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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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An electric kettle is required to heat 0.64 kg of water from 15.4°C to 98.2°C in six

skelet666 [1.2K]

Answer:

Almost done

Explanation:

I am just finishing up my work

7 0

2 years ago

11) (10 points) A large valve is to be used to control water supply in large conduits. Model tests are to be done to determine h

IrinaVladis [17]

Answer:

7.94 ft^3/ s.

Explanation:

So, we are given that the '''model will be 1/6 scale (the modeled valve will be 1/6 the size of the prototype valve)'' and the prototype flow rate is to be 700 ft3 /s. Then, we are asked to look for or calculate or determine the value for the model flow rate.

Note that we are to use Reynolds scaling for the velocity as par the instruction from the question above.

Therefore; kp/ks = 1/6.

Hs= 700 ft3 /s and the formula for the Reynolds scaling => Hp/Hs = (kp/ks)^2.5.

Reynolds scaling==> Hp/ 700 = (1/6)^2.5.

= 7.94 ft^3/ s

7 0

3 years ago

I am trying to make a vacuum cannon but all I can use to get out the air is a speed pump to give air to bicycles. I need to make

Mrrafil [7]

We can actually deduce here that making a airtight seal will take different format. You can:

Use an epoxy-resin to create an airtight seal
Create a glass-metal airtight seal
Make a ceramic-metal airtight seal.

<h3>What is an airtight seal?</h3>

An airtight seal is actually known to be a seal or sealing that doesn't permit air or gas to pass through. Airtight seal are usually known as hermetic seal. They are usually applied to airtight glass containers but the advancement in technology has helped to broaden the materials.

We can see that epoxy-resin can used to create an airtight seal. They create airtight seals to copper, plastics, stainless steels, etc.

When making glass-metal airtight seal, the metal should compress round the solidified glass when it cools.

Learn more about airtight seal on brainly.com/question/14977167

#SPJ1

6 0

1 year ago

Air enters a cmpressor at 20 deg C and 80 kPa and exits at 800 kPa and 200 deg C. The power input is 400 kW. Find the heat trans

aksik [14]

Answer:

The heat is transferred is at the rate of 752.33 kW

Solution:

As per the question:

Temperature at inlet, $T_{i} = 20^{\circ}C$ = 273 + 20 = 293 K

Temperature at the outlet, $T_{o} = 200{\circ}C$ = 273 + 200 = 473 K

Pressure at inlet, $P_{i} = 80 kPa = 80\times 10^{3} Pa$

Pressure at outlet, $P_{o} = 800 kPa = 800\times 10^{3} Pa$

Speed at the outlet, $v_{o} = 20 m/s$

Diameter of the tube, $D = 10 cm = 10\times 10^{- 2} m = 0.1 m$

Input power, $P_{i} = 400 kW = 400\times 10^{3} W$

Now,

To calculate the heat transfer, $Q$ , we make use of the steady flow eqn:

$h_{i} + \frac{v_{i}^{2}}{2} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH' + p_{s}$

where

$h_{i}$ = specific enthalpy at inlet

$h_{o}$ = specific enthalpy at outlet

$v_{i}$ = air speed at inlet

$p_{s}$ = specific power input

H and H' = Elevation of inlet and outlet

Now, if

$v_{i} = 0$ and H = H'

Then the above eqn reduces to:

$h_{i} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH + p_{s}$

$Q = h_{o} - h_{i} + \frac{v_{o}^{2}}{2} + p_{s}$ (1)

Also,

$p_{s} = \frac{P_{i}}{ mass, m}$

Area of cross-section, A = $\frac{\pi D^{2}}{4} =\frac{\pi 0.1^{2}}{4} = 7.85\times 10^{- 3} m^{2}$

Specific Volume at outlet, $V_{o} = A\times v_{o} = 7.85\times 10^{- 3}\times 20 = 0.157 m^{3}/s$

From the eqn:

$P_{o}V_{o} = mRT_{o}$

$m = \frac{800\times 10^{3}\times 0.157}{287\times 473} = 0.925 kg/s$

Now,

$p_{s} = \frac{400\times 10^{3}}{0.925} = 432.432 kJ/kg$

Also,

$\Delta h = h_{o} - h_{i} = c_{p}\Delta T =c_{p}(T_{o} - T_{i}) = 1.005(200 - 20) = 180.9 kJ/kg$

Now, using these values in eqn (1):

$Q = 180.9 + \frac{20^{2}}{2} + 432.432 = 813.33 kW$

Now, rate of heat transfer, q:

q = mQ = $0.925\times 813.33 = 752.33 kW$

4 0

2 years ago

An 800-kg drag racer accelerates from rest to 390 km/hr in 5.8 s. What is the net impulse applied to the racer in the first 5.8

marissa [1.9K]

Answer:

Impulse =14937.9 N

tangential force =14937.9 N

Explanation:

Given that

Mass of car m= 800 kg

initial velocity u=0

Final velocity v=390 km/hr

Final velocity v=108.3 m/s

So change in linear momentum P= m x v

P= 800 x 108.3

P=86640 kg.m/s

We know that impulse force F= P/t

So F= 86640/5.8 N

F=14937.9 N

Impulse force F= 14937.9 N

We know that

v=u + at

108.3 = 0 + a x 5.8

$a=18.66\ m/s^2$

So tangential force F= m x a

F=18.66 x 800

F=14937.9 N

6 0

3 years ago