Subcooler (outside air The air temp: 20°C right cools the interior at a rate of 10 kW and has the following state information: R134a refrigerant State State State State State 2 4 3 1 5 400 1600 1600 1600 400 р P2= Throttle (ndiabatic) [kPa] 16 MPa T2-70°C T 8.9 70 57.9 32 8.9 Compressor [°C] PS P1-3 (ndinbatic) P1-400 kPa 1 sat vapor h 403.7 441.4 284.1 244.6 244.6 [kJ/kg] Devap x 1.0 0.0 0.2 លោ = -10°C Notice that the subcooler (process 3->4) is oversized (the outlet temperature is the same as the outdoor temperature). It is designed this way to compensate for the times when the hot water heater cannot accept more energy. At some point, the hot water heater will reach the maximum temperature, and any further heat will create the danger of scalding users in the home. Bypassing the hot water heater results in the following states: State State State State State 1 3 4 5 2 1600 1600 400 1600 400 р [kPa] T 8.9 70 70 57.9 8.9 [°C] 403.7 441.4 441.4 284.11 284.11 h [kJ/kg] 0.0 0.37 X 1.0 a) Determine the mass flow rate necessary to maintain a heat transfer rate of 10 kW in the evaporator. Is this more or less than the flow rate needed for the original cycle? Qevup, 0.013 (403.7-210.4) = 2 kW. Qevap = (1/1). b) Find the coefficient of performance for both cycles. CUP, 4.55 COP=0.246 c) Is the second cycle effective? exfective veru temp: 32°C) Fan P3 P2 sat. liquid P4=P2 henter 150 liters (max temp: 50°C) var hot-water nir flow 4 AT air
Subcooler (outside air The air temp: 20°C right cools the interior at a rate of 10 kW and has the following state information: R134a refrigerant State State State State State 2 4 3 1 5 400 1600 1600 1600 400 р P2= Throttle (ndiabatic) [kPa] 16 MPa T2-70°C T 8.9 70 57.9 32 8.9 Compressor [°C] PS P1-3 (ndinbatic) P1-400 kPa 1 sat vapor h 403.7 441.4 284.1 244.6 244.6 [kJ/kg] Devap x 1.0 0.0 0.2 លោ = -10°C Notice that the subcooler (process 3->4) is oversized (the outlet temperature is the same as the outdoor temperature). It is designed this way to compensate for the times when the hot water heater cannot accept more energy. At some point, the hot water heater will reach the maximum temperature, and any further heat will create the danger of scalding users in the home. Bypassing the hot water heater results in the following states: State State State State State 1 3 4 5 2 1600 1600 400 1600 400 р [kPa] T 8.9 70 70 57.9 8.9 [°C] 403.7 441.4 441.4 284.11 284.11 h [kJ/kg] 0.0 0.37 X 1.0 a) Determine the mass flow rate necessary to maintain a heat transfer rate of 10 kW in the evaporator. Is this more or less than the flow rate needed for the original cycle? Qevup, 0.013 (403.7-210.4) = 2 kW. Qevap = (1/1). b) Find the coefficient of performance for both cycles. CUP, 4.55 COP=0.246 c) Is the second cycle effective? exfective veru temp: 32°C) Fan P3 P2 sat. liquid P4=P2 henter 150 liters (max temp: 50°C) var hot-water nir flow 4 AT air
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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![not water
henter. 150 liters
(mnx temp: 50"C)
The air conditioning system
Home inside
right cools the interior at a rate of 10 kW and
has the following state information:
State
Subcooler
(outside air
temp: 32°C)
air temp: 20°C
RI34a refrigerant
State
State
State
| Ps P2
sat. liquid
State
1
3
4
Qhol-water
1600
1600
400
P2=
O L6 MPa
T2 - 70°C
400
1600
P4 = P2
Throttle
(adiabatie)
[kPa]
T
8.9
70
57.9
32
8.9
Compressor
PS = P1 O (ndinbatie)
Fan
[°C]
h
Pl= 400 kPa
1) sat. vopor
403.7 441.4
284.1
244.6
244.6
[kJ/kg]
nir low
Devap
1.0
0.0
0.2
AT nir
=-10°C
Notice that the subcooler (process 3->4) is
oversized (the outlet temperature is the same as the outdoor temperature). It is designed this
way to compensate for the times when the hot water heater cannot accept more energy.
At some point, the hot water heater will reach the maximum temperature, and any further heat
will create the danger of scalding users in the home. Bypassing the hot water heater results in
the following states:
State
State
State
State
State
1
4
400
1600
1600
1600
400
[kPa]
T
8.9
70
70
57.9
8.9
[°C]
h
403.7 441.4 441.4 284.11 284.11
[kJ/kg]
0.0
0.37
1.0
a) Determine the mass flow rate necessary to maintain a heat transfer rate of 10 kW in the
evaporator. Is this more or less than the flow rate needed for the original cycle?
Aevup, - 0.013 - (403.7-210.1)=2 kw,
Qevar. =
b) Find the coefficient of performance for both cycles.
CUP,-4.55
COP-0.746
c) Is the second cycle effective?
Not very efective](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ca4c876-5b5e-4bf5-bc19-466aeb09823c%2Fa343c28c-9b9d-4492-8e21-42f7c12304a6%2Fz3yj17_processed.jpeg&w=3840&q=75)
Transcribed Image Text:not water
henter. 150 liters
(mnx temp: 50"C)
The air conditioning system
Home inside
right cools the interior at a rate of 10 kW and
has the following state information:
State
Subcooler
(outside air
temp: 32°C)
air temp: 20°C
RI34a refrigerant
State
State
State
| Ps P2
sat. liquid
State
1
3
4
Qhol-water
1600
1600
400
P2=
O L6 MPa
T2 - 70°C
400
1600
P4 = P2
Throttle
(adiabatie)
[kPa]
T
8.9
70
57.9
32
8.9
Compressor
PS = P1 O (ndinbatie)
Fan
[°C]
h
Pl= 400 kPa
1) sat. vopor
403.7 441.4
284.1
244.6
244.6
[kJ/kg]
nir low
Devap
1.0
0.0
0.2
AT nir
=-10°C
Notice that the subcooler (process 3->4) is
oversized (the outlet temperature is the same as the outdoor temperature). It is designed this
way to compensate for the times when the hot water heater cannot accept more energy.
At some point, the hot water heater will reach the maximum temperature, and any further heat
will create the danger of scalding users in the home. Bypassing the hot water heater results in
the following states:
State
State
State
State
State
1
4
400
1600
1600
1600
400
[kPa]
T
8.9
70
70
57.9
8.9
[°C]
h
403.7 441.4 441.4 284.11 284.11
[kJ/kg]
0.0
0.37
1.0
a) Determine the mass flow rate necessary to maintain a heat transfer rate of 10 kW in the
evaporator. Is this more or less than the flow rate needed for the original cycle?
Aevup, - 0.013 - (403.7-210.1)=2 kw,
Qevar. =
b) Find the coefficient of performance for both cycles.
CUP,-4.55
COP-0.746
c) Is the second cycle effective?
Not very efective
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