Example Of Brayton Cycle – Problem With Solution

5/10/2019

Example of Brayton Cycle – Problem with Solution

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Example of Brayton Cycle – Problem with Solution Brayton Cycle – Problem with Solution

Let assume the closed Brayton cycle, which is the one of most common thermodynamic cycles that can be found in modern gas turbine engines. In this case assume a helium gas turbine with single compressor and single turbine arrangement. One of key parameters of such engines is the maximum turbine inlet temperature and the compressor pressure ratio (PR = p2/p1) which determines the thermal efficiency of such engine. Kate In thisAcademic turbine the high-pressure stage receives gas (point 3 at the figure) from a heat exchanger: Consultant p3 = 6.7 MPa; T3 = 1190 K (917°C)) the isentropic turbine efficiency is ηT = 0.91 (91%) and exhaust it to another heat exchanger, where the outlet pressure is (point 4): p4 = 2.78 MPa T4,is = ? Thus the compressor pressure ratio is equal to PR = 2.41.  Moreover we know, that the compressor receives gas (point 1) at the figure: p1 = 2.78 MPa; T1 = 299 K (26°C) the isentropic compressor efficiency ηK = 0.87 (87%). The heat capacity ratio, , for helium is equal to =cp/cv=1.66

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5/10/2019

Example of Brayton Cycle – Problem with Solution

Calculate: 1. the heat added by the heat exchanger (between 2 → 3) 2. the compressor outlet temperature of the gas (T2,is) 3. the real work done on this compressor, when the isentropic compressor efficiency is ηK = 0.87 (87%) 4. the turbine outlet temperature of the gas (T4,is) 5. the real work done by this turbine, when the isentropic turbine efficiency is ηT = 0.91 (91%) 6. the thermal efficiency of this cycle Solution: 1) + 2) Kate From Academic the first law of thermodynamics, the net heat added is given by Qadd,ex = H3 – H2 [kJ] or Qadd Consultant = Cp.(T3-T2s), but in this case we do not know the temperature (T2s) at the outlet of the compressor. We will solve this problem in intensive variables. We have to rewrite the previous equation (to include ηK) using the term (+h1 – h1) to: Qadd = h3 – h2 = h3 – h1 – (h2s – h1)/ηK  [kJ/kg] Qadd = cp(T3-T1) – (cp(T2s-T1)/ηK) Then we will calculate the temperature, T2s, using p, V, T Relation (from Ideal Gas Law) for adiabatic process between (1 → 2).

In this equation the factor for helium is equal to =cp/cv=1.66. From the previous equation follows that the compressor outlet temperature, T2s, is: We use cookies to ensure that we give you the best experience on our website. If you continue to use this site we  will assum that you agree with it.

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5/10/2019

Example of Brayton Cycle – Problem with Solution

   Specific Heat Calculation

Using this temperature and the isentropic compressor efficiency we can calculate the heat added by the heat exchanger: Qadd = cp(T3-T1) – (cp(T2s-T1)/ηK) = 5200.(1190 – 299) – 5200.(424-299)/0.87 = 4.633 MJ/kg – 0.747 MJ/kg = 3.886 MJ/kg 3) The work done on the gas by the compressor in the isentropic compression process is: WC,s = cp (T2s – T1) = 5200 x (424 – 299) = 0.650 MJ/kg The real work done on the gas by the compressor in the adiabatic compression is then: WC,real = cp (T2s – T1). ηC = 5200 x (424 – 299) / 0.87 = 0.747 MJ/kg 4) Kate The turbine outlet temperature of the gas, T4,is, can be calculated using the same p, V, T Academic Consultant Relation as in 2) but between states 3 and 4:

From the previous equation follows that the outlet temperature of the gas, T4,is, is:

5) The work done by gas turbine in the isentropic expansion is then: W

= c (T – T ) = 5200 x (1190 – 839) = 1.825 MJ/kg

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5/10/2019

Example of Brayton Cycle – Problem with Solution

The real work done by gas turbine in the adiabatic expansion is then: WT,real = cp (T3 – T4s) . ηT = 5200 x (1190 – 839) x 0.91 = 1.661 MJ/kg 6) As was derived in the previous section, the thermal efficiency of an ideal Brayton cycle is a function of pressure ratio and κ:

therefore ηth = 0.295 = 29.5% The thermal efficiency can be also calculated using the work and the heat (without ηK): ηth,s = (WT,s – WC,s) / Qadd,s = (1.825 – 0.650) / 3.983  =  0.295 = 29.5% Finally, the thermal efficiency including isentropic turbine/compressor efficiency is: ηth,real = (WT,real – WC,real) / Qadd = (1.661 – 0.747) / 3.886  =  0.235 = 23.5%  

Kate  References: Academic Consultant

See above: Brayton Cycle

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Example of Brayton Cycle – Problem with Solution

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Example of Brayton Cycle – Problem with Solution

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Example of Brayton Cycle – Problem with Solution

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