# Sizing a Refrigeration system

1. Mar 13, 2012

1. The problem statement, all variables and given/known data

A glass manufacturer operates 11 months of the year and produces 97,000 tons of glass per year that needs to be cooled to ambient temperatures. The melting point of glass is 1673K and ambient temperature is assumed to be 293K. The specific heat capacity of glass is 0.837 kJ/kg.K. How much electricity is required to power a refrigeration system to remove the amount of heat given off by the molton glass?

2. Relevant equations

Qdot=MdotCpΔT

3. The attempt at a solution

Mdot = 97,000,000 kg / (335 days x 24 hours/day x 3600 seconds/hour) = 3.35 kg/s

Qdot = 3.35kg/s x 0.837 kJ/kg.K x (1673K-293K) = 3,871 kW

I know I have to do something with the COP but I'm completely stuck now. Any ideas guys?

Last edited: Mar 13, 2012
2. Mar 13, 2012

### gulfcoastfella

If you assume the coefficient of performance is at the upper limit (and is based on the reversed Carnot Cycle), then the COP is given by

COP = T$_{L}$ / (T$_{H}$ - T$_{L}$)

where T$_{L}$ is the temperature of the low temp reservoir, and T$_{H}$ is the temperature of the high temp reservoir.

3. Mar 13, 2012

### gulfcoastfella

Sorry, that COP definition doesn't apply here. The only refrigeration I'm familiar with is used to move heat from a cold region to a hot region, relatively speaking of course. The reason work is used is because heat will not flow from cold to hot on it's own. Your situation is different, though. If you wanted to, you could cool the molten glass using a heat exchanger. But a heat exchanger doesn't require work in the sense that a refrigeration cycle does.