Answer:
design hour volumes will be 4000 to 6000
Explanation:
given data
AADT = 150000 veh/day
solution
we get here design hour volumes that is express as
design hour volumes = AADT × k × D ..............1
here k is factor and its range is 8 to 12 % for urban
and D is directional distribution i.e traffic equal divided by the direction
so here design hour volumes will be 4000 to 6000
Answer:
Class of fit:
Interference (Medium Drive Force Fits constitute a special type of Interference Fits and these are the tightest fits where accuracy is important).
Here minimum shaft diameter will be greater than the maximum hole diameter.
Medium Drive Force Fits are FN 2 Fits.
As per standard ANSI B4.1 :
Desired Tolerance: FN 2
Tolerance TZone: H7S6
Max Shaft Diameter: 3.0029
Min Shaft Diameter: 3.0022
Max Hole Diameter:3.0012
Min Hole Diameter: 3.0000
Max Interference: 0.0029
Min Interference: 0.0010
Stress in the shaft and sleeve can be considered as the compressive stress which can be determined using load/interference area.
Design is acceptable If compressive stress induced due to selected dimensions and load is less than compressive strength of the material.
Explanation:
Answer:
True, <em>Regeneration is the only process where increases the efficiency of a Brayton cycle when working fluid leaving the turbine is hotter than working fluid leaving the compressor</em>.
Option: A
<u>Explanation:
</u>
To increase the efficiency of brayton cycle there are three ways which includes inter-cooling, reheating and regeneration. <em>Regeneration</em> technique <em>is used when a turbine exhaust fluids have higher temperature than the working fluid leaving the compressor of the turbine. </em>
<em>Thermal efficiency</em> of a turbine is increased as <em>the exhaust fluid having higher temperatures are used in heat exchanger where the fluids from the compressor enters and increases the temperature of the fluids leaving the compressor.
</em>
Answer:
, 
Explanation:
The drag force is equal to:
Where 
is the drag coefficient and 
is the frontal area, respectively. The work loss due to drag forces is:
The reduction on amount of fuel is associated with the reduction in work loss:
Where 
and 
are the original and the reduced frontal areas, respectively.
The change is work loss in a year is:
![\Delta W = (0.3)\cdot \left(\frac{1}{2}\right)\cdot (1.20\,\frac{kg}{m^{3}})\cdot (27.778\,\frac{m}{s})^{2}\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25\times 10^{6}\,m)](https://tex.z-dn.net/?f=%5CDelta%20W%20%3D%20%280.3%29%5Ccdot%20%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%5Cright%29%5Ccdot%20%281.20%5C%2C%5Cfrac%7Bkg%7D%7Bm%5E%7B3%7D%7D%29%5Ccdot%20%2827.778%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%29%5E%7B2%7D%5Ccdot%20%5B%281.85%5C%2Cm%29%5Ccdot%20%281.75%5C%2Cm%29%20-%20%281.50%5C%2Cm%29%5Ccdot%20%281.75%5C%2Cm%29%5D%5Ccdot%20%2825%5Ctimes%2010%5E%7B6%7D%5C%2Cm%29)
The change in chemical energy from gasoline is:
The changes in gasoline consumption is:
Lastly, the money saved is:
