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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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Answer:

Sarah is asking each department head how long they can be without their primary system. Sarah is trying to determine the Recovery Time Objective (RTO) as this is the duration of time within which the primary system must be restored after the disruption.

Recovery Point Objective is basically to determine the age of restoration or recovery point.

Business recovery and technical recovery requirements are to assess the requirements to recover by Business or technically.

Hence, Recovery Time Objective (RTO) is the correct answer.

8 0
2 years ago
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2 years ago
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2 years ago
10. An engineer is designing a total hip implant. She intends to make the femoral stem out of titanium because it forms a good i
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8 0
3 years ago
A fluid of density 900 kg/m3 passes through a converging section of an upstream diameter of 50 mm and a downstream diameter of 2
NISA [10]

Answer:

Q= 4.6 × 10⁻³ m³/s

actual velocity will be equal to 8.39 m/s

Explanation:

density of fluid = 900 kg/m³

d₁ = 0.025 m

d₂ = 0.05 m

Δ P = -40 k N/m²

C v = 0.89

using energy equation

\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g} = \dfrac{P_2}{\gamma}+\dfrac{v_2^2}{2g}\\\dfrac{P_1-P_2}{\gamma}=\dfrac{v_2^2-v_1^2}{2g}\\\dfrac{-40\times 10^3\times 2}{900}=v_2^2-v_1^2

under ideal condition v₁² = 0

v₂² = 88.88

v₂ = 9.43 m/s

hence discharge at downstream will be

Q = Av

Q = \dfrac{\pi}{4}d_1^2 \times v

Q = \dfrac{\pi}{4}0.025^2 \times 9.43

Q= 4.6 × 10⁻³ m³/s

we know that

C_v =\dfrac{actual\ velocity}{theoretical\ velocity }\\0.89 =\dfrac{actual\ velocity}{9.43}\\actual\ velocity = 8.39m/s

hence , actual velocity will be equal to 8.39 m/s

6 0
2 years ago
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