Answer:
i) 25.04 W/m^2 .k
ii) 23.82 minutes = 1429.2 secs
Explanation:
Given data:
Diameter of steel ball = 15 cm
uniform temperature = 350°C
p = 8055 kg/m^3
Cp = 480 J/kg.k
surface temp of ball drops to 250°C
average surface temperature = ( 350 + 250 ) / 2 = 300°C
<u>i) Determine the average convection heat transfer coefficient during the cooling process</u>
<em>Note : Obtain the properties of air at 1 atm at average film temp of 30°C from the table " properties of air " contained in your textbook</em>
average convection heat transfer coefficient = 25.04 W/m^2 .k
<u>ii) Determine how long this process has taken </u>
Time taken by the process = 23.82 minutes = 1429.2 seconds
Δt = Qtotal / Qavg = 683232 / 477.92 = 1429.59 secs
attached below is the detailed solution of the given question
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Answer:
a) 0.4393 mm
b) -0.141 mm
Explanation:
Cylindrical bar : Diameter = 20.3 mm , Length = 205 mm
Force that deforms bar of metal = 46400 N
elastic modulus = 66.9 GPa
Poisson's ratio ( u ) = 0.32
A) Determine the amount by which this specimen will elongate in the direction of applied stress
dl = 
where P = 46400 N
L = 205 mm
A = 
= 323.65
E = 66.9 GPa
dl = ( 46400 * 205 ) / ( 323.65 * 66.9 * 10^3 ) = 0.4393 mm
B) determine the change in diameter of the specimen
change in diameter( compressed due to elongation in length )
= - u * dl
= - 0.32 * 0.4393
= -0.141 mm
