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3 years ago

7

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

1 answer:

denpristay [2]3 years ago

6 0

I’m a concrete mason myself and I can tell you it is a pain in the butt to Roto hammer a hole into the concrete to put the pipe in it’s a lot easier to just pour the concrete around it

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The main function of a router is to

umka2103 [35]

Explanation:

A router is a switching device for networks, which is able to route network packets, based on their addresses, to other networks or devices. Among other things, they are used for Internet access, for coupling networks or for connecting branch offices to a central office via VPN (Virtual Private Network

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3 years ago

Read 2 more answers

You just purchased a 400-L rigid tank for a client who works in the gas industry. The tank is delivered pre-filled with 3 kg of

solniwko [45]

Answer:

the pressure reading when connected a pressure gauge is 543.44 kPa

Explanation:

Given data

tank volume (V) = 400 L i.e 0.4 m³

temperature (T) = 25°C i.e. 25°C + 273 = 298 K

air mass (m) = 3 kg

atmospheric pressure = 98 kPa

To find out

pressure reading

Solution

we have find out pressure reading by gauge pressure

i.e. gauge pressure = absolute pressure - atmospheric pressure

first we find absolute pressure (p) by the ideal gas condition

i.e pV = mRT

p = mRT / V

p = ( 3 × 0.287 × 298 ) / 0.4

p = 641.44 kPa

so

gauge pressure = absolute pressure - atmospheric pressure

gauge pressure = 641.44 - 98

gauge pressure = 543.44 kPa

6 0

3 years ago

Source water pollution in Madagascar

vodka [1.7K]

Answer:

What is the question?

Explanation:

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3 years ago

Air modeled as an ideal gas enters a combustion chamber at 20 lbf/in.2

motikmotik

Answer:

The answer is " $112.97 \ \frac{ft}{s}$ "

Explanation:

Air flowing into the $p_1 = 20 \ \frac{lbf}{in^2}$

Flow rate of the mass $m = 230.556 \frac{lbm}{s}$

inlet temperature $T_1 = 700^{\circ} F$

Pipeline $A= 5 \times 4 \ ft$

Its air is modelled as an ideal gas Apply the ideum gas rule to the air to calcule the basic volume v:

$\to \bar{R} = 1545 \ ft \frac{lbf}{lbmol ^{\circ} R}\\\\ \to M= 28.97 \frac{lb}{\bmol}\\\\ \to pv=RT \\\\\to v= \frac{\frac{\bar{R}}{M}T}{p}$

$= \frac{\frac{1545}{28.97}(70^{\circ}F+459.67)}{20} \times \frac{1}{144}\\\\=9.8 \frac{ft3}{lb}$

$V= \frac{mv}{A}$

$= \frac{230.556 \frac{lbm}{s} \times 9.8 \frac{ft^3}{lb}}{5 \times 4 \ ft^2}\\\\= 112.97 \frac{ft}{s}$

8 0

3 years ago

Steam at 1 MPa, 300 C flows through a 30 cm diameter pipe with an average velocity of 10 m/s. The mass flow rate of this steam i

stealth61 [152]

Answer:

$\dot m = 2.74 kg/s$

Explanation:

given data:

pressure 1 MPa

diameter of pipe = 30 cm

average velocity = 10 m/s

area of pipe $= \frac[\pi}{4}d^2$

$= \frac{\pi}{4} 0.3^2$

A = 0.070 m2

WE KNOW THAT mass flow rate is given as

$\dot m = \rho A v$

for pressure 1 MPa, the density of steam is = 4.068 kg/m3

therefore we have

$\dot m = 4.068 * 0.070* 10$

$\dot m = 2.74 kg/s$

7 0

3 years ago