Thermodynamics Test - 4

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:

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:

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

\(dh = {C_p}dT - T{\left( {\frac{{\partial v}}{{\partial T}}} \right)_p} + v \;dp\)

And,

on equating (3) & (4) we get

\(du = {C_v}dT + \;T{\left( {\frac{{\partial p}}{{\partial T}}} \right)_v}dv - p\;dv \)

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:

\({\left( {\frac{{\delta P}}{{\delta T}}} \right)_{sat}} = \;20.5{\rm{\;kPa}}/{\rm{K}}\)

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,

\({\rm{x}} = {\rm{Dryness\;fraction}} = \frac{{{M_v}}}{{{M_v} + {M_\ell }}}\)

Calculation:

Given:

Mass of water vapour (M_v) = 20 gm

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

Now,

\({\rm{x}} = \frac{{{M_v}}}{{{M_v} + {M_\ell }}}\)

\(\therefore x = \frac{{20}}{{60}}\)

∴ 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,

\({\rm{\rho \;}} = {\rm{\;density\;}} = \frac{1}{V}\)

\({\rm{\rho }} = \frac{1}{{3.34067}}\)

∴ ρ = 0.299 kg/m³