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vivado [14]
2 years ago
7

Accidents occur as a result of ____ and ____.

Engineering
1 answer:
telo118 [61]2 years ago
5 0

BRIGHT HEADLIGHTS

AND SEVERE WEATHER CONDITIONS

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Answer:

enables the representation, analysis and communication of various aspects of an information system. These aspects correspond to varying and incomplete views of information systems and the processes therein.

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Can you guys please introduce yourself​
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Answer: why?

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Steam enters a turbine at 120 bar, 508oC. At the exit, the pressure and quality are 50 kPa and 0.912, respectively. Determine th
Archy [21]

Answer:

The turbine produces <u>955.53 KW</u> power.

Explanation:

Taking the turbine as a system. Applying Law of Conservation of Energy:

m(h₁ - h₂) - Heat Loss = P

where,

m = mass flow rate of steam = 1.31 kg/s

h₁ = enthalpy at state 1 (120 bar and 508°C)

h₂ = enthalpy at state 2 (50 KPa and x = 0.912)

Heat Loss = 225 KW

P = Power generated by turbine = ?

First, we find h₁ from super steam tables:

At,

T = 508°C

P = 120 bar = 12000 KPa = 12 MPa

we find that steam is in super-heated state with enthalpy:

Due to unavailibility of values in chart we approximate the state to 500° C and 12.5 MPa:

h₁ = 3343.6 KJ/kg

Now, for state 2, we have:

P = 50 KPa and x = 0.912

From saturated steam table:

h₂ = hf₂ + x(hfg₂) = 340.54 KJ/kg + (0.912)(2304.7 KJ/kg)

h₂ = 2442.4 KJ/kg

Now, using values in the conservation equation:

(1.31 kg/s)(3343.6 KJ/kg - 2442.4 KJ/kg) - 225 KW = P

<u>P = 955.53 KW</u>

5 0
3 years ago
A sample of sand weighs 490 g in stock and 475 in Oven Dry (OD) condition, respectively. If absorption capability of the sand is
Ivahew [28]

The weight of the specimen in SSD condition is 373.3 cc

<u>Explanation</u>:

a) Apparent specific gravity = \frac{A}{A-C}

Where,

A = mass of oven dried test sample in air = 1034 g

B = saturated surface test sample in air = 1048.9 g

C = apparent mass of saturated test sample in water = 975.6 g

apparent specific gravity = \frac{A}{A-C}

                                         = \frac{1034}{1034-675 \cdot 6}

Apparent specific gravity = 2.88

b) Bulk specific gravity G_{B}^{O D}=\frac{A}{B-C}

G_{B}^{O D}=\frac{1034}{1048.9-675 \cdot 6}

       =  2.76

c) Bulk specific gravity (SSD):

G_{B}^{S S D}=\frac{B}{B-C}

=\frac{1048 \cdot 9}{1048 \cdot 9-675 \cdot 6}

G_{B}^{S S D} = 2.80

d) Absorption% :

=\frac{B-A}{A} \times 100 \%

=\frac{1048 \cdot 9-1034}{1034} \times 100

Absorption = 1.44 %

e) Bulk Volume :

v_{b}=\frac{\text { weight of dispaced water }}{P \omega t}

=\frac{1048 \cdot 9-675 \cdot 6}{1}

= 373.3 cc

5 0
2 years ago
The boy in the wagon begins throwing bricks out of the wagon to simulate rocket propulsion. The wagon begins at rest, and the bo
Digiron [165]

Q:What velocity does the boy attain if he throws the bricks one at a time?

Answer:Linear velocity since it moves back and firth and does not rotate like angular velocity.

5 0
3 years ago
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