An aluminum rod is press fitted onto an aluminum collar. The collar has an inner radius of 1 cm and an outer radius of 2 cm. Giv
1 answer:
The question is incomplete. The complete question is --
An aluminum rod is press fitted onto an aluminum collar. The collar has an inner radius of 1 cm and an outer radius of 2 cm. Given the rod has a diameter of 1.01 cm and the young's modulus of aluminum is 69 GPa, determine the following :
1. the interference value, i
2. the radial pressure at the interference of the collar and the rod
3. the maximum effective stress in the collar
4. if the yield strength of aluminium is 200 MPa and assume a safety factor of 1.5, will the aluminium collar break
Solution:
Given :
Inner radius of the collar = 1 cm
So, inner diameter, = 2 cm
Outer radius of the collar = 12 cm
So, outer diameter, = 4 cm
The aluminium rod diameter, d = 1.01 cm
Now, from the figure, we can see that there will be no interference and so the rod will easily insert inside the collar.
1. So, the interference , i =0
2. The radial pressure is also 0.
3. There will be no stress developed. So the maximum effective stress is 0
4. The collar will not break
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Answer:
The answer to the question is
The heat transferred in the process is -274.645 kJ
Explanation:
To solve the question, we list out the variables thus
R-134a = Tetrafluoroethane
Intitial Temperaturte t₁ = 100 °C
Initial pressure = 3.5 bar = 350 kPa
For closed system we have m₁ = m₂ = m
ΔU = m×(u₂ - u₁) = ₁Q₂ -₁W₂
For constant pressure process we have
Work done = W = = P×ΔV = P × (V₂ - V₁) = P×m×(v₂ - v₁)
From the tables we have
State 1 we have h₁ = (490.48 +489.52)/2 = 490 kJ/kg
State 2 gives h₂ = 206.75 + 0.75 × 194.57= 352.6775 kJ/kg
Therefore Q₁₂ = m×(u₂ - u₁) + W₁₂ = m × (u₂ - u₁) + P×m×(v₂ - v₁)
= m×(h₂ - h₁) = 2.0 kg × (352.6775 kJ/kg - 490 kJ/kg) =-274.645 kJ