Answer: l = 2142.8575 ft
v = 193.99 ft/min.
Explanation:
Given data:
Thickness of the slab = 3in
Length of the slab = 15ft
Width of the slab = 10in
Speed of the slab = 40ft/min
Solution:
a. After three phase
three phase = (0.2)(0.2)(0.2)(3.0)
= 0.024in.
wf = (1.03)(1.03)(1.03)(10.0)
= 10.927 in.
Using constant volume formula
= (3.0)(10.0)(15 x 15) = (0.024)(10.927)Lf
Lf = (3.0)(10.0)(15 x 15)/(0.024)(10.927)
= 6750 /0.2625
= 25714.28in = 2142.8575 ft
b.
vf = (0.2 x 0.2 x 3.0)(1.03 x 1.03 x 10.0)(40)/(0.024)(10.927)
= (0.12)(424.36)/0.2625
= 50.9232/0.2625
= 193.99 ft/min.
Answer:
q = 1.73 W
Explanation:
given data
small end = 5 cm
large end = 10 cm
high = 15 cm
small end is held = 600 K
large end at = 300 K
thermal conductivity of asbestos = 0.173 W/mK
solution
first we will get here side of cross section that is express as
...............1
here x is distance from small end and S1 is side of square at small end
and S2 is side of square of large end and L is length
put here value and we get
S = 5 +
S =
m
and
now we get here Area of section at distance x is
area A = S² ...............2
area A =
m²
and
now we take here small length dx and temperature difference is dt
so as per fourier law
heat conduction is express as
heat conduction q =
...............3
put here value and we get
heat conduction q =
it will be express as
now we intergrate it with limit 0 to 0.15 and take temp 600 to 300 K
solve it and we get
q (30) = (0.173) × (600 - 300)
q = 1.73 W