You're running an engine and seeing that it is exhausting blue smoke. Which is likely to
1 answer:
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Answer:
material remove in 3 min is 16790.4 mm³/s
Explanation:
given data
length L = 80 cm = 800 mm
width W = 30 cm
height H = 15 cm
make grove length = 80 cm
width = 8 cm
depth = 10 cm
mill toll diameter = 4 mm
axial cutting depth = 20 mm
to find out
How much material removed in 3 minutes
solution
first we find time taken for length of advance that is
time =
here advance is given as 0.001166 mts / sec
so time =
time = 686.106 seconds
now we find material remove rate that is
remove rate = mill toll rate × axial cutting depth × advance
remove rate = 4 × 20×0.001166 ×1000
remove rate = 93.28 mm³/s
so
material remove in 3 minute = 3 × 60 = 180 sec
so material remove in 3 min = 180 × 93.28
material remove in 3 min is 16790.4 mm³/s
Answer: Hello the question is incomplete below is the missing part
Question: determine the temperature, in °R, at the exit
answer:
T2= 569.62°R
Explanation:
T1 = 540°R
V2 = 600 ft/s
V1 = 60 ft/s
h1 = 129.0613 ( value gotten from Ideal gas property-air table )
<em>first step : calculate the value of h2 using the equation below </em>
assuming no work is done ( potential energy is ignored )
h2 = [ h1 + ( V2^2 - V1^2 ) / 2 ] * 1 / 32.2 * 1 / 778
∴ h2 = 136.17 Btu/Ibm
From Table A-17
we will apply interpolation
attached below is the remaining part of the solution
