Question

Superheated steam at 20 bar and 450oC flows at a rate of 200 kg/min to an adiabatic turbine, where it expands to 10 bar. The turbine produces 1500 kW of work. The steam then flows to a heat exchanger where it is heated isobarically to its initial temperature. Neglect kinetic energy.

1) Using the energy balance on the turbine, solve for the outlet temperature (in oC).

2) Using the energy balance on the heater, solve for the required heat input (in kW) to the steam.

Answer 1

Step-by-step explanation:

The given data is as follows.

Pressure of steam at inlet of turbine,
`P_(1)` = 20 bar

Temperature at inlet of turbine, T =
`450^(o)C`

Pressure at outlet of turbine,
`P_(2)` = 10 bar

Mass flow rate of steam, m = 200 kg/min

Work produced by the turbine,
`W_(s)` = 1500 kW

Steam is heated at constant pressure to its initial temperature, i.e., temperature at outlet of heat exchanger,
`T_(3)` =
`450^(o)C`.

(1) For an adiabatic turbine, the energy balance is as follows.

`-W_(s) = m({H}_(2) - {H}_(1))`

where
`W_(s)` = work done by the turbine

m = mass flow rate of steam

`H_(1)` and
`H_(2)` are the specific enthalpy of steam at inlet and outlet conditions of turbine.

Obtain the specific enthalpy of steam from Properties of Superheated Steam table

At 20 bar and
`450^(o)C`,
`H_(1)` = 3358 kJ/kg

`H_(2) = H_(1) - (W_(s))/(m)`

`H_(2) = 3358 kJ/kg - (1500kJ/s * 60 s/min)/(200 kg/min)`

`H_(2)` = 2908 kJ/kg

For P = 10 bar,
`H` =2875 kJ/kg for T=
`200^(0)C` and H = 2975 kJ/kg for
`T=250^(o)C`. Interpolate the values.

The temperature corresponding to P = 10 bar and
`H_(2)` = 2908 kJ/kg is T =
`216.5^(o)C`

Therefore, the outlet temperature is
`T_(2) = 216.5^(o)C`.

(2) Energy balance on the heater is as follows.

Q =
`\Delta H` =
`m(H_(3) - H_(2))`

where, Q = heat input required by the steam

`\Delta H` = specific enthalpy change

`H_(3)` = specific enthalpy of steam at the outlet conditions of heat exchanger

At P = 10 bar and
`T_(3)` =
`450^(o)C`,
`H_(3)` = 3371 kJ/kg.

Q =
`(200 kg/min)/(60 s/min) * (3371 - 2908)kJ/kg`

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Q = 1543.33 kJ/s

or, Q = 1543.33 kW

Therefore, the heat input required is Q = 1543.33 kW.