Answer:
18 kJ
Explanation:
Given:
Initial volume of air = 0.05 m³
Initial pressure = 60 kPa
Final volume = 0.2 m³
Final pressure = 180 kPa
Now,
the Work done by air will be calculated as:
Work Done = Average pressure × Change in volume
thus,
Average pressure =
= 120 kPa
and,
Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³
Therefore,
the work done = 120 × 0.15 = 18 kJ
Answer:
Could ask a family member to help
Explanation:
Answer:
(a) Flow rate of vehicles = No of vehicles per mile * Speed
=No of cars per mile * Speed +No of trucks per mile * Speed
= 0.75*50*60 + 0.25*50*40
=2750 vehicles / hour
(b) Let Density of vehicles on grade = x
Density on flat * Speed =Density on grade * Speed
So,( 0.75*50) * 60 + (0.25*50) * 40 = (0.75* x) * 55 + (0.25* x) * 25
So, x= 57.89
So, Density is around 58 Vehicles per Mile.
(c) Percentage of truck by aerial photo = 25%
(d)Percentage of truck bystationary observer on the grade= 25*30/60 * 25/55 =22.73 %
Answer:
slenderness ratio = 147.8
buckling load = 13.62 kips
Explanation:
Given data:
outside diameter is 3.50 inc
wall thickness 0.30 inc
length of column is 14 ft
E = 10,000 ksi
moment of inertia 

Area 


r = 1.136 in
slenderness ratio 

buckling load 


Answer:
The rate of entropy change of the air is -0.10067kW/K
Explanation:
We'll assume the following
1. It is a steady-flow process;
2. The changes in the kinetic energy and the potential energy are negligible;
3. Lastly, the air is an ideal gas
Energy balance will be required to calculate heat loss;
mh1 + W = mh2 + Q where W = Q.
Also note that the rate of entropy change of the air is calculated by calculating the rate of heat transfer and temperature of the air, as follows;
Rate of Entropy Change = -Q/T
Where Q = 30Kw
T = Temperature of air = 25°C = 298K
Rate = -30/298
Rate = -0.100671140939597 KW/K
Rate = -0.10067kW/K
Hence, the rate of entropy change of the air is -0.10067kW/K