All categories

3 years ago

What is differences Between hard shoulder & soft shoulder in civil Engineerin?

Engineering

1 answer:

r-ruslan [8.4K]3 years ago

5 0

Answer:

The 'shoulder' of a road is the land to the edge of the road. On most roads without pavements, the shoulder is a strip of grass or a hedgerow. This is known as a 'soft shoulder'. On a motorway, this strip of land is hardstanding, hence the name 'hard shoulder.'

Mark as brilliant........... 

You might be interested in

NASA SPACE SHUTTLE QUESTION:

vfiekz [6]

Explanation:

The Challenger accident made them aware of the risks. They had been a little naive, since it was never believed that such a thing could happen. This, together with Space Shuttle Columbia disaster revealed the risks of the space shuttle program, which also had very high maintenance costs.

8 0

2 years ago

A cartridge electrical heater is shaped as a cylinder of length L=200mm and outer diameter D=20 mm. Under normal operating condi

lara31 [8.8K]

Answer:

T(water)=50.32℃

T(air)=3052.6℃

Explanation:

Hello!

To solve this problem we must use the equation that defines the transfer of heat by convection, which consists of the transport of heat through fluids in this case water and air.

The equation is as follows!

$Q=ha(Ts-T\alpha )$

Q = heat

h = heat transfer coefficient

Ts = surface temperature

T = fluid temperature

a = heat transfer area

The surface area of a cylinder is calculated as follows

$a=\pi D(\frac{D}{2} +L)$

Where

D=diameter=20mm=0.02m

L=leght=200mm)0.2m

solving

$a=\pi (0.02)(\frac{0.02}{2} +0.2)=0.01319m^2$

For water

Q=2Kw=2000W

h=5000W/m2K

a=0.01319m^2

Tα=20C

$Q=ha(Ts-T\alpha )$

solving for ts

$Ts=T\alpha +\frac{Q}{ha}$

$Ts=20+\frac{2000}{(0.01319)(5000)} =50.32C$

for air

Q=2Kw=2000W

h=50W/m2K

a=0.01319m^2

Tα=20C

$Ts=20+\frac{2000}{(0.01319)(50)}=3052.6C$

3 0

2 years ago

The yield of a chemical process is being studied.The two most important variables are thought to be the pressure and the tempera

anyanavicka [17]

Answer:

note:

solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment

Download docx

7 0

2 years ago

: The interior wall of a furnace is maintained at a temperature of 900 0C. The wall is 60 cm thick, 1 m wide, 1.5 m broad of mat

Snowcat [4.5K]

Answer:

Heat is lost at the rate of 750 J/s or W

The thermal resistance is 1 K/W

Explanation:

interior temperature $T_{2}$ = 900 °C

wall thickness t = 60 cm = 0.6 m

width = 1 m

breadth = 1.5 m

thermal conductivity k = 0.4 W/m-K

outside temperature $T_{1}$ = 150 °C

heat through the wall = ?

The area of the wall A = w x b = 1 x 1.5 = 1.5 m^2

Temperature difference $dt$ = $T_{2}$ - $T_{1}$ = 900 - 150 = 750 °C

note that $dt$ is also equal to 750 K since to convert from °C to K we'll have to add 273 to both temperature, which will still cancel out when we subtract the two temperatures.

To get the heat that escapes through the wall, we use the equation

Q = Ak $\frac{dt}{t}$