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Maksim231197 [3]
2 years ago
11

List and explain the major features that the following building designs must have to relate directly to their functions.

Engineering
1 answer:
Lunna [17]2 years ago
7 0

Answer:

size, design

Explanation:

''.''

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A coal-burning power plant generates electrical power at a rate of 650 megawatts (MW), or 6.50 × 108 J/s. The plant has an overa
Vinvika [58]

Answer:

Energy produce in one year =20.49 x 10¹⁶ J/year

Explanation:

Given that

Plant produce 6.50 × 10⁸ J/s of energy.

It produce  6.50 × 10⁸ J in 1 s.

We know that

1 year = 365 days

1 days = 24 hr

1 hr = 3600 s

1 year = 365 x 24 x 3600 s

1 year = 31536000 s

So energy produce in 1 year = 31536000 x  6.50 × 10⁸ J/year

          Energy produce in one year = 204984 x 10¹² J/year

          Energy produce in one year =20.49 x 10¹⁶ J/year

7 0
3 years ago
A surveyor knows an elevation at Catch Basin to be elev=2156.77 ft. The surveyor takes a BS=2.67 ft on a rod at BM Catch Basin a
fenix001 [56]

Answer:

the elevation at point X is 2152.72 ft

Explanation:

given data

elev = 2156.77 ft

BS = 2.67 ft

FS = 6.72 ft

solution

first we get here height of instrument that is

H.I = elev + BS   ..............1

put here value

H.I =  2156.77 ft + 2.67 ft  

H.I = 2159.44 ft

and

Elevation at point (x) will be

point (x)  = H.I - FS   .............2

point (x)  = 2159.44 ft  - 6.72 ft

point (x)  = 2152.72 ft

3 0
3 years ago
The Stefan-Boltzmann law can be employed to estimate the rate of radiation of energy H from a surface, as in
Mazyrski [523]

Explanation:

A.

H = Aeσ^4

Using the stefan Boltzmann law

When we differentiate

dH/dT = 4AeσT³

dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³

= 8.4085

Exact error = 8.4085x20

= 168.17

H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴

= 1366.376watts

B.

Verifying values

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴

= 1542.468

H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴

= 1205.8104

Error = 1542.468-1205.8104/2

= 168.329

ΔT = 40

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴

= 1735.05

H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴

= 1735.05-1059.83/2

= 675.22/2

= 337.61

5 0
3 years ago
Water is the working fluid in an ideal Rankine cycle. Saturated vapor enters the turbine at 12 MPa, and the condenser pressure i
Brilliant_brown [7]

Answer:

\dot Q_{in} = 372.239\,MW

Explanation:

The water enters to the pump as saturated liquid and equation is modelled after the First Law of Thermodynamics:

w_{in} + h_{in}- h_{out} = 0

h_{out} = w_{in}+h_{in}

h_{out} = 12\,\frac{kJ}{kg} + 191.81\,\frac{kJ}{kg}

h_{out} = 203.81\,\frac{kJ}{kg}

The boiler heats the water to the state of saturated vapor, whose specific enthalpy is:

h_{out} = 2685.4\,\frac{kJ}{kg}

The rate of heat transfer in the boiler is:

\dot Q_{in} = \left(150\,\frac{kg}{s}\right)\cdot \left(2685.4\,\frac{kJ}{kg}-203.81\,\frac{kJ}{kg} \right)\cdot \left(\frac{1\,MW}{1000\,kW} \right)

\dot Q_{in} = 372.239\,MW

3 0
3 years ago
Read 2 more answers
A European car manufacturer reports that the fuel efficiency of the new MicroCar is 48.5 km/L highway and 42.0 km/L city. What a
statuscvo [17]

Answer:

Fuel efficiency for highway = 114.08 miles/gallon

Fuel efficiency for city = 98.79 miles/gallon

Explanation:

1 gallon = 3.7854 litres

1 mile = 1.6093 km

Let's first convert the efficiency to km/gallon:

48.5 km/litre = (48.5 * 3.7854) km/gallon

48.5 km/litre =  183.5919 km/gallon (highway)

42.0 km/litre = (42.0 * 3.7854) km/gallon

42.0 km/litre = 158.9868 km/gallon (city)

Next, we convert these to miles/gallon:

183.5919 km/gallon = (183.5919 / 1.6093) miles/gallon

183.5919 km/gallon = 114.08 miles/gallon (highway)

158.9868 km/gallon = (158.9868 /1.6093) miles/gallon

158.9868 km/gallon = 98.79 miles/gallon (city)

3 0
3 years ago
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