Approximate the volume δV of insulation material required in m3?
V1 = π x 10² x 1400 cm³
V2 = π x (10.6)² x 1400 cm³
V2 - V1 = 1400 π [ 10.6² - 10² ]
V2 - V1 = 1400 π ( 10.6 - 10)(10.6 + 10)
V2 - V1 = 1400 π (0.6)(20.6)
V2 - V1 = 54362 cm³ = 54.362 m³
δv = 54.4 m³ <---------to 1 dec. place
Area=2 pi r l around pipe
Area=2 pi (r+6mm) l outside
Vol=1/2(2 pi r l +2 pi (r+6mm)) l 6mm
Vol=( pi *.1 + pi *.106))*14*.006 all in metres
=22*.206*2*.006 m^3
=.054 m^3
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Verified answer
V1 = π x 10² x 1400 cm³
V2 = π x (10.6)² x 1400 cm³
V2 - V1 = 1400 π [ 10.6² - 10² ]
V2 - V1 = 1400 π ( 10.6 - 10)(10.6 + 10)
V2 - V1 = 1400 π (0.6)(20.6)
V2 - V1 = 54362 cm³ = 54.362 m³
δv = 54.4 m³ <---------to 1 dec. place
Area=2 pi r l around pipe
Area=2 pi (r+6mm) l outside
Vol=1/2(2 pi r l +2 pi (r+6mm)) l 6mm
Vol=( pi *.1 + pi *.106))*14*.006 all in metres
=22*.206*2*.006 m^3
=.054 m^3