# Heat transfer problem,

• Nov 14th 2010, 10:21 AM
Tweety
Heat transfer problem,
Quote:

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.
• Nov 15th 2010, 08:38 AM
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 = $\displaystyle \frac{ln(\frac{r_{o}}{r_{i}})}{2 \pi H \lambda}$

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

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

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

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

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

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

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 $\displaystyle Q = mc \delta T$.
• Nov 15th 2010, 08:40 AM
Unknown008
1. For the first part, you'll need to need to use:

$\displaystyle Q = mc_{water}\theta$

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

Then, use

$\displaystyle Q = mL_{vaporisation}$

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

$\displaystyle Q = mc_{steam}\theta$

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.
• Nov 15th 2010, 08:48 AM
Unknown008
Quote:

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 = $\displaystyle \frac{ln(\frac{r_{o}}{r_{i}})}{2 \pi H \lambda}$

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

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

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

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

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

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

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 $\displaystyle Q = mc \delta T$.

I don't know how to do this... but if what you wrote is correct, then this is not correct:

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

You should be having:

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

So,

$\displaystyle R = \dfrac{6.02 \times 10^{-4}}{50}$
• Nov 15th 2010, 10:27 AM
Tweety
Quote:

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

$\displaystyle Q = mc_{water}\theta$

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

Then, use

$\displaystyle Q = mL_{vaporisation}$

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

$\displaystyle Q = mc_{steam}\theta$

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

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

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

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

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

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
• Nov 15th 2010, 10:35 AM
Unknown008
Yes, you're right (Smile)

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:

$\displaystyle PV = nRT$

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
• Nov 15th 2010, 10:47 AM
Tweety
Thank you, On my steam table, it shows saturated vapour pressure, at 140C to be 0.3614 MPa. Also I know that $\displaystyle pressure_ {absolute} = Pressure_{gauge} + pressure_{barometric}$

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

thanks, you have been a great help.
• Nov 17th 2010, 05:58 AM
bjhopper
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