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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
![\dot{m}=\rho AV_a](https://tex.z-dn.net/?f=%5Cdot%7Bm%7D%3D%5Crho%20AV_a)
Here
- ρ is the density which is given as
![\dfrac{P}{RT}](https://tex.z-dn.net/?f=%5Cdfrac%7BP%7D%7BRT%7D)
- P is the pressure of air at 5500 m from the ISA whose value is 50506.80 Pa
- R is the gas constant whose value is 286.9 J/kg.K
- T is the temperature of the inlet which is given as 253 K
- A is the cross-sectional area of the inlet which is given by using the diameter of 2.0 m
- V_a is the velocity of the aircraft which is given as 250 m/s
So the equation becomes
![\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}](https://tex.z-dn.net/?f=%5Cdot%7Bm%7D%3D%5Crho%20AV_a%5C%5C%5Cdot%7Bm%7D%3D%5Cdfrac%7BP%7D%7BRT%7D%20AV_a%5C%5C%5Cdot%7Bm%7D%3D%5Cdfrac%7B50506.80%7D%7B286.9%5Ctimes%20253%7D%20%5Ctimes%20%28%5Cdfrac%7B%5Cpi%7D%7B4%7D%5Ctimes%202%5E2%29%5Ctimes%20250%5C%5C%5Cdot%7Bm%7D%3D546.4981%5C%20kgs%5E%7B-1%7D)
Now in order to find the flow from the fan, the Bypass ratio is used.
![\dot{m}_f=\dfrac{BPR}{BPR+1}\times \dot{m}](https://tex.z-dn.net/?f=%5Cdot%7Bm%7D_f%3D%5Cdfrac%7BBPR%7D%7BBPR%2B1%7D%5Ctimes%20%5Cdot%7Bm%7D)
Here BPR is given as 8 so the equation becomes
![\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}](https://tex.z-dn.net/?f=%5Cdot%7Bm%7D_f%3D%5Cdfrac%7BBPR%7D%7BBPR%2B1%7D%5Ctimes%20%5Cdot%7Bm%7D%5C%5C%5Cdot%7Bm%7D_f%3D%5Cdfrac%7B8%7D%7B8%2B1%7D%5Ctimes%20546.50%5C%5C%5Cdot%7Bm%7D_f%3D485.77%5C%20kgs%5E%7B-1%7D)
Now the exit velocity is calculated using the total energy balance which is given as below:
![h_4+\dfrac{1}{2}V_a^2=h_5+\dfrac{1}{2}V_e^2](https://tex.z-dn.net/?f=h_4%2B%5Cdfrac%7B1%7D%7B2%7DV_a%5E2%3Dh_5%2B%5Cdfrac%7B1%7D%7B2%7DV_e%5E2)
Here
- h_4 and h_5 are the enthalpies at point 4 and 5 which could be rewritten as
and
respectively. - The value of T_4 is the inlet temperature which is 253 K
- The value of T_5 is the outlet temperature which is 233K
- The value of c_p is constant which is 1005 J/kgK
- V_a is the inlet velocity which is 250 m/s
- V_e is the outlet velocity that is to be calculated.
So the equation becomes
![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](https://tex.z-dn.net/?f=h_4%2B%5Cdfrac%7B1%7D%7B2%7DV_a%5E2%3Dh_5%2B%5Cdfrac%7B1%7D%7B2%7DV_e%5E2%5C%5Cc_pT_4%2B%5Cdfrac%7B1%7D%7B2%7DV_a%5E2%3Dc_pT_5%2B%5Cdfrac%7B1%7D%7B2%7DV_e%5E2)
Rearranging the equation gives
![\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](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7DV_e%5E2%3Dc_pT_4-c_pT_5%2B%5Cdfrac%7B1%7D%7B2%7DV_a%5E2%5C%5C%5Cdfrac%7B1%7D%7B2%7DV_e%5E2%3Dc_p%28T_4-T_5%29%2B%5Cdfrac%7B1%7D%7B2%7DV_a%5E2%5C%5CV_e%5E2%3D2c_p%28T_4-T_5%29%2BV_a%5E2%5C%5CV_e%3D%5Csqrt%7B2c_p%28T_4-T_5%29%2BV_a%5E2%7D%5C%5CV_e%3D%5Csqrt%7B2%5Ctimes%201005%5Ctimes%20%28253-233%29%2B%28250%29%5E2%7D%5C%5CV_e%3D320.46%20m%2Fs)
Now using the cold air approach, the thrust is given as follows
![T=\dot{m}_f(V_e-V_a)\\T=485.77\times (320.46-250)\\T=34227.35\ N](https://tex.z-dn.net/?f=T%3D%5Cdot%7Bm%7D_f%28V_e-V_a%29%5C%5CT%3D485.77%5Ctimes%20%28320.46-250%29%5C%5CT%3D34227.35%5C%20N)
So the thrust of the engine calculated using the cold air is 34227.35 N
Answer:
The answer is VN =37.416 m/s
Explanation:
Recall that:
Pressure (atmospheric) = 100 kPa
So. we solve for the maximum velocity (m/s) to which water can be accelerated by the nozzles
Now,
Pabs =Patm + Pgauge = 800 KN/m²
Thus
PT/9.81 + VT²/2g =PN/9.81 + VN²/2g
Here
Acceleration due to gravity = 9.81 m/s
800/9.81 + 0
= 100/9.81 + VN²/19.62
Here,
9.81 * 2= 19.62
Thus,
VN²/19.62 = 700/9.81
So,
VN² =1400
VN =37.416 m/s
Note: (800 - 100) = 700
<em>Here I hope this is what you need?</em>
<em>When austenite in iron-carbon alloys is transformed isothermally below the eutectoid temperature at small undercooling, it undergoes eutectoid transformation to produce a unique micro- structure termed “pearlite”, which was discovered by Sorby in 1864.</em>
Answer:
Abolition of intermediaries (rent collectors under the pre-Independence land revenue system); Tenancy regulation (to improve the contractual terms including the security of tenure); A ceiling on landholdings (to redistributing surplus land to the landless);