Question

Refrigerant-134a at 700 kPa, 70°C, and 7.2 kg/min is cooled by water in a condenser until it exists as a saturated liquid at the same pressure. The cooling water enters the condenser at 300 kPa and 15°C and leaves at 25°C at the same pressure. Determine the mass flow rate of the cooling water required to cool the refrigerant. The enthalpies of R-134a at the inlet and exit states are 308.34 kJ/kg and 88.82 kJ/kg. The enthalpies of compressed water at both states are 62.98 kJ/kg and 104.83 kJ/kg.

Answer 1

The mass flow rate of cooling water required to cool the refrigerant is
`123.788\,(kg)/(min)`.

Step-by-step explanation:

A condenser is a heat exchanger used to cool working fluid (Refrigerant 134a) at the expense of cooling fluid (water), which works usually at steady state. Let suppose that there is no heat interactions between condenser and surroundings.The condenser is modelled after the First Law of Thermodynamics, which states:

`\dot Q_(ref) - \dot Q_(w) = 0`

`\dot Q_(ref) = \dot Q_(w)`

`\dot m_(ref)\cdot (h_(ref, in) - h_(ref,out)) = \dot m_(w)\cdot (h_(w, out) - h_(w,in))`

The mass flow rate of the cooling water is now cleared:

`\dot m_(w) = \dot m_(ref )\cdot (h_(ref,in)-h_(ref,out))/(h_(w,out)-h_(w,in))`

Given that
`h_(ref,in) = 808.34\,(kJ)/(kg)`,
`h_(ref, out) = 88.82\,(kJ)/(kg)`,
`h_(w,out) = 104.83\,(kJ)/(kg)` and
`h_(w,in) = 62.98\,(kJ)/(kg)`, the mass flow of the cooling water is:

`\dot m_(w) = \left(7.2\,(kg)/(min) \right)\cdot \left((808.34\,(kJ)/(kg)-88.82\,(kJ)/(kg) )/(104.83\,(kJ)/(kg)-62.98\,(kJ)/(kg) ) \right)`

`\dot m_(w) = 123.788\,(kg)/(min)`

The mass flow rate of cooling water required to cool the refrigerant is
`123.788\,(kg)/(min)`.