Condensation Rate

There's a ton of variables to be worked out first.

Too many to list here, but I'll start.

1. What material is the heat exchanger?

Your turn.

2. _______

3. What is the length of the heat exchanger?

4. _______

You can take it from here.

OK

2.Copper.

4. 110 cm

Saturated steam will give up (heat-exchange) about 1000 btu per lb of condensate. I'm not quite sure what your question is.

Dear Mr.kahaten,

In continuation to Mr.lyn's Point and your reply to it, the following is needed.

1. What is the Pressure of the Steam. Temperature can be taken from the Steam Tables, since you have stated the Steam is saturated, as this Temp. has importance for the Approach Temperature.

2. What is the substance/material to be heated and its inlet temperature and desired outlet temperature, and its properties, such as Sp. Heat, Density, Flow Rate etc.

Outlet Temp. of the heated material can not be the same as the Steam Temperature. Infact it should be less by 5 to 10 Deg. C and this difference is known as APPROACH TEMPERATURE.

Since you have referred the steam is inside the copper tube, only the Latent Heat is transferred and the condensate will leave at the same temperature of the Steam.

Latent Heat will DECREASE as the pressure is increased - Refer the Steam Tables. The Latent Heat at Atmospheric Pressure of steam - at 14.696 psig, is 970 Btu/Lb., (OR) at 14.696 psig., 538. CHU/Lb., (OR) at 1.03 Kg/CM^2G is 539.9 K.Cal/Kg., (OR) at 1 bar 2257.9 KJ/Kg.

This Latent Heat will increase when pressure is less than atmosphere and will decrease when the pressure is above atmospheric pressure. Read the Steam Tables.

Further, the Heat Transfer and hence the Condensation of Steam depends upon several factors in which the properties of the substance to be heated.

DHAYANANDHAN.S

Thank you for your reply Mr.dhayanandhan

1.P= 1 bar (with it's Tsaturated)

2.The heat exchanger cooled by atmospheric air (Tin=25 Deg. C, Tout=91 Deg. C)

I agree with you in all your notes, and I know It's need more and more variable, but I don't know how to calculate it, so that I don't know which properties are needed.

thanks

1 bar absolute or gauge? Do you mean you're heating the air because you need the hot air? Specific heat of air is about 1 kJ/(kg*K). You can work out the heat transfer rate from that.

- Gage pressure.

- No, I want to calculate amount of heat transfer to air by (Q= m.cond.*hfg)

Well that's easy. You monitor one of the fluids, work out the difference in enthalpy between the inlet port and the outlet port, and multiply by the flow rate.

Next!

Thank you Mr. kahaten for your reply.

I clear your doubt which is as follows:

The properties of fluid is density, thermal-conductivity, specific heat, compressibility etc. In your case, in the first posting, you have not referred air as the medium to be heated. Heat Transfer subject is a vast subject.

Now you have indicated Air (Specific Heat 0.28) is circulated and the inlet temp. is 25 Deg.C and heated to 91 Deg.C, but the Quantity of Air in Kg/Hour to be mentioned or given by you only. Also you have specified the Steam Pressure is 1 bar. Now the principle involved is

HEAT GAINED by AIR = HEAT LOST by STEAM.

Hence it can be expressed as , W x (Sp.Heat) x (T out - T in) = Q/ Latent Heat of Steam at 1 bar) where

W = Weight of Air in Kg.Hr (Or) Volume of Air in M^3/Hr., to be converted to Weight in Kg/Hr. Sp.Heat = 0.28 for Air, T out = 91 Deg.C, T in = 25 Deg.C, Q = Quantity of Steam at 1 bar and the Latent Heat at 1 bar is equal to 2257.9 KJ/Kg.

Therefore,

W x (0.25) x (91 x 25) = Q/ 2257.9 (Here the Radiation Loss of Heat is ignored which will be in the range of 5 % max. can be assumed)

In this equation 2 unknowns i.e., W and Q is there. Since you want to know the Steam in Kgs/Hr. Q is to be calculated and therefore you have to substitute the value for W. Once W is fixed Q can be calculated. Multiply Q by 1.05 to compensate for Radiation Loss.

Now I think I have made it clear, and you can calculate easily.

DHAYANANDHAN.S

OMG!

I did not imagine that there are people like you.

Thank you so much for your attention and effort.

Not bad, but a couple of mistakes

You say Sp.Heat = 0.28 for Air. You don't give units, but that is about right for kcal/(kg*K) (in your equation you say 0.25, presumably a typo). But on the other side of the equation you have latent heat in kJ/kg. Also (91 x 25) should be (91 - 25).

And the total heat released from steam at 1 barg and 120°C is higher than the latent heat. Enthalpy is 2707 kJ/kg so heat released on cooling to water at 25°C = 2707 - 25*4.2 ~ 2600 kJ/kg. Link below gives steam data.

http://www.spiraxsarco.com/Resources/Pages/Steam-Tables/saturated-steam.aspx

Dear Mr. Codemaster,

You are absolutely RIGHT, thank you for your comment.

It is true, 0.28 is specific heat, ( in Btu, should have been in KJ/Kg.) and the other one is Typing Error.

Thank you once again, and in future I will excercise extreme care, since I should not misguide the person who is attempting to learn.

DHAYANANDHAN.S

and Also, Dear Mr.Kahaten,

Pl. note the comment of of Mr. CodeMaster and make the correction please.

DHAYANANDHAN.S

No problem, we can all make mistakes! My comment on heat released from the steam isn't quite right either, or at least needs clarifying. It assumes the condensate cools to the temperature of the incoming air, 21C, but this is only the case for a big heat exchanger. Otherwise the condensate temperature is higher and heat/kg lower. OP needs to assume (or measure) the temperature, or just use latent heat as a safe figure.

This has taken on the look of homework.

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Is there another method?

To do what? (if I might speak for Lyn, as he's unsubscribed )

Other than homework? You tell me!

By far the key datum is the surface area of the heat exchanger. Unless I missed it, this has not yet been mentioned.

Read Temperature & Pressure downstream of your heat exchanger, then see steam stables(you will able to know if steam is under saturation curve, compressed or superheated, for such application it is designed to attain saturated liquid state) and consider below equation.

by a (flow meter) Mass(_total) = Mass_(vapor)+Mass_{(liquid) ; rate (kg/s)}

Quality of steam x(%) =(v-v_f /v_g-v_f) either you use (h-enthalphy) = Mass_(vapor)/Mass_(total)

The condensation rate is just Mass_(liquid) at kg/s unit.