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snow_lady [41]
2 years ago
8

A magician claims that he/she has invented a novel, super-fantastic heat engine. This engine operates between two reservoirs of

temperatures 250 K and 750 K, respectively. Please verify and explain if this is possible for any of the following QH (heat received from the hot reservoir), QL (heat rejected to the cool reservoir), and W (mechanical work produced):
a. (3a). (3p) QH= 900 kJ, Wmech = 400 kJ, QL = 600 kJ.
b. (3b). (3p) Qu= 900 kJ, W mech = 400 kJ, QL = 500 kJ.
c. (3c). (3p) Qh= 900 kJ, Wmech = 600 kJ, QL = 300 kJ.
d. (3b). (3p) Qh= 900 kJ, Wmech = 800 kJ, QL = 100 kJ.
Engineering
1 answer:
Marysya12 [62]2 years ago
7 0

Answer:

A) Not possible, B) Posible, C) Possible, D) Not possible.

Explanation:

The maximum theoretical efficiency for any thermal engine is defined by Carnot's cycle, whose energy efficiency (\eta), no unit, is expressed below:

\eta = 1-\frac{T_{L}}{T_{H}} (1)

Where:

T_{L} - Cold reservoir temperature, in Kelvin.

T_{H} - Hot reservoir temperature, in Kelvin.

If we know that T_{L} = 250\,K and T_{H} = 750\,K, then the maximum theoretical efficiency for the thermal engine is:

\eta = 1-\frac{T_{L}}{T_{H}}

\eta = 0.667

For real thermal engines, the following inequation is observed:

0 \le \eta_{r} \le \eta (2)

Where \eta_{r} is the efficiency of the real heat engine, no unit.

There are two possible criteria to determine if a given heat engine is real:

Efficiency

\eta_{r} = 1 - \frac{Q_{L}}{Q_{H}} (3)

Where:

Q_{L} - Heat rejected to the cold reservoir, in kilojoules.

Q_{H} - Heat received from the hot reservoir, in kilojoules.

Power output

W = Q_{H}-Q_{L} (4)

Where W is the power output, in kilojoules.

Now we proceed to verify each case:

A) Q_{H} = 900\,kJ, Q_{L} = 600\,kJ, W_{m} = 400\,kJ

\eta_{r} = 0.333

0 \le \eta_{r} \le \eta

W = 300\,kJ

W \ne W_{m}

This engine is not possible.

B) Q_{H} = 900\,kJ, Q_{L} = 500\,kJ, W_{m} = 400\,kJ

\eta_{r} = 0.444

0 \le \eta_{r} \le \eta

W = 400\,kJ

W = W_{m}

The engine is possible.

C) Q_{H} = 900\,kJ, Q_{L} = 300\,kJ, W_{m} = 600\,kJ

\eta_{r} = 0.667

0 \le \eta_{r} \le \eta

W = 600\,kJ

W = W_{m}

The engine is possible.

D) Q_{H} = 900\,kJ, Q_{L} = 100\,kJ, W_{m} = 800\,kJ

\eta_{r} = 0.889

\eta_{r} > \eta

W = 800\,kJ

W = W_{m}

The engine is possible.

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Answer:

  C. Get the names and addresses of witness to the crash

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The best approach is to let your insurance company handle the dispute. Since that is not an option here, the best thing you can do is make sure you know who the witnesses are, so your insurance company can call upon them as needed.

8 0
3 years ago
Water is boiled in a pot covered with a loosely fitting lid at a location where the pressure is 85.4 kPa. A 2.61 kW resistance h
eimsori [14]

Answer:

t = 6179.1 s = 102.9 min = 1.7 h

Explanation:

The energy provided by the resistance heater must be equal to the energy required to boil the water:

E = ΔQ

ηPt = mH

where.

η = efficiency = 84.5 % = 0.845

P = Power = 2.61 KW = 2610 W

t = time = ?

m = mass of water = 6.03 kg

H = Latent heat of vaporization of water = 2.26 x 10⁶ J/kg

Therefore,

(0.845)(2610 W)t = (6.03 kg)(2.26 x 10⁶ J/kg)

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4 0
2 years ago
Which of the following best describes the role of engineers
Fantom [35]

Problem Solvers

Explanation:

Engineers find problems in the world, and then they find solutions for them.

8 0
3 years ago
Identify factors that can cause a process to become out of control. Give several examples of such factors.
Oliga [24]

Answer:

Explained

Explanation:

This situation can occur because of various factors such as:

  • Gradual deterioration of lubrication and coolant.
  • change of environmental condition such as temperature, humidity, moisture, etc.
  • Change in the properties of incoming raw material
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4 0
3 years ago
Steam at 40 bar and 500o C enters the first-stage turbine with a volumetric flow rate of 90 m3 /min. Steam exits the turbine at
a_sh-v [17]

Answer:

(a) 62460 kg/hr

(b) 17,572.95 kW

(c) 3,814.57 kW

Explanation:

Volumetric flow rate, G = 30 m³ / 1 min => 90 / 60 => 1.5

Calculate for h₁ , h₂ , h₃

h₁ is h at P = 40 bar, 500°C => 3445.84 KJ/Kg

Specific volume steam, ц = 0.086441 m³kg⁻¹

h₂ is h at P = 20 bar, 400°C => 3248.23 KJ/Kg

h₃ is h at P = 20 bar, 500°C => 3468.09 KJ/Kg

h₄ is hg at P = 0.6 bar from saturated water table => 2652.85 KJ/Kg

a)

Mass flow rate of the steam, m = G / ц

m = 1.5 / 0.086441

m = 17.35 kg/s

mass per hour is m = 62460 kg/hr

b)

Total Power produced by two stages

= m (h₁ - h₂) + m (h₃ - h₁)

= m [(3445.84 - 3248.23) + (3468.09 - 2652.85)]

= m [ 197.61 + 815.24 ]

= 17.35 [1012.85]

= 17,572.95 kW

c)

Rate of heat transfer to the steam through reheater

= m (h₃ - h₂)

= 17.35 x (3468.09 - 3248.23)

= 17.35 x 219.86

= 3,814.57 kW

8 0
3 years ago
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