Answer:
a) 42.422 KN
b) 44.356 KN
Explanation:
Given data :
Diameter = 20 mm
yield strength = 350 MN/m^2
Torque ( T ) = 100 N.m
Bending moment = 150 N.m
<u>Determine the value of the applied axial tensile force when yielding of rod occurs </u>
first we will calculate the shear stress and normal stress
shear stress ( г ) = Tr / J = [( 100 * 10^3) * 10 ] / 
* ( 20)^4
= 63.662 MPa
Normal stress( Гb + Гa ) = MY/ I + P/A
= [( 150 * 10^3) * 10 ] / * ( 20)^4 + 4P / 
= 190.9859 + 4P / MPa
<u>a) Using MSS theory </u>
value of axial force = 42.422 KN
solution attached below
<u>b) Using MDE theory </u>
value of axial force = 44.356 KN
solution attached below
Answer:
(a) Rate of heat transfer = 34.65 W/m
(b) quality of outlet of pipe x = 0.967
(c) Temperature of outer surface of insulation, T₂ = U1.157°C
Yes it is safe to touch, (But gentle touch)
Explanation:
Detailed explanation is given in the attach document.
Answer:
Q(h=200)=0.35W
Q(h=3000)=5.25W
Explanation:
first part h=200W/Km^2
we must use the convection heat transfer equation for the chip
Q=hA(Ts-T∞)
h=
convective coefficient=200W/m2 K
A=Base*Leght=5mmx5mm=25mm^2
Ts=temperature of the chip=85C
T∞=temperature of coolant=15C
Q=200x2.5x10^-5(85-15)=0.35W
Second part h=3000W/Km^2
Q=3000x2.5x10^-5(85-15)=5.25W
Answer:
b
Explanation:
just took it sondjdjndjjrjridj
Answer: um wuh anyways thxs for the points!
Explanation: ....:/
