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3 years ago

Why might construction crews want to install pipes before the foundation is poured

Engineering

1 answer:

Crazy boy [7]3 years ago

3 0

The answer is choice C

Explanation:

As during construction ,the site is cleared for all debris before laying out the foundation. Even the sewer lines are dug out .

So it will be useful for the construction crews to connect the pipes to the sewer lines before the foundation is poured.

But usually the steps take in construction activity is:- first the site is cleared for the foundation to be poured and once the foundation wall is set , then all utilities , including plumbing and electrical activities are done.,

After this process is over, the city inspector comes to check whether the foundation has been laid down as per the code of construction.

Only after that the rest of the construction activity follows through.

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Air enters a cmpressor at 20 deg C and 80 kPa and exits at 800 kPa and 200 deg C. The power input is 400 kW. Find the heat trans

aksik [14]

Answer:

The heat is transferred is at the rate of 752.33 kW

Solution:

As per the question:

Temperature at inlet, $T_{i} = 20^{\circ}C$ = 273 + 20 = 293 K

Temperature at the outlet, $T_{o} = 200{\circ}C$ = 273 + 200 = 473 K

Pressure at inlet, $P_{i} = 80 kPa = 80\times 10^{3} Pa$

Pressure at outlet, $P_{o} = 800 kPa = 800\times 10^{3} Pa$

Speed at the outlet, $v_{o} = 20 m/s$

Diameter of the tube, $D = 10 cm = 10\times 10^{- 2} m = 0.1 m$

Input power, $P_{i} = 400 kW = 400\times 10^{3} W$

Now,

To calculate the heat transfer, $Q$ , we make use of the steady flow eqn:

$h_{i} + \frac{v_{i}^{2}}{2} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH' + p_{s}$

where

$h_{i}$ = specific enthalpy at inlet

$h_{o}$ = specific enthalpy at outlet

$v_{i}$ = air speed at inlet

$p_{s}$ = specific power input

H and H' = Elevation of inlet and outlet

Now, if

$v_{i} = 0$ and H = H'

Then the above eqn reduces to:

$h_{i} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH + p_{s}$

$Q = h_{o} - h_{i} + \frac{v_{o}^{2}}{2} + p_{s}$ (1)

Also,

$p_{s} = \frac{P_{i}}{ mass, m}$

Area of cross-section, A = $\frac{\pi D^{2}}{4} =\frac{\pi 0.1^{2}}{4} = 7.85\times 10^{- 3} m^{2}$

Specific Volume at outlet, $V_{o} = A\times v_{o} = 7.85\times 10^{- 3}\times 20 = 0.157 m^{3}/s$

From the eqn:

$P_{o}V_{o} = mRT_{o}$

$m = \frac{800\times 10^{3}\times 0.157}{287\times 473} = 0.925 kg/s$

Now,

$p_{s} = \frac{400\times 10^{3}}{0.925} = 432.432 kJ/kg$

Also,

$\Delta h = h_{o} - h_{i} = c_{p}\Delta T =c_{p}(T_{o} - T_{i}) = 1.005(200 - 20) = 180.9 kJ/kg$

Now, using these values in eqn (1):

$Q = 180.9 + \frac{20^{2}}{2} + 432.432 = 813.33 kW$

Now, rate of heat transfer, q:

q = mQ = $0.925\times 813.33 = 752.33 kW$

4 0

3 years ago

A composite wall is to be used to insulate a freezer chamber at -350C. Two insulating materials are to be used with conductiviti

choli [55]

Answer:

thickness1=1.4m

thickness2=2.2m

convection coefficient=0.33W/m^2K

Explanation:

you must use this equation to calculate the thickness:

L=K(T2-T1)/Q

L=thickness

T=temperature

Q=heat

L1=0.04*(0--350)/10=1.4m

L2=0.1(220-0)/10=2.2m

Then use this equation to calculate the convective coefficient

H=Q/(T2-T1)

H=10/(250-220)=0.33W/m^2K

7 0

4 years ago

The most important element of green construction is that it is a(n) __________ approach to building. *

madam [21]

Environmentally friendly

Since it focuses on are sustainable and efficient with and are made with the future in mind.

5 0

3 years ago

For a column that is pinned at both ends, the critical buckling load can be calculated as, Pcr = π2 E I /L^2 where E is Young's

gulaghasi [49]

When a slender member is subjected to an axial compressive load, it may fail by a ... Consider a column of length, L, cross-sectional Moment of Inertia, I, having Young's Modulus, E. Both ends are pinned, meaning they can freely rotate ... p2EI L2 ... scr, is the Euler Buckling Load divided by the columns cross-sectional area

6 0

4 years ago

Calculate the pressure drop in a duct (measured by a differential oil manometer) if the differential height between the two flui

Burka [1]

Answer:

The pressure drop is 269.7N/m^2

Explanation:

∆P = ∆h × rho × g

∆h = 3.2cm = 3.2/100 = 0.032m, rho = 860kg/m^3, g = 9.8m/s^2

∆P = 0.032×860×9.8 = 269.7N/m^2

6 0

3 years ago