Answer:
The time required is 10.078 hours or 605 min
Explanation:
The formula to apply here is ;
K=(d²-d²₀ )/t
where t is time in hours
d is grain diameter to be achieved after heating in mm
d₀ is the grain diameter before heating in mm
Given
d=5.5 × 10^-2 mm
d₀=2.4 × 10^-2 mm
t₁= 500 min = 500/60 =25/3 hrs
t₂=?
n=2.2
First find K
K=(d²-d²₀ )/t₁
K={ (5.1 × 10^-2 mm)²-(2.4 × 10−2 mm)² }/ 25/3
K=(0.051²-0.024²) ÷25/2
K=0.000243 mm²/h
Re-arrange equation for K ,to get the equation for d as;
d=√(d₀²+ Kt) where now t=t₂
![d=\sqrt{0.024^2+0.000243*t} \\\\0.055=\sqrt{0.024^2+0.000243t} \\\\0.055^2=0.024^2+0.000243t\\\\0.055^2-0.024^2=0.000243t\\\\0.002449=0.000243t\\\\0.002449/0.000243=t\\\\10.078=t\\\\t=605min](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B0.024%5E2%2B0.000243%2At%7D%20%5C%5C%5C%5C0.055%3D%5Csqrt%7B0.024%5E2%2B0.000243t%7D%20%5C%5C%5C%5C0.055%5E2%3D0.024%5E2%2B0.000243t%5C%5C%5C%5C0.055%5E2-0.024%5E2%3D0.000243t%5C%5C%5C%5C0.002449%3D0.000243t%5C%5C%5C%5C0.002449%2F0.000243%3Dt%5C%5C%5C%5C10.078%3Dt%5C%5C%5C%5Ct%3D605min)
Answer:
hazardous chemicals leaving the workplace is labeled, tagged or marked with the following information: product identifier; signal word; hazard statement
Explanation:
this is so you know what chemicals are in it
Answer:
M2 = 0.06404
P2 = 2.273
T2 = 5806.45°R
Explanation:
Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.
Therefore from Steam Table, Po1 = (1.028)*(10) = 10.28 atm,
To1 = (1.008)*(1000) = 1008 ºR
R = 1716 ft-lb/slug-ºR cp= 6006 ft-lb/slug-ºR fuel-air ratio (by mass)
F/A =???? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga
For the air q = cp(To2– To1)
(Exit flow – inlet flow) – choked flow is assumed For M1= 0.2
Table A.3 of steam table gives P/P* = 2.273,
T/T* = 0.2066,
To/To* = 0.1736 To* = To2= To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga= FA*(4.5 x 108) ft-lb/slugaFA =
F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit
Answer:
a) ![\mathbf{\sigma _ 1 = 4800 psi}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Csigma%20_%201%20%3D%204800%20psi%7D)
![\mathbf{ \sigma _2 = 0}](https://tex.z-dn.net/?f=%5Cmathbf%7B%20%5Csigma%20_2%20%3D%200%7D)
b)![\mathbf{\sigma _ 1 = 6000 psi}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Csigma%20_%201%20%3D%206000%20psi%7D)
![\mathbf{ \sigma _2 = 3000 psi}](https://tex.z-dn.net/?f=%5Cmathbf%7B%20%5Csigma%20_2%20%3D%203000%20psi%7D)
Explanation:
Given that:
diameter d = 12 in
thickness t = 0.25 in
the radius = d/2 = 12 / 2 = 6 in
r/t = 6/0.25 = 24
24 > 10
Using the thin wall cylinder formula;
The valve A is opened and the flowing water has a pressure P of 200 psi.
So;
![\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}](https://tex.z-dn.net/?f=%5Csigma_%7Bhoop%7D%20%3D%20%5Csigma%20_%201%20%3D%20%5Cfrac%7BPd%7D%7B2t%7D)
![\sigma_{long} = \sigma _2 = 0](https://tex.z-dn.net/?f=%5Csigma_%7Blong%7D%20%3D%20%5Csigma%20_2%20%3D%200)
![\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{200(12)}{2(0.25)}](https://tex.z-dn.net/?f=%5Csigma%20_%201%20%3D%20%5Cfrac%7BPd%7D%7B2t%7D%20%5C%5C%20%5C%5C%20%5Csigma%20_%201%20%3D%20%5Cfrac%7B200%2812%29%7D%7B2%280.25%29%7D)
![\mathbf{\sigma _ 1 = 4800 psi}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Csigma%20_%201%20%3D%204800%20psi%7D)
b)The valve A is closed and the water pressure P is 250 psi.
where P = 250 psi
![\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}](https://tex.z-dn.net/?f=%5Csigma_%7Bhoop%7D%20%3D%20%5Csigma%20_%201%20%3D%20%5Cfrac%7BPd%7D%7B2t%7D)
![\sigma_{long} = \sigma _2 = \frac{Pd}{4t}](https://tex.z-dn.net/?f=%5Csigma_%7Blong%7D%20%3D%20%5Csigma%20_2%20%3D%20%5Cfrac%7BPd%7D%7B4t%7D)
![\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{250*(12)}{2(0.25)}](https://tex.z-dn.net/?f=%5Csigma%20_%201%20%3D%20%5Cfrac%7BPd%7D%7B2t%7D%20%5C%5C%20%5C%5C%20%5Csigma%20_%201%20%3D%20%5Cfrac%7B250%2A%2812%29%7D%7B2%280.25%29%7D)
![\mathbf{\sigma _ 1 = 6000 psi}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Csigma%20_%201%20%3D%206000%20psi%7D)
![\sigma _2 = \frac{Pd}{4t} \\ \\ \sigma _2 = \frac{250(12)}{4(0.25)}](https://tex.z-dn.net/?f=%5Csigma%20_2%20%3D%20%5Cfrac%7BPd%7D%7B4t%7D%20%5C%5C%20%5C%5C%20%20%5Csigma%20_2%20%3D%20%5Cfrac%7B250%2812%29%7D%7B4%280.25%29%7D)
![\mathbf{ \sigma _2 = 3000 psi}](https://tex.z-dn.net/?f=%5Cmathbf%7B%20%5Csigma%20_2%20%3D%203000%20psi%7D)
The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below