Option 1 : Enthalpy

__Explanation:__

Since, the properties like internal energy, enthalpy and entropy of a system cannot be directly measured. They are related to change in the energy of the system.

Hence, we can determine Δu, Δh, Δs but not the absolute values of these properties.

⇒ Therefore, It is necessary to choose a reference state to which, these properties are arbitrary assigned some numerical values.

So for water, **the triple point (T = 0.01°C & P = 611 Pa) is selected as reference a state, where the “Internal energy” (u) and “Entropy” (s) of saturated liquid are assigned a zero value.**

#Note: h = u + Pv

At triple point, u = 0, but p × ν ≠ 0

ThereforeOption 3 : Zero

**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.

Option 4 : A mixture of ice, water and steam

**Concept:**

Homogeneous thermodynamic system:

- A
**homogeneous thermodynamic system**is defined as the one whose chemical composition and physical properties are the same in all parts of the system.

Heterogeneous system:

- A
**heterogeneous system**is defined as one consisting of two or more homogeneous bodies. - The homogeneous bodies of a heterogeneous system are referred to as phases.
- Each phase is separated from other phases by interfaces, or boundaries, and in passing over such a boundary the chemical composition of the substance or its physical properties abruptly change.
- An example of a heterogeneous system is
**ice, water and steam.** - This system has
**three homogeneous bodies, steam, water and ice.** - The
**chemical composition**of the three phases is the**same**, but their**physical properties****differ**drastically.

Option 1 : Temperature drops during throttling

__Concept:__

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.

Option 4 : variable

**Explanation:**

Fusion is solid to a liquid phase transition.

For water, v” < v’ (Indicating contraction) \(\Rightarrow \;\frac{{dp}}{{dT}} \to - ve\)

For most other substances v” > v’ (expansion) \( \Rightarrow \;\frac{{dp}}{{dT}} \to + ve\)

__Important Points__

- The
**fusion curve**has been drawn with a**positive slope**, which is typically the case. - The
**fusion curve of ice/water**has a**negative slope**due to the fact that when ice melt, the**molar volume decreases.** - Ice actually melt at a lower temperature at a higher pressure

**Therefore, The slope of the curve is positive as well as negative hence, it is actually variable in nature.**

Option 4 : 0.611 kPa

__Explanation__

The phase diagram of water is given below.

A substance sublimates on heating If it is kept below its triple point pressure. Hence the lowest pressure at which water can exist in liquid phase in stable equilibrium is triple point pressure.

The critical point properties of water are given below.

**Triple point pressure, P _{tp} = 4.58 mm of Hg = 0.611 kPa**

Triple point temperature, T_{tp} = 273.16 K = 0.01°C

Option 4 : vertical line

__Explanation:__

Mollier chart is drawn between Enthalpy (h) and entropy (s). The turbine output on the Mollier diagram represented by the vertical straight line in ideal conditions.

On Mollier diagram constant pressure line diverge from one another as shown in figure.

Option 1 : 0.29 kg

__Concept:__

Volume of vessel (V) = Volume of liquid (V_{f}) + Volume of vapour (V_{g})

V_{f} = v_{f }× m_{f}

V_{g} = v_{g} × m_{g}

v_{f,g} = Specific volume of liquid and vapour

m_{f,g} = mass of liquid and vapour

__Calculation:__

Given:

V = 0.04 m^{3 }

v_{f }= 0.0011 m^{3}/kg, m_{f} = 5 kg, v_{g} = 0.12 m^{3}/kg

V_{f }= v_{f} × m_{f }= 0.0011 × 5 = 0.0055 m^{3}

V = V_{f} + V_{g} = m_{f}v_{f} + m_{g}v_{g}

0.04 = 5(0.0011) + m_{g}(0.12)

**Concept:**

**Enthalpy** is sum of internal energy plus pV work.

h = u + pv

**Calculation:**

Given data: p = 8 MPa = 8 × 103 kPa, u = 2864 kJ/kg, v = 0.03432 m3/kg

h = u + pv = 2864 + (8 × 10^{3}) × 0.03432 = 3138.56 kJ/kg

**Mistake point:**

- All other quantities are in kPa except pressure. So to convert pressure from MPa to kPa is first step.
- While calculating, take care of all the SI units.
- 1 bar = 10
^{5}Pa - 1 MPa = 1000 kPa = 10 bar

Option 3 : 0.64

**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).

x_{1 }= dryness fraction of steam considering separating calorimeter

x_{2 }= 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 = x _{1 }× x_{2}**

Given:

x_{ 1 }= 0.8, x_{2 }= 0.8

**Calculation: **

x = 0.8 × 0.8

Option 2 : Superheated vapor only

**Explanation:**

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

**The** **Separating calorimeter: **

- It
- 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.

Option 2 : 22.06 MPa

__ Explanation__:

**Critical point**:

At a critical point, the liquid is directly converted into vapor without having a two-phase transition.

**For water at the critical point: **

**Pcr = 220.6 bar = 22.06 MPa**

Tcr = 373.95oC

vcr = 0.005155 m3/kg

Enthalpy of vaporization at a critical point is zero.

The figure below represents the P-V diagram for a pure substance (water).

Option 2 : Extreme points of vaporization curve are triple point and critical point

**Explanation:**

**"Extreme points of fusion curve are triple point and critical point" is an incorrect statement**

**Validation:**

From the phase diagram, it is seen that the triple point and critical point are the extreme points of the vaporization curve.

**"Extreme points of vaporization curve are triple point and critical point" is a correct statement **

**Validation:**

From the phase diagram, it is seen that vaporization starts from the triple point and goes till critical point beyond which liquid is directly flashed into vapour and there is no vaporization.

**"Sublimation curve, fusion curve, and vaporization curve meets at critical point" is an incorrect statement**

**Validation:**

The sublimation curve fusion curve and the vaporization curve meets at the triple point.

**"Fusion curve for water has positive slope" is an incorrect statement**

**Validation: **

The fusion curve of water has a negative slope it can be seen on the phase diagram also, the reason for this negative slope is that water expands after freezing, unlike other substances which contract after freezing.

Region inside the inversion curve is represented by:

[where μ is Joule – Kelvin coefficient]

Option 1 : Cooling region, μ > 0

**Explanation:**

The numerical value of the slope of constant enthalpy curve on a T - P diagram at any point is called Joule - Thomson coefficient and is denoted by μ

\({μ} = \;{\left( {\frac{{\partial T}}{{\partial P}}} \right)_h}\)

Along the inversion curve, μ is zero. The region inside the inversion curve is the cooling region for which μ > 0, and outside the inversion curve, it is the heating region for which μ < 0.

Option 2 : horizontal

**Explanation:**

Mollier diagram is the h-s diagram of pure substance.

- As the pressure increases, the saturation temperature increases, and so the slope of the isobar also increases
- Hence, the constant pressure lines diverge from one another
- The critical isobar is a tangent at the critical point
- Although the slope of an isobar remains continuous beyond the saturated vapour line, the isotherm bends towards the right and its slope decreases asymptotically to zero because in the ideal gas region it becomes horizontal and the constant enthalpy implies a constant temperature

The slope drawn on sublimation curve P – T diagram for all substances gives _____ value.

Option 3 : positive

**Explanation:**

Each single phase of a pure substance is separated by saturation lines.

- The sublimation line separates the solid and vapour regions
- The vaporisation line separates the liquid and vapour regions
- The fusion line separates the solid and liquid regions

The slopes of sublimation and the vaporisation curves for all substances are positive.

The slope of the fusion curve, however may be positive or negative. The fusion curve of most substances has a positive slope. Water is one of the important exceptions. The slope of fusion curve for water is negative.

Option 2 : vertical line

**Explanation:**

- Mollier Diagram: Mollier diagram is enthalpy (h) versus entropy (s) plot.

- It consists of a family of constant pressure lines, constant temperature lines and constant volume lines plotted on enthalpy versus entropy coordinates.
- From the T-ds equation:

Tds = dh – vdp

In the two-phase region, the constant pressure and constant temperature lines coincide.

\({\left( {\frac{{\partial h}}{{\partial s}}} \right)_p} = T\)

- The slope of an isobar on the h-s coordinates is equal to the absolute temperature. If the temperature remains constant the slope will remain constant.
- If the temperature increases the slope of the isobar will increase.
- For the saturated liquid and saturated vapour i.e. within the dome the temperature and pressure remains constant.

Figure: Mollier diagram

**A vertical line in the Mollier diagram represents the isentropic process (s = C).**- A horizontal line in the Mollier diagram represents the isenthalpic process (h = C).

The critical temperature of water in degrees is:

Option 2 : 374.15

__Explanation:__

Critical point:

- The point at which the saturated liquid line and saturated vapour line of a pure substance meet is called the critical point.
- At a critical point, the liquid is directly converted into vapour without having a two-phase transition. So, enthalpy of vaporization at a critical point is zero i.e. At the critical point, saturated liquid and saturated vapour phases are identical.
- The figure below represents the P-V diagram for a pure substance (water).

Below critical point, latent heat is required to convert water into vapour but at this point, there is no need for latent heat i.e. at this point latent heat equals zero.

For **water critical point parameters are:**

Pressure (Pc) = 22 MPa, **Temperature (Tc) = 374°C**

Option 1 : 0.4

**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:**

\(x= \frac{{Mass\;of\;dry~steam}}{{Mass\;of\;dry~steam + Masss\;of\;water\;in\;suspension}} = \frac{{{m_v}}}{{{m_v} + {m_f}}} = \frac{{0.4}}{{0.4 + 0.6}} = \frac{{0.4}}{1} = 0.4\)

Along the ‘triple line’ in a p – v diagram showing all three phase of water, which one of the following statements is correct?

Option 1 : A substance has the same pressure and temperature but different specific volume

**Explanation:**

- The Triple point is a line on the P-V diagram where all the three phases solid, liquid and gases exist in equilibrium. At a pressure below the triple point line, the substance cannot exist in the liquid phase and the substance when heated, transforms from solid to vapour by absorbing the latent heat of sublimation from the surroundings.
- The triple point is merely the point of intersection of the sublimation and vaporization curves. It has been found that on a ‘p-T’ diagram the triple point is represented by a point and on a ‘p-v’ diagram it is a line, and on a ‘u-v’ diagram it is a triangle. In the case of ordinary water, the triple point is at a pressure of 4.58 mm Hg and a temperature of 0.01°C.