Yes, it will. The induction will be the same as long as it’s put back together.
Answer:
e= 50 J/kg
Explanation:
Given that
Speed ,v= 10 m/s
Diameter of the turbine = 90 m
Density of the air ,ρ = 1.25 kg/m³
We know that mechanical energy given as
![E=\dfrac{1}{2}mv^2\ J](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7B1%7D%7B2%7Dmv%5E2%5C%20J)
That is why mechanical energy per unit mass will be
![e=\dfrac{1}{2}v^2\ J/kg](https://tex.z-dn.net/?f=e%3D%5Cdfrac%7B1%7D%7B2%7Dv%5E2%5C%20J%2Fkg)
Now by putting the values in the above equation we get
![e=\dfrac{1}{2}\times 10^2\ J/kg](https://tex.z-dn.net/?f=e%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%2010%5E2%5C%20J%2Fkg)
e= 50 J/kg
That why the mechanical energy unit mass will be 50 J/kg.
Answer:
Not possible.
Explanation:
According to second law of thermodynamics, the maximum efficiency any heat engine could achieve is Carnot Efficiency η defined by:
![\eta=1-\frac{T_{cold}}{T_{hot}}](https://tex.z-dn.net/?f=%5Ceta%3D1-%5Cfrac%7BT_%7Bcold%7D%7D%7BT_%7Bhot%7D%7D)
Where
and
are temperature (in Kelvin) of heat source and heatsink respectively
In our case (I will be using K = 273+°C) :
![\eta=1-\frac{-27+273}{14+273}\\=0.1428](https://tex.z-dn.net/?f=%5Ceta%3D1-%5Cfrac%7B-27%2B273%7D%7B14%2B273%7D%5C%5C%3D0.1428)
In percentage, this is 14.28% efficiency, which is the <em>maximum</em> theoretical efficiency <em>any</em> heat engine could have while working between -27 and 14 °C temperature. Any claim of more efficient heat engine between these 2 temperature are violates the second law of thermodynamics. Therefore, the claim must be false.
Answer:
the rate of heat loss by convection across the air space = 82.53 W
Explanation:
The film temperature
![T_f = \frac{T_1+T_2}{2} \\\\= \frac{20-10}{2}\\\\= \frac{10}{2}\\\\= 5^0\ C](https://tex.z-dn.net/?f=T_f%20%3D%20%5Cfrac%7BT_1%2BT_2%7D%7B2%7D%20%5C%5C%5C%5C%3D%20%5Cfrac%7B20-10%7D%7B2%7D%5C%5C%5C%5C%3D%20%5Cfrac%7B10%7D%7B2%7D%5C%5C%5C%5C%3D%205%5E0%5C%20C)
to kelvin = (5 + 273)K = 278 K
From the " thermophysical properties of gases at atmospheric pressure" table; At
= 278 K ; by interpolation; we have the following
→ v 13.93 (10⁻⁶) m²/s
→ k = 0.0245 W/m.K
→ ∝ = 19.6(10⁻⁶)m²/s
→ Pr = 0.713
![\beta = \frac{1}{T_f} \\=\frac{1}{278} \\ \\ = 0.00360 \ K ^{-1}](https://tex.z-dn.net/?f=%5Cbeta%20%3D%20%5Cfrac%7B1%7D%7BT_f%7D%20%5C%5C%3D%5Cfrac%7B1%7D%7B278%7D%20%5C%5C%20%5C%5C%20%3D%200.00360%20%5C%20K%20%5E%7B-1%7D)
The Rayleigh number for vertical cavity
![Ra_L = \frac{g \beta (T_1-T_2)L^3}{\alpha v}](https://tex.z-dn.net/?f=Ra_L%20%20%3D%20%5Cfrac%7Bg%20%5Cbeta%20%28T_1-T_2%29L%5E3%7D%7B%5Calpha%20v%7D)
= ![\frac{9.81*0.00360(20-(-10))*0.06^3}{19.6(10^{-6})*13.93(10^{-6})}](https://tex.z-dn.net/?f=%5Cfrac%7B9.81%2A0.00360%2820-%28-10%29%29%2A0.06%5E3%7D%7B19.6%2810%5E%7B-6%7D%29%2A13.93%2810%5E%7B-6%7D%29%7D)
= ![8.38*10^5](https://tex.z-dn.net/?f=8.38%2A10%5E5)
![\frac{H}{L}= \frac{1.44}{0.06} \\ \\= 24](https://tex.z-dn.net/?f=%5Cfrac%7BH%7D%7BL%7D%3D%20%5Cfrac%7B1.44%7D%7B0.06%7D%20%5C%5C%20%5C%5C%3D%2024)
For the rectangular cavity enclosure , the Nusselt number empirical correlation:
![Nu_L = 0.42(8.38*10^5)^{\frac{1}{4}}(0.713)^{0.012}(24){-0.3}](https://tex.z-dn.net/?f=Nu_L%20%3D%200.42%288.38%2A10%5E5%29%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%280.713%29%5E%7B0.012%7D%2824%29%7B-0.3%7D)
![NU_L= \frac{hL}{k}= 4.878](https://tex.z-dn.net/?f=NU_L%3D%20%5Cfrac%7BhL%7D%7Bk%7D%3D%204.878)
![\frac{hL}{k}= 4.878](https://tex.z-dn.net/?f=%5Cfrac%7BhL%7D%7Bk%7D%3D%204.878)
![\frac{h*0.06}{0.0245}= 4.878](https://tex.z-dn.net/?f=%5Cfrac%7Bh%2A0.06%7D%7B0.0245%7D%3D%204.878)
![h = \frac{4.878*0.0245}{0.06}](https://tex.z-dn.net/?f=h%20%3D%20%5Cfrac%7B4.878%2A0.0245%7D%7B0.06%7D)
h = 1.99 W/m².K
Finally; the rate of heat loss by convection across the air space;
q = hA(T₁ - T₂)
q = 1.99(1.4*0.96)(20-(-10))
q = 82.53 W
Answer:
A vehicle is considered to be legally parked if it is parked 20 feet (6 m) or more from a pedestrian crosswalk or a marked or unmarked intersection.
Explanation:
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I obtained the provided data from the New York State Driver's Manual. I wish it was useful to help you.
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