Answer:
f = 0.04042
Explanation:
temperature = 0°C = 273k
p = 600 Kpa
d = 40 millemeter
e = 10 m
change in P = 235 N/m²
μ = 2m/s
R = 188.9 Nm/kgk
we solve this using this formula;
P = ρcos*R*T
we put in the values into this equation
600x10³ = ρcos * 188.9 * 273
600000 = ρcos51569.7
ρcos = 600000/51569.7
=11.63
from here we find the head loss due to friction
Δp/pg = feμ²/2D
235/11.63 = f*10*4/2*40x10⁻³
20.21 = 40f/0.08
20.21*0.08 = 40f
1.6168 = 40f
divide through by 40
f = 0.04042
Answer:
true
Explanation:
True, there are several types of polymers, thermoplastics, thermosets and elastomers.
Thermosets are characterized by having a reticulated structure, so they have low elasticity and cannot be stretched when heated.
Because of the above, thermosetting polymers burn when heated.
Answer with Explanation:
Assuming that the degree of consolidation is less than 60% the relation between time factor and the degree of consolidation is
![T_v=\frac{\pi }{4}(\frac{U}{100})^2](https://tex.z-dn.net/?f=T_v%3D%5Cfrac%7B%5Cpi%20%7D%7B4%7D%28%5Cfrac%7BU%7D%7B100%7D%29%5E2)
Solving for 'U' we get
![\frac{\pi }{4}(\frac{U}{100})^2=0.2\\\\(\frac{U}{100})^2=\frac{4\times 0.2}{\pi }\\\\\therefore U=100\times \sqrt{\frac{4\times 0.2}{\pi }}=50.46%](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%20%7D%7B4%7D%28%5Cfrac%7BU%7D%7B100%7D%29%5E2%3D0.2%5C%5C%5C%5C%28%5Cfrac%7BU%7D%7B100%7D%29%5E2%3D%5Cfrac%7B4%5Ctimes%200.2%7D%7B%5Cpi%20%7D%5C%5C%5C%5C%5Ctherefore%20U%3D100%5Ctimes%20%5Csqrt%7B%5Cfrac%7B4%5Ctimes%200.2%7D%7B%5Cpi%20%7D%7D%3D50.46%25)
Since our assumption is correct thus we conclude that degree of consolidation is 50.46%
The consolidation at different level's is obtained from the attached graph corresponding to Tv = 0.2
i)
= 71% consolidation
ii)
= 45% consolidation
iii)
= 30% consolidation
Part b)
The degree of consolidation is given by
![\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.5046\\\\\therefore \Delta H=50.46cm](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20H%7D%7BH_f%7D%3DU%5C%5C%5C%5C%5Cfrac%7B%5CDelta%20H%7D%7B1.0%7D%3D0.5046%5C%5C%5C%5C%5Ctherefore%20%5CDelta%20H%3D50.46cm)
Thus a settlement of 50.46 centimeters has occurred
For time factor 0.7, U is given by
![T_v=1.781-0.933log(100-U)\\\\0.7=1.781-0.933log(100-U)\\\\log(100-U)=\frac{1.780-.7}{0.933}=1.1586\\\\\therefore U=100-10^{1.1586}=85.59](https://tex.z-dn.net/?f=T_v%3D1.781-0.933log%28100-U%29%5C%5C%5C%5C0.7%3D1.781-0.933log%28100-U%29%5C%5C%5C%5Clog%28100-U%29%3D%5Cfrac%7B1.780-.7%7D%7B0.933%7D%3D1.1586%5C%5C%5C%5C%5Ctherefore%20U%3D100-10%5E%7B1.1586%7D%3D85.59)
thus consolidation of 85.59 % has occured if time factor is 0.7
The degree of consolidation is given by
![\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.8559\\\\\therefore \Delta H=85.59cm](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20H%7D%7BH_f%7D%3DU%5C%5C%5C%5C%5Cfrac%7B%5CDelta%20H%7D%7B1.0%7D%3D0.8559%5C%5C%5C%5C%5Ctherefore%20%5CDelta%20H%3D85.59cm)