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3 years ago

The first attempt to strengthen concrete flexural elements by externally bonded steel plates was attempted in?

Engineering

1 answer:

Vikentia [17]3 years ago

7 0

Answer:

Explanation:

The first reported attempts to strengthen concrete flexural elements by externally bonded steel plates were attempted in France around 1964-65 and practical applications date back to 1966-67 in France and South Africa then followed by Japan and Russia.

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Homes may be heated by pumping hot water through radiators. What mass of water (in g) will provide the same amount of heat when

Nitella [24]

Answer:

a mass of water required is mw= 1273.26 gr = 1.27376 Kg

Explanation:

Assuming that the steam also gives out latent heat, the heat provided should be same for cooling the hot water than cooling the steam and condense it completely:

Q = mw * cw * ΔTw = ms * cs * ΔTw + ms * L

where m = mass , c= specific heat , ΔT=temperature change, L = latent heat of condensation

therefore

mw = ( ms * cs * ΔTw + ms * L )/ (cw * ΔTw )

replacing values

mw = [182g * 2.078 J/g°C*(118°C-100°C) + 118 g * 2260 J/g ] /[4.187 J/g°C * (90.7°C-39.4°C)] = 1273.26 gr = 1.27376 Kg

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3 years ago

What is the fastest plane in the world

Aleonysh [2.5K]

Answer:

Lockheed SR-71 Blackbird

Explanation:

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3 years ago

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A cylindrical tank is required to contain a gage pressure 520 kPa . The tank is to be made of A516 grade 60 steel with a maximum

enot [183]

Answer:

t= 4.5 mm

Explanation:

Given that

P = 520 KPa ( gauge)

Maximum allowable normal stress ,σ= 150

d= 2.6 m

Wall thickness = t

The normal stress for pressure vessel given as

$\sigma=\dfrac{Pd}{2t}$ ( hoop stress)

We always take maximum stress for safe design.

$\sigma=\dfrac{Pd}{2t}$

Now by putting the values

$150\times 1000=\dfrac{520\times 2.6}{2t}$

t= 4.5 mm

So the minimum thickness, t, of the wall is 4.5 mm

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3 years ago

A steel rod, which is free to move, has a length of 200 mm and a diameter of 20 mm at a temperature of 15oC. If the rod is heate

kherson [118]

Explanation:

thermal expansion ∝L = (δL/δT)÷L ----(1)

δL = L∝L + δT ----(2)

we have δL = 12.5x10⁻⁶

length l = 200mm

δT = 115°c - 15°c = 100°c

putting these values into equation 1, we have

δL = 200*12.5X10⁻⁶x100

= 0.25 MM

L₂ = L + δ L

= 200 + 0.25

L₂ = 200.25mm

12.5X10⁻⁶ *115-15 * 20

= 0.025

20 +0.025

D₂ = 20.025

as this rod undergoes free expansion at 115°c, the stress on this rod would be = 0

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3 years ago

Examine a process whereby air at 300 K, 100 kPa is compressed in a piston/cylinder arrangement to 600 kPa. Assume the process is

professor190 [17]

Answer:

See attachment and explanation.

Explanation:

- The following question can be solved better with the help of a MATLAB program as follows. The code is given in the attachment.

- The plot of the graph is given in attachment.

- The code covers the entire spectrum of the poly-tropic range ( 1.2 - 1.6 ) and 20 steps ( cases ) have been plotted and compared in the attached plot.

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3 years ago