Concept:

Triple point state:

• It is the point at which it exist in solid, liquid and vapour phase.
• It is a point in P-T diagram, a line in P-V diagram and it is a triangle in U-V diagram.
• Fusion lines in P-T diagram have negative slope and this is due to anomalous behaviour of water.
• Or we can say that for water $\left(\frac{\partial P}{\partial T}\right)$ during fusion is negative and for general substance it is positive that is because density of water from 0°C to 4°C increases.

θ > 90° ⇒ tan θ < 0

∴ slope < 0

$\therefore \frac{\partial P}{\partial T}<0$

From the above diagram and explanation we can clearly see that the slope is negative and hence it will be the correct answer.

Note: There are only three substances which have negative slope for fusion and they are water, antimony and bismuth.

• 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 vapor i.e. within the dome the temperature and pressure remain constant. 

Concept:

The equation that relate partial derivatives of properties of p, v, T and s of a compressible fluid are called Maxwell relations.

The four Gibbsian relations for a unit mass are

1) du = Tds – Pdv

2) dh = Tds + vdP

3) df = - Pdv – sdT

4) dg = -sdT + vdP

Since u,h,f and g are the properties thus point functions and the above relations can be expressed as

dz = Mdx + Ndy

with,

${\left(\frac{\partial T}{\partial v}\right)}_s=+{\left(\frac{\partial p}{\partial s}\right)}_v$

Applying the cyclic relation as

Mdx + Ndy → ${\left(\frac{\delta M}{\delta y}\right)}_x={\left(\frac{\delta N}{\delta x}\right)}_y$

Now,

Replacing M,N,y and x by T,p,v,s of each of the Gibbsian equations in cyclic order, we will get the following four relations.

$1){\left(\frac{\partial T}{\partial p}\right)}_s={\left(\frac{\partial v}{\partial s}\right)}_p$

$2){\left(\frac{\partial p}{\partial T}\right)}_v={\left(\frac{\partial s}{\partial v}\right)}_T$

$3){\left(\frac{\partial T}{\partial v}\right)}_s=-{\left(\frac{\partial p}{\partial s}\right)}_v$

$4){\left(\frac{\partial v}{\partial T}\right)}_p=-{\left(\frac{\partial s}{\partial p}\right)}_T$

These relations are Maxwell’s relations and are extremely important in thermodynamics as they provide the means of determining the change in entropy.

Hence by comparing the above four relations with the options the correct answer will be option 4.

Concept:

• It is the point at which the slope changes from positive to negative and at this point the slope is zero.

• It represents the highest pressure and temperature at which the liquid and vapour phases coexist in equilibrium.
• All the thermodynamic properties of liquid and vapour phases are identical at the critical point.
• The temperature pressure and volume at the critical point are called critical temperature T_c ,

critical pressure P_c and critical volume V_c

• It is a fixed point that’s why degree of freedom will be zero.
• Dryness fraction (x) is not defined at critical point.

Note:

The critical point properties of water are

P_c = 221.2 bar

T_c = 374.15° C

V_c = 0.00317 m³/kg

Concept:

Coefficient of volumetric expansion (β)

It is defined as the ratio of fractional change in volume to change in temperature when pressure is kept constant. It can be expressed as

$\beta =\frac{1}{V}{\left(\frac{\delta v}{\delta T}\right)}_p$

Explanation:

According to the ideal gas equation:

pv = RT (For n = 1)

The differential form of the above equation is

pdv + vdp = RdT      ----(1)

when the pressure p is kept constant, dp = 0

∴ pdv = RdT

∴ $\left(\frac{dv}{dT}\right)=\frac{R}{p}$      ---- (2)

And from the ideal gas equation

$\frac{1}{v}=\frac{R}{p}$       ---- (3)

Now,

If we multiply equation 2 and equation 3 we get the coefficient of volumetric expansion (β)

$\therefore \beta =\frac{1}{v}{\left(\frac{\delta v}{\delta T}\right)}_p$

$\therefore \beta =\frac{p}{RT}\left(\frac{R}{p}\right)$

$\therefore \beta =\frac{1}{T}$

∴ β = T^-1

Concept:

At constant volume

Q = Heat added = change in Internal energy

∴ Q = Δu = u₂ – u₁

Calculation:

As, v₁ = v₂

${\left(v_f+x{v}_{fg}\right)}_1={\left(v_f+x{v}_{fg}\right)}_2$

∴ 0.001 + 0.05 (57.757 – 0.001) = 0.00103 + x₂ (4.129 – 0.00103)

∴ x₂ = 0.699

Now,

$u_2={\left(u_f+x_2{u}_{fg}\right)}_2=1825.88$ kJ/kg

$u_1=\left(u_f+x_1{u}_{fg}\right)=199.82$ kJ/kg

$\therefore Q=u_2-u_1=1626.05$ kJ/kg

Concept:

Using the relation:

Tds = dh - νdp     

Tds = du + pdν    

Calculation:

Now,

Tds = dh - νdp     ---(1)

$Tds=C_{p}dT-T{\left(\frac{\partial v}{\partial T}\right)}_{p}dp$     ---(2)

Tds = du + pdν    ---(3)

$Tds=C_{v}dT+T{\left(\frac{\partial p}{\partial T}\right)}_{v}dv$     ---(4)

Now,

Equating (1) & (2) we get

$dh=C_{p}dT-T{\left(\frac{\partial v}{\partial T}\right)}_p+vdp$

And,

on equating (3) & (4) we get

$du=C_{v}dT+T{\left(\frac{\partial p}{\partial T}\right)}_{v}dv-pdv$

Concept:

According to Clapeyron equation:

${\left(\frac{\delta P}{\delta T}\right)}_{sat}=\frac{h_g-h_f}{T_{sat}\left(v_g-v_f\right)}$

Calculation:

Given:

${\left(\frac{\delta P}{\delta T}\right)}_{sat}=20.5\mathrm{k}\mathrm{P}\mathrm{a}/\mathrm{K}$

h_f = 80 kJ/kg, T_sat = 315 K,v_g = 0.02 m³/kg, v_f = 7.9 × 10^-3 m³/kg = 0.0079 m³/kg.

Now,

${\left(\frac{\delta P}{\delta T}\right)}_{sat}=\frac{h_g-h_f}{T_{sat}\left(v_g-v_f\right)}$

$20.5=\frac{h_g-80}{315\left(0.02-0.0079\right)}$

$h_g-80=78.135$

∴ h_g = 158.13 kJ/kg

Concept:

In throttling process (h₁ = h₂)

And,

h = h_f + x(h_g - h_f)

Calculation:

Given:

h₁ = 820 kJ/kg, h_f2 = 600 kJ/kg, h_g2 = 2550 kJ/kg

P₁ = 1450 kPa, P₂ = 145 kPa

T₁ = 217° C = 290 K, T₂ = 294 K

Now,

In throttling process (h₁ = h₂)

∴ h₁ = h₂ = h_f2 + x(h_g2 - h_f2)

∴ 820 = 600 + x(2550 - 600)

∴ 220 = x(1950)

∴ x = 0.113

Concept:

According to the relation:

$V=V_f+x\left(V_{fg}\right)$

Where,

$\mathrm{x}=\mathrm{D}\mathrm{r}\mathrm{y}\mathrm{n}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{f}\mathrm{r}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}=\frac{M_v}{M_v+M_{\ell }}$

Calculation:

Given:

Mass of water vapour (M_v) = 20 gm

Mass of water in in liquid form (M_l) = 40 gm

Now,

$\mathrm{x}=\frac{M_v}{M_v+M_{\ell }}$

$\therefore x=\frac{20}{60}$

∴ x = 1/3

Now,

$V=V_f+x\left(V_{fg}\right)$

$\therefore V=0.001014+\frac{1}{3}\left(10.02-0.001014\right)$

$\therefore V=3.34067\frac{m^3}{kg}$

And,

$\rho =\mathrm{d}\mathrm{e}\mathrm{n}\mathrm{s}\mathrm{i}\mathrm{t}\mathrm{y}=\frac{1}{V}$

$\rho =\frac{1}{3.34067}$

∴ ρ = 0.299 kg/m³