Answer:
See attached image for diagrams and solution
A motor whose pulley is 80mm in diameter is rotating at 30 revolutions per minute and is connected by a belt to another pulley o
1 answer:
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Explanation:
A.
H = Aeσ^4
Using the stefan Boltzmann law
When we differentiate
dH/dT = 4AeσT³
dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³
= 8.4085
Exact error = 8.4085x20
= 168.17
H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴
= 1366.376watts
B.
Verifying values
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴
= 1542.468
H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴
= 1205.8104
Error = 1542.468-1205.8104/2
= 168.329
ΔT = 40
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴
= 1735.05
H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴
= 1735.05-1059.83/2
= 675.22/2
= 337.61
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