Which of the following statements about PAC storage and housekeeping is correct? (Welding)

Structural engineers use wireless sensor networks to monitor the condition of dams and bridges.

qwelly [4]

Answer: True

Explanation:

Structural engineering is simply referred to as a branch of civil engineering, and they help in the designing of structures like buildings, dams, tunnels, bridges, tunnels, etc. Structural engineers typically work as consultants.

Wireless sensor networks are used by structural engineers to monitor the condition of dams and bridges. Wireless sensor network refers are dedicated sensors that are used in monitoring the physical conditions of the bridges and to know if they're still in good conditions.

8 0

2 years ago

For the same cross-sectional area, which column provides the higher buckling load: a circular bar or a circular tube?

juin [17]

Answer:

Circular tube

Explanation:

Now for better understanding lets take an example

Lets take

Diameter of solid bar= $4\sqrt{2}$ cm

Outer diameter of tube =6 cm

Inner diameter of tube=2 cm

So from we can say that both tubes have equal cross sectional area.

We know that buckling load is given as $P = \dfrac{\pi ^2EI}{L_e^2}$

If area moment of inertia(I) is high then buckling load will be high.

We know that area moment of inertia(I)

For circular tube $I = \dfrac{\pi }{64}(D_o^4-D_i^4)$

For circular bar $I = \dfrac{\pi }{64}D^4$

Now by putting the values

For circular tube $I=62.83 cm^4$

For circular bar $I=50.26 cm^4$

So we can say that for same cross sectional area the area moment of inertia(I) is high for tube as compare to bar.So buckling load will be higher in tube as compare to bar.

3 0

2 years ago

Flow and Pressure Drop of Gases in Packed Bed. Air at 394.3 K flows through a packed bed of cylinders having a diameter of 0.012

devlian [24]

The pressure drop of air in the bed is 14.5 kPa.

<u>Explanation:</u>

To calculate Re:

$R e=\frac{1}{1-\varepsilon} \frac{\rho q d_{p}}{\mu}$

From the tables air property

$\mu_{394 k}=2.27 \times 10^{-5}$

Ideal gas law is used to calculate the density:

ρ = $\frac{2.2}{2.83 \times 10^{-3} \times 394.3}$

ρ = 1.97 Kg / $m^{3}$

ρ = $\frac{P}{RT}$

R = $\frac{R_{c} }{M}$ = 8.2 × $10^{-5}$ / 28.97× $10^{-3}$

R = 2.83 × $10^{-3}$ $m^{3}$ atm / K Kg

q is expressed in the unit m/s

$q=\frac{2.45}{1.97}$

q = 1.24 m/s

Re = $\frac{1}{1-0.4} \frac{1.97 \times 1.24 \times 0.0127}{2.27 \times 10^{-5}}$

Re = 2278

The Ergun equation is used when Re > 10,

$\frac{\Delta P}{L}=\frac{180 \mu}{d_{p}^{2}} \frac{(1-\varepsilon)^{2}}{\varepsilon^{3}} q+\frac{7}{4} \frac{\rho}{d_{p}} \frac{(1-\varepsilon)}{\varepsilon^{3}} q^{2}$

$\frac{\Delta P}{L}=\frac{180 \times 2.27 \times 10^{-5}}{0.0127^{2}} \frac{(1-0.4)^{2}}{0.4^{3}} 1.24$ $+\frac{7}{4} \frac{1.97}{0.0127} \frac{(1-0.4)}{0.4^{3}} 1.24^{2}$