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Elodia [21]
3 years ago
7

If you've wondered about the flushing of toilets on the upper floors of city skyscrapers, how do you suppose the plumbing is des

igned so that there is not an enormous impact of sewage arriving at the basement level?
Engineering
1 answer:
Marina86 [1]3 years ago
3 0

Answer:

<em>The plumbing is designed to reduce the impact of pressure forces due to the height of skyscrapers. This is achieves by narrowing down the pipe down to the basement, using pipes with thicker walls down the basement, and allowing vents; to prevent clogging of the pipes.</em>

<em></em>

Explanation:

<em>Pressure increases with depth and density</em>. In skyscrapers, a huge problem arises due to the very tall height of most skyscrapers. Also, sewage slug coming down has an increased density when compared to that of water, and these two factors can't be manipulated. The only option is to manipulate the pipe design. <em>Pipes in skyscrapers are narrowed down with height, to reduce accumulation at the bottom basement before going to the sewage tank. Standard vents are provided along the pipes, to prevent clogging of the pipes, and pipes with thicker walls are used as you go down the basement of the skyscraper, to withstand the pressure of the sewage coming down the pipes.</em>

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Answer:

particles of a solid object packed together

7 0
3 years ago
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he circular stream of water from a faucet is observed to taper from a diameter of 21 mm to 12 mm in a distance of 52 cm. Determi
Gemiola [76]

Answer: 1.52 × 10^{-3} \frac{m^{3} }{s}

Explanation:

Please kindly find the attached document for the answer.

5 0
3 years ago
A 20-cm-long rod with a diameter of 0.250 cm is loaded with a 5500 N weight. If the diameter decreases to 0.210 cm, determine th
ss7ja [257]

Answer:

1561.84 MPa

Explanation:

L=20 cm

d1=0.21 cm

d2=0.25 cm

F=5500 N

a) σ= F/A1= 5000/(π/4×(0.0025)^2)= 1018.5916 MPa

lateral strain= Δd/d1= (0.0021-0.0025)/0.0025= -0.16

longitudinal strain (ε_l)= -lateral strain/ν = -(-0.16)/0.3

(assuming a poisson's ration of  0.3)

ε_l =0.16/0.3 = 0.5333

b) σ_true= σ(1+ ε_l)= 1018.5916( 1+0.5333)

σ_true = 1561.84 MPa

ε_true = ln( 1+ε_l)= ln(1+0.5333)

ε_true= 0.4274222

The engineering stress on the rod when it is loaded with a 5500 N weight is 1561.84 MPa.

7 0
3 years ago
An example of a transient analysis involving the 1st law of thermodynamics and conservation of mass is the filling of a compress
pickupchik [31]

Answer:

<em>The temperature will be greater than 25°C</em>

Explanation:

In an adiabatic process, heat is not transferred to or from the boundary of the system. The gain or loss of internal heat energy is solely from the work done on the system, or work done by the system. The work done on the system by the environment adds heat to the system, and work done by the system on its environment takes away heat from the system.

mathematically

Change in the internal energy of a system ΔU = ΔQ + ΔW

in an adiabatic process, ΔQ = 0

therefore

ΔU = ΔW

where ΔQ is the change in heat into the system

ΔW is the work done by or done on the system

when work is done on the system, it is conventionally negative, and vice versa.

also W = pΔv

where p is the pressure, and

Δv = change in volume of the system.

In this case,<em> work is done on the gas by compressing it from an initial volume to the new volume of the cylinder. The result is that the temperature of the gas will rise above the initial temperature of 25°C </em>

8 0
3 years ago
Problem 4.079 SI A rigid tank whose volume is 3 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large
salantis [7]

Answer:

Q_{cv} = -1007.86kJ

Explanation:

Our values are,

State 1

V=3m^3\\P_1=1bar\\T_1 = 295K

We know moreover for the tables A-15 that

u_1 = 210.49kJ/kg\\h_i = 295.17kJkg

State 2

P_2 =6bar\\T_2 = 296K\\T_f = 320K

For tables we know at T=320K

u_2 = 228.42kJ/kg

We need to use the ideal gas equation to estimate the mass, so

m_1 = \frac{p_1V}{RT_1}

m_1 = \frac{1bar*100kPa/1bar(3m^3)}{0.287kJ/kg.K(295k)}

m_1 = 3.54kg

Using now for the final mass:

m_2 = \frac{p_2V}{RT_2}

m_2 = \frac{1bar*100kPa/6bar(3m^3)}{0.287kJ/kg.K(320k)}

m_2 = 19.59kg

We only need to apply a energy balance equation:

Q_{cv}+m_ih_i = m_2u_2-m_1u_1

Q_{cv}=m_2u_2-m1_u_1-(m_2-m_1)h_i

Q_{cv} = (19.59)(228.42)-(3.54)(210.49)-(19.59-3.54)(295.17)

Q_{cv} = -1007.86kJ

The negative value indidicates heat ransfer from the system

7 0
3 years ago
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