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andreyandreev [35.5K]
3 years ago
11

Whats the best used for Arch bridge

Engineering
1 answer:
SOVA2 [1]3 years ago
6 0

Answer:

China has also been constructing arch bridges for many years. They built the Zhaozhou Bridge in 605 AD, and it is still around today! Modern arch bridges use materials such as concrete, steel and iron.

Explanation:

You might be interested in
Tech A says that 18 AWG wire is larger than 12 AWG wire. Tech B says that the larger the diameter of the conductor, the more ele
shutvik [7]

Answer:

Both of them are wrong

Explanation:

The two technicians have given the wrong information about the wires.

This is because firstly, a higher rating of AWG means it is smaller in diameter. Thus, the diameter of a 18 AWG wire is smaller than that of a 12 AWG wire and that makes the assertion of the technician wrong.

Also, the higher the resistance, the smaller the cross sectional area meaning the smaller the diameter. A wire with bigger cross sectional area will have a smaller resistance

So this practically makes the second technician wrong too

8 0
3 years ago
The heat required to raise the temperature of m (kg) of a liquid from T1 to T2 at constant pressure is Z T2CpT dT (1) In high sc
a_sh-v [17]

Answer:

(a)

<em>d</em>Q = m<em>d</em>q

<em>d</em>q = C_p<em>d</em>T

q = \int\limits^{T_2}_{T_1} {C_p} \, dT   = C_p (T₂ - T₁)

From the above equations, the underlying assumption is that  C_p remains constant with change in temperature.

(b)

Given;

V = 2L

T₁ = 300 K

Q₁ = 16.73 KJ    ,   Q₂ = 6.14 KJ

ΔT = 3.10 K       ,   ΔT₂ = 3.10 K  for calorimeter

Let C_{cal} be heat constant of calorimeter

Q₂ = C_{cal} ΔT

Heat absorbed by n-C₆H₁₄ = Q₁ - Q₂

Q₁ - Q₂ = m C_p ΔT

number of moles of n-C₆H₁₄, n = m/M

ρ = 650 kg/m³  at 300 K

M = 86.178 g/mol

m = ρv = 650 (2x10⁻³) = 1.3 kg

n = m/M => 1.3 / 0.086178 = 15.085 moles

Q₁ - Q₂ = m C_p' ΔT

C_p = (16.73 - 6.14) / (15.085 x 3.10)

C_p = 0.22646 KJ mol⁻¹ k⁻¹

6 0
3 years ago
A lagoon is designed to accommodate an input flow of 0.10 m^3/s of nonconservative pollutant with concentration 30 mg/L and deca
dexar [7]

Answer:

Volume of the lagoon required for the decay process must be larger than 86580 m³ = 8.658 × 10⁷ L

Explanation:

The lagoon can be modelled as a Mixed flow reactor.

From the value of the decay constant (0.2/day), one can deduce that the decay reaction of the pollutant is a first order reaction.

The performance equation of a Mixed flow reactor is given from the material and component balance thus:

(V/F₀) = (C₀ - C)/((C₀)(-r)) (From the Chemical Reaction Engineering textbook, authored by Prof. Octave Levenspiel)

V = volume of the reactor (The lagoon) = ?

C₀ = Initial concentration of the reactant (the pollutant concentration) = 30 mg/L = 0.03 mg/m³

F₀ = Initial flow rate of reactant in mg/s = 0.10 m³/s × C₀ = 0.1 m³/s × 0.03 mg/m³ = 0.003 mg/s

C = concentration of reactant at any time; effluent concentration < 10mg/L, this means the maximum concentration of pollutant allowed in the effluent is 10 mg/L

For the sake of easy calculation, C = the maximum value = 10 mg/L = 0.01 mg/m³

(-r) = kC (Since we know this decay process is a first order reaction)

This makes the performance equation to be:

(kVC₀/F₀) = (C₀ - C)/C

V = F₀(C₀ - C)/(kC₀C)

k = 0.2/day = 0.2/(24 × 3600s) = 2.31 × 10⁻⁶/s

V = 0.003(0.03 - 0.01)/(2.31 × 10⁻⁶ × 0.03 × 0.01)

V = 86580 m³

Since this calculation is made for the maximum concentration of 10mg/L of pollutant in the effluent, the volume obtained is the minimum volume of reactor (lagoon) to ensure a maximum volume of 10 mg/L of pollutant is contained in the effluent.

The lower the concentration required for the pollutant in the effluent, the larger the volume of reactor (lagoon) required for this decay reaction. (Provided all the other parameters stay the same)

Hope this helps!

5 0
3 years ago
Part A Engineering stress and strain are calculated using the actual cross-sectional area and length of the specimen. True or fa
Galina-37 [17]

Answer: True

Explanation:

Engineering stress is the applied load divided by the original cross-sectional area of a material. It is also known as nominal stress. It can also be defined as the force per unit area of a material. Engineering Stress is usually in large numbers.

While Engineering strain is the amount that a material deforms per unit length in a tensile test.  It can also be defined as extension per unit length. It has no unit as it is a ratio of lengths. Engineering Strain is in small numbers.

5 0
3 years ago
Read 2 more answers
A steam turbine in a power plant receives 5 kg/s steam at 3000 kPa, 500°C. Twenty percent of the flow is extracted at 1000 kPa t
MrMuchimi

Answer:

The temperature of the first exit (feed to water heater) is at 330.15ºC. The second exit (exit of the turbine) is at 141ºC. The turbine Power output (if efficiency is %100) is 3165.46 KW

Explanation:

If we are talking of a steam turbine, the work done by the steam is done in an adiabatic process. To determine the temperature of the 2 exits, we have to find at which temperature of the steam with 1000KPa and 200KPa we have the same entropy of the steam entrance.

In this case for steam at 3000 kPa, 500°C, s= 7.2345Kj/kg K. i=3456.18 KJ/Kg

For steam at 1000 kPa and s= 7.2345Kj/kg K → T= 330.15ºC i=3116.48KJ/Kg

For steam at 200 kPa and s= 7.2345Kj/kg K → T= 141ºC i=2749.74KJ/Kg

For the power output, we have to multiply the steam flow with the enthalpic jump.

The addition of the 2 jumps is the total power output.

4 0
3 years ago
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