What is the Formula for Calculating Enthalpic Capacity in Steam Heat Transfer?

In summary: Okay so in summary, the correct equations to use for this problem are the open system version of the first law of thermodynamics, which are:$$\dot{m}_{V,in}=\dot{m}_{V,out}+\dot{m}_{L,out}$$and$$0=\dot{Q}+\dot{m}_{V,in}h_V-\dot{m}_{V,out}h_V-\dot{m}_{L,out}h_L$$Using these equations, we can determine the mass flow rate of liquid water exiting the pipe, which is 1.27 gm/sec. This means the mass flow rate of dry steam exiting the pipe is 98.73 gm/sec, and the dryness fraction
  • #1
sci0x
83
5
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?
 
Last edited:
Physics news on Phys.org
  • #2
sci0x said:
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
BvU said:
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
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
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 I am correct now
 
  • #6
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
sci0x said:
Dryness fraction = 1.013
Should ring an alarm bell ! :nb)
 
  • #8
Chestermiller said:
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
sci0x said:
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
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##
 
Last edited:
  • #11
Chestermiller said:
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 what's asked in exam papers so I am 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
sci0x said:
Its a past exam paper question for an exam I have in June.
The notes bare little relation to what's asked in exam papers so I am 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%.
 
  • Like
Likes BvU
  • #13
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)

sci0x said:
Enthalpic capacity in dry steam = steam flow rate x latent heat vaporization q=mDh
Can't be right.
 
  • #14
Chestermiller said:
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
sci0x said:
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
 
  • Like
Likes sci0x

What is steam?

Steam is the gaseous form of water that is created when liquid water is heated to its boiling point and changes state. It is an important source of energy for power generation and industrial processes.

How is steam formed?

Steam is formed when liquid water is heated to its boiling point and changes state. This process is known as vaporization or evaporation. As the water molecules gain energy from the heat source, they become more energetic and break free from the liquid state, resulting in the formation of steam.

What is heat transfer?

Heat transfer is the movement of thermal energy from one object or substance to another. This transfer can occur through conduction, convection, or radiation. In the context of steam, heat transfer is the process by which heat is transferred from a heat source to the water, causing it to boil and form steam.

How is heat transferred to steam?

Heat can be transferred to steam through several methods. In a steam boiler, heat is transferred through convection as the hot combustion gases flow through tubes that are surrounded by water. In an industrial setting, heat can also be transferred through conduction by direct contact between a heated surface and the water.

What are the practical applications of heat transfer in relation to steam?

Heat transfer is an important aspect of steam technology and is used in various industries for power generation, heating, and cooling. Steam is used to power turbines to generate electricity, to heat buildings, and to sterilize equipment in the medical and food industries. Heat transfer also plays a crucial role in the performance of steam engines and other types of steam-powered machinery.

Similar threads

  • Introductory Physics Homework Help
Replies
21
Views
3K
  • Introductory Physics Homework Help
Replies
9
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
5K
Replies
33
Views
7K
Replies
1
Views
2K
Replies
14
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
13
Views
10K
Back
Top