Question

Consider a cogeneration power plant that is modified with reheat and that produces \(3 \mathrm{MW}\) of power and supplies \(7 \mathrm{MW}\) of process heat. Steam enters the high-pressure turbine at \(8 \mathrm{MPa}\) and \(500^{\circ} \mathrm{C}\) and expands to a pressure of 1 MPa. At this pressure, part of the steam is extracted from the turbine and routed to the process heater, while the remainder is reheated to \(500^{\circ} \mathrm{C}\) and expanded in the low-pressure turbine to the condenser pressure of 15 kPa. The condensate from the condenser is pumped to 1 MPa and is mixed with the extracted steam, which leaves the process heater as a compressed liquid at \(120^{\circ} \mathrm{C}\). The mixture is then pumped to the boiler pressure. Assuming the turbine to be isentropic, show the cycle on a \(T\) -s diagram with respect to saturation lines, and disregarding pump work, determine \((a)\) the rate of heat input in the boiler and \((b)\) the fraction of steam extracted for process heating.

Short answer

Question: In a cogeneration power plant, determine (a) the rate of heat input in the boiler, and (b) the fraction of steam extracted for process heating. Answer: (a) The rate of heat input in the boiler is 11.92 MW. (b) The fraction of steam extracted for process heating is 69.4%.

Step-by-step solution

Identify the states on the T-s diagram

Begin by labeling the various states involved in the cogeneration cycle: 1. State 1 - Steam entering high-pressure turbine 2. State 2 - Steam exiting the high-pressure turbine and entering the process heater 3. State 3 - Steam exiting the process heater, mixed with condensate, and entering the low-pressure pump 4. State 4 - Steam exiting the low-pressure pump and entering the boiler 5. State 5 - Steam exiting the high-pressure pump and entering the low-pressure turbine 6. State 6 - Steam exiting the low-pressure turbine and entering the condenser Now, draw the T-s diagram and plot these points.

Determine the mass flow rate of steam

We are given that the power output is 3 MW and the process heat output is 7 MW. To determine the mass flow rate of steam entering the high-pressure turbine, we can use the specific enthalpy values at different states in the cycle. Using the steam tables and the given conditions at state 1, we find the specific enthalpy (\(h_1 = 3474.8 \frac{kJ}{kg}\)) and specific entropy (\(s_1 = 7.373 \frac{kJ}{kg \cdot K}\)). Since the high-pressure turbine is isentropic, the specific entropy remains constant (\(s_2 = s_1\)) and we find the specific enthalpy at state 2 is \(h_2 = 2865.1 \frac{kJ}{kg}\). The power output is given by \(W_{out} = m_1 \cdot (h_1 - h_2)\), where \(m_1\) is the mass flow rate of steam entering the high-pressure turbine. Thus, we find \(m_1 = \frac{W_{out}}{(h_1 - h_2)} = \frac{3000}{(3474.8 - 2865.1)} = 4.597 \frac{kg}{s}\).

Determine the fraction of steam extracted for process heating

We can use the process heat output (7 MW) and specific enthalpy values at states 2 and 3 to determine the mass flow rate of steam extracted for process heating, \(m_3\). We know that, for this heat exchange process, the specific enthalpy change in the stream is equal to the process heat supplied, so \(Q_{process} = m_3 \cdot (h_2 - h_3)\), where \(h_3\) is the specific enthalpy at state 3 which is equal to \(h_2\) as the system is isothermal. Using the steam table and the given conditions at state 3, we find the specific enthalpy is \(h_3 = 435.3 \frac{kJ}{kg}\). Thus, we can find \(m_3 = \frac{Q_{process}}{(h_2 - h_3)} = \frac{7000}{(2865.1 - 435.3)} = 3.192 \frac{kg}{s}\). The fraction of steam extracted for process heating is given by the ratio of the mass flow rates: \(\frac{m_3}{m_1} = \frac{3.192}{4.597} = 0.694\). So, the fraction of steam extracted for process heating is 0.694 or 69.4%.

Determine the rate of heat input in the boiler

To determine the heat input in the boiler (q_in), we can use an energy balance for the cycle. The net power output plus the process heat output is equal to the heat input minus the heat rejected in the condenser. We have only to find the heat rejected in the condenser now. The mass flow rate entering the condenser \(m_4 = m_1 - m_3 = 4.597 - 3.192 = 1.405 \frac{kg}{s}\). Using steam tables and the given conditions at state 6, we find the specific enthalpy (\(h_6 = 270.4 \frac{kJ}{kg}\)). The heat rejected in the condenser is given by \(Q_{condenser} = m_4 \cdot (h_6 - h_5)\), where \(h_5 = h_4\) due to isentropic expansion in the low-pressure turbine. Using steam tables and the given conditions at state 4, we find the specific enthalpy (\(h_4 = 1632.5 \frac{kJ}{kg}\)). Then, \(Q_{condenser} = 1.405 \cdot (270.4 - 1632.5) = -1917.6 \mathrm{kW}\). Now, we can calculate the heat input in the boiler using the energy balance equation: \(W_{out} + Q_{process} = q_{in} - Q_{condenser}\) or \(q_{in} = W_{out} + Q_{process} + Q_{condenser} = 3000 + 7000 - (-1917.6) = 11917.6 \mathrm{kW}\). Thus, the rate of heat input in the boiler is 11.92 MW. In summary, (a) the rate of heat input in the boiler is 11.92 MW, and (b) the fraction of steam extracted for process heating is 69.4%.