Question

Water enters the condenser at 30^oC which has a mass of 10 kg/s. If heat added is 100 kW, determine the outlet temperature in KJ/s.

Answer 1

For a condenser, we have the following analysis:

Then, by using [eq. (a)],we have to find the temperature of the water leaving the condenser because we can use the information that we have at the start:

$t_{w_{1}}=30°C=303.15\,K \\ c_w=4.184\frac{kJ}{kg\,K} \\ m_w=10\frac{kg}{s} \\ Q=+100\,kW=100\frac{kJ}{s} \\ Q=h_1-h_2 \\ Q=m_w(h_{w_{2}}-h_{w_{1}})=m_wc_w(t_{w_{2}}-t_{w_{1}}) \\ \implies t_{w_{2}}=t_{w_{1}}+\frac{Q}{m_wc_w}$

After we rearrange, we determine $t_{w_{2}}$ or the temperature of the water leaving the condenser:

$t_{w_{2}}=303.15\,K+2.39\,K \\ \implies t_{w_2}=305.54\,K=32.39\,°C$

With that information we can calculate $h_2=-m_wh_{w_2}$ or the 'outlet temperature' using the requested dimensions:

$h_2=-m_wh_{w_2}=-m_w \times c_{w} \times t_{w_2} \\ h_{2}=-(10\frac{kg}{s})(4.184\frac{kJ}{kg\,K})(305.54\,K) \\ \implies h_{2}=-12783.79\cfrac{kJ}{s}$

In conclusion, we find the 'outlet temperature' of the condenser as h₂ = -12783.79 kJ/s.Reference:

• Rajput, R. K. (2005). A textbook of engineering thermodynamics. Laxmi Publications.