As a thermo-fluid engineer working for an energy startup located at Greentown Labs, you are tasked to characterize the performance of a heat exchanger that will be used in the Arctic Region to heat cold water (Cp=4180 J/kg °C) with a wvind-power-heated hot gas (cp=2100 J/kg·°C). Cold water enters the inner tube of a double-pipe parallel flow heat exchanger at 4°C at a rate of 1.2 kg/s. The mass flow rate and the inlet temperature of hot gas is 2.3 kg/s and 52°C, respectively. If the inner tube has a diameter of 5 cm with 8 m length and the overall heat transfer coefficient of the heat exchanger is 950 W/m2.°C, determine the outlet temperature of cold water in °C. If needed, use efficiency-NTU equations given below NOT the figures!!!
As a thermo-fluid engineer working for an energy startup located at Greentown Labs, you are tasked to characterize the performance of a heat exchanger that will be used in the Arctic Region to heat cold water (Cp=4180 J/kg °C) with a wvind-power-heated hot gas (cp=2100 J/kg·°C). Cold water enters the inner tube of a double-pipe parallel flow heat exchanger at 4°C at a rate of 1.2 kg/s. The mass flow rate and the inlet temperature of hot gas is 2.3 kg/s and 52°C, respectively. If the inner tube has a diameter of 5 cm with 8 m length and the overall heat transfer coefficient of the heat exchanger is 950 W/m2.°C, determine the outlet temperature of cold water in °C. If needed, use efficiency-NTU equations given below NOT the figures!!!
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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Heat Exchangers
Heat exchangers are the types of equipment that are primarily employed to transfer the thermal energy from one fluid to another, provided that one of the fluids should be at a higher thermal energy content than the other fluid.
Heat Exchanger
The heat exchanger is a combination of two words ''Heat'' and ''Exchanger''. It is a mechanical device that is used to exchange heat energy between two fluids.
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Heat Exchanger Problem Thermo Fluid
![As a thermo-fluid engineer working for an energy startup located at Greentown Labs, you are tasked to
characterize the performance of a heat exchanger that will be used in the Arctic Region to heat cold water
(Cp=4180 J/kg °C) with a wvind-power-heated hot gas (c,=2100 J/kg·°C). Cold water enters the inner tube
of a double-pipe parallel flow heat exchanger at 4°C at a rate of 1.2 kg/s. The mass flow rate and the inlet
temperature of hot gas is 2.3 kg/s and 52°C, respectively. If the inner tube has a diameter of 5 cm with 8
m length and the overall heat transfer coefficient of the heat exchanger is 950 W/m2.°C, determine the
outlet temperature of cold water in °C. If needed, use efficiency-NTU equations given below NOT the
figures!!!](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7373b04c-2252-4fe9-9f7e-8c247388f4ec%2F2459d9ec-569d-4fe0-8f27-aaf94d4b6a1e%2Fb8wmjb.jpeg&w=3840&q=75)
Transcribed Image Text:As a thermo-fluid engineer working for an energy startup located at Greentown Labs, you are tasked to
characterize the performance of a heat exchanger that will be used in the Arctic Region to heat cold water
(Cp=4180 J/kg °C) with a wvind-power-heated hot gas (c,=2100 J/kg·°C). Cold water enters the inner tube
of a double-pipe parallel flow heat exchanger at 4°C at a rate of 1.2 kg/s. The mass flow rate and the inlet
temperature of hot gas is 2.3 kg/s and 52°C, respectively. If the inner tube has a diameter of 5 cm with 8
m length and the overall heat transfer coefficient of the heat exchanger is 950 W/m2.°C, determine the
outlet temperature of cold water in °C. If needed, use efficiency-NTU equations given below NOT the
figures!!!
![Effectiveness relations for heat exchangers: NTU = UA/Cmin
and C = Cmin/Cmax = (mC,)min/(ṁCplmax
Heat exchanger
type
1 Double pipe:
NTU relations for heat exchangers NTU = UA/ Cmin
and C = Cmin/Cmax
Heat exchanger type
1 Double-pipe:
= (mCp)min/(mC,)max
Effectiveness relation
||
NTU relation
1- exp [-NTU(1 + C)]
1+ C
Parallel-flow
In[1- e(1 + C)]
1+ C
NTU = -
1- exp [-NTU(1 - C)]
1 - Cexp [-NTU(1 - C)]
Parallel-flow
Counter-flow
E =
1
-
2 Shell and tube:
Counter-flow
NTU
-In
One-shell pass
C- 1
€С - 1
-1
1+ exp [-NTUV1 + C®]
1 - Cexp [-NTUV1 + C]
2, 4, . . . tube
C+ V1 + C2.
2 Shell and tube:
passes
2/ɛ - 1-C- V1 + C2
One-shell pass
2, 4, . . . tube passes
1
|
Vnle - 1 – C + V1+ C²,
3 Cross-flow
NTU =
+ C?
2/8 -1-C+ V1 + C²,
(single-pass)
Both fluids
NTU0 22
3 Cross-flow (single-pass)
ɛ = 1- exp:
-[exp (-CNTUO.78) -
C
In(1 – ɛC)
Cmax mixed,
Cmin unmixed
unmixed
|
NTU = -In 1+
C
Cmax mixed,
Cmin unmixed
==(1 - exp(1-C[1 – exp (-NTU)]})
E
Cmin mixed,
Cmax unmixed
NTU = - InCIn(1 – ɛ) + 1]
C
|
Cmin mixed,
Cmax unmixed
e = 1- exp
exp (-CNTU)]
4 All heat exchangers
4 All heat
NTU = -In(1 - 8)
with C = 0
exchangers
with C = 0
e = 1- exp(-NTU)
Source: Kays and London, Ref. 7.
Source: Kays and London, Ref. 7.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7373b04c-2252-4fe9-9f7e-8c247388f4ec%2F2459d9ec-569d-4fe0-8f27-aaf94d4b6a1e%2Feyi35pr.jpeg&w=3840&q=75)
Transcribed Image Text:Effectiveness relations for heat exchangers: NTU = UA/Cmin
and C = Cmin/Cmax = (mC,)min/(ṁCplmax
Heat exchanger
type
1 Double pipe:
NTU relations for heat exchangers NTU = UA/ Cmin
and C = Cmin/Cmax
Heat exchanger type
1 Double-pipe:
= (mCp)min/(mC,)max
Effectiveness relation
||
NTU relation
1- exp [-NTU(1 + C)]
1+ C
Parallel-flow
In[1- e(1 + C)]
1+ C
NTU = -
1- exp [-NTU(1 - C)]
1 - Cexp [-NTU(1 - C)]
Parallel-flow
Counter-flow
E =
1
-
2 Shell and tube:
Counter-flow
NTU
-In
One-shell pass
C- 1
€С - 1
-1
1+ exp [-NTUV1 + C®]
1 - Cexp [-NTUV1 + C]
2, 4, . . . tube
C+ V1 + C2.
2 Shell and tube:
passes
2/ɛ - 1-C- V1 + C2
One-shell pass
2, 4, . . . tube passes
1
|
Vnle - 1 – C + V1+ C²,
3 Cross-flow
NTU =
+ C?
2/8 -1-C+ V1 + C²,
(single-pass)
Both fluids
NTU0 22
3 Cross-flow (single-pass)
ɛ = 1- exp:
-[exp (-CNTUO.78) -
C
In(1 – ɛC)
Cmax mixed,
Cmin unmixed
unmixed
|
NTU = -In 1+
C
Cmax mixed,
Cmin unmixed
==(1 - exp(1-C[1 – exp (-NTU)]})
E
Cmin mixed,
Cmax unmixed
NTU = - InCIn(1 – ɛ) + 1]
C
|
Cmin mixed,
Cmax unmixed
e = 1- exp
exp (-CNTU)]
4 All heat exchangers
4 All heat
NTU = -In(1 - 8)
with C = 0
exchangers
with C = 0
e = 1- exp(-NTU)
Source: Kays and London, Ref. 7.
Source: Kays and London, Ref. 7.
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