Thermodynamics Test - 5

Concept:

Reversible work and irreversibility

• Reversible work (W_rev) is the maximum amount of useful work that can be produced as a system undergoes a process between the specified initial and final states.
• For a process that require work, reversible work represents the minimum amount of work necessary to carry out that process.
• The difference between the reversible work and the useful work (W) is due to irreversibility.
• The irreversibility is equivalent to the exergy destroyed.
• For a total reversible process, the actual and reversible work terms are identical, and thus the irreversibility is zero.
• Irreversibility represents the energy that could have been converted to work but could not be converted because of the effects like friction losses, heat transfer through the finite temperature difference, etc.
• If Heat transfer takes place from a reservoir which is at the almost same temperature to the system then sgen is zero. If we are using infinite reservoirs then it will always be the case as written above, therefore overall sgen will be zero and the whole process will be reversible.

The irreversibility (I) of the process is defined as

I = W_r – W

Where,

W_r = Reversible work

W = Actual work

It can also be expressed as:

I = T₀ΔS_universe

Where,

T₀ = surrounding temperature

ΔS_universe = entropy change of the universe 

Concept:

For a polytropic process,

\(\frac{{{P_1}}}{{{P_2}}} = {\left( {\frac{{{V_2}}}{{{V_1}}}} \right)^n} = {\left( {\frac{{{T_1}}}{{{T_2}}}} \right)^{\frac{n}{{n - 1}}}}\)

\({s_2} - {s_1} = {c_v}\ln \frac{{{{\rm{T}}_2}}}{{{{\rm{T}}_1}}} + R\ln \frac{{{{\rm{V}}_2}}}{{{{\rm{V}}_1}}} = {c_v}\ln \frac{{{{\rm{T}}_2}}}{{{{\rm{T}}_1}}} + \;R\ln {\left( {\frac{{{T_1}}}{{{T_2}}}} \right)^{\frac{1}{{n - 1}}}} = {c_v}\ln \frac{{{{\rm{T}}_2}}}{{{{\rm{T}}_1}}} + \frac{R}{{n - 1}}\ln \frac{{{T_1}}}{{{T_2}}}\)

\({s_2} - {s_1} = {c_v}\ln \frac{{{{\rm{T}}_2}}}{{{{\rm{T}}_1}}} - \frac{R}{{n - 1}}\ln \frac{{{{\rm{T}}_2}}}{{{{\rm{T}}_1}}} = {c_v}\ln \frac{{{{\rm{T}}_2}}}{{{{\rm{T}}_1}}} - \frac{{{c_v}\left( {\gamma - 1} \right)}}{{n - 1}}\ln \frac{{{{\rm{T}}_2}}}{{{{\rm{T}}_1}}} = {c_v}\ln \frac{{{{\rm{T}}_2}}}{{{{\rm{T}}_1}}}\left( {1 - \frac{{\left( {\gamma - 1} \right)}}{{n - 1}}} \right)\)

\({s_2} - {s_1} = {c_v}\ln \frac{{{{\rm{T}}_2}}}{{{{\rm{T}}_1}}}\left( {\frac{{n - \gamma }}{{n - 1}}} \right)\)

The entropy change is given by:

\(\Delta {\rm{S}} = {{\rm{c}}_{\rm{v}}}{\rm{\;}}\left[ {\frac{{{\rm{n}} - {\rm{\gamma }}}}{{{\rm{n}} - 1}}} \right]\ln \frac{{{{\rm{T}}_2}}}{{{{\rm{T}}_1}}}\)

\(\Delta {\rm{S}} = \frac{{{\rm{n}} - {\rm{\gamma }}}}{{\left( {{\rm{n}} - 1} \right)\left( {\gamma - 1} \right)}}R\ln \frac{{{{\rm{T}}_2}}}{{{{\rm{T}}_1}}}\)

Concept:

According to Clausius Inequality:

\(\oint \frac{{\delta Q }}{T} = 0\)

Calculation:

Given:

T₁ = 500 K, T₂ = 1000 K

Q₁ = 50 kW, Q₂ = 30 kW, Q₃ = 40 kW

Now,

\(\frac{{{Q_1}}}{{{T_1}}} + \frac{{{Q_2}}}{{{T_2}}} - \frac{{{Q_3}}}{T} = 0\)

\(\frac{{50}}{{1000}} + \frac{{30}}{{500}} - \frac{{40}}{T} = 0\)

Concept:

Available energy:

• The portion of the energy supplied as heat which can be converted into work by a reversible engine is called available energy.
• It is also called as exergy.
• It is a property that determines the useful work potential at a given amount of energy at some specified state. It is also known as availability.
• Since work is a function of the initial state, the path of the process, and the final state we can say that work is a path function. It is not a property of the system but exergy is a property of a system and surroundings or we can say that it is the function of the total energy of the system.
• The work potential of the energy contained in the system at a specified state is maximum work that can be obtained from the system.
• It is a property of the system and surrounding combined means that we can improve the efficiency of the system itself or the other way round is that we can change the surrounding condition to increase the exergy.

Concept:

\({\left( {{\rm{\Delta }}s} \right)_{total}} = {\left( {{\rm{\Delta }}s} \right)_{block\;1}} + {\left( {{\rm{\Delta }}s} \right)_{block\;2}}\)

Calculation:

Now,

Heat lost by block 1 = Heat gained by block 2

∴ \(mC_p\left( {{T_1} - {T_f}} \right) = mC_p\;\left({{T_f} - {T_2}} \right) \)

∴ \({T_f} = \frac{{{T_1} + {T_2}}}{2} = 750 K\)

Now,

\({\left( {{\rm{\Delta }}s} \right)_{total}} = {\left( {{\rm{\Delta }}s} \right)_{block\;1}} + {\left( {{\rm{\Delta }}s} \right)_{block\;2}}\)

\({\left( {{\rm{\Delta }}s} \right)_{total}} = mC_p\;\ln\;\left( {\frac{{{T_f}}}{{{T_1}}}} \right) + mC_p\;\ln\;\left( {\frac{{{T_f}}}{{{T_2}}}} \right)\)

\({\left( {{\bf{\Delta }}s} \right)_{total}} = mC_p\left( {\ln\;\left( {\frac{{{T_f}}}{{{T_1}}}} \right) + \ln\left( {\frac{{{T_f}}}{{{T_2}}}} \right)} \right)\)

\({\left( {{\rm{\Delta }}s} \right)_{total}} = 2 \times 1 \times \left[ {\ln\;\left( {\frac{{{{750}^2}}}{{500 \times 1000}}} \right)} \right]\)

Concept:

Using the concept of availability function of a flowing gas

\({\rm{Change\;in\;A}}.{\rm{E}} = {\phi _1} - {\phi _2}\)

And,

\({\phi _1} = {h_1} - {T_0}{s_1} + \frac{{V_1^2}}{{2000}}\)

\({\phi _2} = {h_2} - {T_0}{s_2} + \frac{{V_2^2}}{{2000}}\)

Calculation:

Now,

\({\rm{Change\;in\;A}}.{\rm{E}} = {\phi _1} - {\phi _2}\)

\({\rm{Change\;in\;A}}.{\rm{E}} = {h_1} - {h_2} + {T_0}\left( {{s_2} - {s_1}} \right) + \;\frac{{V_1^2 - V_2^2}}{{2000}}\)

\(\therefore {\rm{Change\;in\;A}}.{\rm{E}} = {C_p}\left( {{T_2} - {T_1}} \right) + 300\;\left( {{C_p}\;\ln\;\left( {\frac{{{T_2}}}{{{T_1}}}} \right) - 0.287\;\ln\;\left( {\frac{{{P_2}}}{{{P_1}}}} \right)} \right) + \frac{{V_1^2 - V_2^2}}{{2000}}\)

\(\therefore {\rm{Change\;in\;A}}.{\rm{E}} = 1\left( {500 - 400} \right) + 300\;\left( {1\;\ln\;\left( {\frac{{400}}{{500}}} \right) - 0.287\;\ln\;\left( {\frac{1}{5}} \right)} \right) + \frac{{\left( {100 - 25} \right)}}{{2000}}\)

∴ Change in A.E = 171.667 kJ/kg

Concept:

Irreversibility = W_actual - W_min

Calculation:

For the case of minimum work

(Δs)_univ = 0

\({s_2} - {s_1} = 0 \Rightarrow {C_p}\;\ln\;\frac{{{T_2}}}{{{T_1}}} - R\;\ln\frac{{{p_2}}}{{{p_1}}} = 0\)

∴\(\ln \frac{{{T_2}}}{{300}} - 0.287\ln 4 = 0\)

∴ T₂ = 446.59 K

Now,

Given:

flow rate (ṁ) = 1 kg/s

Inlet temperature (T₁) = 300 K, Outlet temperature (T₂) = 446.59 K (calculated above)

Inlet velocity (V₁) = 400 m/s, Outlet velocity (V₂) = 900 m/s

Using Steady flow energy equation (SFEE)

\(W + \dot m{h_1} + \frac{{\dot m{V^2_1}\;}}{{2000}} = \dot m{h_2} + \frac{{\dot mV_2^2}}{{2000}} + Q\)

\(W = \dot m{h_2} + \frac{{\dot m{V^2_2}\;}}{{2000}} - \dot m{h_1} - \frac{{\dot mV_1^2}}{{2000}} + Q\)

\(W = \left( {446.59 - 300} \right) + \frac{1}{{2000}}\left( {900 - 400} \right)\)   (∵ h₂ - h₁ = C_p (T₂ - T₁))

∴ W_min = 146.84 kW

Now,

In actual condition.

Given:

Inlet temperature (T1)  = 300 K, Actual Outlet temperature (T2) = 500 K

Inlet velocity (V1)= 400 m/s, Outlet velocity (V2)= 900 m/s

Using Steady flow energy equation (SFEE)

\(W + \dot mh + \frac{{\dot mV_1^2}}{{2000}} = \dot m{h_2} + \frac{{\dot mV_2^2}}{{2000}}\)

\(W = \dot m{h_2} + \frac{{\dot m{V^2_2}\;}}{{2000}} - \dot m{h_1} - \frac{{\dot mV_1^2}}{{2000}} + Q\)

\(W = \left( {500 - 300} \right) + \frac{1}{{2000}}\left( {900 - 400} \right)\)   (∵ h2 - h1 = Cp (T2 - T1))

\(W = 1\left( {500 - 300} \right) + 0.25 \)

∴ Wactual = 200.25 kW

Now,

Irreversibility = Wactual - Wmin

∴ Irreversibility = 200.25 - 146.84

∴ Irreversibility = 53.41 kW