Answer:
power = 49.95 W
and it is self locking screw
Explanation:
given data
weight W = 100 kg = 1000 N
diameter d = 20mm
pitch p = 2mm
friction coefficient of steel f = 0.1
Gravity constant is g = 10 N/kg
solution
we know T is
T = w tan(α + φ )
...................1
here dm is = do - 0.5 P
dm = 20 - 1
dm = 19 mm
and
tan(α) =
...............2
here lead L = n × p
so tan(α) =
α = 3.83°
and
f = 0.1
so tanφ = 0.1
so that φ = 5.71°
and now we will put all value in equation 1 we get
T = 1000 × tan(3.83 + 5.71 )
T = 1.59 Nm
so
power =
.................3
put here value
power =
power = 49.95 W
and
as φ > α
so it is self locking screw
Answer Explanation:
SPRING STIFFNESS :The stiffness of a body is a measure of resistance offered by an elastic body to deformation.it is denoted by K every object has some stiffness for spring the spring stiffness is the force required to cause unit deflection
DAMPING CONSTANT: the damping constant is a number decided by manufacturer that describes the material property. Damping is an influence within an oscillatory system that has the effect of restricting its oscillation
Answer: True
Explanation:
An Infinite loop occurs in a computer programming when a sequence of instructions run endlessly without stopping until there is an external intervention, this intervention could be pull or plug. It is also known as endless loop. It occurs due to using of variables that are not properly updated or when there is an error in looping condition.
Answer:
1) q=18414.93 W
2) C=12920$
Explanation:
Given data:
pipe length L=25m
pipe diameter D=100mm =0.1 m
air temperature
=
=25
°C.....= 298.15k
pipe surface temp
=150
°C.....=423.15k
surface emissivity e= 0.8
boiler efficiency η=0.90
natural gas price Cg=$0.02 per MJ
1) Total heat loss and radiation heat loss combined
q=
q=
б(
^4-
^4)]....... (1)
б=5.67×10^-8 W/m^2K^4 (boltzmann constant)
area A =L.Dπ=25×0.1π=7.85 m^2
putting all these values in eq (1)
q=18414.93 W
2) suppose boiler is operating non stop annual energy loss will be
E=q.t
=18414.93.3600.24.365
=5.81×10^11 J
to find furnace energy consumption
Ef =E/η
=6.46×10^5 MJ
annual cost
C=Cg. Ef
=12920$