Thermo Question Steady-Flow System

Question

An air-conditioning system involves the mixing of cold air and warm outdoor air before the mixture is routed to the conditioned room in steady operation. Cold air enters the mixing chamber at 7°C and 105 kPa at a rate of 0.55 m3/s while warm air enters at 34°C and 105 kPa. The air leaves the room at 24°C. The ratio of the mass flow rates of the hot to cold airstreams is 1.6. Using variable specific heats, determine (a) the mixture temperature at the inlet of the room and (b) the rate of heat gain of the room.

Answer 1

(a) Determining the Mixture Temperature at the Inlet of the Room We are given:

Cold air enters at 7°C and 105 kPa with a volumetric flow rate of 0.55 m³/s

Warm air enters at 34°C and 105 kPa.

The ratio of the mass flow rates of hot to cold air is 1.6.

The volumetric flow rate of the cold air is given, so we can find its mass flow rate if we assume air behaves as an ideal gas and use the specific volume at the given state.

For air at 7^oC and 105KPa the density is 1.3075 Kg/m³.

So the Mass flow rate of cold air is 0.7191 Kg/s and the Mass flow rate of Hot air is 1.1506 Kg/s.

Now, using the balance of enthalpies, we can find the mixture enthalpy and then find the mixture temperature.

In the following I used M instead of mdot.

M_ch_c+M_hh_h=(M_c+M_h)h_m

0.719*280.44+1.1506*307.66=1.8697*h_m so h_m=297.172 KJ/Kg

which is very close to the air specific enthalpy at 24^oC

(b) Determining the Rate of Heat Gain of the Room

To find the room heat gain, we can use the exact enthalpy of air at 24^oC which is 297.58 KJ/Kg. so the heat rate is total mass flow rate times the difference in enthalpies.

Heat rate=(M_c+M_h)*(h_m-h_room)=1.8697*(297.172-297.58)=-0.762 Watts