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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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Is a water proof material used around tubs and

Leviafan [203]

I think it’s hard board

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3 years ago

Read 2 more answers

What are the 5 different components of the four-stroke engine?

Dmitrij [34]

Answer: Piston.

Crankshaft.

Camshaft.

Spark plug.

Cylinder.

Valves.

Carburetor.

Flywheel.

Explanation: What Are the Strokes of a 4-Cycle Engine? At the end of the compression (previous) stroke, the spark plug fires and ignites the compressed air/fuel mixture. This ignition/explosion forces the piston back down the cylinder bore and rotates the crankshaft, propelling the vehicle forward.

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3 years ago

Decompose the signal (1+ 0.1 cos5t) cos100t into a linear combination of sinusoidal functions, and find the amplitude, frequency

frutty [35]

Answer:

as answered in the attachment

Explanation:

The detailed step by step derivation and the use of trigonometric identities is as shown in the attachment.

8 0

4 years ago

A converging-diverging nozzle with an exit to throat area ratio of 4.0 is designed to expand air isentropically to atmospheric p

34kurt

Answer

0.9, 1172.35kPa

Explanation:

<em>Question (in proper order) Attached below</em>

Air is flowing inside the throat has following inlet conditions

$P_{0}=1000 kPa$

$T_{0}=500 K$

$M=1.8$

$M=\frac{u}{c}=1.8$

'u' is the speed of sound in the air

$\Rightarrow u=1.8\times c$

$=1.8\times 340.29$

$=612.522\frac{m}{sec}$

Therefore volumetric flow rate entering,

$Q=612.522\times 0.0008$

$=0.4900176\frac{m^{3}}{sec}$

Using ideal gas equation

PV=nRT

$n=\frac{PV}{RT}$

$=\frac{1000\times 0.4900176}{8.314\times 500}$

=0.117878 gmoles/sec

Therefore , mass flow rate

$Mass = 0.117878\times 29$

$=3.4184 grams/sec$

Given

$\frac{A}{A_{0}}=2$

$\Rightarrow A=0.0016.m^{2}$

Using continuity equation

$A_{1}V_{1}=A_{2}V_{2}$

$\Rightarrow V_{2}=\frac{A_{1}V_{1}}{A_{2}}$

$=\frac{0.0008\times 612.522}{0.0016}$

$=306.261\frac{m}{sec}$

Hence exit velocity = 306.261 m/sec

Exit Mach number

$M=\frac{u}{c}=\frac{306.261}{340.29}=0.9$

Temperature will remain same as 500 K

Now

Using Bernoulli's equation

$\frac{P_{1}}{\rho g}+\frac{v_{1}^{2}}{2g}+z_{1}=\frac{P_{2}}{\rho g}+\frac{v_{2}^{2}}{2g}+z_{2}$

Here

$z_{1} = z_{2}$

$\frac{P_{1}}{\rho g}+\frac{v_{1}^{2}}{2g}-\frac{v_{2}^{2}}{2g}=\frac{P_{2}}{\rho g}$

$\Rightarrow \frac{1000000}{\rho g}+\frac{612.522^{2}}{2g}-\frac{306.261^{2}}{2g}=\frac{P_{2}}{\rho g}$

$\Rightarrow \frac{1000000}{1.225}+\frac{612.522^{2}}{2}-\frac{306.261^{2}}{2}=\frac{P_{2}}{1.225}$

$\Rightarrow P_{2}=1172.35kPa$

4 0

3 years ago

A concrete building slab, the temperature is normally on the upper surface of the slab (the inside), andit is-on the lower surfa

Crank

Complete Question:

A concrete building slab, on a basement floor is 5 m long, 3 m wide and 0.6 m thick. During the winter, the temperature is normally 18°C on the upper surface of the slab (the inside), and it is -7°C on the lower surface, with a linear temperature profile in between. If the concrete has a thermal conductivity of 1.4 W/m-K, what is the rate of heat loss through the slab? If the basement is heated by a gas furnace operated with an efficiency of 66%, how much energy must be utilized to maintain the basement temperature for 90 days assuming that nearly all heat losses occur through the slab?

Answer:

a) Rate of heat loss, $\dot{Q} = 875 W$

b) Energy that must be utilized to maintain the basement temperature,

Em = 10309 MJ

Explanation:

Length of the slab, l = 5 m

Width of the slab, w = 3 m

Thickness of the slab, t = 0.6 m

Cross Sectional Area of the slab, A = l x b

A = 5 x 3

A = 15 m²

Upper Surface Temperature, T₁ = 18°C = 18 + 273 = 291 K

Lower Surface Temperature, T₂ = -7°C = -7 + 273 = 266 K

a) The rate of heat loss is given by the formula:

$\dot{Q} = KA \frac{dT}{dx} \\\dot{Q} = 1.4 * 15 \frac{291 - 266}{0.6}\\\dot{Q} = 1.4 * 15 \frac{25}{0.6}\\\dot{Q} = 875 W$

b) Energy, $E = \dot{Q}t$

t = 90 days = 90 * 24 * 3600 = 7776000 s

E = 875 * 7776000

E = 6804000000J

E = 6804 MJ

If Efficiency = 66%, Energy that must be utilized to maintain the basement temperature.

6804 = 66% * Em

6804 = 0.66 * Em

Em = 6804/0.66

Em = 10309 MJ

4 0

4 years ago