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saw5 [17]
2 years ago
9

HELP PLEASE

Engineering
1 answer:
Sveta_85 [38]2 years ago
8 0

..........23÷357=0.0644257703........

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On a cold winter day, wind at 55 km/hr is blowing parallel to a 4-m high and 10-m long wall of a house. If the air outside is at
koban [17]

Answer:

16.21 kW

Explanation:

Solution

Given that,

The velocity of wind = 55 km/hr

The length of the wall L = 10m

The height of the wall w = 4m

The surface temperature at wall Ts = 12° C

Temperature of air T∞ = 5°C

Now,

The properties  of the air at atm and average film temperature =( 12 + 5)/2 = 8.5°C, which is taken from the air table properties.

k= 0.02428 W/m°C

v= 1.413 *10 ^⁻5

Pr =0.7340

Now,

Recall Reynolds number when air flow parallel to 10 m side

[ 55 * 1000/3600) m/s (10 m)/1.413 *10^⁻5 m²/s

Rel =1.081 * 10⁷

This value is greater than Reynolds number.

The nusselt number is computed as follows:

Nu =hL/k

(0.037Rel^0.08 - 871)Pr^1/3

Nu =1.336 * 10 ^4

The heat transfer coefficient is

h = k/L Nu

= 0.2428 W/m°C /10 m (1.336 * 10 ^4)

h = 32.43 W/m°C

The heat transfer area of surface,

As = 40 m²

= ( 4 m) (10 m)

As = 40 m²

The rate of heat transfer is determined as follows:

Q = hAs( Ts - T∞)

= (32.43 W/m²°C) (40 m) (12 - 5)°C

=9081 W

Q = 9.08 kW

When the velocity is doubled,

let say V = 110km/hr

The Reynolds number is

Rel = VL/v

= [110 * 100/3600) m/s] (10 m)/ 1.413 *10^⁻5 m²/s

Rel = 2.163 * 10 ^7

This value is greater for critical Reynolds number

The nusselt number is computed as follows:

Nu =hL/k

(0.037Rel^0.08 - 871)Pr^1/3

[0.037 ( 2.163 * 10 ^7)^0.08 - 871] (0.7340)^1/3

Nu =2.344 * 10^4

The heat transfer coefficient is

h = k/L Nu

= 0.2428 W/m°C /10 m (2.384 * 10 ^4)

h= 57.88 W/m²°C

The heat transfer area of surface,

As =  wL

= ( 10 m) (4 m)

As = 40 m²

he rate of heat transfer is determined as follows:

Q = hAs( Ts - T∞)

= (57.88 W/m²°C) (40 m²) (12 - 5)°C

= 16,207 W

= 16.21 kW

3 0
3 years ago
One employee was climbing a metal ladder to hand an electric drill to the journeyman installer on a scaffold about five feet abo
lbvjy [14]

Answer:

ACCIDENT PREVENTION RECOMMENDATIONS  

Fatal Fact made us to understand some of the prevention techniques as stated below.

1. Use approved ground fault circuit interrupters or an assured equipment grounding conductor program to protect employees on construction sites.  

2. Use equipment that provides a permanent and continuous path from circuits, equipment, structures, conduit or enclosures to ground.

3. Inspect electrical tools and equipment daily and remove damaged or defective equipment from use until it is repaired.

Explanation:

In order to gain a better understanding of the answer above let explain some terms

Ground Fault Circuit Interrupters :

    A ground fault circuit interrupter (GFCI), or Residual Current Device (RCD) is a type of circuit breaker which shuts off electric power when it senses an imbalance between the outgoing and incoming current. The main purpose is to protect people from an electric shock caused when some of the current travels through a person's body due to an electrical fault such as a short circuit, insulation failure, or equipment malfunction.

So the first statement is implying that in order to prevent this accident that  this device (GFCI) should have  been used in that  construction site, or as  an alternative before the  construction commenced  the company should have drafted a lay down conductor program(i.e. a step by step conductor program) that assured equipment grounding  in order to protect employees on construction sites

5 0
3 years ago
You are in charge of ordering the concrete for a basement wall concrete pour. The wall forms are all set up and ready. The wall
choli [55]

Answer:

189.15cy

Explanation:

To understand this problem we need to understand as well the form.

It is clear that there is four wall, two short and two long.

The two long are \rightarrow 120ft5in+2(10ft)

The two long are \rightarrow 122ft1in=122.08ft

The two shors are \rightarrow 86ft4.5in = 86.375ft

The height and the thickness are 14ft and 0.83ft respectively.

So we only calculate the Quantity of concrete,

Q_c = [(2*122.08)+(2*86-375)]*14*0.833\\Q_c=4864.02ft^3

That in cubic yards is equal to 180.15 (1cy=27ft^3)

Hence, we need order 5% plus that represent with the quantity

Q_{ordered}=1.05*180.15=189.15cy

8 0
3 years ago
Which type of modeling can create virtual designs that can save clients thousands of dollars?
swat32

Answer:

VR Prototyping

VR Prototyping Can Save you Thousands of Dollars.

Explanation:

there you go lad

8 0
2 years ago
What are the success factors for mechanical engineering?
Eva8 [605]

Answer:

-effective technical skills.

-the ability to work under pressure.

-problem-solving skills.

-creativity.

-interpersonal skills.

-verbal and written communication skills.

-commercial awareness.

-teamworking skills.

Explanation:

is this what ur looking for? if so there ya go lol

7 0
2 years ago
Read 2 more answers
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