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Deffense [45]
3 years ago
9

A 600 MW power plant has an efficiency of 36 percent with 15

Engineering
2 answers:
ololo11 [35]3 years ago
8 0

Answer:

401.3 kg/s

Explanation:

The power plant has an efficiency of 36%. This means 64% of the heat form the source (q1) will become waste heat. Of the waste heat, 85% will be taken away by water (qw).

qw = 0.85 * q2

q2 = 0.64 * q1

p = 0.36 * q1

q1 = p /0.36

q2 = 0.64/0.36 * p

qw = 0.85 *0.64/0.36 * p

qw = 0.85 *0.64/0.36 * 600 = 907 MW

In evaporation water becomes vapor absorbing heat without going to the boiling point (similar to how sweating takes heat from the human body)

The latent heat for the vaporization of water is:

SLH = 2.26 MJ/kg

So, to dissipate 907 MW

G = qw * SLH = 907 / 2.26 = 401.3 kg/s

larisa [96]3 years ago
5 0

Answer:

m = 367.753 kg.s

Explanation:

GIVEN DATA:

Power plant capacity W=600 MW

Plant efficiency is \eta = 36\%

efficiency = \frac{W}{QH}

0.36 = \frac{600}{QH}

QH = 1666.6 MW

from first law of thermodynamics we hvae

QH -QR = W

Amount of heat rejection is QR = 1066.66 MW

As per given information we have 15% heat released to atmosphere

QR = 0.15 \tiimes 1066.66 = 159.99 MW

AND 85% to cooling water

Q = 0.85 \times 1066.66 = 906.66 MW

from saturated water table

at temp 150 degree c we have Hfg = 2465.4 kJ/kg

rate of cooiling water is given as  =  mhfg

906.66 \times 1000  KW = m \times 2465.4

m = 367.753 kg.s

where m is rate of makeup water that is added to offset

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adelina 88 [10]
All of the above. Answer.
7 0
3 years ago
What is the primary responsibility of ABET?
nalin [4]

Answer:

C

Explanation:

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3 years ago
A single square-thread screw has an input power of 3 kW at a speed of 1 rev/s. The screw has a diameter of 40 mm and a pitch of
Sonbull [250]

Answer:

Axial Resisting Load, F = 31.24kN

Efficiency = 16.67%

Explanation:

Given

Input Power = P,in = 3kW = 3,000W

Speed, S = 1rev/s

Pitch, p = 8mm

Thread frictional coefficient = μt = 0.18

Collar frictional coefficient = μc = 0.09

Friction radius of collar, Rc = 50mm

First, we calculate the torque while the load is being lifted in terms of 'F'.

This is calculated by

T = ½FDm[1 + πDmμt]/[πDm - μtp]

By substituton.

T = ½F(40-4)[1 + π(40-4)0.18]/[π(40-4) - 0.18 * 8]

T = 18F(1 + 6.48π)/(36π - 1.44)

T = 3.44F.Nmm

T = 3.44 * 10^-3F Nm

Then we calculate the torque due to friction from the collar

T = Fμc * Rc

T = F * 0.09 * 50

T = 4.5F. Nmm

T = 4.5 * 10^-3F Nm

Then, we calculate the axial resisting load 'F' by using the the following power input relation.

P,in = Tw

P,in = (T1 + T2) * 2πN

Substitute each value

3,000 = (3.44 + 4.5) * 10^-3 * F * 10^-3 * 2 * π * 2

F = 3000/((3.44 + 4.5) * 10^-3 * 10^-3 * 2 * π * 2

F = 31,247.69N

F = 31.24kN

Hence, the axial resisting load is

F = 31.24kN

Calculating Efficiency

Efficiency = Fp/2πP

Efficiency = 2Fp/P,in

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Efficiency = 2 * 31,247.69 * 8 * 10^-3/3000

Efficiency = 0.166654346666666

Efficiency = 16.67%

8 0
3 years ago
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Anna11 [10]

Answer:

The answer is "conditionally unstable"

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Classification dependent on ELR:

Larger than 10 \frac{^{\circ}C}{1000} m Around 10 and 6 \frac{^{\circ}C}{1000} m or less 6 \frac{^{\circ}C}{1000} m volatile implicitly unreliable Therefore ELR is implicitly unreliable 9 \frac{^{\circ}C}{1000} m, that's why it is "conditionally unstable".

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