Challenge Question: Specific Heats [Solved]

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QUESTION: Your city in northern Montana is holding a town council meeting on installing heated sidewalks to cut down on the costs of shoveling and plowing snow every winter. As a chemical engineer, you are brought in to consult on the feasibility and cost effectiveness of this design.

An average winter day can bring temperatures of -20?F and 1.5 feet of snow. The town council would like to install the heated portion of the sidewalk in the main downtown stretch of the city. This stretch of sidewalk is 2 miles long and 9 feet wide on each side of the street. The proposal calls for two pipes 8 in. in diameter to run underneath each sidewalk. There is already a steam plant in your town, which provides steam to the city for various utility purposes. The steam, after it has been used, returns back to the plant at 350?C. An idea has been suggested to use the waste steam at the heat for the sidewalk system. The benefit to this is that no extra steam is produced. The initial costs for installation of the pipes is going to be around $50,000, and to pump the steam the additional distance under the streets costs $1.50/kW*hr. The current snow removal system costs $750 per job and takes at least an hour to implement. This current system is used at least 80 times per winter.

To calculate the feasibility of this idea, make the following assumptions: the thermal conductivity of the cement is not important, to idealize the system, the cement is perfectly conductive; the steam in the pipes is at 10 bar of pressure and has a flow rate of 100 kg/s; the final pressure of the steam out is less than 1 bar; the specific heat of snow is 2.108 kJ/kg*?C and the snow has to be heated to 40?F; the snow is 30% by volume water and the density of the air in town is 1.225 kg/m3.

Is the steam pipe proposal a cost-effective way to remove the snow from your city's sidewalks? Explain.

Postscript: This type of snow removal is a commonly used method in northern European countries such as Sweden and Norway.


Congratulations to Vinith Bejugam, who is a PhD student and graduate assistant at the University of Arkansas, Fayetteville. He provided the best answer to this challenge. Vinith is currently working with Dr. Keith Roper, who is researching electromagnetic interactions with nanometer-scale architectures to find applications that will serve healthcare, energy, and the environment. Vinith's research interests are plasmonics, nanomaterials, membranes separations, and alternative fuels.

The proposed technique of snow melting would be marginally cost-effective. First, the installation charges of pipes have been rounded off to $50,000. In addition, it would cost $1.5/kw*hr for pumping steam across additional distances. However, we do not know the exact figure. On the other hand, the current snow removal system costs around $60,000 every winter for a single job, which is a huge amount, yet we do not know the number of people who would be assigned to do such task. From this viewpoint, our game is full of improbability.

The energy output to the snow in this system could be analyzed quantitatively by using Chapman and Katunich's equation(courtesy:
qo= qs + qm + Ar (qe+ qh)
where qs = sensible heat transferred to the snow
qm = heat of fusion
Ar = ratio of snow-free area to total area (dimensionless),
qe= heat of evaporation
qh= heat transfer by convection and radiation
qo,qs,qm,qe and qh are in KJ/h.m^2

Further, qs= s .cp. D (Td- Ta)
s = rate of snowfall (inches of water equivalent per hour),
D = density of water equivalent of snow
cp= specific heat of snow
Td=40 F
Ta = air temperature (20 F)

Let us take s as 1.5 feet per day
=0.75 in/hr
Density of water equivalent(D)= Snow depth/density
Density of snow can be calculated as follows:
?snow/?water=weight of snow(30%)/weight of water
This gives ?snow=0.3675kg/m^3
And D= 0.007
Upon sustituting these values in appropriate units and solving for the above equation yields qs=0.004KJ/Kg-Kelvin

qm is the heat of fusion to melt the snow
qm = s. hf. D
hf=enthalphy of fusion for water(333.77KJ/Kg)

Ar is the Snow Free Area Ratio is defined by:
Ar = Af/At, where
Af = equivalent snow-free area, ft2
As = equivalent snow-covered area, ft2
At = Af + As = total area, ft2
For simplicity, let us assume Ar=1(no accumulation of snow)

qe is the heat of evaporation which can be neglected since we would want to bring snow to 40 F and that is all!

qh is the heat transfer by convection and radiation.
qh= 11.4 (0.0201 V + 0.055) (tf - ta)
where tf = water film temperature (F), usually taken as 33F.
where v is the speed of wind

However, all the above equations are generally used for pavement heating through geothermal energy and not steam. The term qh might vary considerably for steam heating system. The sum total of qs and qm =0.0485KJ/h-m^2 which is quite low , and the steam at 350 C will reach more than this requirement beyond any doubt. Thus, there's no dearth in providing energy to the system(pavement).

Now, the type of material employed for the pipe is very significant for determining costs in the long run once this model is considered. Although initial expenses would be low, factors such as pipe material, corrosion problems and fouling problems have to be considered before designing an effective snow-melting system. Having said this, further information has to be gathered in order to reach accurate calculations for the cost of this system.

Image: Borstal Boy via Flickr


Jeff's picture

"...the specific heat of snow is 2.108 kJ/kg*°C and the snow has to be heated to 40°F; the snow is 30% by volume water and the density of the air in town is 1.225 kg/m3." .... In the preceding comment are we to assume that the 30% of the snow that is water is 30% liquid water with 70% being ice, or that the composition is 30% water/ice with the balance being "trapped" air, in which we can use the air density to get a average snow density. If it's the latter, then I'm correct that the specific heater provided is a bulk material property for the snow.

Hi Jeff, Sorry for the confusion. The snow is 30% by weight water and the remainder is air. For this problem, you can assume that the energy is all going to the water and does not affect the air. This idealizes the system and makes for an easier calculation. Gool luck and thanks for the question. Cheers, Kilean

David's picture

I'd use the steam to heat an antifreeze solution rather than using steam directly. Even when not snowing, the steam system would have to be kept warm; trap failure could result in freezing and destruction of the system.