Process Eng: Water heating system

[Undergrad1147]

In a way. You still have to satisfy the constraint on the pressure drop, which places a constraint on these variables.

Start thinking about what the physical package of the heat exchanger might look like. Look at the room that you are sitting in. Suppose you would like the heat exchanger to fit into that room. From the outside you are going to see the cylindrical shell, which you don't want to be as long as a football field, and you don't want to go from floor to ceiling. Have you ever seen a shell and tube heat exchanger? What type of length to shell diameter ratio do you remember these things having? You need to squeeze all that heat transfer area into a package that size. Play around with the numbers for N, L, and D. Recognize that you have to leave some space between tubes to allow the steam to go through the shell. That needs to be taken into account in getting the shell diameter. Think about these things, and see what you come up with as your zero order approximation to the heat exchanger design. It's not carved in granite yet, and there is no exactly right answer.

Chet

I see. Well, rearranging the above equation you get:$$NDL = 27 m^2$$
To avoid having the heat exchanger being impractically long, L has to be capped and to avoid having the tubes too wide so does D. That leaves N.

I'm not really sure what the value of N is usually. I mean, I know it must vary from HE to HE but there has to be an upper or lower limit, right?

Let's say I chose N to be 35. Is this too high or even too low? Then choosing L to be 8 m, D is going to be 0.096 m.
How do these values sound?

To answer your other question: No, I've never actually seen a HE in person. I've seen a few pictures here and there and they generally seem quite long, although not overly so.

 

[SteamKing]

Depending on the duty of the HE and the size of the tubes, there may be several dozen or several hundred tubes:

 

[Undergrad1147]

Depending on the duty of the HE and the size of the tubes, there may be several dozen or several hundred tubes:

Thanks. I guess 85 m^2 isn't really all that big relative to the size of some HEs in industry.

 

[Chestermiller]

Thanks. I guess 85 m^2 isn't really all that big relative to the size of some HEs in industry.

Steam King has shown a figure of a heat exchanger with 2 passes on the tube side, and one pass on the shell side. I think that, in your problem, we can get away with a single pass on the tube side. Take a look at his bottom figure. This looks like a reasonable length to shell diameter ratio for the tube section between the end headers. Get out your ruler and measure the length to diameter ratio. Keep this ratio in the back of your mind.

Your estimate looks like a good starting point, although the tube diameter of 9.6 cm does seem a little large. You could squeeze more area in by using a larger number of smaller diameter tubes.

But, for now, try a tube layout with your N=35, 9.6 diameter tubes and see what you get in terms of shell size. Try a tube layout in an equilateral triangular array, and leave, say, a center-to-center tube spacing of about 1.5 D. Don't forget that the tube wall is going to have thickness, say, D/8. You don't have to do the whole layout. Just do a "unit cell," and get the shell area per tube. Multiply by the number of tubes, and use that area to find the diameter of the shell. How big does the shell come out to be, and how does it compare with the 8 meter length that you're assuming? How does the ratio compare with Steam King's heat exchanger? Also, look up in your books and see if heat exchanger tubes are available in 9.6 cm size. Start looking up commercially available heat exchanger tube sizes.

I must admit that I was starting to think in terms of 3-5 cm tubes, N = 100 or more, and L = about 5 - 6 meters. So we were not so far apart in our thinking. But let's see how the design evolves.

Chet

 

[Undergrad1147]

Steam King has shown a figure of a heat exchanger with 2 passes on the tube side, and one pass on the shell side. I think that, in your problem, we can get away with a single pass on the tube side. Take a look at his bottom figure. This looks like a reasonable length to shell diameter ratio for the tube section between the end headers. Get out your ruler and measure the length to diameter ratio. Keep this ratio in the back of your mind.

Your estimate looks like a good starting point, although the tube diameter of 9.6 cm does seem a little large. You could squeeze more area in by using a larger number of smaller diameter tubes.

But, for now, try a tube layout with your N=35, 9.6 diameter tubes and see what you get in terms of shell size. Try a tube layout in an equilateral triangular array, and leave, say, a center-to-center tube spacing of about 1.5 D. Don't forget that the tube wall is going to have thickness, say, D/8. You don't have to do the whole layout. Just do a "unit cell," and get the shell area per tube. Multiply by the number of tubes, and use that area to find the diameter of the shell. How big does the shell come out to be, and how does it compare with the 8 meter length that you're assuming? How does the ratio compare with Steam King's heat exchanger? Also, look up in your books and see if heat exchanger tubes are available in 9.6 cm size. Start looking up commercially available heat exchanger tube sizes.

I must admit that I was starting to think in terms of 3-5 cm tubes, N = 100 or more, and L = about 5 - 6 meters. So we were not so far apart in our thinking. But let's see how the design evolves.

Chet

I see. I'll have a look into that now and then try play around with the possible values of N, D and L that may be a little more suitable.

Could we leave the HE for now and focus on the final part, the pump, briefly? I feel like I may be going into a little too much detail with the HE as we really haven't covered anything close to this in our current module. Maybe come next term after our module on heat exchange, I'd have more of an idea but I think I'll have to go ahead and clear up with my lecturer, what he expects with regards to the detail of this design.

With regards to the pump, I'm not really too sure where to start. I'm not given too much information about it apart from the fact that the fluid has to be pumped from the atmosphere to 1.69 MPa. Could you give me an idea of where to start? Thanks.

 

[Chestermiller]

Have you taken thermo yet? If so, do you remember the form of the first law for a flow system, involving shaft work and change in enthalpy of the flow stream? How much of a temperature rise do you expect through the pump? (I expect negligible).

chet

 

[Undergrad1147]

Have you taken thermo yet? If so, do you remember the form of the first law for a flow system, involving shaft work and change in enthalpy of the flow stream? How much of a temperature rise do you expect through the pump? (I expect negligible).

chet

I'm currently taking thermo. I think our terminology may be different to yours, or else I'm not following. A flow system, as in a control volume at steady state? One form of the first law which comes to mind that may be applicable to the pump is is: $$Q - W + \Sigma m_{in}h_{in} - \Sigma m_{out}h_{out} = \Delta U$$
The pump would be steady flow so $\Delta U$ would be 0, yes? Is this the equation that you're referring to or have I misunderstood you?

 

[Chestermiller]

I'm currently taking thermo. I think our terminology may be different to yours, or else I'm not following. A flow system, as in a control volume at steady state? One form of the first law which comes to mind that may be applicable to the pump is is: $$Q - W + \Sigma m_{in}h_{in} - \Sigma m_{out}h_{out} = \Delta U$$
The pump would be steady flow so $\Delta U$ would be 0, yes? Is this the equation that you're referring to or have I misunderstood you?

Yes, this is exactly the equation I was referring to, and yes, ΔU = 0. Also, Q is equal to zero (if you assume that the pump is adiabatic). So, what do you get for the shaft work?

 

[Undergrad1147]

Yes, this is exactly the equation I was referring to, and yes, ΔU = 0. Also, Q is equal to zero (if you assume that the pump is adiabatic). So, what do you get for the shaft work?

It's just one stream so $m_{in}$ will equal $m_{out}$, yes? If so, then: $$W = m(h_{in} - h_{out})$$
We're given the mass flow rate so we'll end up with the equation: $$\dot{W} = \dot{m} (h_{in} - h_{out})$$ Is this correct? The water is initially at 20 C and atmospheric presure, thus it's enthalpy is 83.95 kJ/kg and after leaving the pump it's at 20 C and 1.69 MPa with an enthalpy of 85.5 kJ/kg. Thus the Shaft Work Rate is equal to: $$\frac{36(1000) (83.95 - 85.5)}{3600} = -15.5 J/s = -15.5 W$$

Does this sound correct?

 

[Chestermiller]

It's just one stream so $m_{in}$ will equal $m_{out}$, yes? If so, then: $$W = m(h_{in} - h_{out})$$
We're given the mass flow rate so we'll end up with the equation: $$\dot{W} = \dot{m} (h_{in} - h_{out})$$ Is this correct?

OK so far.

The water is initially at 20 C and atmospheric presure, thus it's enthalpy is 83.95 kJ/kg and after leaving the pump it's at 20 C and 1.69 MPa with an enthalpy of 85.5 kJ/kg. Thus the Shaft Work Rate is equal to: $$\frac{36(1000) (83.95 - 85.5)}{3600} = -15.5 J/s = -15.5 W$$

Does this sound correct?

No. The change in specific enthalpy for an incompressible fluid is Δh=C_pΔT+vΔP, where v is the specific volume = 1/ρ, where ρ is the density. Try again. (15.5 Watts is a little low for a pump that is pumping all that water to a high pressure).

 

[Undergrad1147]

OK so far.


No. The change in specific enthalpy for an incompressible fluid is Δh=C_pΔT+vΔP, where v is the specific volume = 1/ρ, where ρ is the density. Try again. (15.5 Watts is a little low for a pump that is pumping all that water to a high pressure).

I was thinking it was a little low alright. Why isn't it $C_v$ as opposed to $C_p$ in the above equation though? h = u + Pv, so $\Delta h = \Delta u + \Delta Pv$ and $\Delta u = C_v \Delta T$ no?

 

[Chestermiller]

I was thinking it was a little low alright. Why isn't it $C_v$ as opposed to $C_p$ in the above equation though? h = u + Pv, so $\Delta h = \Delta u + \Delta Pv$ and $\Delta u = C_v \Delta T$ no?

For an incompressible liquid, C_p and C_v are the same.

Incidentally, that result you got probably was correct, except off by a factor of 1000. Those specific enthalpies were in kJ, not J. So the work ought to be 15.5 kW, not 15.5 W. Try doing it the way I suggested also to see that the result comes out the same.

The 15.5 kW is the mechanical work that must be delivered by the pump. But the pump will not be 100% efficient in converting electrical energy from the motor to mechanical energy. Pick what you consider to be a reasonable efficiency, and divide the 15.5 kW by that to get the actual pump requirement.

 

[Undergrad1147]

For an incompressible liquid, C_p and C_v are the same.

Incidentally, that result you got probably was correct, except off by a factor of 1000. Those specific enthalpies were in kJ, not J. So the work ought to be 15.5 kW, not 15.5 W. Try doing it the way I suggested also to see that the result comes out the same.

The 15.5 kW is the mechanical work that must be delivered by the pump. But the pump will not be 100% efficient in converting electrical energy from the motor to mechanical energy. Pick what you consider to be a reasonable efficiency, and divide the 15.5 kW by that to get the actual pump requirement.

Yeah. Using your way you get -15.9 kW. So, they're fairly close. If you were to say the pump is about 80% efficient then, 19.875 kJ/s needs to be supplied to the pump for the water to be pumped from $P_{atm}$ to 1.69 MPa.

 

[Chestermiller]

Yeah. Using your way you get -15.9 kW. So, they're fairly close. If you were to say the pump is about 80% efficient then, 19.875 kJ/s needs to be supplied to the pump for the water to be pumped from $P_{atm}$ to 1.69 MPa.

Very nice. I would round that result off to 20 kW.

I'm hoping you might consider continuing with the heat exchanger design. My suggestion: assume 100 tubes of 4 cm diameter, and use the equations in Chapter 14 of BSL to determine the heat transfer coefficients on the water side and on the steam side of the tubes, and then to determine the required length L. It's pretty much cook book. At the beginning of the chapter, they tell you to use the "film temperature" to determine the required physical properties of water. The film temperature is arithmetic average of the inlet and the outlet water temperatures, then averaged with the steam temperature. You need the water viscosity at this temperature, the heat capacity, the density, and the thermal conductivity. You need to determine the Reynolds number in each tube and the Prantdl number for water at the film temperature. There is then a forced convection graph from which you can determine the heat transfer coefficient on the water side. There is also an equation later in the chapter for getting the heat transfer coefficient for condensing vapor on a horizontal tube. From this, you can get a much much more accurate estimate of the overall heat transfer coefficient than the 680 value which you used earlier, and that it was too crude an approximation.

 

[Undergrad1147]

Very nice. I would round that result off to 20 kW.

I'm hoping you might consider continuing with the heat exchanger design. My suggestion: assume 100 tubes of 4 cm diameter, and use the equations in Chapter 14 of BSL to determine the heat transfer coefficients on the water side and on the steam side of the tubes, and then to determine the required length L. It's pretty much cook book. At the beginning of the chapter, they tell you to use the "film temperature" to determine the required physical properties of water. The film temperature is arithmetic average of the inlet and the outlet water temperatures, then averaged with the steam temperature. You need the water viscosity at this temperature, the heat capacity, the density, and the thermal conductivity. You need to determine the Reynolds number in each tube and the Prantdl number for water at the film temperature. There is then a forced convection graph from which you can determine the heat transfer coefficient on the water side. There is also an equation later in the chapter for getting the heat transfer coefficient for condensing vapor on a horizontal tube. From this, you can get a much much more accurate estimate of the overall heat transfer coefficient than the 680 value which you used earlier, and that it was too crude an approximation.

Thanks for the help. I'll have to hold off on the heat exchanger for now until I free up some time. I'll definitely get back to it at some point and I'll post about it. Thanks for the help with the problem in general.

 

[Chestermiller]

Thanks for the help. I'll have to hold off on the heat exchanger for now until I free up some time. I'll definitely get back to it at some point and I'll post about it. Thanks for the help with the problem in general.

My pleasure. It was very nice working with you. You showed great determination and tenaciousness.

Chet