Find the toughness (or energy to cause fracture) for a metal that experiences both elastic and plastic deformation. Assume Equat
ion 6.5 for elastic deformation, that the modulus of elasticity is 172 GPa (25 ´ 106 psi), and that elastic deformation terminates at a strain of 0.01. For plastic deformation, assume that the relationship between stress and strain is described by Equation 6.19, in which the values for K and n are 6900 MPa (1 ´ 106 psi) and 0.30, respectively. Furthermore, plastic deformation occurs between strain values of 0.01 and 0.75, at which point fracture occurs.
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Answer:
a)
b)
c)
d)
Explanation:
Non horizontal pipe diameter, d = 25 cm = 0.25 m
Radius, r = 0.25/2 = 0.125 m
Entry temperature, T₁ = 304 + 273 = 577 K
Exit temperature, T₂ = 284 + 273 = 557 K
Ambient temperature,
Pipe length, L = 10 m
Area, A = 2πrL
A = 2π * 0.125 * 10
A = 7.855 m²
Mass flow rate,
Rate of heat transfer,
a) To calculate the convection coefficient relationship for heat transfer by convection:
Note that we cannot calculate the heat loss by the pipe to the environment without first calculating the surface temperature of the pipe.
c) The surface temperature of the pipe:
Smear coefficient of the pipe,
b) Heat loss from the pipe to the environment:
d) The required fan control power is 25.125 W as calculated earlier above
Answer:
b). The same for all pipes independent of the diameter
Explanation:
We know,
From the above formulas we can conclude that the thermal resistance of a substance mainly depends upon heat transfer coefficient,whereas radius has negligible effects on heat transfer coefficient.
We also know,
Factors on which thermal resistance of insulation depends are :
1. Thickness of the insulation
2. Thermal conductivity of the insulating material.
Therefore from above observation we can conclude that the thermal resistance of the insulation is same for all pipes independent of diameter.