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GrogVix [38]
3 years ago
10

A commercial refrigerator with refrigerant -134a as the working fluid is used to keep the refrigerated space at -30C by rejectin

g its waste heat to cooling water that enters the condenser at 18C at a rate of 0.25 kg/s and leaves at 26C. The refrigerant enters the condenser at 1.2 MPa and 65C and leaves at 42C. The inlet state of the compressor is 60 kPa and -34C and the compressor is estimated to gain a net heat of 450 W from the surroundings. Determine
a. the quality of the refrigerant at the evaporator inlet,



b.the refrigeration load,



c. the COP of the refrigerator, and



d. the theoretical maximum refrigeration load for the same power input to the compressor.

Engineering
Bahaa
2 years ago
full answer plz
3 answers:
Leto [7]3 years ago
7 0

Answer:

a. the quality of the refrigerant at the evaporator inlet, x = 0.487

b.the refrigeration load, 5.46 kw

c. the COP of the refrigerator, 2.24

d. maximum refrigeration load 12.43 kw

Explanation:

Mariana [72]3 years ago
6 0

Answer:

a) 0.487

b) refrigeration load = 5.46w

c) cop = 2.24

d)ref load max = 12.43kw

Explanation:

Bahaa2 years ago
0 0

A commercial refrigerator with refrigerant-134a as the working fluid is used to
keep the refrigerated space at -30°C by rejecting its waste heat to cooling water
that enters the condenser at 18°C at a rate of 0.25 kg/s and leaves at 26°C. The
refrigerant enters the condenser at 1.2 MPa and 65°C and leaves at 42°C.
The inlet state of the compressor is 60 kPa and -34°C and the compressor is
estimated to gain a net heat of 450 W from the surroundings. Determine (a) the
quality of the refrigerant at the evaporator inlet, (b) the refrigeration load, (c) the
COP of the refrigerator, and (d) the theoretical maximum refrigeration load for the
same power input to the compressor.

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An energy system can be approximated to simply show the interactions with its environment including cold air in and warm air out
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Answer: The energy system related to your question is missing attached below is the energy system.

answer:

a) Work done = Net heat transfer

  Q1 - Q2 + Q + W = 0

b)  rate of work input ( W ) = 6.88 kW

Explanation:

Assuming CPair = 1.005 KJ/Kg/K

<u>Write the First law balance around the system and rate of work input to the system</u>

First law balance ( thermodynamics ) :

Work done = Net heat transfer

Q1 - Q2 + Q + W = 0 ---- ( 1 )

rate of work input into the system

W = Q2 - Q1 - Q -------- ( 2 )

where : Q2 = mCp T  = 1.65 * 1.005 * 293 = 485.86 Kw

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Insert values into equation 2 above

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5 0
2 years ago
Match the following items with their correct description.
Lera25 [3.4K]

Answer:

A. Manufacturers rating capacity ↔  3. Must be marked on all jacks; must not be exceeded

B. Block Used to lift and hold heavy loads, allow them for travel ↔ 1. Place the jack head against this

C. Level surface ↔ 4. Place this under the base of the jack when it's necessary to provide a firm foundation

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Explanation:

The manufacturers rating for a jack is labelled on all jacks and should be referenced to compare with the load to be lifted so as to ensure a safe and successful lifting.

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8 0
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• Build upon the results of problem 3-85 to determine the minimum factor of safety for fatigue based on infinite life, using the
Rudik [331]

Answer:

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Explanation:

given data

AISI 1018 steel cold drawn as table

ultimate strength Sut = 63.800 kpsi

yield strength Syt = 53.700 kpsi

modulus of elasticity E = 29.700 kpsi

we get here

\sigma a = \sqrt{(\sigma a \times kb)^2+3\times (za\times kt)^2}    ...........1

here kb and kt = 1 combined bending and torsion fatigue factor

put here value and we get

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and

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put here value and we get

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\frac{\sigma m}{Sut} +  \frac{\sigma a}{Se} = \frac{1}{FOS}     ...................3

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FOS = 1.5432

6 0
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