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# Copper bearing 80.10 mm at 20 *C needs to fit into a hole 80.00 mm. What temp does it have to cool?
# assume alpha for copper is 17 x 10^-6 /*C.
# if clearence, note it.
# cl(0.02) will calculate the temp with 0.02 mm of clearence
def cl(clnc=0):
L0 = 80
T2 = 20
dL = 0.1 + clnc
alpha = 17e-6
T1 = T2 - ( (dL) / (alpha*L0))
T1 = round(T1,2)
return(T1)
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cl()
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In [7]:
cl(0.02)
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In [8]:
cl(0.01)
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In [9]:
cl(0.05)
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In [10]:
cl(.1)
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In [11]:
cl(0.01)
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At room temperature (20 °C), copper bearing with an outside diameter of 80.10 mm does not fit into an aluminum case with a hole diameter of 80.00 mm. In order for this to be accomplished, a furnace or hotplate can be used to heat the case. What temperature must the case be heated up to in order for the bearing to fit in the case? (assume the bearing stays at room temperature.) If assuming a clearance, note the clearance amount as part of your solution on your answer sheet below the answer. (assume the coefficient of thermal expansion of aluminum is 24 × 10-6 °C-1 and the coefficient of thermal expasion of copper is 17 × 10-6 °C-1)
In [3]:
def cl(clnc=0):
L0 = 80
#T2 = ?
T1 = 20
dL = 0.1 + clnc
alpha = 24e-6
T2 = T1 + ( (dL) / (alpha*L0))
T2 = round(T2,2)
return(T2)
In [4]:
cl()
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In [5]:
cl(0.05)
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