Which type of memory is programmed at the factory? RAM ROM Cache or Virtual memory
1 answer:
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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:
net boiler heat = 301.94 kW
Explanation:
given data
saturated steam = 6.0 bars
temperature = 18°C
flow rate = 115 m³/h = 0.03194 m³/s
heat use by boiler = 90 %
to find out
rate of heat does the boiler output
solution
we can say saturated steam is produce at 6 bar from liquid water 18°C
we know at 6 bar from steam table
hg = 2756 kJ/kg
and
enthalpy of water at 18°C
hf = 75.64 kJ/kg
so heat required for 1 kg is
=hg - hf
= 2680.36 kJ/kg
and
from steam table specific volume of saturated steam at 6 bar is 0.315 m³/kg
so here mass flow rate is
mass flow rate =
mass flow rate m = 0.10139 kg/s
so heat required is
H = h × m
here h is heat required and m is mass flow rate
H = 2680.36 × 0.10139
H = 271.75 kJ/s = 271.75 kW
now 90 % of boiler heat is used for generate saturated stream
so net boiler heat =
net boiler heat =
net boiler heat = 301.94 kW