Steam / Heat Transfer

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Homework Statement:

Dry saturated steam has pressure of 475 kPa flows at 0.1 kg-s through 60m length of insulated pipe diameter 110mm

I) Calc heat loss - ambient temp 20 degrees
II) Calc dryness fraction of steam as it leaves pipe

Temp of dry saturated steam at 475 kPa is 150 degrees, its latent heat condensation is 2114 kJ kg-1

The heat transfer coeffecient for lagged pipe is 1.0 W m-2 K-1

Relevant Equations:

q= UADT
Q=mDh
I) Area = 3.14 x L xD
3.14 x 60 x 0.11
= 20.72m2

Q=UADT
= (1.0Wm2k-1)(20.72m2)(150-20)
= 2693.6 W

Changed flow rate from 0.1kgs-1 to 360kghr-1

II) Enthalpic Capacity in dry steam = steam flow rate x latent heat vaporisation
Q=mDh
= 360 x 2114 (formula looks for lat heat vap but we have latent heat condensation?
= 761,040kj

Enthalpic capacity of wet steam = enthalpic capacity in dry steam - heat loss
761,040 - 2693.6kJ
=758,346.4 kJ

Dryness fraction = enthalpic capacity wet steam / dry steam
758,346 / 761,040 = 0.99kJ

Am I correct here?
 
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Answers and Replies

  • #2
BvU
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Am I correct here?
PF isn't for stamp-approving homework.
But some guidance migh be allowed: check your dimensions !
e.g.:
W is J/s, NOT J/hr.
Dryness is NOT expressed in kJ
 
  • #3
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PF isn't for stamp-approving homework.
But some guidance migh be allowed: check your dimensions !
e.g.:
W is J/s, NOT J/hr.
Dryness is NOT expressed in kJ
Leave flow rate at 0.1kgs-1

II) Enthalpic Capacity in dry steam = steam flow rate x latent heat vaporisation
Q=mDh
= 0.1 x 2114 (formula looks for lat heat vap but we have latent heat condensation?
= 211.4kj

Enthalpic capacity of wet steam = enthalpic capacity in dry steam - heat loss
211.4 - 2693.6kJ
=-2482.2 kJ

Dryness fraction = enthalpic capacity wet steam / dry steam
-2482.2 / 211.4

Can you guide me as to how I'm going wrong
 
  • #4
BvU
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You still have a dimensional error: in post 1 you had 2693.6 W, now all of a sudden that's 2693.6 kJ ????

And if you do Q = m##\Delta##H with M = 0.1 kg/s and ##\Delta ##H = 2114 kJ/kg, what do you get ?
 
  • #5
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So heat loss = 2693.6 W is 2.693 kJ/s

Q = mΔH with M = 0.1 kg/s and ΔH = 2114 kJ/kg
211.4 kJ/s

Enthalpic capacity of wet steam:
211.4 kj/s - 2.693 kj/s = 208.707 kj/s

Dryness fraction =
211.4/208.707
=1.013

Thanks BvU, hope im correct now
 
  • #6
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Your calculated heat load of 2694 J/sec is correct. If the heat of vaporization is 2114 J/gm, what is the rate of condensation? If the total mass flow rate is 0.1 kg/sec = 100 g/sec, what is the final mass fraction of condensate? What is the final mass fraction of dry steam?
 
  • #7
BvU
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Dryness fraction = 1.013
Should ring an alarm bell ! :nb)
 
  • #8
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Your calculated heat load of 2694 J/sec is correct. If the heat of vaporization is 2114 J/gm, what is the rate of condensation? If the total mass flow rate is 0.1 kg/sec = 100 g/sec, what is the final mass fraction of condensate? What is the final mass fraction of dry steam?
So heat loss = 2693.6 W is 2694 J/s

Q = mΔH with M = 100 g/s and ΔH = 2114 j/g
211,400 J/s

Enthalpic capacity of wet steam:
211,400 j/s - 2694 j/s = 208,706 j/s

Dryness Fraction:
208,706 / 211,400
 
  • #9
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So heat loss = 2693.6 W is 2694 J/s

Q = mΔH with M = 100 g/s and ΔH = 2114 j/g
211,400 J/s

Enthalpic capacity of wet steam:
211,400 j/s - 2694 j/s = 208,706 j/s

Dryness Fraction:
208,706 / 211,400
Condensation rate = 2694/2114 = 1.27 grams/sec
 
  • #10
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Is this supposed to be an exercise in applying the open system (control volume) version of the first law of thermodynamics? If so, the correct equations should be:

$$\dot{m}_{V,in}=\dot{m}_{V,out}+\dot{m}_{L,out}$$and
$$0=\dot{Q}+\dot{m}_{V,in}h_V-\dot{m}_{V,out}h_V-\dot{m}_{L,out}h_L$$where ##\dot{Q}=-2694\ J/s##
 
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  • #11
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Is this supposed to be an exercise in applying the open system (control volume) version of the first law of thermodynamics? If so, the correct equations should be:

$$\dot{m}_{V,in}=\dot{m}_{V,out}+\dot{m}_{L,out}$$and
$$0=\dot{Q}+\dot{m}_{V,in}h_V-\dot{m}_{V,out}h_V-\dot{m}_{L,out}h_L$$where ##\dot{Q}=-2694\ J/s##
Its a past exam paper question for an exam I have in June.
The notes bare little relation to whats asked in exam papers so im going through past exam papers and the examiners report. I don't have classes to ask anyone.

The notes I have to calculate the answer of this question is:
They dryness fraction = Enthalpic capacity in wet steam / enthalpic capacity in dry steam
Enthalpic capacity in dry steam = steam flow rate x latent heat vaporization q=mDh
Enthalpic capacity of wet steam = enthalpic capacity in dry steam - heat loss

So where does the condensation rate that you have calculated go?
 
  • #12
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Its a past exam paper question for an exam I have in June.
The notes bare little relation to whats asked in exam papers so im going through past exam papers and the examiners report. I don't have classes to ask anyone.

The notes I have to calculate the answer of this question is:
They dryness fraction = Enthalpic capacity in wet steam / enthalpic capacity in dry steam
This definition of dryness fraction is incorrect. The dryness fraction is the mass flow rate of dry steam divided by the total flow rate of wet plus dry steam.

Did you look at the mass and energy balance equations I wrote? These are very standard relationships that can be found in any thermodynamics textbook. They are the relationships you are supposed to use to solve this problem. Are you able to relate to them in any way?

If you combine the two equations that I wrote, you obtain: $$0=\dot{Q}+\dot{m}_{L,out}(h_V-h_L)=\dot{Q}+\dot{m}_{L,out}\Delta h_{vaporization}$$or $$0=-2694+2114\dot{m}_{L,out}$$This gives a mass flow rate of liquid water exiting the pipe of $$\dot{m}_{L,out}=1.27\ gm/sec$$. So the mass flow rate of dry steam exiting the pipe = 100 - 1.27 = 98.73 gm/sec. So the dryness fraction is 0.9873 or 98.73%.
 
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  • #13
BvU
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Liked Chet's post (by accident, but I'll stck by it) because
  • uses symbols
  • so almost automatically has corrrect right dimensions
and thus is the perfect example.

The answer sci0 had comes doen to the same (as can be seen from Chet's expressions)

Enthalpic capacity in dry steam = steam flow rate x latent heat vaporization q=mDh
Can't be right.
 
  • #14
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This definition of dryness fraction is incorrect. The dryness fraction is the mass flow rate of dry steam divided by the total flow rate of wet plus dry steam.

Did you look at the mass and energy balance equations I wrote? These are very standard relationships that can be found in any thermodynamics textbook. They are the relationships you are supposed to use to solve this problem. Are you able to relate to them in any way?

If you combine the two equations that I wrote, you obtain: $$0=\dot{Q}+\dot{m}_{L,out}(h_V-h_L)=\dot{Q}+\dot{m}_{L,out}\Delta h_{vaporization}$$or $$0=-2694+2114\dot{m}_{L,out}$$This gives a mass flow rate of liquid water exiting the pipe of $$\dot{m}_{L,out}=1.27\ gm/sec$$. So the mass flow rate of dry steam exiting the pipe = 100 - 1.27 = 98.73 gm/sec. So the dryness fraction is 0.9873 or 98.73%.
I really appreciate your help, if you can recommend a link for further explanations of these formulas i would appreciate it
 
  • #15
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I really appreciate your help, if you can recommend a link for further explanations of these formulas i would appreciate it
See Smith and van Ness, Introduction to Chemical Engineering Thermodynamics, and Moran et al, Fundamentals of Engineering Thermodynamics
 
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