A concrete mix design calls for 6.5 sacks of cement, a water/cement ratio of 0.45, and an air content of 2.5%. 1. Complete the m
2 answers:
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Answer:
the torque capacity is 30316.369 lb-in
Explanation:
Given data
OD = 9 in
ID = 7 in
coefficient of friction = 0.2
maximum pressure = 1.5 in-kip = 1500 lb
To find out
the torque capacity using the uniform-pressure assumption.
Solution
We know the the torque formula for uniform pressure theory is
torque = 2/3 × × coefficient of friction × maximum pressure ( R³ - r³ ) .....................................1
here R = OD/2 = 4.5 in and r = ID/2 = 3.5 in
now put all these value R, r, coefficient of friction and maximum pressure in equation 1 and we will get here torque
torque = 2/3 × × 0.2 × 1500 ( 4.5³ - 3.5³ )
so the torque = 30316.369 lb-in
Answer:
Amount of air left in the cylinder=m=0.357 Kg
The amount of heat transfer=Q=0
Explanation:
Given
Initial pressure=P1=300 KPa
Initial volume=V1=0.2
Initial temperature=T=20 C
Final Volume==0.1
Using gas equation
m1==(300*0.2)/(.287*293)
m1=0.714 Kg
Similarly
m2=(P2*V2)/R*T2
m2=(300*0.1)/(0.287*293)
m2=0.357 Kg
Now calculate mass of air left,where me is the mass of air left.
me=m2-m1
me=0.715-0.357
mass of air left=me=0.357 Kg
To find heat transfer we need to apply energy balance equation.
Where me=m1-m2
And as the temperature remains constant,hence the enthalpy also remains constant.
h1=h2=he=h
Q=(me-(m1-m2))*h
me=m1-me
Thus heat transfer=Q=0