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snow_lady [41]
3 years ago
6

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 tur

bine 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.
Chemistry
1 answer:
sergejj [24]3 years ago
8 0

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} - \frac{W_{s}}{m}

      H_{2} = 3358 kJ/kg - \frac{1500kJ/s \times 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 = \frac{200 kg/min}{60 s/min} \times (3371 - 2908)kJ/kg

[/tex]

          Q = 1543.33 kJ/s

or,    Q = 1543.33 kW

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

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The new temperature of the nitrogen gas is 516.8 K or 243.8 C.

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Gay-Lussac's law can be expressed mathematically as follows:

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Where P = pressure, T = temperature, K = Constant

You want to study two different states, an initial state and a final state. You have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment. By varying the temperature to a new value T2, then the pressure will change to P2, and the following will be fulfilled:

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In this case:

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<u><em>The new temperature of the nitrogen gas is 516.8 K or 243.8 C.</em></u>

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