_2}}}{{{A_2}}}$

$\frac{{1000}}{{\frac{{\pi {D^2}}}{4}}} = \frac{{{F_2}}}{{\frac{{\pi {d^2}}}{4}}};$

∴ ${F_2} = \frac{{{d^2}}}{{{D^2}}} \times 1000 = \frac{{{{1.5}^2}}}{{{{15}^2}}} \times 1000 = 10 ~kg$

Here, F2 = force required on the plunger

The head loss for flow in an open channel can be expressed in similar to that for pipe the flow, using the hydraulic radius and the length of the channel and these are given by:

${\rm{h}} = {\rm{f}} \times \frac{{\rm{L}}}{{4{\rm{;R}}}} \times \frac{{{{\rm{V}}^2}}}{{2{\rm{g}}}}$

Where, R = Hydraulic radius, L = Length of channel, f = friction factor, and V = Velocity of flow

It is clear from the above that for a given value of L, R, and f losses in open channel are proportional to square of velocity.

∴ The losses is open channel vary as proportional to V².

Explanation:

Nozzle: A steam nozzle is a passage of varying cross-sections, which converts the heat energy of steam into kinetic energy. The main use of a steam nozzle is in a steam turbine, which produces a jet of steam with a high velocity.

Convergent-Divergent nozzle:  Generally the configuration of the nozzle is of the decreasing area because we deal with the subsonic flow but when the flow becomes sonic or supersonic we have to use the convergent-divergent configuration of the nozzle.

Mach no: It is defined as the ratio of the velocity of flow to the velocity of sound in the same fluid medium under the same conditions.

\({\rm{Ma;}} = \frac{{{{\rm{V}}{{\rm{flow}}}}}}{{{{\rm{V}}{{\rm{sound}}}}}}\)

• Subsonic flow: when the velocity of flow is lower than the velocity of sound in that medium then this flow is called subsonic flow. for subsonic flow Ma < 1
• Sonic flow: when the velocity of flow becomes equal to the velocity of sound. for sonic flow Ma = 1
• Supersonic flow: when the velocity of flow becomes greater than the velocity of sound. for supersonic flow Ma > 1. When the mach number between 1 and 5, it is called supersonic flow.
• Hypersonic flow: When the mach number is more than 5, it is called hyper sonic flow

Concept:

Kinematic viscosity: A quantity representing the dynamic viscosity of a fluid per unit density. 

The kinematic viscosity is expressed as:

$\nu=\frac{μ}{ρ}$

where μ = absolute or dynamic viscosity, ρ = density of the fluid.

Calculation:

Given:

μ = 1.2 × 10-4 Ns/m2, ρ = 600 kg/m3.

The kinematic viscosity is expressed as:

$\nu=\frac{μ}{ρ}$

$\nu=\frac{1.2\times10^{-4}}{600}=20\times10^{-8};m^2/sec.$

Explanation:

Reynold's number: 

• It is a dimensionless number that determines the nature of the flow of liquid through a pipe.
• It is defined as the ratio of the inertial force to the viscous force for a flowing fluid.
• Reynold's number is written as Re.
• ${Re} = \frac{{{\rm{Inertial;force}}}}{{{\rm{Viscous;force}}}}$

Additional Information

• If  Reynold's number lies between 0 - 2000, then the flow of liquid is streamlined or laminar.
• If Reynold's number lies between 2000 - 3000, the flow of liquid is unstable and changing from streamline to turbulent flow.
• If Reynold's number is above 3000, the flow of liquid is turbulent.

Explanation:

R-12, Dichloro-difluoro-methane (CCI₂F₂).

• The R-12 is a very popular refrigerant. It is a colorless, almost odorless liquid with a boiling point of -29°C at atmospheric pressure.
• It is non-toxic, non-corrosive, non-irritating, and non-flammable.
• It has a relatively low latent heat value which is an advantage in small refrigerating machines.
• A large amount of refrigerant circulated will permit the use of less sensitive and more positive operating and regulating mechanisms.
• It operates at a low but positive head and backpressure and with good volumetric efficiency.
• This refrigerant is used in many different types of industrial and commercial applications such as refrigerators, freezers, water coolers, room and window air-conditioning units, etc.
• Its principal use is found in reciprocating and rotary compressors, but its use in centrifugal compressors for large commercial air-conditioning is increasing.

Explanation:

Pitot Tube 

• Pitot Tube is a device used calculating the velocity of flow at any point in a pipe or a channel.
• The flow of water in a pipe about 3 metres in diameter can be measured by Pitot tube.
• It is based on the principle that if the velocity of flow at a point becomes zero, the pressure there is increased due to the conversion of the kinetic energy into pressure energy.

Working:

• The liquid flows up the tube and when equilibrium is attained, the liquid reaches a height above the free surface of the water stream
• Since the static pressure, under this situation, is equal to the hydrostatic pressure due to its depth below the free surface, the difference in level between the liquid in the glass tube and the free surface becomes the measure of dynamic pressure$p_0-p=\frac{\rho V^2}{2} = h\rho g$  where p0, p and V are the stagnation pressure, static pressure and velocity respectively at point A
• Such a tube is known as a Pitot tube and provides one of the most accurate means of measuring the fluid velocity

• For an open stream of liquid with a free surface, this single tube is sufficient to determine the velocity, but for a fluid flowing through a closed duct, the Pitot tube measures only the stagnation pressure and so the static pressure must be measured separately.

Additional Information

Rotameter:

• Rotameter works as a constant pressure drop variable area meter that measures the flow rate of fluid in a closed tube.

Concept:

Ideal-gas equation is given as, PV = nRoT

The characteristic gas equation is given as, PV = mRT

where R is known as Characteristic Gas Constant (or simply Gas Constant) and Ro is Universal Gas Constant.

$R = \frac{{{R_0}}}{M}$where M is molecular weight.

And Characteristic gas constant is written as, $R={C_p} - {C_v}$

where, C_p = heat capacity at constant pressure, C_v = heat capacity at constant volume

Explanation:

Boyle’s law:

According to Boyle’s law for a fixed mass of gas at a constant temperature, the volume is inversely proportional to the pressure. That means, for example, if you double the pressure, you will halve the volume. This can express mathematically as,

PV = constant

The relation PV = constant holds good only for a small range of pressure. The reason is that at high pressure the repulsive forces among the molecules increase which don’t allow the volume to decrease in the proportion it should decrease according to Boyle’s Law.

So the PV = constant doesn’t hold at high pressure.

Explanation: 

Avogadro's law:

• Avogadro law states that equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules.
• V∝ n, where n is the number of moles of the gas
• $V=k_{4}n$
• The number of molecules in one mole of a gas is 6.022×1023, which is known as the Avogadro constant.

Additional Information

• Dalton's theory was based on the law of conservation of mass and the law of constant composition.
• It is also known as the law of constant composition and the law of multiple proportions.
• Dalton’s atomic theory states that all matter is composed of atoms and all compounds are composed of atomic combinations in defined ratios.

Concept:

As the gas is heated at constant pressure, so by applying Charles’s law

$\frac{{{V_2}}}{{{V_1}}} = \frac{{{T_2}}}{{{T_1}}}$

Where V & T are volume and temperature respectively.

Calculation:

Given:

T1 = 27°C = 300 K, V1 = V, V2 = 2V

Therefore, $2= \frac{{{T_2}}}{{300}}$

T2 = 600 K = 600 - 273

= 327°C

Explanation:

Heat Pump:

A heat pump is a device that works on a reversed Carnot cycle and transfers heat from a lower temperature body to a higher temperature body.

A heat pump maintains a body at a temperature higher than the surrounding temperature or a lower temperature body.

\({\rm{CO}}{{\rm{P}}{{\rm{heat;pump}}}} = \frac{{{\rm{Refrigeration;Effect}}}}{{{\rm{Work;Input}}}} = \frac{{{{\rm{Q}}{\rm{1}}}}}{{{{\rm{Q}}{\rm{1}}} - {{\rm{Q}}{\rm{2}}}}} = \frac{{{{\rm{T}}{\rm{1}}}}}{{{{\rm{T}}{\rm{1}}} - {{\rm{T}}_{\rm{2}}}}}\)

\({\rm{CO}}{{\rm{P}}{{\rm{heat;pump}}}} = \frac{{{{\rm{T}}{\rm{1}}}}}{{{{\rm{T}}{\rm{1}}} - {{\rm{T}}{\rm{2}}}}} = \frac{{{{\rm{T}}{\rm{1}}} - {{\rm{T}}{\rm{2}}}}}{{{{\rm{T}}{\rm{1}}} - {{\rm{T}}{\rm{2}}}}} + \frac{{{{\rm{T}}{\rm{2}}}}}{{{{\rm{T}}{\rm{1}}} - {{\rm{T}}_{\rm{2}}}}}\)

\({\rm{CO}}{{\rm{P}}{{\rm{heat;pump}}}} = 1 + {\rm{;CO}}{{\rm{P}}{{\rm{Refrigerator}}}}\)

Explanation:

The thermal efficiency of the Otto Cycle:

${\eta _{otto}} = 1 - \frac{1}{{{r^{\gamma - 1}}}}$

Compression ratio: r = v1/v2

Explanation:

Heat and work, both are path functions; Their magnitude depends on the path followed during a process as well as the end state.

• Work and heat are modes of energy transfer.
• Heat is energy transferred due to temperature differences only.
• Work is the energy transfer associated with a force acting through a distance.

Relationship between heat and work:

• Both are recognized at the boundaries of a system as they cross the boundaries; That is, both heat and work are boundary phenomena
• Systems possess energy, but not heat or work
• Both are associated with a process, not a state; Unlike properties, heat or work has no meaning at a state

Properties are classified into two types.

• Intensive property: The properties of the system which are independent of the mass of a system are called Intensive property.
• E.g. pressure, density, temperature, all specific properties, etc.
• Extensive property: The properties of the system which are dependent on the mass of a system are called Extensive property.
• E.g. energy, volume, entropy, etc.

Explanation:

• For cold starting (simply when your engine parts are not warmed) - 9 : 1 (rich on fuel).
• For Idling (when you stop at signal without turning off engine) - 10 : 1 (rich mixture).
• For cruising (when you run the engine at constant medium/economy speed) - 16:1(lean mixture).
• When you accelerate bike/car - 13 : 1 (Rich mixture but require less fuel than idling).

To achieve maximum power and quick speeding/overtaking, the engine needs a ‘rich’ mixture. It is about 12 - 13 parts air by weight to 1 part of fuel (12-13 : 1 air-fuel ratio).

For chemically correct, or stoichiometric combustion λ (λ = 1), is the ratio of air supplied to air needed for complete combustion.

Explanation:

If there is a relation between x, y & z, then z may be expressed as a function of x & y.

An equation of type dz = Mdx + Ndy  is exact differentials, it can be written as ${\left( {\frac{{\partial M}}{{\partial y}}} \right)_x} = {\left( {\frac{{\partial N}}{{\partial x}}} \right)_y}$

For a pure substance undergoing a reversible process, these equations hold true:

1. dU = T.ds – p.dv
2. dH = T.ds + v.dp
3. dF = -p.dV – s.dT
4. dG = v.dP –s.dT

They can be written as

1. ${\left( {\frac{{\partial T}}{{\partial v}}} \right)_s} = - {\left( {\frac{{\partial p}}{{\partial s}}} \right)_v}$
2. ${\left( {\frac{{\partial T}}{{\partial p}}} \right)_s} = ;{\left( {\frac{{\partial v}}{{\partial s}}} \right)_p}$
3. ${\left( {\frac{{\partial p}}{{\partial T}}} \right)_v} = ;{\left( {\frac{{\partial s}}{{\partial v}}} \right)_T}$
4. ${\left( {\frac{{\partial v}}{{\partial T}}} \right)_p} = - {\left( {\frac{{\partial s}}{{\partial p}}} \right)_T}$

These are known as Maxwell’s Equation.

Out of these options, 4 does not match with Maxwell’s Equation.

Other Maxwell relations are:

if x =  ϕ(y, z)

$dx = {\left( {\frac{{\partial x}}{{\partial y}}} \right)_z}dy + {\left( {\frac{{\partial x}}{{\partial z}}} \right)_y}dz$

$or,;{\left( {\frac{{\partial x}}{{\partial y}}} \right)_z}{\left( {\frac{{\partial z}}{{\partial x}}} \right)_y}{\left( {\frac{{\partial y}}{{\partial z}}} \right)_x} = - 1$

If F = ϕ(x, y, z)

${\left( {\frac{{\partial x}}{{\partial y}}} \right)_F}{\left( {\frac{{\partial y}}{{\partial z}}} \right)_F}{\left( {\frac{{\partial z}}{{\partial x}}} \right)_F} = 1$

Concept:

• Diesel engines or (C.I. Engines) are heavier than petrol engines (S.I. Engines) because of the large compression ratio in the case of diesel engines.
• The reason for the large compression ratio is compression ignition i.e. in diesel engines, ignition takes place by the high temperature developed by the compression since there is no spark plug like in petrol or S.I. Engine.
• Also, the flashpoint at which fuel gets ignited is more in the case of diesel engines so the compression ratio needs to be sufficient enough to produce that self-ignition temperature of the fuel for the combustion to take place.
• Compression ratio (r): It is defined as the ratio of volume before compression to the volume after compression.
• $r = \frac{{({V_c} + {V_s})}}{{{V_c}}};$
• where Vc and Vs are clearance and swept volume respectively.
• The value of the compression ratio for diesel engines lies between 16 : 1 to 22 : 1 whereas for the petrol engine it lies between 6 : 1 to 12 : 1.
• A 5 BHP engine running at full load would consume diesel of the order of 1 kg/hr

Important Points

Difference between C.I. Engine and S.I. Engine:

Explanation:

• High carbon content in diesel oil used for diesel engines leads to deposition on engine parts.
• High carbon content leads to incomplete combustion of carbon atoms, as a result, the engine produces soot particles. Carbon deposits on hot chamber components generally occur because of incomplete combustion of fuel. Such deposits enhance hot corrosion.
• The dissolution of the oxides scale is accelerated and sulfur penetration into the base alloy is promoted by the presence of carbon.
• There are two main types of deposition mechanisms: Decomposition of hydrocarbons to elemental carbon and hydrogen, and polymerization of hydrocarbon species into poly-nuclear aromatic hydrocarbons.
• The carbon deposits are distinguished by their origin between carbon and coke. Carbon is typically a product of carbon mono-oxide dissociation, whereas coke is produced by decomposition or condensation of hydrocarbons. By further dehydrogenation, coke can be transformed into carbon.
• The amount of carbon dioxide produced is a function of the carbon content of the fuel.
• The heat content produced when a fuel is burned is mainly determined by the carbon and hydrogen content of the fuel.

The high content of carbon in diesel fuel causes harmful deposits to form in fuel injectors, intake valves, and combustion chambers and can cause: Lower gas mileage, poor acceleration and lost power, hard starts, auto-ignition (knock and ping), increased exhaust emissions.

Explanation:

• In diesel engines, only air is sent into the combustion chamber during induction. This air is compressed during the compression stroke and towards the end of the compression stroke, fuel is injected by the fuel injection system into the cylinder - just before the desired start of combustion. Liquid fuel is injected at high velocities as one or more jets through small orifices or nozzles in the injector tip.
• Since the air temperature and pressure are above the fuel’s ignition point, spontaneous ignition of portions of already mixed fuel and air occurs after a delay period of a few crank angle degrees.
• So, in the case of a diesel engine, the mixing of fuel and air takes place in the Engine cylinder.

Important Points

• There are certain basic differences existing between the combustion process in the SI and CI engines.
• In SI Engines, a homogeneous carburetted mixture of gasoline and air, in a certain proportion, is compressed (Compression ratio 6 : 1 to 10 : 1) and the mixture is ignited at one place before the end of the compression stroke by mean of an electric spark (Spark plug). A single flame front progresses through the air-fuel mixture after ignition.
• In the CI engine, only air is compressed through a high compression ratio (16 : 1 to 20 : 1), raising its temperature and pressure to a high value. Fuel is injected through one or more jets into this highly compressed air in the combustion chamber.

Explanation:

• Although Nitrogen is the most abundant gas in the atmosphere, it does not take part in the combustion of IC Engines because Nitrogen is a very inert gas.

• In nitrogen gas, i.e. N₂, there are three bonds between the nitrogen atoms.

• Therefore it is very difficult to break these bonds.

• Since a chemical reaction occurs by breaking and making chemical bonds, nitrogen is relatively very inert and therefore it gets exhausted without combustion and transformation.

• Below is the combustion of propane in atmospheric air with perfect combustion:

${C_3}{H_8} + ;5\left[ {{O_2} + ;\frac{{79}}{{21}}{N_2}} \right]; \to 4{H_2}O + 3C{O_2} + ;\frac{{79 \times 5}}{{21}}{N_2}$

Explanation:

• Petrol ignites by a spark from the spark plugs, diesel in a petrol engine clogs up the spark plugs and fuel system. That means the engine will not run.
• When we put diesel in petrol engine the engine will not work because the flashpoint of diesel is very high ( 50°C ) as compared to that of petrol ( -43°C ) so vapour will not form of diesel fuel.
• As petrol engine has carburetor which mixes fuel and air and supply to the engine but here there will be no combustion and vapour formed hence the engine will not run.

Concept:

Sensitivity of governor: 

• The extent of sleeve movement for a change in speed of the governor is known as the Sensitivity of the governor.
• In the other words, for the same sleeve movement, the governor in which less change in speed is required will be more sensitive.
• For the same sleeve movement, $Sensitivity = ;\frac{{Range;of;speed}}{{Mean;speed}}$ = $\frac{{{N_2} - ;{N_1}}}{{\frac{{{N_1} + ;{N_2}}}{2}}}$

Calculation:

Given: 

N₁ = 970 rpm N₂ = 980 rpm

Sensitivity = $\frac{{{N_2} - ;{N_1}}}{{\frac{{{N_1} + ;{N_2}}}{2}}}$ = $\frac{{980 - 970}}{{\frac{{980 + 970}}{2}}} = ;.01 = 1;\%$

So, the sensitivity of the governor is 1 %.

Explanation:

Dry steam, or saturated steam, is characterized by the vapor quality or dryness fraction(x), which is equal to unity.

When the vapor quality is equal to 0, it is referred to as the saturated liquid state (single-phase).

On the other hand, when the vapor quality is equal to 1, it is referred to as the saturated vapor state or dry steam (single-phase).

Between these two states, vapor-liquid mixture or wet steam (two-phase mixture) exist. At constant pressure, the addition of energy does not change the temperature of the mixture, but the vapor quality and specific volume change.

In the case of dry steam (100% quality), it contains 100% of the latent heat available at that pressure. Saturated liquid water, which has no latent heat and therefore 0% quality, will therefore only contain sensible heat.

Dryness fraction is defined as the ratio of the mass of dry steam and the combined mass of dry steam & mass of water in the mixture. It is denoted by x.

$Dryness;fraction;\left( x \right) = \frac{{mass;of;vapour;\left( {{m_v}} \right)}}{{mass;of;vapour;\left( {{m_v}} \right) + mass;of;liquid;\left( {{m_l}} \right)}}$

For saturated liquid x = 0

For saturated vapour = 1

The value of the dryness fraction lies between 0 and 1.

Note:

The term quality is used in the wet vapor zone and is of no significance in superheated or compressed liquid regions.

Heating of dry steam above saturation temperature is known as Superheat.

Explanation:

• Dryness fraction is defined as the ratio of the mass of dry steam and the combined mass of dry steam & mass of water vapour in the mixture. It is denoted by x.

$Dryness;fraction;\left( x \right) = \frac{{mass;of;vapour;\left( {{m_v}} \right)}}{{mass;of;vapour;\left( {{m_v}} \right) + mass;of;liquid;\left( {{m_l}} \right)}}$

• For saturated liquid x = 0
• For saturated vapour x = 1
• The value of the dryness fraction lies between 0 and 1.

Calculation:

Given:

m_v = 0.4 kg, m = 1 kg

$x= \frac{{mass;of;vapour;\left( {{m_v}} \right)}}{{mass;of;vapour;\left( {{m_v}} \right) + mass;of;liquid;\left( {{m_l}} \right)}} = \frac{{{m_v}}}{{{m_v} + {m_f}}} = \frac{{0.4}}{1} = 0.4$

Explanation:

Boiler Feed Waters (Water for Steam Making)

• Water is used directly or indirectly in many industrial processes.
• The problem of preparing feed water for boilers is common in many industries. Its importance and complexity are obvious from the large number of technical investigations made in this field.

The treatment of boiler feed waters may be broadly divided into two categories:

1. External treatment. In this, the scale-forming and corrosive impurities are removed from the water before it enters the boiler.
2. Internal treatment. In this, dosages of various materials are added to the boiler to react with the non-carbonate hardness, reduce the sticking tendency of scale-forming materials, and neutralize or overcome corrosive tendencies.

In many cases, both the above forms of treatment are employed.

In the external treatment of boiler feed water, one or more of the following processes are usually employed:

1. Removal of "hardness" or scale-forming salts

• Lime-soda process
• Zeolite process
• Demineralization
• Distillation.

2. Removal of silica:

• Magnesia process, ferric sulfate process, anion-exchange process, Fluosilicate process.

3. Removal of the suspended matter:

• pre-treatment: sedimentation, coagulation, settling, and filtration.
• softening processes.

4. Removal of dissolved gases: CO₂ and O₂

• Decarbonation by aeration
• Deaeration
• Chemical treatment.

5. Removal of oil

• Mechanical oil-separators
• Sodium aluminate treatment.

Most of these processes have been discussed earlier and others will be discussed later, along with the methods for internal treatment.

Explanation:

• Axial flow compressors usually consist of a series of stages
• Each stage comprises of a row of rotor blades followed by a row of stator blades
• The working fluid is initially accelerated by the rotor blades and then decelerated in the stator passages
• In the stator, the kinetic energy transferred in the rotor is converted to static pressure
• This process is repeated in several stages to yield the necessary overall pressure ratio
• Degree of reaction provides a measure of the extent to which the rotor contributes to the overall pressure rise in the stage
• For all types of axial flow compressors, the degree of reaction is usually 0.5.

${R_x} = \frac{{{\rm{Static;enthalpy;rise;in;the;rotor}}}}{{{\rm{Stagnation;enthalpy;rise;in;the;stage}}}}=\frac{\Delta h_m}{\Delta h_m+\Delta h_f}$

Explanation:

Stage efficiency $= {\eta _s} = \frac{{{\rm{Work done by the rotor}}}}{{{\rm{Isentropic enthalpy drop}}}}$

or, = $\frac{{Blade,speed}}{{Fluid,,velocity,,at,the,,blade,,inlet}}$

∴ The maximum value of blade efficiency

\({({\eta b}){\max }} = \frac{{{{\cos }^2}\alpha }}{2},(1 + kc)\)

For equation blades, \({({\eta b}){\max }} = \frac{{{{\cos }^2}\alpha }}{2}(1 + kc)\) 

If the friction over blade surface is neglected

\({({\eta b}){\max }} = {\cos ^2}\alpha \)

Explanation:

• Compounding of steam turbines is the method in which energy from the steam is extracted in a number of stages rather than a single stage in a turbine.
• A compounded steam turbine has multiple stages i.e. it has more than one set of nozzles and rotors, in series, keyed to the shaft or fixed to the casing, so that either the steam pressure or the jet velocity is absorbed by the turbine in number of stages.
• Compounding of steam turbines are done, where rotor speed (flow speed) is reduced in steps, this prevent over speeding of turbine, which could lead to mechanical failures of turbine blades.

Concept:

The work done during compression of air is given by 

dw = ∫ vdP

That is the area under curve projected on P–axis (on PV plot)

Now, between two same pressure limits P1 and P2 (P2 > P1), the final volume of the air under various processes is

Say $\frac{P_2}{P_1}=r$,

Process-1 (Isothermal):

P1V1 = P2V2

${V_2} = \frac{{{V_1}}}{r}$

Process-2 (Adiabatic):

$P_1V_1^{\gamma}=P_2V_2^{\gamma}$

${V_2} =\frac{V_1}{r^{\frac{1}{\gamma}}}$

Comparing denominators, (V2) Isothermal < (V2) Adiabatic

Now plotting the processes on the PV plot, we get

From the figure, it is clearly evident that the area under the curve (work done) is minimum for isothermal compression and maximum for isentropic compression.

∴ work input is maximum for Adiabatic compression

Concept: 

Processes: -

1. 1-2: isentropic compression
2. 2-3: constant pressure energy addition
3. 3-4: isentropic expansion
4. 4-1: constant pressure energy rejection

Energy added, Q1 = m Cp (T3 - T2)

Energy rejected, Q2 = m Cp (T4 - T1)

Thermal efficiency, 

$\eta = \frac{{{Q_1} - {Q_2}}}{{{Q_1}}} = 1 - \frac{{{T_4} - {T_1}}}{{{T_3} - {T_2}}}$

$\eta = 1 - \frac{{{T_1}\left[ {\left( {\frac{{{T_4}}}{{{T_1}}}} \right) - 1} \right]}}{{{T_2}\left[ {\left( {\frac{{{T_3}}}{{{T_2}}}} \right) - 1} \right]}}$

The pressure ratio of the Brayton cycle, rp is defined as, 

${r_P} = \frac{{{P_2}}}{{{P_1}}}$

Processes 1-2 and 3-4 are isentropic. Hence,

$\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^{\frac{{\left( {\gamma - 1} \right)}}{\gamma }}}$

$\frac{{{T_3}}}{{{T_4}}} = {\left( {\frac{{{P_3}}}{{{P_4}}}} \right)^{\frac{{\left( {\gamma - 1} \right)}}{\gamma }}}$

We get,

$\frac{{{T_2}}}{{{T_1}}} = \frac{{{T_3}}}{{{T_4}}}$

$\eta = 1 - \frac{{{T_1}}}{{{T_2}}} = 1 - {\left( {\frac{{{P_1}}}{{{P_2}}}} \right)^{\frac{{\gamma - 1}}{\gamma }}}$

$\eta = 1 - {\left( {\frac{1}{{{r_p}}}} \right)^{\frac{{\gamma - 1}}{\gamma }}}$

Hence, the Ideal Efficiency of simple gas turbine cycle depends on Pressure ratio Only.

Mistake Points

• In the Practical cycle case, the cycle efficiency depends on the Maximum temperature as well as the Pressure ratio.