A heat exchanger is set up so that saturated liquid water en

A heat exchanger is set up so that saturated liquid water entering at 100°C is cooled by a stream of air that enters at 0°C and 100 kPa. The water is flowing at a rate of 0.05 kg/s and the air enters at a rate of 0.1 kg/s. (This is a lot of air.) The air exits at 50°C. (

a) What is the final temperature of the water stream? (You can assume it is a saturated liquid at the exit.)

(b) What is the volumetric flow rate (m3/s) of the air at the inlet and outlet?

Solution

mass flow rate of water, m_w= 0.05 kg/s

inlet temp of water. t_w1=100 ⁰ C

let outlet temp of water be t_w2.

mass flow rate of air. m_a= 0.1 kg/s

inlet temp of air t_a1= 0 ⁰C

pressure of air, P=100 kPa

out let temperature of air, t_a2=50⁰C

In absence of data on effectiveness of heat exchanger, we assume that it is 100%.

let specific heat of water s=4.187 kJ/kg/K,

specific heat of air at constant pressure=C_{p= 1.005 kJ/kg/K}

Universal gas constant for air, R=0.287 kJ/kg/k

Then in a heat exchanger,

Heat lost by hot fluid = heat gained by cold fluid

m_wxSx(t_w1-t_w2)=m_axC_px(t_2-t₁)

0.05x4.187x(100-t_w2)=0.1x1.005x(50-0)

t_w2=76⁰C

a) The final temperature of the water stream =76⁰ C

Now for volumetric flow rates:

Using perfect gas equation assuminf air to be a perfect gas, PV=m RT

V=mRT/P

Volumetric flow rate of air at inlet =V₁=m_aRT_a1/P₁

T_a1=t_a1+273=0+273=273 K

V₁= 0.1X0.287X(0+273)/100

= 0.0783 m³/s

Similarly, Volumetric flow rate of air at outlet (T_a2=t_a2+273=50+273=323 K)

V₂₌ m_aRT_a2/P2

Here P_a2=P_a1 because it is a constant pressure process.

V₂=0.1x0.287x323/100

   =0.0927 m³/s.