Answer:
(B) FALSE
Explanation:
view factor depends on the surface emissivity and the surface of geometry view factor is the term used in radiative heat transfer. View factor is depends upon the radiation which leave the surface and strike the surface.View factor is also called shape factor configuration factor it is denoted by
Answer:
The stress in the bar is 34.72 MPa.
The design factor (DF) for each case is:
A) DF=0.17
B) DF=0.09
C) DF=0.125
D) DF=0.12
E) DF=0.039
F) DF=1.26
G) DF=5.5
Explanation:
The design factor is the relation between design stress and failure stress. In the case of ductile materials like metals, the failure stress considered is the yield stress. In the case of plastics or ceramics, the failure stress considered is the breaking stress (ultimate stress). If the design factor is less than 1, the structure or bar will endure the applied stress. By the opposite side, when the DF is higher than 1, the structure will collapse or the bar will break.
we will calculate the design stress in this case:
The design factor for metals is:
The design factor for plastic and ceramics is:
We now need to know the yield stress or the ultimate stress for each material. We use the AISI and ASTM charts for steels, materials charts for non-ferrous materials and plastics safety charts for the plastic materials.
For these cases:
A) The yield stress of AISI 120 hot-rolled steel (actually is AISI 1020) is 205 MPa, therefore:
B) The yield stress of AISI 8650 OQT 1000 steel is 385 MPa, therefore:
C) The yield stress of ductile iron A536-84 (60-40-18) is 40Kpsi, this is 275.8 MPa, therefore:
D) The yield stress of aluminum allot 6061-T6 is 290 MPa, therefore:
E) The yield stress of titanium alloy Ti-6Al-4V annealed (certified by manufacturers) is 880 MPa, therefore:
F) The ultimate stress of rigid PVC plastic (certified by PVC Pipe Association) is 4Kpsi or 27.58 MPa, therefore:
In this case, the bar will break.
F) You have to consider that phenolic plastics are used as matrix in composite materials and seldom are used alone with no reinforcement. In this question is not explained if this material is reinforced or not, therefore I will use the ultimate stress of most pure phenolic plastics, in this case, 6.31 MPa:
This material will break.
Answer:
And if we solve for T we got:
The answer for this case would be at 5cm from the base of the fin.
Explanation:
Data given
For this case we have the following data given:
represent the heat transfer coefficient.
represent the perimeter for this case and would be given by:
represent the thermal conductivity
represent the width
represent the thickness
Solution to the problem
For this case we assume that we have steady conditions, the temperature of the fins varies just in one direction, the heat transfer coefficient not changes with the time and the thermal properties of the fin not change.
We can determine the temperature if the fin at x=5 cm=0.05 m from the base with the following formula:
Where m is a coefficient given by:
The value of x for this case represent the distance
represent the base temperature
represent the temperature of the sorroundings or the ambient.
If we replace we have this:
And if we solve for T we got:
The answer for this case would be at 5cm from the base of the fin.
Answer:
h=1.122652m
Explanation:
Assuming density of air = 1.2kg/m³
the differential pressure is given by: