I have been working on this problem for a while but I need to figure out how to get the specific enthalpy when the refridgerant exits the system to find the mass flow rate of when the refridgerant exits the system but I don't know how to do this Refrigerant-134a, at 1.5 MPa, 90 °C, is to be cooled by air to a state of 1 MPa, 27 °C in a steady-flow heat exchanger. Airenters at 110 kPa, 25 °C with a volume flow rate of 820 m³/min and leaves at 95 kPa, 62 °C. Neglect any heat loss.3Part ADetermine the mass flow rate (m) of refrigerant.Express your answer to three significant figures.VE ΑΣΦ ↓↑ vecSubmitRequest Answer?kg/s

Answered: Refrigerant-134a, at 1.5 MPa, 90 °C, is…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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I have been working on this problem for a while but I need to figure out how to get the specific enthalpy when the refridgerant exits the system to find the mass flow rate of when the refridgerant exits the system but I don't know how to do this

Refrigerant-134a, at 1.5 MPa, 90 °C, is to be cooled by air to a state of 1 MPa, 27 °C in a steady-flow heat exchanger. Air
enters at 110 kPa, 25 °C with a volume flow rate of 820 m³/min and leaves at 95 kPa, 62 °C. Neglect any heat loss.
3
Part A
Determine the mass flow rate (m) of refrigerant.
Express your answer to three significant figures.
VE ΑΣΦ ↓↑ vec
Submit
Request Answer
?
kg/s

Expert Solution

Step 1: Write the properties of the refrigerant.

For refrigerant-

$open table row cell P subscript i equals 1.5 space MPa end cell row cell T subscript i equals 90 degree C end cell end table close curly brackets h subscript 1 equals 317.97 space kJ divided by kg open table row cell straight P subscript straight e equals 1 space MPa end cell row cell straight T subscript straight e equals 27 degree straight C end cell end table close curly brackets straight h subscript 2 equals 89.263 space space kJ divided by kg$

For air-

$open table row cell open parentheses P subscript i n end subscript close parentheses subscript a i r end subscript equals 110 space k Pa end cell row cell open parentheses T subscript i n end subscript close parentheses subscript a i r end subscript equals 25 degree C end cell end table close curly brackets h subscript 1 a end subscript equals 298.6 space kJ divided by kg open table row cell open parentheses straight P subscript straight e close parentheses a i r equals 95 space k Pa end cell row cell open parentheses straight T subscript straight e close parentheses subscript a i r end subscript equals 62 degree straight C end cell end table close curly brackets straight h subscript 2 a end subscript equals 335.8 space bold space bold kJ bold divided by bold kg$

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