Answer:
I think reduce your following distance
Answer:
shoe box covered in aluminum foil
Answer:
Flow rate is ![1.82\times 10^{-8} m^{3}/s](https://tex.z-dn.net/?f=1.82%5Ctimes%2010%5E%7B-8%7D%20m%5E%7B3%7D%2Fs)
Explanation:
Given information
Density of oil, ![\rho_{oil}= 850 Kg/m^{3}](https://tex.z-dn.net/?f=%5Crho_%7Boil%7D%3D%20850%20Kg%2Fm%5E%7B3%7D)
kinematic viscosity, ![v= 0.00062 m^{2} /s](https://tex.z-dn.net/?f=v%3D%200.00062%20m%5E%7B2%7D%20%2Fs)
Diameter of pipe, D= 5 mm= 0.005 m
Length of pipe, L=40 m
Height of liquid, h= 3 m
Volume flow rate for horizontal pipe will be given by
where
is dynamic viscosity and
is pressure drop
At the bottom of the tank, pressure is given by
![P_{bottom}=\rho_{oil} gh=850 Kg/m^{3}\times 9.81 m/s^{2}\times 3 m= 25015.5 N/m^{2}](https://tex.z-dn.net/?f=P_%7Bbottom%7D%3D%5Crho_%7Boil%7D%20gh%3D850%20Kg%2Fm%5E%7B3%7D%5Ctimes%209.81%20m%2Fs%5E%7B2%7D%5Ctimes%203%20m%3D%2025015.5%20N%2Fm%5E%7B2%7D)
Since at the top pressure is zero, therefore change in pressure a difference between the pressure at the bottom and the top. It implies that change in pressure is still ![25015.5 N/m^{2}](https://tex.z-dn.net/?f=25015.5%20N%2Fm%5E%7B2%7D)
Dynamic viscosity, ![\mu=\rho_{oil}v= 850 Kg/m^{3}\times 0.00062 m^{2}/s=0.527 Kg/m.s](https://tex.z-dn.net/?f=%5Cmu%3D%5Crho_%7Boil%7Dv%3D%20850%20Kg%2Fm%5E%7B3%7D%5Ctimes%200.00062%20m%5E%7B2%7D%2Fs%3D0.527%20Kg%2Fm.s)
Now the volume flow rate will be
![\bar v=\frac {25015.5 N/m^{2}\times \pi \times 0.005^{4}}{128\times 0.527 Kg/m.s \times 40}=1.82037\times 10^{-8} m^{3}/s\approx 1.82\times 10^{-8} m^{3}/s](https://tex.z-dn.net/?f=%5Cbar%20v%3D%5Cfrac%20%7B25015.5%20N%2Fm%5E%7B2%7D%5Ctimes%20%5Cpi%20%5Ctimes%200.005%5E%7B4%7D%7D%7B128%5Ctimes%200.527%20Kg%2Fm.s%20%5Ctimes%2040%7D%3D1.82037%5Ctimes%2010%5E%7B-8%7D%20m%5E%7B3%7D%2Fs%5Capprox%201.82%5Ctimes%2010%5E%7B-8%7D%20m%5E%7B3%7D%2Fs)
Proof of flow being laminar
The velocity of flow is given by
![V_{flow}=\frac {\bar v}{A}=\frac {1.82\times 10^{-8} m^{3}/s}{0.25\times \pi\times 0.005^{2}}=0.000927104 m/s](https://tex.z-dn.net/?f=V_%7Bflow%7D%3D%5Cfrac%20%7B%5Cbar%20v%7D%7BA%7D%3D%5Cfrac%20%7B1.82%5Ctimes%2010%5E%7B-8%7D%20m%5E%7B3%7D%2Fs%7D%7B0.25%5Ctimes%20%5Cpi%5Ctimes%200.005%5E%7B2%7D%7D%3D0.000927104%20%20m%2Fs)
Reynolds number, ![Re=\frac {\rho_{oil} v_{flow} D}{\mu}=\frac {850 Kg/m^{3}\times 0.000927104 m/s\times 0.005}{0.527 kg/m.s}=0.007476648](https://tex.z-dn.net/?f=Re%3D%5Cfrac%20%7B%5Crho_%7Boil%7D%20v_%7Bflow%7D%20D%7D%7B%5Cmu%7D%3D%5Cfrac%20%7B850%20Kg%2Fm%5E%7B3%7D%5Ctimes%200.000927104%20m%2Fs%5Ctimes%200.005%7D%7B0.527%20kg%2Fm.s%7D%3D0.007476648)
Since the Reynolds number is less than 2300, the flow is laminar and the assumption is correct.
Answer:
Fire resistance is measured in hour.
Explanation:
The fire resistance of the structure is estimated in a number of hours during which material and a piece of the structure can oppose a fire. The fire resistance of a joining structure is dictated by the type of structure, the supporting dividers, the segments, the bars, and the supports. The table shows the fire resistance level of most of the structure's segments.
While deciding the level of free opposition, the cost will be higher. In this way, the structure is planned with the least tightness at the fire level authorized by the building law.