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
what is geo technical
Question 1. If a fiber weight 3.0 g and composite specimen weighing 4.g. The composite specimen weighs 2.0 g in water. If the specific gravity of the fiber and matrix is 2.4 and 1.3, respectively, find the 1. Theoretical density of composite 2. Experimental density 3. Void fraction
Answer:
Explanation:
From the given information:
weight of fiber [tex]w_f[/tex] = 3.0 g
weight of composite specimen [tex]w_c[/tex] = 4.0 g
specimen composite weight in water [tex]C_{wm}[/tex] = 2.0 g
specific gravity of fiber [tex]S_f[/tex] = 2.4
specific gravity of matrix [tex]S_m[/tex] = 1.3
The weight of the matrix = weight of the composite - the weight of fiber
⇒ (4.0 - 3.0) g
= 1.0 g
The theoretical density of the composite [tex]\rho_{ct}[/tex] can be determined by using the formula:
[tex]\dfrac{1}{\rho_{ct}} = \dfrac{w_f}{w_cS_f}+ \dfrac{w_m}{w_cS_m}[/tex]
[tex]\dfrac{1}{\rho_{ct}} = \dfrac{3.0}{(4.0 \times 2.4)}+ \dfrac{1.0}{(4.0\times 1.3)}[/tex]
[tex]\dfrac{1}{\rho_{ct}} = \dfrac{3.0}{9.6}+ \dfrac{1.0}{5.2}[/tex]
[tex]\dfrac{1}{\rho_{ct}} =0.505\\[/tex]
[tex]\rho_{ct} =\dfrac{1}{0.505}[/tex]
[tex]\mathbf{\rho_{ct} = 1.980 \ g/cm^3}[/tex]
The experimental density [tex]\rho _{ce}[/tex] is determined by using the equation:
[tex]\rho _{ce} = \dfrac{w_f + w_c}{\dfrac{w_f }{S_f} + \dfrac{w_c }{S_m} }[/tex]
[tex]\rho _{ce} = \dfrac{3.0 + 4.0}{\dfrac{3.0 }{2.4} + \dfrac{4.0 }{1.3} }[/tex]
[tex]\rho _{ce} = \dfrac{3.0 + 4.0}{1.250 +3.077 }[/tex]
[tex]\mathbf{\rho _{ce} = 1.620 \ g/cm^3}[/tex]
The void fraction is: [tex]= \dfrac{\rho_{ct}-\rho_{ce}}{\rho_{ct}}[/tex]
[tex]= \dfrac{1.980-1.620}{1.980}[/tex]
= 0.1818
Your organization recently purchased 20 Android tablets for use by the organization's management team. To increase the security of these devices, you want to ensure that only specific apps can be installed. Which of the following would you implement?
A. Credential Manager.
B. App whitelisting.
C. App blacklisting.
D. Application Control.
Design a ductile iron pumping main carrying a discharge of 0.35 m3/s over a distance of 4 km. The elevation of the pumping station is 140 m and that of the exit point is 150 m. The required terminal head is 10 m. Estimate the pipe diameter and pumping head using the explicit design procedure g
Answer:
[tex]D=0.41m[/tex]
Explanation:
From the question we are told that:
Discharge rate [tex]V_r=0.35 m3/s[/tex]
Distance [tex]d=4km[/tex]
Elevation of the pumping station [tex]h_p= 140 m[/tex]
Elevation of the Exit point [tex]h_e= 150 m[/tex]
Generally the Steady Flow Energy Equation SFEE is mathematically given by
[tex]h_p=h_e+h[/tex]
With
[tex]P_1-P_2[/tex]
And
[tex]V_1=V-2[/tex]
Therefore
[tex]h=140-150[/tex]
[tex]h=10[/tex]
Generally h is give as
[tex]h=\frac{0.5LV^2}{2gD}[/tex]
[tex]h=\frac{8Q^2fL}{\pi^2 gD^5}[/tex]
Therefore
[tex]10=\frac{8Q^2fL}{\pi^2 gD^5}[/tex]
[tex]D=^5\frac{8*(0.35)^2*0.003*4000}{3.142^2*9.81*10}[/tex]
[tex]D=0.41m[/tex]