You have access to a flow of saturated steam (i.e., saturated vapor) that is con

Question

You have access to a flow of saturated steam (i.e., saturated vapor) that is continuously vented from an open feedwater heater in a coal-fired heating plant. The mass flow rate of the steam is = 0.35 kg/s and the steam is at Pstm = 50 psi. The environment is at pressure Pamb = 1 atm and Tamb = 20ºC. You are interested in trying to utilize the steam to provide some power. Initially, you would like to estimate the maximum possible rate at which power could be produced using this resource in this environment. The maximum possible rate of power that could be produced occurs if the steam is provided to a set of reversible equipment that exchanges heat with the atmosphere. The steam must be brought to the dead state as it exhausts from the equipment. This hypothetical system is shown in Figure 6.B-14(a).

a.)  Analyze the system shown in Figure 6.B-14(a) in order to determine the maximum possible rate at which power can be produced using the steam.

 In order to use the steam to provide some power, you have decided to feed it to a turbine which exhausts to the atmosphere, as shown in Figure 6.B-14(b).  The efficiency of the turbine is ht = 0.82.  Note that the steam leaving the turbine is at atmospheric pressure but not atmospheric temperature.  Therefore, the steam will eventually re-equilibrate thermally with the atmosphere and finally come to the dead state temperature.

 b.)  Analyze the system shown in Figure 6.B-14(b) and determine the power produced by the turbine.

Explanation / Answer

http://www.waterproperties.eu/

This site can give thermodynamic properties.

Now in first figure its ideal process.

Since Pressure P1 = 50psi = 50*6.895=344.75kPa, X1 = 1.00

h1 (enthalpy) = 2731.83 kJ/kg

Properties of dead state, P2 = 1atm = 101.325 kPa, Tamb = T2 = 20 celsius

h2 = 83.95 kJ/kg

Now W =mass flow rate X(h1 -h2) = 0.35(2731.83-83.95) = 926.758 kW

In second case, turbine work. If the turbine is ideal, process will be isentropic, i.e s1 = s2

Now s1 is known as state is defined by P1 = 344.75 Kpa, X1 =1

s1 = 6.9465 kJ/kgK

Now P2 = 1atm = 101.325 kPa, s2 = 6.9465 kJ/kgK

using this h2 = 2523.37 kJ/kg through interpolation

Now efficiency of turbine = 0.82

Hence Work = 0.82*0.35*(2731.83-2523.37) = 59.828kW

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