Brayton Cycle MCQ Quiz - Objective Question with Answer for Brayton Cycle - Download Free PDF

Explanation:

Steam rate:

• The capacity of a steam plant is often expressed in terms of steam rate or specific steam consumption.
• It is defined as the rate of steam flow (kg/s) required to produce unit shaft output (1 kW).

\(Steam\;Rate = \frac{{3600}}{{{W_{net}}}}\;kg/kW - hr\)

Concept:

The Brayton cycle is an ideal gas turbine cycle that consists of isentropic compression, constant-pressure heat addition, isentropic expansion, and constant-pressure heat rejection. The net work output is the difference between turbine and compressor work.

Given:

Pressure ratio, \( r_p = 5 \)

Inlet temperature, \( T_1 = 300~K \)

Maximum temperature, \( T_3 = 928~K \)

\( C_p = 1.0~\text{kJ/kg·K},~\gamma = 1.41 \)

Calculation:

\( T_2 = T_1 \times r_p^{\frac{\gamma - 1}{\gamma}} = 300 \times 5^{\frac{0.41}{1.41}} \approx 300 \times 1.6 = 480~K \)

\( T_4 = T_3 \times r_p^{-\frac{\gamma - 1}{\gamma}} = 928 \div 1.6 = 580~K \)

Net work output per kg of air:

\( W_{net} = C_p \left[ (T_3 - T_4) - (T_2 - T_1) \right] \)

\( W_{net} = 1.0 \times \left[ (928 - 580) - (480 - 300) \right] = 348 - 180 = 168~\text{kJ/kg} \)

Concept:

We use isentropic expansion relations in a turbine to determine the power developed when supplying air to a cabin.

Given:

• Mass flow rate, \( \dot{m} = 0.1 \, \text{kg/s} \)
• Inlet pressure, \( P_1 = 0.4 \, \text{MN/m²} = 400 \, \text{kPa} \)
• Exit pressure, \( P_2 = 0.1 \, \text{MN/m²} = 100 \, \text{kPa} \)
• Exit temperature, \( T_2 = 285 \, \text{K} \)
• Specific heat at constant pressure, \( C_p = 1 \, \text{kJ/kg·K} \)
• Ratio of specific heats, \( \gamma = 1.41 \)
• Given relation: \( 4^{\frac{(\gamma - 1)}{\gamma}} = 1.5 \)

Step 1: Determine the inlet temperature (T₁)

For an isentropic process:

\( \frac{T_1}{T_2} = \left(\frac{P_1}{P_2}\right)^{\frac{\gamma - 1}{\gamma}} = 1.5 \)

\( T_1 = 285 \times 1.5 = 427.5 \, \text{K} \)

Step 2: Calculate the work output (power)

The power developed by the turbine is given by:

\( W = \dot{m} \times C_p \times (T_1 - T_2) \)

\( W = 0.1 \times 1 \times (427.5 - 285) = 14.25 \, \text{kW} \)

Concept:

We use the isentropic relations for an ideal Brayton cycle to determine the final temperatures at the end of the compression and expansion processes.

Given:

• Minimum pressure, \( P_1 = 1 \, \text{bar} \)
• Maximum pressure, \( P_2 = 6 \, \text{bar} \)
• Minimum temperature, \( T_1 = 300 \, \text{K} \)
• Maximum temperature, \( T_3 = 1500 \, \text{K} \)
• Ratio of specific heats, \( \gamma = 1.4 \)

Step 1: Calculate temperature at the end of compression (T₂)

The compression process is isentropic. The relation between temperature and pressure is:

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

Substitute the given values:

\( T_2 = 300 \left( \frac{6}{1} \right)^{\frac{1.4 - 1}{1.4}} = 300 \times 6^{0.2857} \approx 300 \times 1.668 \approx 500 \, \text{K}\)

Step 2: Calculate temperature at the end of expansion (T₄)

The expansion process is also isentropic. The relation between temperature and pressure is:

\( T_4 = T_3 \left( \frac{P_1}{P_2} \right)^{\frac{\gamma - 1}{\gamma}} \)

Substitute the given values:

\( T_4 = 1500 \left( \frac{1}{6} \right)^{\frac{1.4 - 1}{1.4}} = 1500 \times 6^{-0.2857} \approx 1500 \times 0.599 \approx 900 \, \text{K}\)

Explanation:

Given:

T₁ = 300 K, T₂ = 550 K, T₄ = 1250 K, T₅ = 800 K, T₆ = 600 K,

From energy balance across regenerator

T₃ - T₂ = T₅ - T₆

T₃ - 550 = 800 - 600

Тз = 750 K

Now, 

\(\eta=1-\frac{Q_{R}}{Q_{s}}=1-\frac{\left(T_{6}-T_{1}\right)}{\left(T_{4}-T_{3}\right)}=1-\frac{(600-300)}{(1250-750)}\)

\(\eta=1-\frac{300}{500}=0.4 \text { or } 40 \%\)

Concept:

The working of closed cycle gas turbine is as follows

Process 1-2: Isentropic compression of gas takes place in the compressor.

Process 2-3: It denotes the heating of gas in the heating chamber at constant pressure.

Process 3-4: In this process, the expansion of gas takes place isentropically.

Process 4-1: This process shows the cooling of gas at constant pressure in the cooling chamber.

• In a closed-cycle gas turbine, the same working fluid is recirculated again and again
• In an open cycle gas turbine, the working fluid is used only one time.

Concept:

Work ratio: 

• It is defined as the ratio of net work to turbine work.
• The gas turbine is work on the Brayton cycle.
• This cycle consist of  two adiabatic process and two isobaric process.

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\(Work Ratio=\frac{W_{net}}{W_T}=\frac{W_T-W_C}{W_T}\)

Turbine work is given by,

WT = Cp(T3 - T4)

Compressor work is given by,

WC = Cp(T2 - T1)

Therefore, \(Work ~Ratio=\frac{C_p(T_3-T_4)-C_p(T_2-T_1)}{C_p(T_3-T_4)}\)

\(\Rightarrow1-\frac{T_2-T_1}{T_3-T_4}\)

\(Work~ratio=1-\frac{T_1(\frac{T_2}{T_1}-1)}{T_4(\frac{T_3}{T_4}-1)}\)

Here, \(\frac{T_2}{T_1}=\frac{T_3}{T_4}=r_p^{\frac{\gamma -1}{\gamma}}\)

Therefore, \(Work~ratio=1-\frac{T_1}{T_4}\)

\(Work~ratio=1-\frac{T_1}{T_4}\times\frac{T_3}{T_3}\)

\(Work~ratio=1-\frac{T_1}{T_3}\times\frac{T_3}{T_4}\)

\(Work~ratio=1 - \frac {T_1}{T_3} (r_p)^\frac {\gamma - 1}{\gamma}\)

Explanation:

The basic differences between the gas turbine and steam turbine are:

Explanation:

An open-cycle or closed-cycle gas turbine is a turbine that uses a gas (e.g. air, nitrogen, helium, argon, etc.) for the working fluid as part of a closed thermodynamic system. Heat is supplied from an external source. Such recirculating turbines follow the Brayton cycle.

Gas turbines operate on the Brayton cycle/Joule cycle. The Joule cycle consists of four internally reversible processes:

• Isentropic compression (in a compressor)
• Constant-pressure heat addition
• Isentropic expansion (in a turbine)
• Constant-pressure heat rejection

Otto cycle: It is the standard air cycle used in the spark ignition (SI) engines or petrol engines.

The Carnot cycle can be thought of as the most efficient heat engine cycle.

Stirling engine is used in a mechanical propulsion system and electrical generation systems.

Concept:

Work ratio is defined as net work output and work done by turbine

Work ratio = \(\frac{{Net\;work\;output}}{{positive\;workdone\;\left( {Turbine\;work} \right)}}\)

\(WR = \frac{{{C_p}\left[ {{T_3} - {T_4}} \right] - {C_p}\left[ {{T_2} - {T_1}} \right]}}{{{C_p}\left[ {{T_3} - {T_4}} \right]}}\)

\(WR = 1-\frac{{{}{{} {}} {}\left[ {{T_2} - {T_1}} \right]}}{{{}\left[ {{T_3} - {T_4}} \right]}}\)

Efficiency is defined as the ratio of net work output to the energy supplied to the cycle.

Heat added in system \( = {C_p}\left[ {{T_3} - {T_2'}} \right]\)

Head rejected = C_p [T_4' – T₁]

For ideal Regenerator

T₂ = T_4' and T2' = T4 

\(\eta = 1 - \frac{{{Q_{Rejected}}}}{{{Q_{added}}}}\)

\(\eta = 1 - \frac{{{C_p}\left[ {{T_4'} - {T_1}} \right]}}{{{C_p}\left[ {{T_3} - {T_2'}} \right]}}\)

\(\eta = 1 - \frac{{{C_p}\left[ {{T_2} - {T_1}} \right]}}{{{C_p}\left[ {{T_3} - {T_4}} \right]}}\)

\(\eta = 1 - \frac{{\left[ {{T_2} - {T_1}} \right]}}{{\left[ {{T_3} - {T_4}} \right]}}\)

Hence, Efficiency is equal to work ratio

Concept:

The Brayton cycle is a theoretical cycle for simple gas turbine. Ideal Brayton cycle consists of two isentropic and two constant pressure processes.

As the plot depicts for a fixed turbine inlet temperature T3, the net work output per cycle increases with the pressure ratio initially, it reaches a maximum value and then it starts to decrease.                                                        

For a fixed values of Tmax and Tmin pressure ratio for maximum work done by gas turbine:

\({\left( {\frac{{{T_{\max }}}}{{{T_{\min }}}}} \right)^{\gamma /2\left( {\gamma - 1} \right)}}\)

where, γ = specific heat ratio, Tmax = Maximum temperature in gas turbine cycle, Tmin = Minimum temperature in the gas turbine cycle.

Concept:

Gas turbines operate on the Brayton cycle/Joule cycle. The Brayton cycle consists of four internally reversible processes:

• Isentropic compression (in a compressor)
• Constant-pressure heat addition
• Isentropic expansion (in a turbine)
• Constant-pressure heat rejection

Work ratio: Work ratio in the gas turbine is defined as the ratio of net-work to the turbine work.

The net-work in the turbine is given by Wturbine – Wcompressor, because the compressor is work consuming device therefore it consumes the work.

\(Work - ratio = \frac{{{{\rm{W}}_{{\rm{net}}}}}}{{{{\rm{W}}_{{\rm{turbine}}}}{\rm{\;}}}} = \;\frac{{{{\rm{W}}_{{\rm{turbine}}}}{\rm{\;}} - {\rm{\;}}{{\rm{W}}_{{\rm{compressor}}}}}}{{{{\rm{W}}_{{\rm{turbine}}}}}} = 1 - \;\frac{{{{\rm{W}}_{{\rm{compressor}}}}}}{{{{\rm{W}}_{{\rm{turbine}}}}}}\)

\( \Rightarrow WR = 1 - \frac{{m{C_P}\left( {{T_2} - {T_1}} \right)}}{{m{C_P}\left( {{T_3} - {T_4}} \right)}}\;\)

\( \Rightarrow WR = 1 - \frac{{\left( {{T_2} - {T_1}} \right)}}{{\left( {{T_3} - {T_4}} \right)}}\)

As \(\frac{{{T_2}}}{{{T_1}}} = \frac{{{T_3}}}{{{T_4}}} = {\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^{\frac{{\gamma - 1}}{\gamma }}} = r_P^{\frac{{\gamma - 1}}{\gamma }}\)

\( \Rightarrow WR = 1 - \frac{{{T_1}\left( {r_P^{\frac{{\gamma - 1}}{\gamma }}} \right) - {T_1}\;}}{{{T_3} - \frac{{{T_3}}}{{\left( {r_P^{\frac{{\gamma - 1}}{\gamma }}} \right)}}}}\)

\( \Rightarrow WR = 1 - \frac{{{T_1}}}{{{T_3}}}r_P^{\frac{{\gamma - 1}}{\gamma }}\)

∴ Work ratio depends on r_P (Pressure ratio), γ (ratio of specific heats), temperature ratio (T₁/T₃). 

Concept:

Brayton Cycle:

The Brayton cycle (or Joule cycle) represents the operation of a gas turbine engine.

The cycle consists of four processes

• Process a – b: Reversible adiabatic compression in the inlet and compressor
• Process b – c: Isobaric heat addition (fuel combustion)
• Process c – d: Reversible adiabatic expansion in the turbine and exhaust nozzle
• Process d – a: Cool the air at constant pressure back to its initial condition.

The efficiency of the Brayton cycle:

\({\eta _{\rm{b}}} = 1 - \;\frac{1}{{r{_p^{\frac{γ - 1}{γ}}}}}\)

Where, r_p = pressure ratio = \(\frac{{{P_b}}}{{{P_a}}} = \frac{{{P_c}}}{{{P_d}}}\)

Otto cycle:

Otto cycle consists of four processes. They are as follows:

Process 1 – 2: Reversible adiabatic or Isentropic compression

Process 2 – 3: Constant Volume Heat Addition

Process 3 – 4: Isentropic (reversible adiabatic) expansion

Process 4 – 1: Constant Volume Heat Rejection

The efficiency of the Otto cycle:

\({\eta _{otto}} = 1 - \frac{1}{{{r^{γ - 1}}}}\)

Where r = compression ratio \(= \,\,\frac{{V_1}}{{V_2}}\)

We know that compression in both the Otto cycle and Brayton cycle is an isentropic process, so we may write

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

Thus for the same pressure ratio or compression ratio in Otto and Brayton, their efficiency is equal.

Concept:

In a turbine, the thermal efficiency is given by 

\(\eta=\frac{W_{net}}{Q_{supplied}}\)

The net work in a gas turbine is the difference of work done by the turbine and work consumed by the compressor 

W_net = W_T - W_C

Efficiency is 20%

∴ \(\eta=\frac{W_{net}}{Q_{in}}=\frac{W_{T}-W_C}{Q_{supplied}}\)

Calculation:

Given, WT = 600 kJ/kg, WC = 400 kJ/kg, Q_supplied = 1000 kJ/kg

\(\eta=\frac{600-400}{1000}=0.2\)

Concept:

A gas turbine work on Brayton cycle, Thermal Efficiency of Brayton cycle is given by 

 \({\eta _{th}} = 1 - \frac{1}{{{{\left( {{r_p}} \right)}^{\frac{{\gamma - 1}}{\gamma }}}}}\)where rp = pressure ratio, \(r_p=\frac{P_2}{P_1}\)

For Maximum work : \({r_p} = {\left( {\frac{{{T_3}}}{{{T_1}}}} \right)^{\frac{\gamma }{{2\left( {\gamma - 1} \right)}}}}\)

Where T3 is the maximum temperature in the cycle which is at the inlet of the turbine,

and T1 is the minimum temperature in the cycle which is at the inlet of the compressor now if we put the maximum compression ratio in the thermal efficiency formula then 

\({\eta _{th}} = 1 - \sqrt {\frac{{{T_1}}}{{{T_3}}}} = 1 - \sqrt {\frac{{{T_{min}}}}{{{T_{max}}}}} \)

Calculation:

Given, temperature limits, T3 = 1200 K, and T1 = 300 K

So \({\eta _{th}} = 1 - \sqrt {\frac{{{T_1}}}{{{T_3}}}} = 1 - \sqrt {\frac{{{T_{min}}}}{{{T_{max}}}}} \)

\({\eta _{th}} = 1 - \sqrt {\frac{{300}}{{1200}}} = 0.5\;or\;50\% \)