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CalcTree
This calculator evaluates ratio equations, efficiency and effectiveness for the Brayton Cycles.


The following are the variables for the equations.

  1. V = the volume. i.e. V(1) is the volume at state 1.
  2. r(k) is the compression ratio
  3. r(e) is the expansion ratio
  4. r(c) is the cut-off ratio
  5. MEP = mean effective pressure. The constant theoretical pressure would produce the same network in one complete cycle if it acted on the piston. And can be defined as the following:

MEP = Net Work For One CycleDisplacement VolumeMEP\ =\ \frac{Net\ Work\ For\ One\ Cycle}{Displacement\ Volume}
  1. η = the efficiency of the cycle
  2. k = the adiabatic index
  1. P = the pressure. i.e. P(1) is the pressure at state 1.
  1. Where: P(1)[D(eff)-N(eff)] = Q(h)-Q(c), Q(h) is the amount of heat initially extracted, and Q(c) is the heat expelled.

WT=mCPΔTW_T=mC_P{\Delta}T

T2T1=T3T4\frac{T_2}{T_1}=\frac{T_3}{T_4}

QA=mCPΔTQ_A=mC_P{\Delta}T

Work Calculations

Inputs


m
:100.00g



C(P)
:4.18J/g*K



T (change in temp)
:20.00degC


Output


W
:8,360.00kJ


WP=mCPΔTW_P=mC_P{\Delta}T

Efficiency Calculations


Inputs


W(C)
:20.00kJ



W(T)
:30.00kJ


Output


r(BW)
:0.67


rBW=WCWTr_{BW}=\frac{W_C}{W_T}


η
:-0.10


η=11rpk1k\eta=1-\frac{1}{r_p\frac{k-1}{k}}

Net Work and Combustor Calculations


Inputs


P(1)
:40.00atm



P(2)
:100.00atm



T(1)
:25.00degC



T(3)
:35.00degC



T(max)
:25.00degC



T(min)
:30.00degC



Q(air)
:80.00J



Q (fuel)
:100.00J



k
:7.00


Maximum Net Work Results



P(x)
:63.25atm


Px=P1P2P_x=\sqrt{P_1P_2}


T(2)
:29.58degC


T2=T1T3T_2=\sqrt{T_1T_3}


r(p)
:0.90


rp=(TmaxTmin)k2k2r_p=(\frac{T_{max}}{T_{min}})^{\frac{k}{2k-2}}

Combustor Efficiency Results



e(c)
:0.80


ec=QairQfuele_c=\frac{Q_{air}}{Q_{fuel}}


r(B)
:0.65


rB=rpk1k(T1T3)r_B=r_p^{\frac{k-1}{k}}(\frac{T_1}{T_3})

Brayton Cycle

In using the calculators and inputting values above, refer to Figure 1 and the different states of the Brayton cycle.
Figure 1: Temperature vs Entropy Graph for a Brayton Cycle


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