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Ilia_Sergeevich [38]
3 years ago
10

An HVAC contracting company sent out three work crews to complete three

Engineering
1 answer:
Doss [256]3 years ago
6 0

Answer: 156 hours

Explanation:

Crew One worked 10-, 9-, 11-, 12-, and 9-hours for the week. The total hours worked by Crew One will be:

= 10 + 9 + 11 + 12 + 9

= 51 hours.

Crew Two worked 9, 12, 12, 9 and 9 hours. The total hours worked by Crew Two for the week will be:

= 9 + 12 + 12 + 9 + 9

= 51 hours

Crew Three worked 12-, 12-, 10-, 9-, and 11-hour daily. The total hours worked by Crew Three for the week will be:

= 12 + 12 + 10 + 9 + 11

= 54 hours

The total hours worked by the three crews for the week will be:

= 51 hours + 51 hours + 54 hours

= 156 hours

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What does WCS stand for? A. Western CAD System B. Worldwide Coordinate Sectors C. World Coordinate System D. Wrong CAD Settings
Ray Of Light [21]

Answer:

The correct answer is C. World Coordinate System

Explanation:

The World Coordinate System has to do with that coordinate system which is fixed in the activities of the CADing. There is a default system in which we can refer to them as soon as we want to manipulate the objects and add new elements.

7 0
3 years ago
Joinn my zo om lets play some blookets<br> 98867 708157<br> 9dPQPW
pav-90 [236]

Answer:k

Explanation:

6 0
2 years ago
Air is contained in a vertical piston–cylinder assembly such that the piston is in static equilibrium. The atmosphere exerts a p
oee [108]

Answer:

a) 24 kg

b) 32 kg

Explanation:

The gauge pressure is of the gas is equal to the weight of the piston divided by its area:

p = P / A

p = m * g / (π/4 * d^2)

Rearranging

p * (π/4 * d^2) = m * g

m = p * (π/4 * d^2) / g

m = 1200 * (π/4 * 0.5^2) / 9.81 = 24 kg

After the weight is added the gauge pressure is 2.8kPa

The mass of piston plus addded weight is

m2 = 2800 * (π/4 * 0.5^2) / 9.81 = 56 kg

56 - 24 = 32 kg

The mass of the added weight is 32 kg.

5 0
2 years ago
How does the two-stroke Otto cycle differ from the four-stroke Otto cycle?
Digiron [165]

Answer:

Two stroke cycle                                               Four stroke cycle

1.Have on power stroke in one revolution.   1.have one power  

                                                                   stroke in two  revolution                                                                            

2.Complete the cycle in 2 stroke                 2.Complete the cycle in 4 stroke    

3.It have ports                                                3.It have vales

                                                                         

4.Greater requirement of cooling              4.Lesser requirement of cooling  

5.Less thermal efficiency                            5.High thermal efficiency

6.Less volumetric efficiency                       6.High volumetric efficiency    

7.Size of flywheel is less.                           7.Size of flywheel is more.

3 0
2 years ago
A stationary gas-turbine power plant operates on a simple ideal Brayton cycle with air as the working fluid. The air enters the
ololo11 [35]

Answer:

A) W' = 15680 KW

B) W' = 17113.87 KW

Explanation:

We are given;

Temperature at state 1; T1 = 290 K

Temperature at state 3; T3 = 1100 K

Rate of heat transfer; Q_in = 35000 kJ/s = 35000 Kw

Pressure of air into compressor; P_c = 95 kPa

Pressure of air into turbine; P_t = 760 kPa

A) The power assuming constant specific heats at room temperature is gotten from;

W' = [1 - ((T4 - T1)/(T3 - T2))] × Q_in

Now, we don't have T4 and T2 but they can be gotten from;

T4 = [T3 × (r_p)^((1 - k)/k)]

T2 = [T1 × (r_p)^((k - 1)/k)]

r_p = P_t/P_c

r_p = 760/95

r_p = 8

Also,k which is specific heat capacity of air has a constant value of 1.4

Thus;

Plugging in the relevant values, we have;

T4 = [(1100 × (8^((1 - 1.4)/1.4)]

T4 = 607.25 K

T2 = [290 × (8^((1.4 - 1)/1.4)]

T2 = 525.32 K

Thus;

W' = [1 - ((607.25 - 290)/(1100 - 525.32))] × 35000

W' = 0.448 × 35000

W' = 15680 KW

B) The power accounting for the variation of specific heats with temperature is given by;

W' = [1 - ((h4 - h1)/(h3 - h2))] × Q_in

From the table attached, we have the following;

At temperature of 607.25 K and by interpolation; h4 = 614.64 KJ/K

At T3 = 1100 K, h3 = 1161.07 KJ/K

At T1 = 290 K, h1 = 290.16 KJ/K

At T2 = 525.32 K, and by interpolation, h2 = 526.12 KJ/K

Thus;

W' = [1 - ((614.64 - 290.16)/(1161.07 - 526.12))] × 35000

W' = 17113.87 KW

4 0
2 years ago
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