Concept:

For constant volume process

\(W = \smallint PdV\)

Since

dV = 0

so W = 0 for constant volume process.

For free expansion work done is zero. W = 0

The process in which high-pressure fluid is converted to low pressure by using a throttle valve is Throttling.

In the throttling process enthalpy remains constant, hence work done is zero.

Explanation:

This is a P-V curve of a pure substance (water) where we can see different types of lines as per their state.

The dome has been divided into two regions by two lines GH and HD.

GH line represents a saturated liquid which differentiate a liquid and liquid-vapour mixture i.e. when a liquid just starts boiling it moves from liquid region to liquid-vapour region crossing saturated liquid line.

HD line represents a saturated vapour which differentiate a vapour and liquid-vapour mixture i.e. when a liquid just starts condensing it moves from vapour to liquid-vapour line crossing saturated vapour line.

The two lines i.e. saturated liquid line (GH) and saturated vapour line (HD) meets at a point (H) which is known as critical point.

Explanation:

Enthalpy of a substance is defined as:

h = u + pV

According to the First Law of Thermodynamics for a process 

δQ = du + δW

δQ = du + p.dV     [process involving just pdV work]

At constant pressure,

p.dV = d(pV)

(δQ)_p = du + d(pV)

(δQ)_p = d(u + pV)

(δQ)_p = dh

i.e. heat transferred at constant pressure increases the enthalpy of the system.

For an ideal gas pV = RT

h = u + pV

∴ h = u + RT

The internal energy (u) of an ideal gas is a function of temperature i.e. u = f(T) since there are no intermolecular forces of attraction and repulsion.

∴ the enthalpy of an ideal gas also depends on the temperature i.e. h = f(T).

Explanation:

Entropy change for an ideal gas can be calculated by:

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

where m = mass of the gas, cp = 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

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

\({S_2} - {S_1} = Rln\left( {\frac{{{P_1}}}{{{P_2}}}} \right)\)     {∵ 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 cv = specific heat at constant volume, V1 = initial volume, V2 = final volume

For an isothermal process (T₁ = T₂)

Specific entropy change will be:

\({S_2} - {S_1} = Rln\left( {\frac{{{V_2}}}{{{V_1}}}} \right) = (C_P - C_V) ln(\frac {V_2 }{V_1}) \)

Concept:

This will be the case of sublimation. As the weather is sub-zero, so the water present in clothes will freeze and will sublimate directly and we feel the clothes as dry.

Sublimation

• Sublimation is a process in which solid changes into the gaseous state without passing through the liquid state.
• Sublimation is endothermic in nature and occurs at temperature and pressure below the one at which all three states of the substance exist simultaneously.
• The energy required for sublimation is known as the enthalpy of sublimation.
• The sublimation process can be seen in the case of camphor, naphthalene balls, wet cloths turn into dry clothes, etc.

​Vaporization

• Vaporization can be defined as the process in which liquid state changes into the vapor state.

Melting

• ​Melting, or fusion, is a physical process that results in the phase transition of a substance from a solid to a liquid.
• This occurs when the internal energy of the solid increases, typically by the application of heat or pressure, which increases the substance's temperature to the melting point.

Condensation

• Condensation is the deposition of a liquid or a solid from its vapor, generally upon a surface that is cooler than the adjacent gas.

Explanation:

Entropy-

It can be defined as a measure of the degree of molecular disorder/randomness existing in the system.

We know that,

\(\rho = \frac{P}{{RT}}\)

Density and Temperature are inversely related at constant Pressure.

Initially, hot water has a lower density and cold water has a higher density. Due to the density differences, hot water floats above cold water. Due to temperature differences heat transfer takes place within the mixture from hot water to cold water. Molecules of hot water start moving downwards and molecules of cold water start moving upwards as they lose and gain heat respectively. Thus, the mixture comes in motion as a result of buoyancy force originated from variation in fluid density.

∵ This motion increases the molecule's randomness and thus increases the entropy of the mixture.

Concept:

From Steady Flow Energy Equation (S.F.E.E.)

\(m\left( {{h_1} + \;\frac{{V_1^2}}{2} + g{z_1}} \right) + Q = m\;\left( {{h_2} + \frac{{V_2^2}}{2} + g{z_2}} \right) + W\)

For Nozzles and diffusers:

• Q = 0, as the nozzle is perfectly insulated,
• W = 0, no work is done by the nozzle, V1 <<< V2 and potential energy change is neglected
• Horizontal nozzles, no change in Z

∴\({h_1} + \frac{{V_1^2}}{2} = {h_2} + \frac{{V_2^2}}{2}\)

Calculation:

Given:

Δh = h₁ - h₂ = 100 kJ/kg, V₁ = 0 m/s;

From the above equation,

\({h_1-h_2} = \frac{{V_2^2}}{2}\)

2 × 100 × 1000 = (V₂)²

⇒ V₂ = 447.2 m/s

Concept:

Reversible heat engine cycle will reject least rate of heat (QL) per kW net output (W) of engine.

For reversible heat engine,

\(\eta = \frac{W}{{{Q_H}}} = \frac{{{T_H} - {T_L}}}{{{T_H}}} = 1 - \frac{{{T_L}}}{{{T_H}}}\)

Calculation:

Given T_H = 800°C = 1073 K, T_L = 30°C = 303 K, W = 1 kW,

\(\frac{W}{{{Q_H}}} = 1 - \frac{{{T_L}}}{{{T_H}}} \Rightarrow \frac{1}{{{Q_H}}} = 1 - \frac{{303}}{{1073}}\)

⇒ Q_H = 1.39 kW

Concept:

The numerical value of the slope of an isenthalpe on a T-p diagram at any point is called the Joule-Thomson coefficient and is denoted by μ.

And \({\mu } = {\left( {\frac{{\partial T}}{{\partial p}}} \right)_h}\)

\({\mu }\left\{ {\begin{array}{*{20}{c}} { < 0}&{temperature\;increases}\\ { = 0}&{temperature\;remains\;constant}\\ { > 0}&{temperature\;decreases} \end{array}} \right.\)

Concept:

According to the First Law of thermodynamics, “For a closed system undergoing a cycle, net heat transfer is equal to network transfer.”

When heat energy is supplied to a gas, two things may occur:

• The internal energy of the gas may change
• The gas may do some external work by expanding

According to the first law of Thermodynamics:

δQ = δW + ΔU

When a process is executed by a system, the change in stored energy of the system is numerically equal to the net heat interaction minus the net work interaction during the process:

ΔU = δQ – δW where U is the internal energy that is introduced by this law.

For the cyclic process: ΔU = 0

\(\oint \delta Q = \oint \delta W\)

The first law of thermodynamics states that for a cyclic process, the cyclic integral of heat is equal to the cyclic integral of work.

ΣQ = ΣW

So, heat and work are mutually convertible.

The second law of thermodynamics introduces the concept of entropy.

Explanation:

Steady Flow Energy Equation:

When there is mass transfer across the system boundary, the system is called an open system.

When there is an involvement of heat, work and the rate of flow of mass and energy across the control surface are constant, the equation used is Steady Flow Energy Equation.

In steady flow, the ratio of heat transfer, work transfer and mass flow at inlet and outlet is the same.

\({h_1} + \frac{{V_1^2}}{2} + {z_1}g + \frac{{dQ}}{{dm}} = {h_2} + \frac{{V_2^2}}{2} + {z_2}g + \frac{{dW}}{{dm}}\;\;\;\;(1)\)

Eq (1) is the Steady flow energy equation in mass form.

\({\dot m_1}({h_1} + \frac{{V_1^2}}{2} + {z_1}g) + \dot Q = {\dot m_2}\left( {{h_2} + \frac{{V_2^2}}{2} + {z_2}g} \right) + \dot W\;\;\;\;\;(2)\)

Eq (2) is the Steady flow energy equation in rate form.

This equation can be applied to a wide variety of processes like pipeline flows, heat transfer processes, mechanical power generation in engines and turbines, flow through nozzle and diffusers etc.

Eg. nozzle and diffuser having converging and diverging passage.

​Euler's Equation:

In fluid flow, numerous forces acting on the fluid element. When the forces due to gravity F_g and pressure force F_p are considered then it is known as Euler's equation of motion.

\(\frac{{dp}}{ρ } + gdz + vdv = 0\;\;\;\;\;(1)\)

Euler's equation is derived considering the fluid element along a streamline.

Bernoulli's equation:

Bernoulli's equation is obtained by integrating the Euler's equation of motion.

\(\frac{p}{{ρ g}} + \frac{{{v^2}}}{{2g}} + z = C\)

Following assumptions are made in the derivation of Bernoulli's equation:

1. Flow is ideal i.e inviscous.
2. Flow is steady i.e. time variation is zero.
3. Flow is incompressible i.e. ρ is constant.
4. Flow is irrotaional i.e. ωx = ωy = ωz = 0.

Laplace Equation:

If ϕ(x, y, z) represents a function, then the Laplace equation is used to check the possibility of the existence of such function which is given by:

\(\frac{\partial ^2\phi}{\partial x^2}+\frac{\partial ^2\phi}{\partial y^2}+\frac{\partial ^2\phi}{\partial z^2}=0\)

Explanation:

Polytropic Process is represented by:

PVn = C

where P is pressure, V is volume, and n is the polytropic index.

• The isothermal process is governed by Boyle’s law.
• The temperature remains constant in this process. Pressure and volume are inversely proportional to each other. Hence, when P increases V decreases, and when V increases P decreases.
• Thus the curve developed is a rectangular hyperbola. So, the graph for the Isothermal process shows a hyperbolic curve.

T = C

i.e. PV = C (from ideal gas equation)
where P = Pressure, T = Temperature, V = Volume

Concept:

• The first law of thermodynamics is a restatement of the law of conservation of energy. 
  • It states that energy cannot be created or destroyed in an isolated system; energy can only be transferred or changed from one form to another.

When heat energy is supplied to a gas, two things may occur:

• The internal energy of the gas may change
• The gas may do some external work by expanding

According to the first law of Thermodynamics:

δQ = δW + ΔU

Where δQ is change in heat, δW is work done, and ΔU is internal energy change.

CALCULATION:

Given that:

Heat given to system = +30 kJ

Internal energy reduction = -91 kJ

From the First law of thermodynamics
dQ = dW + ΔU

∴ dW = dQ – ΔU

= 30 + 91 = 121 kJ

Magnitude of work is positive, therefore, 121 kJ of work is done by the system

Explanation:

Here is a figure of the PV diagram of the Carnot cycle.

Process 1-2 represents reversible isothermal heat addition,

Process 2-3 represents reversible adiabatic expansion or work output.

Process 3-4 represents reversible isothermal heat rejection 

Process 4-1 represents reversible adiabatic work input.

• Now, if we are dealing with the ideal gas cycle, then the isothermal process or transferring heat energy at constant temperature should be extremely slow because only then the process will be reversible and hence it is very much difficult to attain. Hence statement (1) is correct.
• Again without ideal insulation, which is impossible to achieve in real life, the reversible adiabatic or isentropic process must be extremely fast. Hence statement (2) is correct.
• The thermal efficiency of the Carnot cycle is a function of source and sink temperature limits only. There is no need to maintain lower or higher pressure ratios for the Carnot cycle. Hence statement (3) is not correct.
• The thermal efficiency of the Carnot cycle is given by η = 1 \(- \frac{{{{\rm{T}}_2}}}{{{{\rm{T}}_1}}}\) where T₂ is the absolute temperature of the sink and T₁ is the absolute temperature of the source. Hence the thermal efficiency of a Carnot cycle is a function of source and sink temperatures. So, statement (4) is not correct.

So, the correct option is an option (1).

Concept:

\(W = \mathop \smallint \limits_{{v_1}}^{{v_2}} pdV = P\left( {{V_2} - {V_1}} \right)\)

Work Done = Area under P – V diagram

Calculation:

WD = Area of rectangle ABCDA = AB × BC = 2V_o × P_o = 2V_oP_o

Explanation:

Celsius scale 

• In this scale, LFP (ice point) is taken 0° and UFP (steam point) is taken 100°.
• The temperature measured on this scale all in degree Celsius (° C).

Fahrenheit scale 

• This scale of temperature has LFP as 32° F and UFP as 212° F .
• The change in temperature of 1° F corresponds to a change of less than 1° on the Celsius scale.

Kelvin scale 

• The Kelvin temperature scale is also known as the thermodynamic scale. The triple point of water is also selected to be the zero of the scale of temperature.
• The temperatures measured on this scale are in Kelvin (K).

Rankine scale

• This scale of temperature has LPF as 492° R and UFP as 672° R.
• Interval of this scale is according to Fahrenheit.
• The temperature measured on this scale are in Rankine (R)

All these temperatures are related to each other by the following relationship

\(\frac C{100}= \frac {F-32}{180}=\frac {R-492}{180}= \frac {K -273}{100}\)

Relationship between the Celsius and Fahrenheit scale is –

\(\frac{F-32}{9}=\frac{C}{5}\)

\(\Rightarrow C=\frac{5}{9}\left( F-32 \right)\)

Explanation:

We have,

F + P = C + 2

For invariant point ⇒ F = 0

0 + P = 2 + 2

∴ P = 4

Explanation:

A control volume is also known as an open system. In this, both heat and mass can cross the boundary of the system.

A control volume can be seen as a fixed region across which mass and energy transfers are studied.

Control Volume System (Open System)

• It is a system of fixed volume
• This type of system is usually referred to as an "open system” or a "control volume"
• Mass transfer can take place across a control volume
• Energy transfer may also occur into or out of the system
• A control volume can be seen as a fixed region across which mass and energy transfers are studied
• Control Surface - It is the boundary of a control volume across which the transfer of both mass and energy takes place
• The mass of a control volume (open system) may or may not be fixed.

Control Mass System (Closed System)

• It is a system of fixed mass with a fixed identity
• This type of system is usually referred to as a "closed system"
• There is no mass transfer across the system boundary
• Energy transfer may take place into or out of the system

Explanation:

A separating and throttling calorimeter is a combination of separating and a throttling calorimeter.

The Separating calorimeter:

• It consists of two concentric chambers, the inner chamber and the outer chamber, which communicates with each other through an opening at the top.
• As the steam discharges through the metal basket, which has a large number of holes, the water particles due to their heavier momentum get separated from the steam and collect in the chamber.
• The comparatively dry steam in the inner chamber moves up and then down aging through the annular space between the two chambers and enters the Throttling Calorimeter.
• It is a vessel used initially to separate some of the moisture from the steam, to ensure superheat conditions after throttling. The steam is made to change direction suddenly; the moisture droplets, being heavier than the vapor, drop out of suspension and are collected at the bottom of the vessel.

The Throttling Calorimeter:

• It is a vessel with a needle valve fitted on the inlet side.
• The steam is throttled through the needle valve and exhausted to the condenser. The state of steam for the separating and throttling calorimeter should be in superheated vapor just after throttling.
• Then only pressure and temperature can be measured by a manometer and thermometer and then using the steam table enthalpy of superheated vapor is determined.
• From the steady flow energy equation, we can find the enthalpy after throttling and dryness fraction.

Concept:

A perfect gas is also an Ideal gas, which follows the Ideal gas equation of states i.e. PV = mRT all temperature.

where, P = pressure of gas, V = volume occupied, m = mass of a gas, R = universal gas constant. 

The universal gas constant (R) is the difference between specific heat constants for constant pressure (C_p) and constant volume (C_v)

i.e.R = C_p - C_v

A real gas behaves as an Ideal gas at low pressure and very high temperature. Air is a perfect gas.

Gases that obey the gas laws (Charles law, Boyles law, and Universal Gas Law) are called ideal gases.

Boyle’s, Charles’, and Gay Lussac's Laws describe the basic behavior of fluids with respect to volume, pressure, and temperature.

The Combined gas law or General Gas Equation is obtained by combining Boyle's Law, Charles's law, and Gay-Lussac's Law. It shows the relationship between the pressure, volume, and temperature for a fixed mass (quantity) of gas:

\(\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\)

\({C_P} - {C_V} = \frac{R}{J}\) is valid for a perfect gas only.