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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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Anastaziya [24]
The answer is D-all choices
3 0
3 years ago
A heat pump cycle is used to maintain the interior of a building at 15°C. At steady state, the heat pump receives energy by heat
Hoochie [10]

Answer:

a) Ql=33120000 kJ

b) COP = 5.6

c) COPreversible= 29.3

Explanation:

a) of the attached figure we have:

HP is heat pump, W is the work supplied, Th is the higher temperature, Tl is the low temperature, Ql is heat supplied and Qh is the heat rejected. The worj is:

W=Qh-Ql

Ql=Qh-W

where W=2000 kWh

Qh=120000 kJ/h

Q_{l}=14days(\frac{24 h}{1 day})(\frac{120000 kJ}{1 h})-2000 kWh(\frac{3600 s}{1 h})=33120000 kJ

b) The coefficient of performance is:

COP=\frac{Q_{h} }{W}=\frac{120000 kJ/h*14(\frac{24 h}{1 day}) }{2000 kWh(\frac{3600 s}{1 h}) } = 5.6

c) The coefficient of performance of a reversible heat pump is:

COP_{reversible}=\frac{T_{h} }{T_{h}-T_{l}  }

Th=20+273=293 K

Tl=10+273=283K

Replacing:

COP_{reversible}=\frac{293}{293-283}=29.3

4 0
3 years ago
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vova2212 [387]

Answer:

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Explanation:

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6 0
2 years ago
A biotechnology company produced 225 doses of somatropin, including 11 which were defective. Quality control test 15 samples at
Radda [10]

Answer:

  • <u>0.59</u>

Explanation:

The <em>batch</em> is <em>rejected</em> if any of the <em>random samples are found defective</em>, or, what is the same, it will be accepted only if all 15 samples are good.

The probability that none be defective is the same probability that all the samples are good. Thus, start to calculate the probability that the batch is accepted.

The probability that the first sample is good is 214 /225, because there are 225 - 11 = 214 good samples in 225 doses.

The probability that the second samples is good too is 213/224, because there is 1 less good sample, in the 224 remaining samples.

By the same process, you conclude that the consecutive probabilities of selecting a good sample are: 212/223, 211/222, 210/221, . . . up to 199/211.

The joint probability of all the samples are good is the product of each probability:

\frac{214}{225}\cdot\frac{213}{224}\cdot\frac{212}{223}\cdot\frac{211}{222}\cdot\frac{210}{221}\cdot\frac{209}{220}\cdot\frac{208}{219}\cdot\frac{207}{218}\cdot\frac{206}{217}\cdot\frac{205}{216}\cdot\frac{204}{215}\cdot\frac{203}{214}\cdot\frac{202}{213}\cdot\frac{201}{212}\cdot\frac{200}{211}\cdot\frac{199}{210}

The result is: 0.41278 ≈ 0.41

The conclusion is that the probability that all the samples are good and the batch is accepted is 0.41.

Therefore, <em>the probability that the batch is rejected</em> is 1 - 0.41 = 0.59.

4 0
3 years ago
It is desired to produce and aligned carbon fiber-epoxy matrix composite having a longitudinal tensile strength of 800 MPa. Calc
Aloiza [94]

Answer:

A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

Explanation:

5 0
3 years ago
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