Question

Turbine bleed steam enters an open feed water heater of a regenerative Rankine cycle at 40 psia and \(280^{\circ} \mathrm{F}\) while the cold feed water enters at \(110^{\circ} \mathrm{F}\). Determine the ratio of the bleed steam mass flow rate to the inlet feed water mass flow rate required to heat the feed water to \(250^{\circ} \mathrm{F}\)

Short answer

Answer: The required mass flow rate ratio of bleed steam to inlet feed water is approximately 0.0728.

Step-by-step solution

Find the enthalpy of the bleed steam

Using the pressure and temperature of the bleed steam, we can look up the enthalpy values in a steam table or use an online tool like the NIST webbook to find the enthalpy of the bleed steam. We find \(h_s = 1174.3\, Btu/lbm\).

Find the enthalpy of the cold feed water

Next, we need to find the enthalpy of the cold feed water. As feed water is in the compressed liquid state, we can approximate the specific enthalpy as \(h_w = h_{f}\) at \(110^{\circ} \mathrm{F}\) using the steam table. We find \(h_w = 76.56\, Btu/lbm\).

Find the enthalpy of the outlet mixture

The outgoing mixture has to be heated to \(250^{\circ} \mathrm{F}\). At this temperature, we can again approximate the specific enthalpy as \(h_{mix} = h_{f}\), using the steam table. We find \(h_{mix} = 150.563\, Btu/lbm\).

Apply energy balance equation

Now we can apply the energy balance equation over the feed water heater: $$m_s \cdot h_s + m_w \cdot h_w = m_{mix} \cdot h_{mix}$$ Substituting the mass balance equation yields: $$m_s \cdot h_s + m_w \cdot h_w = (m_s + m_w) \cdot h_{mix}$$

Solve for the mass flow rate ratio

We are interested in finding the ratio of the bleed steam mass flow rate to the inlet feed water mass flow rate, i.e., \(\frac{m_s}{m_w}\). To find this, we can rearrange the energy balance equation: $$m_s(h_s - h_{mix}) = m_w(h_{mix} - h_w)$$ Dividing both sides by \(m_w\) gives: $$\frac{m_s}{m_w} = \frac{h_{mix} - h_w}{h_s - h_{mix}}$$ Now, we can plug in the values for the enthalpies: $$\frac{m_s}{m_w} = \frac{150.563 - 76.56}{1174.3 - 150.563}$$ Calculating the ratio, we get: $$\frac{m_s}{m_w} = 0.0728$$ So, the required mass flow rate ratio of bleed steam to inlet feed water is approximately \(0.0728\).