# The Volumetric Flowrate Under Suction Condition

For a National Board Exam Review:

An air conditioning plant with a capacity of $400$ $kW$ of refrigeration has an evaporating and condensing temperature of $3$ $^oC$ and $37$ $^oC$ respectively. If it uses Refrigerant 12, what volumetric rate of flow under suction condition?

Answer is $0.172$ $m^3/s$

So I try:

$${ V = mv}$$

$${ COP = \frac{T_{min}}{T_{max} - T_{min}} = \frac{Q_A}{W} = \frac{h_1 - h_4}{h_2 - h_1}}$$

$${ COP = \frac{ (3 + 273) }{37 - 3} = \frac{400}{W} = \frac{400}{h_2 - h_1}}$$

$${ W = h_2 - h_1 = 49.24859678 kJ/kg }$$

I've viewed every formula in the book and many other books. I cannot find a way to find m, for mass. even so... it's also hard to find the specific volume of freon without a given temperature. How am I supposed to go about this?

• Are you allowed to research all values that are not given? And what exactly is meant by"under suction conditions". Unfortunatly I don't have a formula at hand as I never came across a question phrased like this. Aug 4, 2015 at 20:29
• @idkfa Yes, this is for a philippine board exam and does not tie to any academic institution. the downside though that whatever we use, we should be able to memorize it.. Suction conditions = Compressor Inlet in the cycle. Aug 4, 2015 at 23:21