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xeze [42]
3 years ago
10

The town of Mustang, TX is concerned that waste heat discharged from a new up- stream power plant will decimate the minnow popul

ation in its local stream. The power plant is 35% efficient with a rated output of 350 MW, and 75% of the waste heat is discharged into the stream, which flows at 150 m3/s. Before the power plant starts operations, the stream temperature was 20◦C. If the minnows can handle max- imum stream temperatures of 22◦C, will the water be safe at the point of cooling water discharge? (Assume that the waste heat is discharged to the flow of the whole stream.)
Engineering
1 answer:
vitfil [10]3 years ago
7 0

Answer:

Yes the water will be safe at the point of cooling water discharge

Explanation:

Power losses in plant= 350- 350×0.35=227.5MW

Rate of heat rejection to stream= 0.75× 227.5= 170.625MW

Rate of heat rejection= rate of flow of water× c × ΔT

170625000= 150000000× 4.186 × (Final temperature- 20)

Final temperature= 20.3 ◦C

The final temperature of stream will be 20.3 ◦C. Thechange is very small so the minnows will be able to handle this temperature.

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If welding is being done in the vertical position, the torch should have a travel angle of?
siniylev [52]

Answer:

Between 35°– 45°

Explanation:

In the vertical position, Point the flame in the direction of travel. Keep the flame tip at the correct height above the base metal. An angle of 35°–45° should be maintained between the torch tip and the base metal. This angle may be varied up or down to heat or cool the weld pool if it is too narrow or too wide

4 0
2 years ago
A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the mot
Thepotemich [5.8K]

78950W the answer

Explanation:

A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor

A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor

5 0
3 years ago
A house that was heated by electric resistance heaters consumed 1500 kWh of electric
gladu [14]

Answer:

2.5=1500/Whp=> Whp=600 kWh

delWgain=1500-600=900 kWh

Money saved= 900* 6tk*=5400 tk

5 0
3 years ago
A car is traveling at sea level at 78 mi/h on a 4% upgrade before the driver sees a fallen tree in the roadway 150 feet away. Th
Dmitrij [34]

Answer: V = 47.7 mi/hr

Explanation:

first we calculate elements of aero-dynamic resistance

Ka = p/2 * CD * A.f

p is the density of air(0.002378 slugs/ft^3) for zero altitude, CD is the drag coefficient(0.35) and A.f is the front region of the vehicle

so we substitute

Ka = 0.002378/2 * 0.35 * 18

Ka = 0.00749

Now we calculate the final speed of the vehicle (V2) using the relation;

S = (YbW/2gKa)In[ (UW + KaV1^2 + FriW ± Wsinθg) / (UW + KaV2^2 + FriW ± Wsinθg)

so

WE SUBSTITUTE

150 = (1.04 * 2700 / 2 * 32.2 * 0.0075) In [(0.8 * 2700 + 0.0075 *(78mil/hr * 5280ft/1min * 1hr/3600s)^2 + 0.017 * 2700 ± 2700 * 0.04) / (0.8 * 2700 + 0.0075 * V2^2 + 0.017 * 2700 ± 2700 * 0.04)]

150 = (2808/0.483) In [(2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

150 = 5813.66 In [ (2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

divide both sides by 5813.66

0.0258 = In [ (2412.06) / ( 0.0075V2^2 + 2313.9)]

take the e^ of both side

e^0.0258 = (2412.06) / ( 0.0075V2^2 + 2313.9)

1.0261 = (2412.06) / ( 0.0075V2^2 + 2313.9)]

(0.0075V2^2 + 2313.9) = 2412.06 / 1.0261

(0.0075V2^2 + 2313.9) = 2350.7

0.0075V2^2 = 2350.7 - 2313.9

0.0075V2^2 = 36.8

V2^2 = 36.8 / 0.0075

V2^2 = 4906.6666

V2 = √4906.6666

V2 = 70.0476 ft/s

converting to miles per hour

V2 = 70.0476 ft/s * 1 mil / 5280 ft * 3600s / 1hr

V = 47.7 mi/hr

8 0
3 years ago
Write short notes on: (any four) a) Suspended ground floor b) Soil exploration c) Baulking of sand d) Bearing capacity of soil e
vredina [299]

Answer:

a) A suspended floor is a ground floor with a void underneath the structure. The floor can be formed in various ways, using timber joists, precast concrete panels, block and beam system or cast in-situ with reinforced concrete. However, the floor structure is supported by external and internal walls.

b) Soil exploration consists of determining the profile of the natural soil deposits at the site, taking the soil samples and determining the engineering properties of soils using laboratory tests as well as in-situ testing methods

c) Bulking in sand Occurs When dry sand interacts with the atmospheric moisture. Presence of moisture content forms a thin layer around sand particles. This layer generates the force which makes particles to move aside to each other. This results in the increase of the volume of sand.

d) In a nutshell, bearing capacity is the capacity of soil to support the loads that are applied to the ground above. It depends primarily on the type of soil, its shear strength and its density. It also depends on the depth of embedment of the load – the deeper it is founded, the greater the bearing capacity.

Explanation:

<h2>please follow me</h2>
6 0
3 years ago
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