Answer:
a)supplying the intake of an engine with air at a density greater than the density of the surrounding atmosphere
Explanation:
Supercharging is the process of supplying the intake of an engine with air at a density greater than the density of the surrounding atmosphere.
By doing this , it increases the power out put and increases the brake thermal efficiency of the engine.It also increases the volumetric efficiency of the engine.
So the our option a is right.
Answer: a) 1.05kW b) 3.78MJ c) 5.3 bars
Explanation :
A)
Conversions give 900 kcal as 900000 x 4.2 J/cal {4.2 J/cal is the standard factor}
= 3780kJ
And 1 hour = 3600s
Therefore, Power in watts = 3780/3600 = 1.05kW = 1050W
B)
At 15km/hour a 15km run takes 1 hour.
1 hour is 3600s and the runner burns 1050 joule per second.
Energy used in 1 hour = 3600 x 1050 J/s
= 3780000 J or 3.78MJ
C)
1 mile = 1.61km so 13.1 mile is 13.1 x 1.61 = 21.1km
15km needs 3.78 MJ of energy therefore 21.1km needs 3.78 x 21.1/15 = 5.32MJ =5320 kJ
Finally,
1 Milky Way = 240000 calories = 4.2 x 240000 J = 1008000J or 1008kJ
This means that the runner needs 5320/1008 = 5.3 bars
Answer:
t = 2244.3 sec
Explanation:
calculate the thermal diffusivity
Temperature at 28 mm distance after t time = = 50 degree C
we know that
![\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)
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)
solving fot t we get
t = 2244.3 sec
