Answer:
Determine the added thrust required during water scooping, as a function of aircraft speed, for a reasonable range of speeds.= 132.26∪
Explanation:
check attached files for explanation
Answer:
1) The exergy of destruction is approximately 456.93 kW
2) The reversible power output is approximately 5456.93 kW
Explanation:
1) The given parameters are;
P₁ = 8 MPa
T₁ = 500°C
From which we have;
s₁ = 6.727 kJ/(kg·K)
h₁ = 3399 kJ/kg
P₂ = 2 MPa
T₂ = 350°C
From which we have;
s₂ = 6.958 kJ/(kg·K)
h₂ = 3138 kJ/kg
P₃ = 2 MPa
T₃ = 500°C
From which we have;
s₃ = 7.434 kJ/(kg·K)
h₃ = 3468 kJ/kg
P₄ = 30 KPa
T₄ = 69.09 C (saturation temperature)
From which we have;
h₄ =
+ x₄×
= 289.229 + 0.97*2335.32 = 2554.49 kJ/kg
s₄ =
+ x₄×
= 0.94394 + 0.97*6.8235 ≈ 7.563 kJ/(kg·K)
The exergy of destruction,
, is given as follows;
= T₀ ×
= T₀ ×
× (s₄ + s₂ - s₁ - s₃)
= T₀ ×
×(s₄ + s₂ - s₁ - s₃)/(h₁ + h₃ - h₂ - h₄)
∴
= 298.15 × 5000 × (7.563 + 6.958 - 6.727 - 7.434)/(3399 + 3468 - 3138 - 2554.49) ≈ 456.93 kW
The exergy of destruction ≈ 456.93 kW
2) The reversible power output,
=
+
≈ 5000 + 456.93 kW = 5456.93 kW
The reversible power output ≈ 5456.93 kW.
Answer:
a)Δs = 834 mm
b)V=1122 mm/s
![a=450\ mm/s^2](https://tex.z-dn.net/?f=a%3D450%5C%20mm%2Fs%5E2)
Explanation:
Given that
![s = 15t^3 - 3t\ mm](https://tex.z-dn.net/?f=s%20%3D%2015t%5E3%20-%203t%5C%20mm)
a)
When t= 2 s
![s = 15t^3 - 3t\ mm](https://tex.z-dn.net/?f=s%20%3D%2015t%5E3%20-%203t%5C%20mm)
![s = 15\times 2^3 - 3\times 2\ mm](https://tex.z-dn.net/?f=s%20%3D%2015%5Ctimes%202%5E3%20-%203%5Ctimes%202%5C%20mm)
s= 114 mm
At t= 4 s
![s = 15t^3 - 3t\ mm](https://tex.z-dn.net/?f=s%20%3D%2015t%5E3%20-%203t%5C%20mm)
![s = 15\times 4^3- 3\times 4\ mm](https://tex.z-dn.net/?f=s%20%3D%2015%5Ctimes%204%5E3-%203%5Ctimes%204%5C%20mm)
s= 948 mm
So the displacement between 2 s to 4 s
Δs = 948 - 114 mm
Δs = 834 mm
b)
We know that velocity V
![V=\dfrac{ds}{dt}](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7Bds%7D%7Bdt%7D)
![\dfrac{ds}{dt}=45t^2-3](https://tex.z-dn.net/?f=%5Cdfrac%7Bds%7D%7Bdt%7D%3D45t%5E2-3)
At t= 5 s
![V=45t^2-3](https://tex.z-dn.net/?f=V%3D45t%5E2-3)
![V=45\times 5^2-3](https://tex.z-dn.net/?f=V%3D45%5Ctimes%205%5E2-3)
V=1122 mm/s
We know that acceleration a
![a=\dfrac{d^2s}{dt^2}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bd%5E2s%7D%7Bdt%5E2%7D)
![\dfrac{d^2s}{dt^2}=90t](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5E2s%7D%7Bdt%5E2%7D%3D90t)
a= 90 t
a = 90 x 5
![a=450\ mm/s^2](https://tex.z-dn.net/?f=a%3D450%5C%20mm%2Fs%5E2)
When a slender member is subjected to an axial compressive load, it may fail by a ... Consider a column of length, L, cross-sectional Moment of Inertia, I, having Young's Modulus, E. Both ends are pinned, meaning they can freely rotate ... p2EI L2 ... scr, is the Euler Buckling Load divided by the columns cross-sectional area