Question

Steam is to be condensed in the condenser of a steam power plant at a temperature of \(50^{\circ} \mathrm{C}\) with cooling water from a nearby lake that enters the tubes of the condenser at \(12^{\circ} \mathrm{C}\) at a rate of \(240 \mathrm{kg} / \mathrm{s}\) and leaves at \(20^{\circ} \mathrm{C}\). Assuming the condenser to be perfectly insulated, determine (a) the rate of condensation of the steam and ( \(b\) ) the rate of energy destruction in the condenser.

Short answer

Answer: The rate of condensation of the steam is 64 kg/s and the rate of energy destruction in the condenser is 4,589,839.81 W.

Step-by-step solution

Equate the heat transfer in and out of the condenser

As the condenser is perfectly insulated, the heat transfer into the system should equal the heat transfer out of the system. For cooling water, the change in temperature is ΔT_c = 20°C - 12°C = 8°C. For steam, the change in temperature is ΔT_s = 50°C - 20°C = 30°C. The equation can be written as: $$m_s c_{s} ΔT_s = m_c c_{c} ΔT_c $$

Plug in the known values and solve for m_s

We need to find the mass flow rate for the steam. Let's plug in the known values and constants into the equation: For water/steam, the specific heat capacity is \(c_s = c_c = 4186 J/(kg\cdot K)\). Given \(m_c = 240 kg/s\). The equation now becomes: $$m_s \cdot 4186 \cdot 30 = 240 \cdot 4186 \cdot 8$$ Solve for m_s: $$m_s = \frac{240 \cdot 4186 \cdot 8 }{4186 \cdot 30}$$ $$m_s = 64 kg/s$$ The rate of condensation of the steam is 64 kg/s.

Calculate the rate of energy destruction in the condenser

For the rate of energy destruction in the condenser, we can use the formula: $$\dot E_{destruction} = T_0 \cdot (s_{out} - s_{in})$$ where T_0 is the reference temperature (usually the ambient temperature), s_out is the entropy of the exit stream, and s_in is the entropy of the inlet stream. Assuming that the specific heat capacity remains constant for both inlet and exit streams, we can write the entropy change as follows: $$Δs = s_{out} - s_{in} = c \cdot ln\left({\frac{T_{out}}{T_{in}}}\right)$$ For cooling water, \(c = 4186 J/(kg\cdot K)\), \(T_{in} = 12 + 273.15 = 285.15 K\), and \(T_{out} = 20 + 273.15 = 293.15 K\). $$Δs_c = 4186 \cdot ln\left({\frac{293.15}{285.15}}\right)$$ $$Δs_c = 81.486 J/(kg\cdot K)$$ For steam, \(c_s = 2010 J/(kg\cdot K)\), \(T_{in} = 50 + 273.15 = 323.15 K\), and \(T_{out} = 273.15 K\) (since the steam is condensed and leaves at \(0^{\circ} C\)). $$Δs_s = 2010 \cdot ln\left({\frac{273.15}{323.15}}\right)$$ $$Δs_s = -320.255 J/(kg\cdot K)$$ Now, we can calculate the rate of energy destruction: $$\dot E_{destruction} = 285.15\cdot(64\cdot (-320.255) + 240\cdot81.486)$$ $$\dot E_{destruction} = -4,589,839.81 W$$ As the value is negative, it indicates that energy is being destroyed in the condenser, at a rate of 4,589,839.81 W. To summarize: (a) The rate of condensation of the steam is 64 kg/s. (b) The rate of energy destruction in the condenser is 4,589,839.81 W.