BPSC AE Paper 5 (Mechanical) 2019 Official Paper (Previous Year Paper)

Concept:

COP of Heat Pump is given as, $COP=\frac{T_H}{T_H-T_L}$

Calculation:

Given:

T_L = -3°C = 270 K, T_H = 37°C = 310 K

$COP = \frac{{{T_H}}}{{{T_H} - {T_L}}} = \frac{{310}}{{310-270}} = \frac{{310}}{{40}} = 7.75$

Concept:

• The process in which the moisture or water vapor or the humidity is removed from the air keeping its dry bulb (DB) temperature constant is called the dehumidification process.
• This process is represented by a straight vertical line on the psychrometric chart starting from the initial value of relative humidity, extending downwards and ending at the final value of the relative humidity.
• The process in which the moisture or water vapor or humidity is added to the air without changing its dry bulb (DB) temperature is called a humidification process.

Explanation:

Knocking in CI engine:

• In CI engine, the knocking tendency increases with a decrease in compression ratio.
• If the compression ratio is low in CI engine, there will not be enough temperature generation for ignition of the fuel-air mixture.
• So fuel will get accumulated in the chamber.
• And when ignition occurs, due to this fuel accumulation, explosive combustion will take place leading to noise and vibration, technically known as detonation

Explanation:

Specific Fuel Consumption (sfc):

• The fuel consumption characteristics of an engine are generally expressed in terms of specific fuel consumption in kg of fuel per kW-hr. It is an important parameter that reflects how good engine performance is. It is inversely proportional to the thermal efficiency of the engine.
• Brake specific fuel consumption (bsfc) and indicated specific fuel consumption (isfc) is the specific fuel consumptions on the basis of brake power (bp) and indicated power (ip).
• $sfc = \frac{{Fuel;consumption;per;unit;time}}{{power}}$
• The specific fuel consumption per kW-hr for a diesel engines is very close to 0.27 kg / kW-hr.

Concept:

Equivalent Evaporation:

• Generally, the output or evaporative capacity of the boiler is given as kg of water evaporated per hour but as different boilers generate steam at different pressures and temperature (from feed water at different temperatures) and as such have different amounts of heat, the number of kg of water evaporated per hour in no way provides the exact means for comparison of the performance of the boilers.
• Hence to compare the evaporative capacity or performance of different boilers working under different conditions it becomes imperative to provide a common base so that water is supposed to be evaporated under standard conditions.
• The standard conditions adopted are: Tem­perature of feed water is 100°C and converted into dry and saturated steam at 100°C.
• As per these standard conditions, 1 kg of water at 100°C necessitates 2267 kJ (539 kcal in MKS units) to get converted to steam at 100°C.
• Thus the equivalent evaporation may be defined as the amount of water evaporated from water at 100°C to dry and saturated steam at 100°C per kg of coal burnt.

Explanation:

Compressor Surge:

• Compressor surge is associated with the stalling of the compressor blades due to high positive incidence. The following is a simplified explanation of surge:
• The amount of diffusion has to be controlled and an excessive amount of diffusion will cause the flow to separate from the blade contours, resulting in stalling. Stalling gives rise to a significant loss in efficiency due to high profile losses.
• Closing the throttle valve, can produce such conditions. For a given temperature rise across the compressor, the pressure ratio will fall as the efficiency decreases due to stalling.
• If the efficiency loss due to stalling is sufficiently large and rapid, the required discharge pressure cannot be delivered by the compressor and the flow will reverse in the suction side of the compressor.
• This flow reversal results in a reduction in pressure in the discharge volume downstream of the compressor.
• When the pressure in the discharge volume has decayed sufficiently, the flow will progress in the normal direction and the discharge pressure will build up (normal compression).
• Since the conditions that gave rise to the flow reversal in the first instance still prevail, the flow can again break down (separate) and reverse.
• Such a reversal can occur at quite a high frequency, referred to as surge, and can be very destructive to a compressor. This recompression of what is effectively the compressor discharge air, already at a high temperature, results in very high compressor discharge temperatures during the surge.

Fig. 1

Figure showing theoretical characteristics showing regions and surge operations.

Explanation:

Neutron Reproduction Factor (K) and Critical Size:

All the neutrons produced may not cause further fission. The neutrons may be lost due to:

• Leakage from the system (or fissionable matter) and
• Capture by impurities or non-fissionable uranium atoms.

Whether chain reaction once started will remain steady, increase or decrease depends on a parameter called the neutron reproduction factor (K). 

Mathematically,

\(K = \frac{Rate~ of~ production~ of neutrons}{Rateoflossof~neutrons}\)

• If K = 1, the chain reaction will proceed steadily, i.e. it will be sustained. The size of the fissionable material corresponding to K = 1 is called Critical Size and the reactor is called a critical reactor.
• If K > 1, the rate of chain reaction keeps on increasing resulting in an explosion. The size of the fissionable material is greater than the critical size called supercritical size. This is the working principle of atomic bombs.
• If K < 1, the no. of fissions rapidly decreases and the process dies down.

The value of K should be kept slightly greater than 1 i.e. around 1.04 to overcome the neutron loss due to leakage. Maintaining K value exact is the most difficult problem in reactor control.

Explanation:

Volumetric efficiency of an engine

• Volumetric efficiency is a measure of the success with which the air supply and thus the charge is inducted into the engine
• It is a very important parameter since it indicates the breathing capacity of the engine.
• Volumetric efficiency is defined as the ratio of mass of air actually drawn into the cylinder to the mass of air that ideally could be drawn into cylinder.
• \({\eta v} = \frac{{m{air}^.\left( {measured} \right)}}{{{\rho _{air}}{V_d} \times \frac{N}{n}}}\) where V_d = volume of the cylinder, N = rotation per sec, n = no. of rotation per cycle
• For 4-stroke engine, n = 2 and for 2-stroke engine, n = 1

Important Points

• The volumetric efficiency of a SI engine is generally lies in the range of 71% - 90%.
• The volumetric efficiency of a CI engine is generally lie in the range of 85 – 90%

Concept:

Centrifugal compressor

Operating Principle:

A centrifugal compressor works by converting kinetic energy into pressure energy. The gas enters the impeller, where it gains kinetic energy through the impeller's high-speed rotation. This kinetic energy is then converted into pressure energy in the diffuser.

Performance Parameters:

Pressure Ratio:

• Centrifugal compressors typically achieve a low to moderate pressure ratio (ranging from 1.1 to 4) per stage.
• This makes them suitable for applications where the required pressure increase is not very high.

High Mass Flow:

• Due to their radial flow design, centrifugal compressors can handle high volumetric flow rates effectively.
• They are often used in situations requiring high mass flow rates at moderate pressures.

Applications:

They are commonly used in applications where moderate pressure ratios and lower mass flow rates are required. Examples include refrigeration systems, small gas turbines, turbochargers, and HVAC systems.

Advantages:

• Compact and robust design.
• Less sensitive to variations in flow.
• Capable of handling a wide range of gases and operating conditions.
• Simpler and more economical maintenance compared to axial compressors.

Pressure ratio for centrifugal compressor = 4 : 1

Pressure ratio for axial flow compressor = 1.2 : 1

Explanation:

In convection studies, it is common practice to nondimensional the governing equations and combines the variables which group together into dimensionless numbers in order to reduce the number of total variables.

Nusselt number is a parameter to non-dimensional heat transfer coefficient.

Non-dimensional groupings

• Nusselt No. Nu = hLc/k = (convection heat transfer strength)/ (conduction heat transfer strength)
• Prandtl No. Pr = n/a = (momentum diffusivity)/ (thermal diffusivity)
• Reynolds No. Re = Ux/ν = (inertia force)/ (viscous force)

The dimensionless parameter which represents the natural convection effects and is called the Grashof number.

Grashof number, Gr, as the ratio between the buoyancy force and the viscous force:

$Gr = \frac{{g\beta \left( {{T_s} - {T_\infty }} \right)L_c^3}}{{{\nu ^2}}}$

Nusselt number is a function of the Grashof number and the Prandtl number alone. Nu = f (Gr, Pr)

Concept:

• Thermal radiation is the transmission of heat in the form of radiant energy from one body to another body by the thermal motion of charged particles of matter.
• The mechanism of radiation is divided into 3 phases
• Conversion of thermal energy of the hot source into electromagnetic waves
• Passage of wave motion through intervening medium
• Transformation of waves into heat
• Thermal radiation is limited to wavelengths ranging from 0.1 to 100 µm of the electromagnetic spectrum.

It includes some portion of UV radiation (0.1 to 0.4 µm), entire visible radiation (0.4 to 0.7 µm), and entire infrared radiation (0.7 to 100 µm)

Explanation:

Adiabatic process: 

When a thermodynamic system undergoes a change in such a way that no exchange of heat takes place between the system and surrounding, the process is known as an adiabatic process.

ΔQ = 0

Entropy (S): 

It is a measure of the disorder of the molecular motion of a system.

Greater is the disorder, greater is the entropy.

The change in entropy is,

${\rm{\Delta }}S = \frac{{{\rm{\Delta }}Q}}{T}$

where, ΔQ = heat exchange, T = temperature of the system.

For a reversible adiabatic process

$dS = \frac{{dQ}}{T} = 0$

Entropy remains constant in an adiabatic process which is also reversible.

For the sake of simplicity consider the case of a closed system, i.e. a control mass, which does not exchange any mass with the surroundings.

There are two ways in which the entropy of such a system can change, firstly through heat transfer at the system boundary and secondly through entropy generation inside the system.

Now the process is adiabatic, so the heat transfer is zero and so the entropy change is zero through heat transfer.

Also, the process is reversible, so there is no entropy generated inside the system and the entropy change due to entropy generation is zero.

In this way, the entropy change is zero for an adiabatic process which is also reversible.

Explanation:

Joule – Thomson coefficient:- When the gas in steady flow passes through a constriction, e.g. in an orifice or valve, it normally experiences a change in temperature. From the first law of thermodynamics, such a process is isenthalpic and one can usefully define a Joule – Thomson coefficient as

$\mu = {\left( {\frac{{\partial T}}{{\partial P}}} \right)_H}$

As a measure of the change in temperature which results from a drop in pressure across the construction.

• For an ideal gas, μ = 0, because ideal gases neither warm not cool upon being expanded at constant enthalpy.
• If μ is +ve, then the temperature will fall during throttling.
• If μ is -ve, then the temperature will rise during throttling.

Thickness of Laminar Boundary layer:

$δ = \frac{{5x}}{{\sqrt {R{e_x}} }}$

where, x = distance from the leading edge

Rex = local Reynold's Number = $R{e_x} = \frac{{ρ Vx}}{μ } = \frac{{Vx}}{ν }$

where, ρ = density of fluid in kg/m3, V = average velocity in m/s

μ = dynamic viscosity in N-s/m2 and ν = kinematic viscosity in m2/s.

$δ = \frac{{5x}}{{\sqrt {\frac{{\rho Vx}}{\mu }} }} ∝ \frac{x}{{\sqrt x }}$ thus, δ ∝ x1/2

Explanation:

Shielding of Nuclear Reactor:

• A nuclear reactor is a prolific source of potentially dangerous nuclear radiation. It is unavoidable since most of the radiation released originates with the fission process itself.
• In addition to the highly energetic neutrons and gamma rays that are emitted simultaneously with the fission event the fission fragments formed are highly radioactive nuclides that emit α and β radiations. The essential shielding problem is the attenuation of the penetrative fast neutrons and high energy gamma rays that are released in the reactor core and reactor shield.
• The charged particle released, such as α and β radiation, and the heavily ionising fission fragments are readily stopped within at most few centimeters of dense material so that the charged particle does not penetrate beyond the fuel elements and certainly not beyond the reactor core.
• The shielding of charged particle radiation is therefore much less severe problem and shielding is done to protect against highly energetic neutrons and gamma rays.

Explanation

The significance of Reynolds’s analogy is that we would be able to predict the local convective heat transfer coefficient hx at any location x on the flat plate just by knowing the local skin friction coefficient at the same location of x even when there is no heat transfer between the plate and the fluid.

In short, Reynolds’s analogy shows the interrelationship between fluid friction and Newton’s Law of viscosity.

According to this. 

$\frac{{{N_u}_x}}{{{R_e}_x{P_r}}} = {S_t}_x = \frac{{{C_f}_x}}{2}$

where

Nux = Local Nusselt number, Rex = Local Reynolds number, Stx = Local Stanton number, Pr = Prandtl number, Cfx = Skin friction coefficient

Explanation:

Gas turbine: 

• An open cycle or closed-cycle gas turbine is a turbine that uses a gas (e.g. air, nitrogen, helium, argon, etc.) for the working fluid as part of a closed thermodynamic system.

• Heat is supplied from an external source.

• Such recirculating turbines follow the Brayton cycle.

• Gas turbines operate on the Brayton cycle/Joule cycle. The Joule cycle consists of four internally reversible processes:

• Isentropic compression (in a compressor)

• Constant-pressure heat addition

• Isentropic expansion (in a turbine)

• Constant-pressure heat rejection

• Heat is added by combustion in the Gas turbines.

• In this process, energy is put into the system by burning fuel in flowing gas.

• The combustion is maintained in a stream of air moving with a high velocity in the region of 30-60 m/s.

• The air-fuel ratio vary from 60 : 1 to 120 : 1.

• The stoichiometric ratio is approximately 15 : 1.

Concept:

Compressor: 

• It is a device used to increase the pressure of a gas.
• The volumetric efficiency of a compressor is defined as the ratio of actual volume sucked by the compressor at the inlet to the swept volume.
• volumetric efficiency is given by,
• ${\eta _v} = 1 + c - c{\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^{\frac{1}{n}}}$
• ⇒ ${\eta _{vol}} = 1 - C\left[ {{{\left( {\frac{{{p_2}}}{{{p_1}}}} \right)}^{\frac{1}{n}}} - 1} \right]$
• where C is clearance ratio it is the ratio of clearance volume to swept volume.
• P₂ is the delivery pressure and P₁  is the suction pressure.
• So from the formula, it is shown with an increase in clearance ratio and pressure ratio volumetric efficiency is decreased.

Volumetric efficiency decreases when

1. Outlet pressure Increases
2. Clearance ratio Increases

Volumetric efficiency increases when

1. 'n' increases
2. Inlet temperature increases

As the clearance ratio increases, the volumetric efficiency of the reciprocating compressor decreases as the pressure ratio increases. Options 2 and 4 both are correct but here, we have to select the most appropriate answer among the given options. So, here option 4 is the correct one.

Concept:

Work ratio: 

• It is defined as the ratio of net-work to turbine work.
• The gas turbine works on the Brayton cycle.
• This cycle consists of two adiabatic processes and two isobaric processes.

​

$Work Ratio=\frac{W_{net}}{W_T}=\frac{W_T-W_C}{W_T}$

where, Turbine work, W_T = C_p(T₃ - T₄), Compressor work, W_C = C_p(T₂ - T₁)

$\therefore Work ~Ratio=\frac{C_p(T_3-T_4)-C_p(T_2-T_1)}{C_p(T_3-T_4)}\Rightarrow1-\frac{T_2-T_1}{T_3-T_4}$​

$Work~ratio=1-\frac{T_1(\frac{T_2}{T_1}-1)}{T_4(\frac{T_3}{T_4}-1)}$

Here, $\frac{T_2}{T_1}=\frac{T_3}{T_4}=r_p^{\frac{\gamma -1}{\gamma}}$

$\therefore Work~ratio=1-\frac{T_1}{T_4}$

$Work~ratio=1-\frac{T_1}{T_4}\times\frac{T_3}{T_3}$

$Work~ratio=1-\frac{T_1}{T_3}\times\frac{T_3}{T_4}$

$Work~ratio=1 - \frac {T_1}{T_3} (r_p)^\frac {\gamma - 1}{\gamma}$

Common uses of some Radioactive elements:

Enriched uranium:

• U-235 can start a nuclear reaction and sustain it. But that 0.7 percent in naturally occurring uranium isn’t enough to make a bomb or even a nuclear reactor for a power plant.
• A power plant requires uranium with 3 to 4% U-235 (this is known as low-enriched or reactor-grade uranium), and a bomb needs uranium with a whopping 90 percent U-235 (highly enriched uranium).

Uranium enrichment, then, is the process by which a sample of uranium has its proportion of U-235 increased through the process of isotope separation.

Boiling water reactor: BWRs actually boil the water. Water is converted to steam, and then recycled back into the water by a part called the condenser, to be used again in the heat process. We can understand the overall working of the Boiling Water Reactor by the following diagram.

Concept:

In a turbine, the thermal efficiency is given by 

$\eta=\frac{W_{net}}{Q_{supplied}}$

The net work in a gas turbine is the difference of work done by the turbine and work consumed by the compressor 

W_net = W_T - W_C

Efficiency is 20%

∴ $\eta=\frac{W_{net}}{Q_{in}}=\frac{W_{T}-W_C}{Q_{supplied}}$

Calculation:

Given, WT = 600 kJ/kg, WC = 400 kJ/kg, Q_supplied = 1000 kJ/kg

$\eta=\frac{600-400}{1000}=0.2$

Explanation:

Clausius-Clapeyron equation ​

• The Clausius–Clapeyron relation is a way of characterizing a discontinuous phase transition between two phases of matter of a single constituent.
• On a pressure-temperature (P – T) diagram, the line separating the two phases is known as the coexistence curve.
• The Clausius–Clapeyron relation gives the slope of the tangents to this curve.
• The Clausius–Clapeyron relation can be used to find the relationship between pressure and temperature along phase boundaries. The dotted green line gives the anomalous behaviour of water.

Clausius Clapeyron equation which defines the slope of the vapor pressure curve:

$\frac{{dP}}{{dT}} = \frac{{{h_{fg}}}}{{T\left( {{V_g};-;{V_f}} \right)}} = \frac{L}{{T\left( {{V_2};-;{V_1}} \right)}}$

Explanation:

Entropy change for an ideal gas can be calculated by:

${S_2} - {S_1} = m{c_p}\ln \left( {\frac{{{T_2}}}{{{T_1}}}} \right) - mRln\left( {\frac{{{P_2}}}{{{P_1}}}} \right)$

where m = mass of the gas, c_p = specific heat at constant pressure, R = gas constant

T₁ = initial temperature, T₂ = final temperature, P₁ = initial pressure, P₂ = final pressure

 As the process is isothermal i.e. T₁ = T₂

∴ ${S_2} - {S_1} = m{c_p}\ln \left( 1 \right) - Rln\left( {\frac{{{P_2}}}{{{P_1}}}} \right)$  

${S_2} - {S_1} = Rln\left( {\frac{{{P_1}}}{{{P_2}}}} \right)$     {∵ ln(1) = 0} 

Entropy change for an ideal gas can also be calculated by:

${S_2} - {S_1} = m{c_v}\ln \left( {\frac{{{T_2}}}{{{T_1}}}} \right) + mRln\left( {\frac{{{V_2}}}{{{V_1}}}} \right)$

where c_v = specific heat at constant volume, V₁ = initial volume, V₂ = final volume

Explanation:

• Heat transfer in the constant temperature process by first law of thermodynamics is given as:
• Q = U + W
• where, Q = heat supplied. U = 0 = Internal energy, W = work done
• So, the heat transfer equation is, Q = W = 100 kW

Important Points

Concept:

Loss of head due expansion,

${h_L} = \frac{{{{\left( {{V_1} - {V_2}} \right)}^2}}}{{2g}}$

Calculation:

Given:

D₁ = 6 cm, D₂ = 12 cm

We know,

A₁V₁ = A₂V₂

$\Rightarrow \frac{\pi }{4}D_1^2{V_1} = \frac{\pi }{4}D_2^2{V_2}$

$\Rightarrow {V_2} = {\left( {\frac{{{D_1}}}{{{D_2}}}} \right)^2}{V_1}$

$\Rightarrow {V_2} = \frac{1}{4}{V_1}$

Loss of head due expansion,

$\therefore {h_L} = {\left( {\frac{({{V_1} - {V_2}})^2}{{2g}}} \right)} = {\left( {\frac{({{V_1} - \frac{1}{4}{V_1}})^2}{{2g}}} \right)}$

$\Rightarrow {h_L} = \frac{9}{{16}}\frac{{V_1^2}}{{2g}}$

Concept:

We have, heat conduction through the cylindrical wall is,

$Q=\frac{2\pi kL(\Delta T)}{ln(\frac{r_2}{r_1})}$

where, k = Thermal conductivity, L = length of the cylinder

From the above equation,

$Q\propto \frac{1}{ln(\frac{r_2}{r_1})}$

Hence, heat conduction is inversely proportional to $\ln \left( {\frac{{{r_2}}}{{{r_1}}}} \right)$

Explanation:

The two finite bodies will attain a common temperature Tf. Then the change in energy of the two finite bodies accordingly will be

U_i = C_pT₁ + C_pT₂

U_f = C_pT_f + C_pT_f

$\Delta U$ = U_f - U_i

$\Delta U = 2{C_p}{T_f} - {C_p}\left( {{T_1} + {T_2}} \right)$

And the work output will be

$W = {C_p}\left( {{T_1} + {T_2} - 2{T_f}} \right)$

The change in total entropy will occur entirely in the two finite bodies, for which

$\Delta S = {C_p}\ln \frac{{{T_f}}}{{{T_1}}} + ;{C_p}\ln \frac{{{T_f}}}{{{T_2}}} = 2{C_p}\ln \frac{{{T_f}}}{{\sqrt {{T_1} {T_2}} }};$

To maximize W, we have to minimize Tf, and by the third equation, we have to minimize the change in entropy. 

The minimum possible value of change in entropy is zero, corresponding to a reversible process. Hence the maximum work output will be one for which

${T_f} = \sqrt {{T_1} \ \times {T_2}}$

and finally,

${W_{max}} = {C_p}\left( {{T_1} + {T_2} - 2\sqrt {{T_1} \times {T_2}} ;} \right)$

It can be rearranged as

${W_{max}} = {C_p}{\left( {\sqrt {{T_1}} - \sqrt {{T_2}} ;} \right)^2}$

Explanation:

Joule – Thomson coefficient

• When the gas in steady flow passes through a constriction, e.g. in an orifice or valve, it normally experiences a change in temperature.
• From the first law of thermodynamics, such a process is isenthalpic and one can usefully define a Joule – Thomson coefficient as
• $\mu = {\left( {\frac{{\partial T}}{{\partial P}}} \right)_H}$
• As a measure of the change in temperature which results from a drop in pressure across the construction.
• For an ideal gas, μ = 0, because ideal gases neither warm not cool upon being expanded at constant enthalpy. ex. air
• If μ is +ve, then the temperature will fall during throttling. ex. oxygen, nitrogen, etc
• If μ is -ve, then the temperature will rise during throttling. ex. hydroge, helium, neon, etc

Concept:

Throttling Calorimeter: A throttling calorimeter is based on the principle of throttling the wet steam so that it becomes superheated.

Separating Calorimeter: It is a vessel used initially to separate some of the moisture from the steam, to ensure superheat conditions after throttling.

Combined Separating and Throttling Calorimeter

A combined separating and throttling calorimeter is therefore found most suitable for accurate measurement of dryness of steam reducing the drawback of Separating Calorimeter (i.e. water particles from wet steam are not fully separated) and the drawback of Throttling Calorimeter (i.e. not suitable for very wet steam).

x₁ = dryness fraction of steam considering separating calorimeter

x₂ = dryness fraction of steam considering the throttling calorimeter

x = actual dryness fraction of steam in the sample

Calculation:

Actual dryness fraction of steam in the sample is:

X = x₁ × x₂

Given:             

x ₁ = 0.8, x₂ = 0.8   

Calculation:       

x = 0.8 × 0.8

∴ X = 0.64

Blade or diagram efficiency is defined as the ratio of work done on the blades to the energy supplied to the blades.

\({\eta b} = \frac{{Work;done;on;blades}}{{Energy;supplied;to;blades}} = \frac{{2\left( {{V{w1}} + {V_{w2}}} \right)u}}{{V_1^2}}\)

Blade speed ratio is the ratio of blade speed to a steam speed

$\rho = \frac{u}{{{V_1}}}$

η_b will be maximum when:

$\frac{{d{\eta _b}}}{{d\rho }} = 0 \Rightarrow \rho = \frac{{\cos {\alpha _1}}}{2} = \frac{{\cos 30^\circ }}{2} = 0.433$

α₁ is of the order 18° to 22°

${\eta _{b,max}} = {\cos ^2}{\alpha _1};$

Concept:

Shape Factor is defined as the fraction of radiation energy leaving a surface that reaches another surface, for example, F12 means the fraction of energy leaving the surface 1 reaches surface 2.

Summation Rule: F11 + F12 = 1

Reciprocity theorem: A1 F12 = A2 F21 (A is the surface area)

Calculation:

Given:

For the infinite parallel surface, F₁₁ = 0

∴ F₁₂ = 1

Efficiency of 1^st engine

${\eta _1} = 1 - \frac{{{T_2}}}{{{T_1}}} = 1 - \frac{{{T_2}}}{{900}}$

Efficiency of 2^nd engine

${\eta _2} = 1 - \frac{{{T_3}}}{{{T_2}}} = 1 - \frac{{400}}{{{T_2}}}$

As ${\eta _1} = {\eta _2}$

$\begin{array}{l} \Rightarrow 1 - \frac{{{T_2}}}{{900}} = 1 - \frac{{400}}{{{T_2}}}\ \Rightarrow {T_2} = \sqrt {900 \times 400} = 600;K \end{array}$

Short Trick:

For efficiency to be equal

$\begin{array}{l} {T_2} = \sqrt {{T_1}{T_3}} \ {T_2} = \sqrt {900 \times 400} = 600K \end{array}$

Concept:

Gibbs statement concerning equilibrium

Gibbs based his entire treatment of heterogeneous equilibrium on the following statements.

The criteria of equilibrium of a material system which is isolated from all external influences may be expressed in either of the following entirely equivalent forms.

1. For equilibrium of any isolated system, it is necessary and sufficient that in all possible variations of the state of the system which do not alter its energy, the variations of its entropy shall either vanish or be negative.

• (δS)_E ≤ 0

2. For equilibrium of any isolated system it is necessary and sufficient that in all possible variations of the state of the system which do not alter its entropy,  the variations of its energy shall either vanish or be positive.

• (δS)_E ≥ 0

Concept:

Stream Function (ψ):

$v = \frac{{\partial \psi }}{{\partial x}};u = - \frac{{\partial \psi }}{{\partial y}}$

Velocity Potential Function(ϕ):

$u = - \frac{{\partial \phi }}{{\partial x}};v = - \frac{{\partial \phi }}{{\partial y}};w = - \frac{{\partial \phi }}{{\partial z}}$

Calculation:

$\begin{array}{l} \psi = {x^2} - {y^2} ;u =- \frac{{\partial \psi }}{{\partial y}} =- \frac{\partial }{{\partial y}}\left[ {{x^2} - {y^2}} \right] \end{array}$

u = 2y

$v = \frac{{\partial \psi }}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {{x^2} - {y^2}} \right]$

v = 2x

Resultant velocity at (1, 1)

$\begin{array}{l} V= \sqrt {{u^2} + {v^2}} = \sqrt {{{\left( { 2y} \right)}^2} + {{\left( { 2x} \right)}^2}} = \sqrt {4{y^2} + 4{x^2}} = \sqrt {4 + 4} = 2\sqrt 2 ;units \end{array}$

Concept:

If P₁, P_2, . . . . . ._, P_N, P_N+1 is the pressure from the beginning stage to last stage respectively of a multistage compressor then,

$\frac{{{P_2}}}{{{P_1}}} = \frac{{{P_3}}}{{{P_2}}} = \ldots \ldots \ldots \frac{{{P_{N + 1}}}}{{{P_N}}} = K$

P_N+1 =K^N *P₁

where K is constantly known as pressure ratio between any two-stage. And N is the number of stages

Calculation:

Given:

P₁ = 1 bar, P₃ = 4 bar, N = 2

$\frac{{{P_3}}}{{{P_1}}} = {k^2} \Rightarrow 4 = {k^2}$

K = 2

$\frac{{{P_2}}}{{{P_1}}} = k = 2 \Rightarrow {P_2} = 2{P_1} = 2;bar$

Explanation:

Basic Vapour Absorption Refrigeration System

• The basic absorption cycle employs two fluids, the absorbate or refrigerant, and the absorbent
• The most common fluids are Water/Ammonia as the refrigerant and lithium bromide/ water as the absorbent
• These fluids are separated and recombined in the absorption cycle
• In the absorption cycle, the low-pressure refrigerant vapour is absorbed into the absorbent releasing a large amount of heat (Absorber pressure is equal to evaporator pressure)
• The liquid refrigerant/absorbent solution is pumped to a high-operating pressure generator using significantly less electricity than that for compressing the refrigerant for an electric chiller
• Heat is added at the high-pressure generator from a gas burner, steam, hot water or hot gases
• The added heat causes the refrigerant to desorb from the absorbent and vaporize
• The vapours flow to a condenser, where heat is rejected and condense to a high-pressure liquid
• The liquid is then throttled through an expansion valve to the lower pressure in the evaporator where it evaporates by absorbing heat and provides useful cooling
• The remaining liquid absorbent, in the generator, passes through a valve, where its pressure is reduced, and then is recombined with the low-pressure refrigerant vapours returning from the evaporator so the cycle can be repeated

Explanation:

We know that,

Radiation heat exchange between parallel plates

$Q = \frac{{\sigma \left( {;T_1^4;-;T_2^4;} \right)}}{{\left( {;\frac{1}{{{\varepsilon _1}}};+;\frac{1}{{{\varepsilon _2}}};-;1;} \right)}}$

Radiation heat exchange when radiation shield is inserted between plates 

$Q = \frac{{\sigma \left( {;T_1^4;-;T_2^4;} \right)}}{{\left( {;\frac{1}{{{\varepsilon _1}}};+;\frac{1}{{{\varepsilon _2}}};+;\frac{2}{{{\varepsilon _3}}};-;2;} \right)}}$

Calculation:

Given:

${\varepsilon _1} = ;{\varepsilon _2} = \varepsilon = 0.5;\left( {{\rm{;emissivity;of;plates;}}} \right)$

${\varepsilon _3} = 0.25;\left( {{\rm{emissivity;of;radiation;shield}}} \right);$

${Q_1} = \frac{{\sigma \left( {;T_1^4;-;T_2^4;} \right)}}{{\left( {\frac{1}{{{\varepsilon _1}}};+;\frac{1}{{{\varepsilon _2}}};-;1} \right)}}$

${Q_1} = \frac{{\sigma \left( {;T_1^4;-;T_2^4;} \right)}}{{\left( {;\frac{1}{{0.5}};+;\frac{1}{{0.5}};-;1;} \right)}}$

${Q_1} = \frac{{\sigma \left( {;T_1^4;-;T_2^4;} \right)}}{3}$

Now,

If radiation shield inserted between plates

${Q_2} = \frac{{\sigma \left( {;T_1^4;-;T_2^4;} \right)}}{{\left( {;\frac{1}{{{\varepsilon _1}}};+;\frac{1}{{{\varepsilon _2}}};+;\frac{2}{{{\varepsilon _3}}};-;2;} \right)}}$

${Q_2} = ;\frac{{\sigma\left( {;T_1^4;-;T_2^4;} \right)}}{{\left( {;\frac{1}{{0.5}};+;\frac{1}{{0.5}};+;\frac{2}{{0.25}};-;2;} \right)}}$

${Q_2} = \frac{{\sigma \left( {;T_1^4;-;T_2^4;} \right)}}{{10}}$

$\therefore;\frac{{{Q_2}}}{{{Q_1}}} = \frac{3}{{10}}$

Concept:

A nozzle is a passage of smoothly varying cross-section by means of which the pressure energy of working fluid is converted into kinetic energy.

Convergent-Divergent nozzle: In a convergent-divergent nozzle the cross-section of the nozzle first decreases and then increases.

The relationship between the area, velocity, and the Mach number for a nozzle is governed by the equation

$\frac{{dA}}{A} = - \frac{{dV}}{V}\left( {1 - {M^2}} \right)$

So, from the figure, it is shown that the flow velocity in the convergent section of the nozzle is subsonic flow.

Concept:

Wall shear stress is given by:

${\rm{}}{{\rm{\tau }}_{\rm{w}}} = - \frac{{\partial {\rm{P}}}}{{\partial {\rm{x}}}}\times\frac{{\rm{R}}}{2}{\rm{}}$

where $\frac{{\partial {\rm{P}}}}{{\partial {\rm{x}}}} = pressure~gradient$, R = radius of pipe = D/2

for length L, the pressure drop is ΔP

∴ Pressure gradient is $\frac{{\partial {\rm{P}}}}{{\partial {\rm{x}}}} = \frac{{{\rm{Δ P}}}}{{{\rm{L}}}}~$

Wall shear stress is 

${\rm{}}{{\rm{\tau }}_{\rm{w}}} = - \frac{{\partial {\rm{P}}}}{{\partial {\rm{x}}}}\times\frac{{\rm{R}}}{2}{\rm{}}= \frac{{{\rm{Δ P}}}}{{\rm{L}}}\times \frac{{\rm{D}}}{{2 \times 2}} = \frac{{{\rm{Δ PD}}}}{{4{\rm{L}}}}$

Concept:

Hydraulic or blading efficiency of Pelton turbine

${\eta _h} = \frac{{2u\left( {{V_1} - u} \right)\left( {1 + k\cos{\beta _2}} \right)}}{{{V_1}^2}} = 2\left( {ρ - {ρ ^2}} \right)\left( {1 + k\cos{\beta _2}} \right)$

where u = bucket speed, V₁ = jet speed

For maximum efficiency: 

$\frac{d}{{dρ }}{\eta _h} = 0; \to ;ρ = \frac{u}{{{V_1}}} = \frac{1}{2}$

where ρ is bucket to jet-speed ratio.

So for maximum power:

$ρ = \frac{u}{V_1} = \frac{1}{2} = 0.5$

Explanation:

• The phenomena in which air stream fails to follow the blade contour is called stalling.
• Stalling and surging both are unstable operation condition in an axial compressor, a stall occurs normally when airflow separates from the convex sides of the compressor blades.
• This means that the airflow and the speed of the rotor do not match.
• If stall develops further, it may cause a compressor surge.
• When surging, the airflow through the compressor will reverse and comes out from the front of the compressor.
• Compressor surging can cause vibration and compressor blades damage.

Adiabatic or chemical dehumidification:

This process is mainly used in industrial air conditioning. In this process, air is passed once through a chemical contact. The moisture gets condensed out of the air.

In this adiabatic or chemical dehumidification, the parameters vary as follows

i) Dry bulb temperature increases

ii) Specific Humidity decreases

iii) Dew point temp decreases

iv) Relative humidity decreases

v) Enthalpy and wet bulb temp remain constant

vi) Sp. Volume increases

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:

LH = 0.25 SH

$SHF = \frac{{SH}}{{SH + LH}} = \frac{{SH}}{{SH + ;0.25;SH}} = 0.8$

Concept:

Sensible cooling is the process of decreasing DBT at a constant specific humidity.

Sensible cooling:

During this process, the moisture content of air remains constant, but its temperature decreases as it flows over a cooling coil.

Other parameters are varied as given

• Dry bulb temperature decreases
• Dew point temp remains constant
• Sp. humidity remains constant
• Relative humidity increases
• Enthalpy decreases
• Wet-bulb temperature decreases
• Sp. Volume decreases

Concept:

• Let 1 kg of air at temperature t_d1 is passed over the coil having its temperature t_d3 as shown in the figure.

• A little consideration will show that when air passes over a coil, some of it (say x kg) just by-passes, unaffected while the remaining (1 - x) kg comes in direct contact with the coil.

• This by-pass process of air is measured in terms of a by-pass factor.

• The BPF depends upon the following factors :

• The number of fins provided in a unit length i.e. the pitch of the cooling coil fins.

• The number of rows in a coil in the direction of flow.

• The velocity of flow of air.

• It may be noted that the by-pass factor of a cooling coil decreases with a decrease in fin spacing and an increase in the number of rows.

Concept:

The radiation energy emitted by a body per unit time is given by:

Eb = ϵAσT4

Where ϵ is emissivity of the body.

σ = The Stefan – Boltzmann constant = 5.67 × 10-8 W m-2K-4

Calculation:

For a given area  E α T4

$\frac{{{E_2}}}{{{E_1}}} = \frac{{T_2^4}}{{T_1^4}}$

$\frac{{{E_2}}}{{{E_1}}} = \frac{{{{\left( {2{T_1}} \right)}^4}}}{{T_1^4}} \Rightarrow {E_2} = 16E$,

E₁ = P(given)

E₂ = 16 P

Concept:

Diesel cycle:

Processes in compression engine (diesel cycle) are:

Process 1-2: Reversible adiabatic compression

Process 2-3: Constant pressure heat addition

Process 3-4: Reversible adiabatic expansion

Process 4-1: Constant volume of heat rejection

cut-off ratio:

The cut-off ratio is the ratio of the volume after combustion to the volume before combustion.

Cut-off ratio: ${r_c} = \frac{{{V_3}}}{{{V_2}}}$

Compression ratio: ${r} = \frac{{{V_1}}}{{{V_2}}}$

The efficiency of the diesel cycle is given by

$\eta = 1 - \frac{1}{{{r^{\gamma - 1}}}}\left[ {\frac{{r_c^\gamma - 1}}{{\gamma \left( {{r_c} - 1} \right)}}} \right]$

The mean effective pressure (p_m) which is an indication of the internal work output increases with a pressure ratio at a fixed value of compression ratio and the ratio of specific heats.

The expression for mean effective pressure for diesel cycle,

${p_m} = \frac{{{p_1}\left[ {\gamma {r^\gamma }\left( {{r_c} - 1} \right) - r\left( {r_c^\gamma - 1} \right)} \right]}}{{\left( {\gamma - 1} \right)\left( {r - 1} \right)}}$

From the expression,

The mean effective pressure of the diesel cycle having a fixed compression ratio will increase if the cut-off ratio increases.

Concept:

1 – 2 – 3 – 4 : Otto cycle

1’ – 2’ – 3’ – 4’ : Diesel cycle

The maximum pressure is same and same heat input is given.

It is clear from T-S diagram that heat rejected by diesel cycle (4’ – 1) is less than heat rejected by Otto cycle (4 – 1).

Hence the diesel cycle is more efficient than Otto cycle. Also for dual cycle, the compression ratio is greater than Otto cycle, hence more efficient than Otto but less efficient than diesel cycle.

Concept:

In a converging-diverging nozzle, when the back pressure ratio reaches a critical value, the flow will become choked with the subsonic flow in the converging section, sonic flow at the throat, and supersonic flow in the diverging section.

A further decrease in the back pressure ratio results in the formation of a shock wave within the diverging section.

The location of the shock is such that the pressure at the diverging section exit will equal the back pressure. As the back pressure ratio is decreased further, the shock wave moves downstream of the throat toward the exit of the device as shown below.