Answer:
12 N-m
Explanation:
The dc motor is operating at 24 V that is its terminal voltage V =24 V
Armature resistance = 0.2 ohm
No load speed = 240 radian /sec
For motor we know that as the motor is on no load so so
Power developed in the motor
Now we know that power = torque× angular speed
So
Answer:
Some types of aggregate are susceptible to damage from repeated freezing and thawing due to their porosity. An aggregate being porous allows water molecules to enter in between the rocks.
When freezing occurs, water is known to expand. The expansion of this in the rocks creates a type of pressure which results in the fracture of the rocks. Subsequent freezing and thawing will allow for more fracture between the rock particles which will lead to its disintegration.
Answer:
see explaination for all the answers and full working.
Explanation:
deflection=8P*DN/Gd^4
G(for steel)=70Gpa=70*10^9N/m^2=70KN/mm^2
for outer spring,
deflection=8*3*50^3*5/(70*9^4)=32.66mm
for inner spring
deflection=8*3*30^3*10/(70*5^4)=148.11mm
max stress=k*8*P*C/(3.14*d^2)
for outer spring
c=50/9=5.55
k=(4c-1/4c-4)+.615/c=1.2768
max stress=1.2768*8*3*5.55/(3.14*9^2=.66KN.mm^2
for inner spring
c=6
k=1.2525
max stress=2.29KN/mm^2
Answer:
a. 81 kj/kg
b. 420.625K
c. 101.24kj/kg
Explanation:
![\frac{t2}{t1} =[\frac{p2}{p1} ]^{\frac{y-1}{y} }](https://tex.z-dn.net/?f=%5Cfrac%7Bt2%7D%7Bt1%7D%20%3D%5B%5Cfrac%7Bp2%7D%7Bp1%7D%20%5D%5E%7B%5Cfrac%7By-1%7D%7By%7D%20%7D)
t1 = 360
p1 = 0.4mpa
p2 = 1.20
y = 1.13
substitute these values into the equation
![\frac{t2}{360} =[\frac{1.20}{0.4} ]^{\frac{1.13-1}{1.13} }](https://tex.z-dn.net/?f=%5Cfrac%7Bt2%7D%7B360%7D%20%3D%5B%5Cfrac%7B1.20%7D%7B0.4%7D%20%5D%5E%7B%5Cfrac%7B1.13-1%7D%7B1.13%7D%20%7D)
![\frac{t2}{360} =[\frac{1.2}{0.4} ]^{0.1150}\\\frac{t2}{360} =1.1347](https://tex.z-dn.net/?f=%5Cfrac%7Bt2%7D%7B360%7D%20%3D%5B%5Cfrac%7B1.2%7D%7B0.4%7D%20%5D%5E%7B0.1150%7D%5C%5C%5Cfrac%7Bt2%7D%7B360%7D%20%3D1.1347)
when we cross multiply
t2 = 360 * 1.1347
= 408.5
a. the work required in the firs compressor
w=c(t2-t1)
c=1.67x10³
t1 = 360
t2 = 408.5
w = 1670(408.5-360)
= 1670*48.5
= 80995 J
= 81KJ/kg
b. 
n = 80%
t2 = 408.5
t1 = 360
0.80 = 408.5-360 ÷ t'2-360
cross multiply to get the value of t'2
0.80(t'2-360) = 48.5
0.80t'2 - 288 = 48.5
0.8t'2 = 48.5+288
0.8t'2 = 336.5
t'2 = 336.5/0.8
= 420.625
this is the temperature at the exit of the first compressor
c. cooling requirement
w = c(t2-t1)
= 1.67x10³(420.625-360)
= 1670*60.625
= 101243.75
= 101.24kj/kg
