Answer:
the approximate temperature is equal to 533 K
Step-by-step explanation:
the conditions given by the exercise are as follows:
P1=pressure at state 1 as saturated vapor=1 bar
m1=m2=mass flow rate at state 1=10 kg/s
P3=pressure of air at state 3=1 bar
T3=temperature of air at state 3=1200 K
m3=m4=mass flow rate of air=5 kg/s
The energy balance heat gained is equal to heat loss:
m1(h2-h1)=m3(h3-h4)=m3Cp(T3-T2) (eq. 1)
From saturated water tables:
h1=hg=2675.5 kJ/kg
Cp=1.004 kJ/kgK
if T2=240°C=513 K
From the superheated water tables we have at 513 K and 1 bar:
h2=2954.5 kJ/kg
Replacing values in equation 1:
10*(2954.5-2675.5)=5*1.004*(1200-513)
2790≠3448.74
We need to try another temperature
if T=260°C=533K
at 533K and 1 bar h2=2994.35 kJ/kg
Replacing in equation 1:
10*(2994.35-2675.5)=5*1.004*(1200-533)
3188.5≠3348.34
We observe that the difference is increasing. At this temperature, is approximately equal, thus the approximate temperature is equal to 533 K