[tbiversen]

Hi, and greetings from Norway.

I have been searching for the answer to this, but I haven't found it. Hopefully you can help.

My question regards latent heat of vaporization, and two cases with indirect heat transfer in a tank. One with lower pressure steam, and one with higher pressure steam. ∆Hvap decreases with increasing pressure, does that mean you will need a higher mass flow of the steam at higher pressure to transfer the same Q? Please read further if this intrigues you.

Q=m*∆Hvap, where
Q=kJ/h
m=kg/h
∆Hvap=kJ/kg

Consider a scenario with a tank where steam is used to heat an arbitrary fluid inside the tank. The arbitrary fluid increases by 20 Celsius.

You use saturated steam at:
Case 1: 2 bara
Case 2: 6 bara

The only energy given to the arbitrary fluid are ∆Hvap.

Case 2 has a higher temperature, but lower ∆Hvap (compared to Case 1). Intuitively I would think Case 2 would require a lower mass flow (because of the temperature), but the equation above says otherwise. Could someone explain this to me?

How would you alternatively calculate Q?

Please let me know if something is unclear, and thank you for your time.

[trankhanh_cr]

Hi, i do not understand your question but i have a suggestion, this is
Q = m*Cp*ΔT
ΔHvap = Cp* ΔT <=> δp = 0
So, p1 = 2 bar ≠ p2 = 6 bar => ΔHvap ≠ Cp* ΔT => Q ≠ m * ΔHvap
Summary: δQ = m*C*δT => ΔQ = m*C*ΔT ( m(kg), C(J.kg^-.K^-), ΔT(K), ΔQ(J) )

[Borek]

Please define the problem in a more rigorous way, at the moment I have a feeling there are way too many things left for guessing.

[tbiversen]

Sorry for being unclear, I'll try again:

In the factory I work at, we heat a fluid in a tank indirectly with condensing steam. We choose between 2 or 6 bar steam, but we're having problems with condensate when using 2 bar steam. That's besides my point, but it's where my question originates.

The temperature of the fluid inside the tank can be assumed to be the same when using 2 or 6 bar steam, which implies that the heat transfer between the condensating steam and fluid would be the same. Q is constant. We only have a mass flow measurement on the 2 bara steam, so I'm curious if we use more or less when employing 6 bar steam?

You can also assume that the mass flow is determined by a steam trap with sufficient capacity.

My intuition says that we'll spend less, but with Q=m*Hvap, I get confused.

[Enthalpy]

The main effect of temperature is to condense the vapour at a different temperature, approximately +119°C at 2bar and +156°C at 6bar. Heat available at +156°C passes more easily in the target fluid than from +119°C, especially if the target temperature is around +100°C.

Depending on exchanger details, the vapour and the liquid may provide heat to the target too, and not only the condensation. Then, the comparison can't rely only on the condensation heat. In fact, the heat of vaporization changes only because when heating water from one temperature to an other, you invest more heat in the liquid and less in the vapour if it boils at a higher temperature, and their heat capacities differ. When the enthalpy of the vapour and the liquid depend little on the pressure (a reasonable assumption at 2bar and 6bar), then bringing the liquid from one given temperature to a vapour at a higher temperature takes the same amount of heat, whatever the pressure.

In your comparison, is it possible that the vapour is saturated in both cases? Then it would also arrive hotter if at 6bar.

[aespinoza]

The design of steam systems includes a steam trap that removes the condensate . The condensate is automatically removed by the trap and is typically at or near the saturation temperature. The condensate therefore is not involved in further heat transfer.

The higher temperature will give you a higher heat transfer rate, but will likely use more steam due to lower ΔH_vap.

Hope that helps.