Answer:
a) 5.2 kPa
b) 49.3%
Explanation:
Given data:
Thermal efficiency ( л ) = 56.9% = 0.569
minimum pressure ( P1 ) = 100 kpa
<u>a) Determine the pressure at inlet to expansion process</u>
P2 = ?
r = 1.4
efficiency = 1 - [ 1 / (rp)
]
0.569 = 1 - [ 1 / (rp)^0.4/1.4
1 - 0.569 = 1 / (rp)^0.285
∴ (rp)^0.285 = 0.431
rp = 0.0522
note : rp = P2 / P1
therefore P2 = rp * P1 = 0.0522 * 100 kpa
= 5.2 kPa
b) Thermal efficiency
Л = 1 - [ 1 / ( 10.9 )^0.285 ]
= 0.493 = 49.3%
Answer:
304.13 mph
Explanation:
Data provided in the question :
The Speed of the flying aircraft = 300 mph
Tailwind of the true airspeed = 50 mph
Now,
The ground speed will be calculated as:
ground speed = ![\sqrt{300^2+50^2}](https://tex.z-dn.net/?f=%5Csqrt%7B300%5E2%2B50%5E2%7D)
or
The ground speed = ![\sqrt{92500}](https://tex.z-dn.net/?f=%5Csqrt%7B92500%7D)
or
The ground speed = 304.13 mph
Hence, the ground speed is 304.13 mph
Answer:
(absolute).
Explanation:
Given that
Pressure ratio r
r=8
![r=\dfrac{P_2_{abs}}{P_1_{abs}}](https://tex.z-dn.net/?f=r%3D%5Cdfrac%7BP_2_%7Babs%7D%7D%7BP_1_%7Babs%7D%7D)
-----1
P₁(gauge) = 5.5 psig
We know that
Absolute pressure = Atmospheric pressure + Gauge pressure
Given that
Atmospheric pressure = 14.5 lbf/in²
P₁(abs) = 14.5 + 5.5 psia
P₁(abs) =20 psia
Now by putting the values in the above equation 1
Therefore the exit gas pressure will be 160 psia (absolute).
Answer:
t = 2244.3 sec
Explanation:
calculate the thermal diffusivity
![\alpha = \frac{k}{\rho c}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7Bk%7D%7B%5Crho%20c%7D)
![= \frac{50}{7800\times 480} = 1.34 \times 10^{-5} m^2/s](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B50%7D%7B7800%5Ctimes%20480%7D%20%3D%201.34%20%5Ctimes%2010%5E%7B-5%7D%20m%5E2%2Fs)
Temperature at 28 mm distance after t time = = 50 degree C
we know that
![\frac[ T_{28} - T_s}{T_i -T_s} = erf(\frac{x}{2\sqrt{at}})](https://tex.z-dn.net/?f=%5Cfrac%5B%20T_%7B28%7D%20-%20T_s%7D%7BT_i%20-T_s%7D%20%3D%20erf%28%5Cfrac%7Bx%7D%7B2%5Csqrt%7Bat%7D%7D%29)
![\frac{ 50 -25}{300-25} = erf [\frac{28\times 10^{-3}}{2\sqrt{1.34\times 10^{-5}\times t}}]](https://tex.z-dn.net/?f=%5Cfrac%7B%2050%20-25%7D%7B300-25%7D%20%3D%20erf%20%5B%5Cfrac%7B28%5Ctimes%2010%5E%7B-3%7D%7D%7B2%5Csqrt%7B1.34%5Ctimes%2010%5E%7B-5%7D%5Ctimes%20t%7D%7D%5D)
![0.909 = erf{\frac{3.8245}{\sqrt{t}}}](https://tex.z-dn.net/?f=0.909%20%3D%20erf%7B%5Cfrac%7B3.8245%7D%7B%5Csqrt%7Bt%7D%7D%7D)
from gaussian error function table , similarity variable w calculated as
erf w = 0.909
it is lie between erf w = 0.9008 and erf w = 0.11246 so by interpolation we have
w = 0.08073
![erf 0.08073 = erf[\frac{3.8245}{\sqrt{t}}]](https://tex.z-dn.net/?f=erf%200.08073%20%3D%20erf%5B%5Cfrac%7B3.8245%7D%7B%5Csqrt%7Bt%7D%7D%5D)
![0.08073 = \frac{3.8245}{\sqrt{t}}](https://tex.z-dn.net/?f=0.08073%20%3D%20%5Cfrac%7B3.8245%7D%7B%5Csqrt%7Bt%7D%7D)
solving fot t we get
t = 2244.3 sec