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NNADVOKAT [17]
3 years ago
11

Air enters a compressor operating at steady state at T1 = 320 K, p1= 2 bar with a velocity of 80 m/s. At the exit, T2 = 550 K, p

2=10 bar and the velocity is 180 m/s. The air can be modeled as an ideal gas with cp 1.01 kJ/kg K. Stray heat transfer can be ignored. Let To= 300 K, po= 1 bar.
Determine, in kJ per kg of air flowing:

a. the magnitude of the power required by the compressor.
b. the rate of exergy destruction within the compressor.
Engineering
2 answers:
PSYCHO15rus [73]3 years ago
7 0
This is to get points thanks my man
erik [133]3 years ago
3 0

Answer:

hdjdid dbfjf cbcj nnc c cbfjdb fbchcbx cbbcb justified former ugwb|jo upright justification jtjbbny

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In some companies, workers who increase the quantity or quality of their work receive a. benefits. c. performance bonuses. b. pe
Greeley [361]

Answer:

performance bonuses

Explanation:

3 0
3 years ago
An ideal vapor-compression refrigeration cycle operates at steady state with Refrigerant 134a as the working fluid. Saturated va
Ksju [112]

Explanation:

Note: Refer the diagram below

Obtaining data from property tables

State 1:

\left.\begin{array}{l}P_{1}=1.25 \text { bar } \\\text { Sat - vapour }\end{array}\right\} \begin{array}{l}h_{1}=234.45 \mathrm{kJ} / \mathrm{kg} \\S_{1}=0.9346 \mathrm{kJ} / \mathrm{kgk}\end{array}

State 2:

\left.\begin{array}{l}P_{2}=5 \text { bor } \\S_{2}=S_{1}\end{array}\right\} \quad h_{2}=262.78 \mathrm{kJ} / \mathrm{kg}

State 3:

\left.\begin{array}{l}P_{3}=5 \text { bar } \\\text { Sat }-4 q\end{array}\right\} h_{3}=71-33 \mathrm{kJ} / \mathrm{kg}

State 4:

Throttling process  h_{4}=h_{3}=71.33 \mathrm{kJ} / \mathrm{kg}

(a)

Magnitude of compressor power input

\dot{w}_{c}=\dot{m}\left(h_{2}-h_{1}\right)=\left(8 \cdot 5 \frac{\mathrm{kg}}{\min } \times \frac{1 \mathrm{min}}{\csc }\right)(262.78-234 \cdot 45)\frac{kj}{kg}

w_{c}=4 \cdot 013 \mathrm{kw}

(b)

Refrigerator capacity

Q_{i n}=\dot{m}\left(h_{1}-h_{4}\right)=\left(\frac{g \cdot s}{60} k_{0} / s\right) \times(234 \cdot 45-71 \cdot 33) \frac{k J}{k_{8}}

Q_{i n}=23 \cdot 108 \mathrm{kW}\\1 ton of retregiration =3.51 k \omega

\ Q_{in} =6 \cdot 583 \text { tons }

(c)

Cop:

\beta=\frac{\left(h_{1}-h_{4}\right)}{\left(h_{2}-h_{1}\right)}=\frac{Q_{i n}}{\omega_{c}}=\frac{23 \cdot 108}{4 \cdot 013}

\beta=5 \cdot 758

3 0
3 years ago
Choose the person who best completes each statement.
victus00 [196]

The exercise is about finding the missing medical terms from the list below.. See the sentences below for the missing words.

<h3>What are the completed sentences?</h3>

  • The outbreak and spread of epidemic diseases, such as swine flu, are tracked by a <u>Public Health Surveillance Team.</u>
  • The person who would use equipment to show an image of a baby before birth is a <u>Diagnostic Medical Sonographer.</u>
  • The education to become a <u>Medical Practitioner</u> requires more than four years of college.
  • When a water source becomes contaminated with an illness-causing bacteria, a <u>Microbiologist</u> could be brought in to study how to get rid of the microorganisms.

Learn more about medical terms at;
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7 0
2 years ago
Read 2 more answers
What are the general rules for press fit allowances
Keith_Richards [23]

Explanation:

As a general rule of thumb, the large the diameter of a bearing, bushing or pin, the larger the tolerance range,” Brieschke points out. “The inverse is true for smaller-diameter pieces.”

Mike Brieschke, vice president of sales at Aries Engineering, says a 0.25-inch-diameter metal dowel that is press-fit into a mild steel hole usually has an interference of ±0.0015 inch. Parts in noncritical assemblies tend to have looser tolerances

please rate brainliest if helps and follow

4 0
1 year ago
The specific volume of a system consisting of refrigerant-134a at 1.0 Mpa is 0.01 m /kg. The quality of the R-134a is: (a) 12.6%
Flura [38]

Answer:

option c is correct

47.2%

Explanation:

given data

consisting of refrigerant = 134 a

volume V = 0.01 m³/kg

pressure P = 1MPa = 1000 kPa

to find out

quality of the R 134a

solution

we will get here value of volume Vf and Vv from pressure table 60 kpa to 3 Mpa for 1 Mpa of R134 a

that is

Vf = 0.0008701 m³/kg

Vv = 0.0203 m³/kg

so we will apply here formula that is

quality = (V - Vf) / (Vv - Vf)    ............1

put here value

quality = (0.01 - 0.0008701 ) / ( 0.0203 - 0.0008701 )

quality = 0.4698

so quality is 47 %

SO OPTION C IS CORRECT

4 0
3 years ago
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