Where are revolved sections placed in a print? A) in between break lines B) cutting planes are used to identify their locations
1 answer:
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Answer:
diameter is 14 mm
Explanation:
given data
power = 15 kW
rotation N = 1750 rpm
factor of safety = 3
to find out
minimum diameter
solution
we will apply here power formula to find T that is
power = 2π×N×T / 60 .................1
put here value
15 × = 2π×1750×T / 60
so
T = 81.84 Nm
and
torsion = T / Z ..........2
here Z is section modulus i.e = πd³/ 16
so from equation 2
torsion = 81.84 / πd³/ 16
so torsion = 416.75 / / d³ .................3
so from shear stress theory
torsion = σy / factor of safety
so here σy = 530 for 1020 steel
so
torsion = σy / factor of safety
416.75 / d³ = 530 × / 3
so d = 0.0133 m
so diameter is 14 mm
Answer:
Explanation:
Let assume that heating and boiling process occurs under an athmospheric pressure of 101.325 kPa. The heat needed to boil water is:
The heat liberated by the LP gas is:
A kilogram of LP gas has a minimum combustion power of . Then, the required mass is:
Answer:
Relative density = 0.7 or 70%
Explanation:
The following information was provided by this question
Pd = 1.72mg/mg³
Pd max = 1.81 mg/mg³
Pd min = 1.54 mg/mg³
We substitute into the formula. This formula is contained in the attachment.
[(1/1.54)-(1/1.72)]/[1/1.54 - 1/1.81]
= 0.649350 - 0.581395 / 0.649350 - 0.552486
= 0.067955/0.096864
= 0.7015
= 0.7
The relative density is Therefore 0.7 or 70% when converted to percentage
