1. ## Heat transfer problem,

A heat exchanger produces steam by boiling water within vertical tubes, as shown in Figure 3.1

i)Water is supplied at a rate of 2 kg s–1 and an initial temperature of 80°C. Calculate the rate of heat transfer required to produce saturated steam at 140°C. What would the pressure in the top of the heat exchanger be (in kPa gauge)?

ii)The water/steam passes through a bundle of 50 vertical tubes, each of 3 cm internal diameter, 4 cm outer diameter, and 4 m in length, made of stainless steel with thermal conductivity 19 W m–1 K–1. Calculate the overall resistance to heat transfer across the tube walls. Hence calculate the required temperature driving force to deliver the heat transfer rate calculated in part (i).
I am really stuck, don't know what to do.

Any help appreciated.

2. Okay I asked my teacher and he said I have to use this equation for the second part,

thermal resistance for a single pipe = $\frac{ln(\frac{r_{o}}{r_{i}})}{2 \pi H \lambda}$

$H = 4m$
so $r_{o} = 2 cm = 0.02 m$

$r_{i} = 1.5cm = 0.015 m$

$\lambda = 19 W m^{-1} K^{-1}$

pluggiing these numbers in I get $6.02\times 10^{-4}$

so the overall resistance, I am not sure but Can I just use the same formula as in electrical circuits, e.g $\frac{1}{R} = \frac{1}{R} + \frac{1}{R} ...$

and times that by 50, since there are 50 tubes
so R = $\frac{1}{ 6.02 x 10^{-4}} \times 50$

Is this correct?

I really need help, with the first part, since we are turning water into steam will I have to use Q = ml , for latent heat? or is it $Q = mc \delta T$.

3. 1. For the first part, you'll need to need to use:

$Q = mc_{water}\theta$

That'll give you the energy required to raise the temperature of water from 80 to 100 C

Then, use

$Q = mL_{vaporisation}$

That'll give you the energy required to boil the water.

$Q = mc_{steam}\theta$

Again to get the heat required to heat the steam at 100 to 140 C.

Use m = 2 kg, this will give the heat required in 1 second.

4. Originally Posted by Tweety
Okay I asked my teacher and he said I have to use this equation for the second part,

thermal resistance for a single pipe = $\frac{ln(\frac{r_{o}}{r_{i}})}{2 \pi H \lambda}$

$H = 4m$
so $r_{o} = 2 cm = 0.02 m$

$r_{i} = 1.5cm = 0.015 m$

$\lambda = 19 W m^{-1} K^{-1}$

pluggiing these numbers in I get $6.02\times 10^{-4}$

so the overall resistance, I am not sure but Can I just use the same formula as in electrical circuits, e.g $\frac{1}{R} = \frac{1}{R} + \frac{1}{R} ...$

and times that by 50, since there are 50 tubes
so R = $\frac{1}{ 6.02 x 10^{-4}} \times 50$

Is this correct?

I really need help, with the first part, since we are turning water into steam will I have to use Q = ml , for latent heat? or is it $Q = mc \delta T$.
I don't know how to do this... but if what you wrote is correct, then this is not correct:

$R = \dfrac{1}{ 6.02 \times 10^{-4}} \times 50$

You should be having:

$\dfrac{1}{R} = \dfrac{1}{ 6.02 \times 10^{-4}} \times 50$

So,

$R = \dfrac{6.02 \times 10^{-4}}{50}$

5. Originally Posted by Unknown008
1. For the first part, you'll need to need to use:

$Q = mc_{water}\theta$

That'll give you the energy required to raise the temperature of water from 80 to 100 C

Then, use

$Q = mL_{vaporisation}$

That'll give you the energy required to boil the water.

$Q = mc_{steam}\theta$

Again to get the heat required to heat the steam at 100 to 140 C.

Use m = 2 kg, this will give the heat required in 1 second.
Thank you very much,

so for the first one

$Q = 2 kg \times 4.198 KJ Kg^{-1} K^{-1} \times 100-80 C^{o} = 167.92$

for the next one I looked up the latent heat of vaporization for water, in my steam table, at 100C and found it to be 2257 KJ/Kg

$Q = 2kg \times 2257 = 4514 KJ$

$Q = 2kg \times 2.034 KJ Kg^{-1} K^{-1} \times 140-100C^{o} = 162.72$

Is this correct?

So now do I just add all of my values up to get the rate of heat required to produce saturated steam at 140?

Thanks so much

6. Yes, you're right

For the pressure, I'm sorry but I don't know what you are supposed to be using. Personally, I would have use the ideal gas equation, but this might be seen mostly as chemistry...

2 kg of water -> 111.1 moles of water.
Assuming that steam behaves as an ideal gas, the pressure is given by:

$PV = nRT$

I would say that the volume is kept constant from liquid to gas, hence a volume of 2 litres, or 2 dm^3 = 0.002 m^3
R = 8.31,
T = 273.15 + 140 = 412.15 K
n = 111.11

So, we can get P.

However, this gives an incredibly large value... 190275916.7 Pa = 190 MPa

7. Thank you, On my steam table, it shows saturated vapour pressure, at 140C to be 0.3614 MPa. Also I know that $pressure_ {absolute} = Pressure_{gauge} + pressure_{barometric}$

$1 MPa = 10^{6} Nm^{-2} = 10bar$

thanks, you have been a great help.

8. ## Heat Transfer Problem

Hi Tweety,
Part 1 is an enthalpy balance as shown by others. The pressure of saturated steam at 140 C is 53 psi absolute
Part 2 requires the heating media ,hot oil . high pressure steam, or other fluids

bjh