Engineering College

Consider a turbofan engine installed on an aircraft flying at an altitude of 5500m. The CPR is 12 and the inlet diameter of this engine is 2.0m The bypass ratio of this engine 8. The bypass ratio (BPR) of a turbofan engine is the ratio between the mass flow rate of the bypass stream to the mass flow rate entering the core. The inlet temperature is 253K and the outlet temperature is 233K. Determine the thrust of this engine in order to fly at the velocity of 250 m/s. Assume cold air approach. The engine is ideal.

Answers

Answer 1

Answer:

The thrust of the engine calculated using the cold air is 34227.35 N

Explanation:

For the turbofan engine, firstly the overall mass flow rate is considered. The mass flow rate is given as

[tex]\dot{m}=\rho AV_a[/tex]

Here

ρ is the density which is given as [tex]\dfrac{P}{RT}[/tex]P is the pressure of air at 5500 m from the ISA whose value is 50506.80 PaR is the gas constant whose value is 286.9 J/kg.KT is the temperature of the inlet which is given as 253 KA is the cross-sectional area of the inlet which is given by using the diameter of 2.0 mV_a is the velocity of the aircraft which is given as 250 m/s

So the equation becomes

[tex]\dot{m}=\rho AV_a\\\dot{m}=\dfrac{P}{RT} AV_a\\\dot{m}=\dfrac{50506.80}{286.9\times 253} \times (\dfrac{\pi}{4}\times 2^2)\times 250\\\dot{m}=546.4981\ kgs^{-1}[/tex]

Now in order to find the flow from the fan, the Bypass ratio is used.

[tex]\dot{m}_f=\dfrac{BPR}{BPR+1}\times \dot{m}[/tex]

Here BPR is given as 8 so the equation becomes

[tex]\dot{m}_f=\dfrac{BPR}{BPR+1}\times \dot{m}\\\dot{m}_f=\dfrac{8}{8+1}\times 546.50\\\dot{m}_f=485.77\ kgs^{-1}[/tex]

Now the exit velocity is calculated using the total energy balance which is given as below:

[tex]h_4+\dfrac{1}{2}V_a^2=h_5+\dfrac{1}{2}V_e^2[/tex]

Here

h_4 and h_5 are the enthalpies at point 4 and 5 which could be rewritten as [tex]c_pT_4[/tex] and [tex]c_pT_5[/tex] respectively.The value of T_4 is the inlet temperature which is 253 KThe value of T_5 is the outlet temperature which is 233KThe value of c_p is constant which is 1005 J/kgKV_a is the inlet velocity which is 250 m/sV_e is the outlet velocity that is to be calculated.

So the equation becomes

[tex]h_4+\dfrac{1}{2}V_a^2=h_5+\dfrac{1}{2}V_e^2\\c_pT_4+\dfrac{1}{2}V_a^2=c_pT_5+\dfrac{1}{2}V_e^2[/tex]

Rearranging the equation gives

[tex]\dfrac{1}{2}V_e^2=c_pT_4-c_pT_5+\dfrac{1}{2}V_a^2\\\dfrac{1}{2}V_e^2=c_p(T_4-T_5)+\dfrac{1}{2}V_a^2\\V_e^2=2c_p(T_4-T_5)+V_a^2\\V_e=\sqrt{2c_p(T_4-T_5)+V_a^2}\\V_e=\sqrt{2\times 1005\times (253-233)+(250)^2}\\V_e=320.46 m/s[/tex]

Now using  the cold air approach, the thrust is given as follows

[tex]T=\dot{m}_f(V_e-V_a)\\T=485.77\times (320.46-250)\\T=34227.35\ N[/tex]

So the thrust of the engine calculated using the cold air is 34227.35 N

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