Question

Reserve Problem 4.006 Tutorial SI

A feedwater heater in a vapor power plant operates at steady state with liquid water entering at inlet 1 with 45°C and 5.0 bar. Water vapor at 320°C and 5.0 bar enters at inlet 2. Saturated liquid water exits with a pressure of 5.0 bar. Ignore heat transfer with the surroundings and all kinetic and potential energy effects. The mass flow rate of the liquid entering at inlet 1 is 3.2 × 105 kg/h. Determine the temperature at the exit, , in °C, and the mass flow rate at inlet 2, in kg/h.

Answer 1

a)
`T = 151.83^(\textdegree)C`, b)
`\dot m_(2) = 58.61* 10^(3)\,(kg)/(h)`

Step-by-step explanation:

A feedwater heater is a mixing chamber that pre-heats water prior to compressor, improving the efficiency of the power cycle. This system is modelled after the First Law of Thermodynamics:

`\dot m_(1)\cdot h_(1) + \dot m_(2)\cdot h_(2) - (\dot m_(1)+ \dot m_(2))\cdot h_(3) = 0`

The properties at inlets and outlets are, respectively:

Inlet 1 (Subcooled Liquid)

`h_(1) = 188.44\,(kJ)/(kg)`

Inlet 2 (Superheated Vapor)

`h_(2) = 3106\,(kJ)/(kg)`

Outlet 3 (Saturated Liquid)

`h_(3) = 640.09\,(kJ)/(kg)`

a) The temperature at the exit is:

`T = 151.83^(\textdegree)C`

b) The mass flow rate at inlet 2 is:

`\left(3.2* 10^(5)\,(kg)/(h) \right)\cdot \left(188.44\,(kJ)/(kg) \right) + \dot m_(2)\cdot \left(3106\,(kJ)/(kg) \right) - \left(3.2* 10^(5)\,(kg)/(h)+\dot m_(2)\right)\cdot \left(640.09\,(kJ)/(kg) \right)= 0`
`2465.91\cdot \dot m_(2) -144528000 = 0`

`\dot m_(2) = 58.61* 10^(3)\,(kg)/(h)`