Answer:
where P is density of air and v flow is volume flow rate, so in order to find velocity of air and mass flow rate formula is given by :
so velocity is:
An uninsulated, thin-walled pipe of 100-mm diameter is used to transport water to equipment that operates outdoors and uses the
water as a coolant. During particularly harsh winter conditions, the pipe wall achieves a temperature of −15°C and a cylindrical layer of ice forms on the inner surface of the wall. If the mean water temperature is 3°C and a convection coefficient of 2000 W/m2 · K is maintained at the inner surface of the ice, which is at 0°C, what is the thickness of the ice layer?
1 answer:
You might be interested in
Answer:
number of pulses produced = 162 pulses
Explanation:
give data
radius = 50 mm
encoder produces = 256 pulses per revolution
linear displacement = 200 mm
solution
first we consider here roll shaft encoder on the flat surface without any slipping
we get here now circumference that is
circumference = 2 π r .........1
circumference = 2 × π × 50
circumference = 314.16 mm
so now we get number of pulses produced
number of pulses produced = × No of pulses per revolution .................2
number of pulses produced = × 256
number of pulses produced = 162 pulses
Answer:
a)σ₁ = 265.2 MPa
b)σ₂ = -172.8 MPa
c)
d)Range = 438 MPa
Explanation:
Given that
Mean stress ,σm= 46.2 MPa
Stress amplitude ,σa= 219 MPa
Lets take
Maximum stress level = σ₁
Minimum stress level =σ₂
The mean stress given as
2 x 46.2 = σ₁ + σ₂
σ₁ + σ₂ = 92.4 MPa --------1
The amplitude stress given as
2 x 219 = σ₁ - σ₂
σ₁ - σ₂ = 438 MPa --------2
By adding the above equation
2 σ₁ = 530.4
σ₁ = 265.2 MPa
-σ₂ = 438 -265.2 MPa
σ₂ = -172.8 MPa
Stress ratio
Range = 265.2 MPa - ( -172.8 MPa)
Range = 438 MPa