Question

Four hundred kg/h of Steam from a low pressure turbine at 12.5 atm. and 250 C goes into a feed water heater which preheats the water entering the economizer in a boiler. The steam is comes out of the feed water at 200 C. The water at a flow rate of 500 kg/h enters the feed water heater at 20C. Calculate the feed water exit temperature.

Answer 1

Explanation:

According to the heat relationship,

    Heat given by steam = heat taken by water

Hence,    $(mC_{p} \Delta T)_{steam}$ = $(mC_{p} \Delta T)_{water}$

where, $C_{p}$ = specific heat

Value of $C_{p}_{steam}$ at 250^{o}C and 12.5 atm is 2.326 $kJ/Kg^{o}C$.

And, $C_{p}_{steam}$ at 200^{o}C and 12.5 atm is 2.743 $kJ/Kg^{o}C$.

Therefore, average $C_{p}_{steam}$ = $\frac{2.326 kJ/Kg^{o}C + 2.743 kJ/Kg^{o}C}{2}$

                                          = 2.5345 kJ/Kg^{o}C

As the given data is as follows.

      $m_{steam}$ = 400 kg/hr, $\Delta T_{steam}$ = $250^{o}C - 200^{o}C$ = $50^{o}C$, $m_{water}$ = 500 kg/hr

            $C_{p}_{water}$ = 4.186 $kJ/kg^{o}C$

Hence, putting the given values into the above formula as follows.

           $(mC_{p} \Delta T)_{steam}$ = $(mC_{p} \Delta T)_{water}$

  $(400 \times 2.5345 kJ/kg^{o}C \times 50^{o}C)_{steam}$ = $(500 kg \times 4.186 kJ/kg^{o}C \Delta T)_{water}$

                $\Delta T_{water} = 24.2^{o}C$

As we know that, $\Delta T_{water} = T_{2} - T_{1}$

                             $T_{2} - 20^{o}C = 24.2^{o}C$

                            $T_{2} = 44.2^{o}C$

Thus, we can conclude that the new temperature of water is $44.2^{o}C$.