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4 years ago

14

The ___________ section of the bid package might include information concerning provisions for water and other utilities, sanita

tion equipment, or storage areas.

1 answer:

dexar [7]4 years ago

3 0

Answer:

Procurement Process

Explanation:

Procurement Process describes the series of activities that an organization partakes in to get products or services in order to achieve their goals. The choice of the procurement process is very important for the success of a construction project.

So during a bidding process, the procurement process is section where the organization will need to get water and other utilities, sanitation equipment or storage needed for the success of a construction project.

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A fluid at 300 K flows through a long, thin-walled pipe of 0.2-m diameter. The pipe is enclosed in a concrete casing that is of

andrew-mc [135]

Answer:

The correct answer is "1341.288 W/m".

Explanation:

Given that:

T₁ = 300 K

T₂ = 500 K

Diameter,

d = 0.2 m

Length,

l = 1 m

As we know,

The shape factor will be:

⇒ $SF=\frac{2 \pi l}{ln[\frac{1.08 b }{d} ]}$

By putting the value, we get

⇒ $=\frac{2 \pi l}{ln[\frac{1.08\times 1}{0.2} ]}$

⇒ $=3.7258 \ l$

hence,

The heat loss will be:

⇒ $Q=SF\times K(T_2-T_1)$

$=3.7258\times 1\times 1.8\times (500-300)$

$=3.7258\times 1.8\times (200)$

$=1341.288 \ W/m$

3 0

3 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,

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

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

= 4089.748 Pa/m

ΔP = 4089.748 × 3.66

ΔP = 14.5 kPa

4 0

3 years ago

a metal rod 24mm diameter and 2m long is subjected to an axial pull of 40 kN. If the rod is 0.5mm, then find the stress-induced

ozzi

Answer:

i dont know but i will take the points tho hahah

Explanation:

8 0

3 years ago

A(n)<br> is a safety device commonly<br> used with a slotted nut.

liraira [26]

A safety device called a cotter pin. The cotter pin fits through a hole in the bolt or part. This keeps the nut from turning and possibly coming off.

5 0

3 years ago

The state of plane strain on an element is:

balu736 [363]

Answer:

a. ε₁=-0.000317

ε₂=0.000017

θ₁= -13.28° and θ₂=76.72°

b. maximum in-plane shear strain =3.335 *10^-4

Associated average normal strain ε(avg) =150 *10^-6

θ = 31.71 or -58.29

Explanation:

$\epsilon _{1,2} =\frac{\epsilon_x + \epsilon_y}{2} \pm \sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\epsilon _{1,2} =\frac{-300 \times 10^{-6} + 0}{2} \pm \sqrt{(\frac{-300 \times 10^{-6}+ 0}{2}) ^2 + (\frac{150 \times 10^-6}{2})^2}\\\\\epsilon _{1,2} = -150 \times 10^{-6} \pm 1.67 \times 10^{-4}$

ε₁=-0.000317

ε₂=0.000017

To determine the orientation of ε₁ and ε₂

$tan 2 \theta_p = \frac{\gamma_xy}{\epsilon_x - \epsilon_y} \\\\tan 2 \theta_p = \frac{150 \times 10^{-6}}{-300 \times 10^{-6}-\ 0}\\\\tan 2 \theta_p = -0.5$

θ= -13.28° and 76.72°

To determine the direction of ε₁ and ε₂

$\epsilon _{x' }=\frac{\epsilon_x + \epsilon_y}{2} + \frac{\epsilon_x -\epsilon_y}{2} cos2\theta + \frac{\gamma_xy}{2}sin2\theta \\\\\epsilon _{x'} =\frac{-300 \times 10^{-6}+ \ 0}{2} + \frac{-300 \times 10^{-6} -\ 0}{2} cos(-26.56) + \frac{150 \times 10^{-6}}{2}sin(-26.56)\\\\$

=-0.000284 -0.0000335 = -0.000317 =ε₁

Therefore θ₁= -13.28° and θ₂=76.72°

b. maximum in-plane shear strain

$\gamma_{max \ in \ plane} =2\sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\gamma_{max \ in \ plane} = 2\sqrt{(\frac{-300 *10^{-6} + 0}{2} )^2 + (\frac{150 *10^{-6}}{2})^2}$

=3.335 *10^-4

$\epsilon_{avg} =(\frac{\epsilon_x + \epsilon_y}{2} )$

ε(avg) =150 *10^-6

orientation of γmax

$tan 2 \theta_s = \frac{-(\epsilon_x - \epsilon_y)}{\gamma_xy} \\\\tan 2 \theta_s = \frac{-(-300*10^{-6} - 0)}{150*10^{-6}}$

θ = 31.71 or -58.29

To determine the direction of γmax

$\gamma _{x'y' }= - \frac{\epsilon_x -\epsilon_y}{2} sin2\theta + \frac{\gamma_xy}{2}cos2\theta \\\\\gamma _{x'y' }= - \frac{-300*10^{-6} - \ 0}{2} sin(63.42) + \frac{150*10^{-6}}{2}cos(63.42)$

= 1.67 *10^-4

4 0

4 years ago