Question

# The gas-turbine portion of a combined gas-steam power plant has a pressure ratio of 16. Air enters the compressor at 300 K at a rate of 14 kg/s and is heated to 1500 K in the combustion chamber. The combustion gases leaving the gas turbine are used to heat the steam to $400^{\circ} \mathrm{C}$ at 10 MPa in a heat exchanger. The combustion gases leave the heat exchanger at at 420 K. The steam leaving the turbine is condensed at 15 kPa. Assuming all the compression and expansion processes to be isentropic, determine (a) the mass flow rate of the steam. (b) the net power output, and (c) the thermal efficiency of the combined cycle. For air, assume constant specific heats at room temperature.

Solution

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To solve this problem we will need the enthalpies at all points of the steam cycle.

For the $\textbf{enthalpy}$ $h_1$ we will use the saturated liquid water tables and the given pressure $p_1=15\text{ kPa}$.

$\begin{equation*} h_1=226\,\frac{\text{kJ}}{\text{kg}} \end{equation*}$

For the $\textbf{enthalpy}$ $h_2$ we will add the work done by the pump $w_{p1}$ to the enthalpy $h_1$. For the work $w_{p1}$ we will need the specific volume of the water $v_1=0.00101\text{ m}^3\text{/kg}$ and the pressure $p_1$ and $p_2=10000\text{ kPa}$ .

\begin{align*} h_2&=h_1+ v_1\cdot (p_2 - p_1) \\ h_2&=226\,\frac{\text{kJ}}{\text{kg}}+ 0.00101\,\frac{\text{m}^3}{\text{kg}}\cdot (10000\text{ kPa}- 15\text{ kPa}) \\ h_2&=236.08\,\frac{\text{kJ}}{\text{kg}} \end{align*}

To determine the $\textbf{enthalpy }h_3$ we will use the given pressure $p_3=10000\text{ kPa}$ and the temperature $T_3=400\text{\textdegree}\text{C}$ in the appropriate software.

$\begin{equation*} h_3=3100\,\frac{\text{kJ}}{\text{kg}} \end{equation*}$

The entropy $s_3=6.210\text{ kJ}\text{/kg K}$ is equal to entropy $s_4$ and we can use that and the pressure $p_4=15\text{ kPa}$ in the appropriate software to determine the $\textbf{enthalpy}$ $h_4$.

$\begin{equation*} h_4=2010\,\frac{\text{kJ}}{\text{kg}} \end{equation*}$

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