Question

A well-insulated heat exchanger has one line with 2 kg/s of air at 125 kPa and 1000 K entering, and leaving at 100 kPa and 400 K. The other line has 0.5 kg/s water entering at 200 kPa and 20 °C, and leaving at 200 kPa. Calculate the exit temperature of the water and the total rate of entropy generation?

Answer 1

Answer:

120°C

Explanation:

Step one:

given data

T_{wi} = 20^{\circ}C

T_{Ai}=1000K

T_{Ae}= 400kPa

P_{Wi}=200kPa

P_{Ai}=125kPa

P_{We}=200kPa

P_{Ae}=100kPa

m_A=2kg/s

m_W=0.5kg/s

We know that the energy equation is

$m_Ah_{Ai}+m_Wh_W=m_Ah_{Ae}+m_Wh_{We}$

making $h_{We}$ the subject of formula we have

$h_{We}=h_{Wi}+\frac{m_A}{mW}(h_A-h_{Ae})$

from the saturated water table B.1.1 , corresponding to  $T_{wi}= 20c$

$h_{Wi}=83.94kJ/kg$

from the ideal gas properties of air table B.7.1 , corresponding to T=1000K

the enthalpy is:

$h_{Ai}=1046.22kJ/kg$

from the ideal gas properties of air table B.7.1 corresponding to T=400K

$h_{Ae}=401.30kJ/kg$

Step two:

substituting into the equation we have

$h_{We}=h_{Wi}+\frac{m_A}{mW}(h_A-h_{Ae})$

$h_{We}=83.94+\frac{2}{0.5}(2046.22-401.30)\\\\h_{We}=2663.62kJ/kg$

from saturated water table B.1.2 at $P_{We}=200kPa$  we can obtain the specific enthalpy:

$h_g=2706.63kJ/kg$

we can see that $h_g>h_{Wi}$, hence there are two phases

from saturated water table B.1.2 at $P_{We}=200kPa$

$T_{We}=120 ^{\circ} C$