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iragen [17]
3 years ago
14

Determine the nature of the following cycle (reversible, irreversible, or impossible): a refrigeration cycle draws heat from a c

old reservoir at 250 K and rejects 950 KJ to a hot reservoir at 300 K while receiving 70 kJ of work to operate. Draw a schematic of the cycle clearly indicating the hot and cold reservoir and the direction of heat and work transfers. (10 pts.)
Engineering
1 answer:
vlabodo [156]3 years ago
6 0

Answer:

Impossible.

Explanation:

The ideal Coefficient of Performance is:

COP_{i} = \frac{250\,K}{300\,K-250\,K}

COP_{i} = 5

The real Coefficient of Performance is:

COP_{r} = \frac{950\,kJ-70\,kJ}{70\,kJ}

COP_{r} = 12.571

Which leads to an absurds, since the real Coefficient of Performance must be equal to or lesser than ideal Coefficient of Performance. Then, the cycle is impossible, since it violates the Second Law of Thermodynamics.

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The enforcement of OSHA standards is provided by federal and state
Gnoma [55]

Answer:

Explanation:

Enforcing OSHA, Occupational Safety and Health Administration, standards is not a job for electricians, lawmakers or tax collectors. The right answer is safety inspectors.

3 0
2 years ago
Air at 26 kPa, 230 K, and 220 rn/s enters a turbojet engine in flight. The air mass flow rate is 25 kg/s. The compressor pressur
Paha777 [63]

Answer:

Explanation:

Answer:

Explanation:

Answer:  

Explanation:  

This is a little lengthy and tricky, but nevertheless i would give a step by step analysis to make this as simple as possible.  

(a). here we are asked to determine the Temperature and Pressure.  

Given that the properties of Air;  

ha = 230.02 KJ/Kg  

Ta = 230 K  

Pra = 0.5477  

From the energy balance equation for a diffuser;  

ha + Va²/2 = h₁ + V₁²/2  

h₁ = ha + Va²/2 (where V₁²/2 = 0)  

h₁ = 230.02 + 220²/2 ˣ 1/10³  

h₁ = 254.22 KJ/Kg  

⇒ now we obtain the properties of air at h₁ = 254.22 KJ/Kg  

from this we have;  

Pr₁ = 0.7329 + (0.8405 - 0.7329)[(254.22 - 250.05) / (260.09 - 250.05)]  

Pr₁ = 0.77759  

therefore T₁ = 254.15K  

P₁ = (Pr₁/Pra)Pa  

= 0.77759/0.5477 ˣ 26  

P₁ = 36.91 kPa  

now we calculate Pr₂  

Pr₂ = Pr₁ (P₂/P₁) = 0.77759 ˣ 11 = 8.55349  

⇒ now we obtain properties of air at  

Pr₂ = 8.55349 and h₂ = 505.387 KJ/Kg  

calculating the enthalpy of air at state 2  

ηc = h₁ - h₂ / h₁ - h₂  

0.85 = 254.22 - 505.387 / 254.22 - h₂  

h₂ = 549.71 KJ/Kg  

to obtain the properties of air at h₂ = 549.71 KJ/Kg  

T₂ = 545.15 K

⇒ to calculate the pressure of air at state 2

P₂/P₁ = 11

P₂ = 11 ˣ 36.913  

p₂ = 406.043 kPa

but pressure of air at state 3 is the same,

i.e. P₂ = P₃ = 406.043 kPa

P₃ = 406.043 kPa

To obtain the properties of air at  

T₃ = 1400 K, h₃ = 1515.42 kJ/Kg and Pr = 450.5

for cases of turbojet engine,

we have that work output from turbine = work input to the compressor

Wt = Wr

(h₃ - h₄) = (h₂ - h₁)

h₄ = h₃ - h₂ + h₁  

= 1515.42 - 549.71 + 254.22

h₄ = 1219.93 kJ/Kg

properties of air at h₄ = 1219.93 kJ/Kg

T₄ = 1140 + (1160 - 1140) [(1219.93 - 1207.57) / (1230.92 - 1207.57)]

T₄ = 1150.58 K

Pr₄ = 193.1 + (207.2 - 193.1) [(1219.93 - 1207.57) / (1230.92 - 1207.57)]

Pr₄ = 200.5636

Calculating the ideal enthalpy of the air at state 4;

Лr = h₃ - h₄ / h₃ - h₄*

0.9 = 1515.42 - 1219.93 / 1515.42 - h₄  

h₄* = 1187.09 kJ/Kg

now to obtain the properties of air at h₄⁻ = 1187.09 kJ/Kg

P₄* = 179.7 + (193.1 - 179.7) [(1187.09 -1184.28) / (1207.57 - 1184.28)]

P₄* = 181.316

P₄ = (Pr₄/Pr₃)P₃       i.e. 3-4 isentropic process

P₄ = 181.316/450.5 * 406.043

P₄ = 163.42 kPa

For the 4-5 process;

Pr₅ = (P₅/P₄)Pr₄

Pr₅ = 26/163.42 * 200.56 = 31.9095

to obtain the properties of air at Pr₅ = 31.9095

h₅= 724.04 + (734.82 - 724.04) [(31.9095 - 3038) / (32.02 - 30.38)]

h₅ = 734.09 KJ/Kg

T₅ = 710 + (720 - 710) [(31.9095 - 3038) / (32.02 - 30.38)]

T₅ = 719.32 K

(b) Now we are asked to calculate the rate of heat addition to the air passing through the combustor;

QH = m(h₃-h₂)

QH = 25(1515.42 - 549.71)

QH = 24142.75 kW

(c). To calculate the velocity at the nozzle exit;

we apply steady energy equation of a flow to nozzle

h₄ + V₄²/2 = h₅ + V₅²/2

h₄  + 0  = h₅₅ + V₅²/2

1219.9 ˣ 10³ = 734.09 ˣ 10³ + V₅²/2

therefore, V₅ = 985.74 m/s

cheers i hope this helps

6 0
4 years ago
Liquefied Natural Gas (LNG) is a natural gas in its liquid form that is clear, colorless, odorless, non-corrosive, and non-toxic
ZanzabumX [31]

Liquefied Natural Gas (LNG) can be defined as a natural gas which in liquid form appear clear and colorless. It is odorless, non-toxic, and non-corrosive. Therefore, the given statement is A) True.

  • LNG or Liquified Natural Gas is a fossil fuel that is produced after the compression of organic matter in the form of algae and phytoplankton.
  • LNG consists of 95% methane gas.
  • The combustion of LNG produces carbon dioxide and water vapors.
  • It burns with a least pollution thus called as cleanest fossil fuel.
  • The liquefaction of the natural gas takes place at -160 degree Celsius. The liquefaction of the gas causes it to transport easily in gas tanks.
  • LNG is colorless, and clear.
  • LNG does not possess any smell and it is non-corrosive to metallic tanks.
  • LNG is also non-toxic.

Learn more about natural gas:

brainly.com/question/12200462

6 0
3 years ago
Thermosets burn upon heating. a)-True b)- false?
hram777 [196]

Answer:

true

Explanation:

True, there are several types of polymers, thermoplastics, thermosets and elastomers.

Thermosets are characterized by having a reticulated structure, so they have low elasticity and cannot be stretched when heated.

Because of the above, thermosetting polymers burn when heated.

5 0
3 years ago
An inductor and resistor are connected in parallel to a 120-V, 60-Hz line. The resistor has a resistance of 50 ohms, and the ind
BARSIC [14]

Answer:

hi

Explanation:

the answer would be I dont know

6 0
3 years ago
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