To solve for the thermal efficiency of the Rankine cycle, we need to follow these steps:

Step 1: Identify relevant values

From the image, the following information is given:

• Boiler pressure: $60 \, \text{bar}$
• Condenser pressure: $0.1 \, \text{bar}$
• Boiler exit enthalpy: $h_1 = 2785.10 \, \text{kJ/kg}$
• Boiler exit entropy: $s_1 = 5.8891 \, \text{kJ/kg-K}$
• At 0.1 bar:
  • Saturated liquid enthalpy: $h_f = 191.81 \, \text{kJ/kg}$
  • Latent heat of vaporization: $h_{fg} = 2392.82 \, \text{kJ/kg}$
  • Entropy of saturated liquid: $s_f = 0.6492 \, \text{kJ/kg-K}$
  • Entropy of vaporization: $s_{fg} = 7.5010 \, \text{kJ/kg-K}$
• Saturated liquid volume at 0.1 bar: $v_f = 0.001010 \, \text{m}^3/\text{kg}$

Step 2: Calculate turbine work (W_t)

Since the steam leaving the boiler is dry and saturated, we assume the expansion in the turbine is isentropic (constant entropy process). Therefore, the entropy at the condenser (state 2) is equal to the entropy at the boiler exit (state 1):

$s_2 = s_1 = 5.8891 \, \text{kJ/kg-K}$

Now, we calculate the dryness fraction $x_2$ of the steam at the condenser (state 2) using the entropy relation:

$s_2 = s_f + x_2 \cdot s_{fg}$

Substituting the values:

$5.8891 = 0.6492 + x_2 \cdot 7.5010$

Solve for $x_2$:

$x_2 = \frac{5.8891 - 0.6492}{7.5010} = 0.6988$

Next, calculate the enthalpy at state 2:

$h_2 = h_f + x_2 \cdot h_{fg}$

Substitute the values:

$h_2 = 191.81 + 0.6988 \cdot 2392.82 = 1858.85 \, \text{kJ/kg}$

Thus, the turbine work per unit mass is:

$W_t = h_1 - h_2 = 2785.10 - 1858.85 = 926.25 \, \text{kJ/kg}$

Step 3: Calculate pump work (W_p)

The pump work can be calculated as:

$W_p = v_f \cdot (P_1 - P_2)$

Where:

• $v_f = 0.001010 \, \text{m}^3/\text{kg}$
• $P_1 = 60 \, \text{bar} = 6000 \, \text{kPa}$
• $P_2 = 0.1 \, \text{bar} = 10 \, \text{kPa}$

Substituting values:

$W_p = 0.001010 \cdot (6000 - 10) = 0.001010 \cdot 5990 = 6.05 \, \text{kJ/kg}$

Step 4: Calculate heat input (Q_in)

The heat input is the difference in enthalpy at the boiler:

$Q_{in} = h_1 - h_4$

Since the fluid is saturated at state 4:

$h_4 = h_f = 191.81 \, \text{kJ/kg}$

Thus:

$Q_{in} = 2785.10 - 191.81 = 2593.29 \, \text{kJ/kg}$

Step 5: Calculate thermal efficiency

Finally, the thermal efficiency is given by:

$\eta = \frac{W_t - W_p}{Q_{in}} \cdot 100$

Substitute the values:

$\eta = \frac{926.25 - 6.05}{2593.29} \cdot 100 = \frac{920.20}{2593.29} \cdot 100 = 35.49\%$

Rounding off to one decimal place:

$\eta \approx 35.5\%$

Thus, the thermal efficiency of the cycle is 35.5%.

Would you like further clarification on any steps?

Related Questions:

1. How does isentropic efficiency affect the overall thermal efficiency of a Rankine cycle?
2. What would be the efficiency if the pump work was neglected in this calculation?
3. How do different condenser pressures influence the thermal efficiency of the Rankine cycle?
4. What are the common methods used to improve the efficiency of the Rankine cycle?
5. How does superheating the steam affect the efficiency of the Rankine cycle?

Tip:

To increase the efficiency of a Rankine cycle, methods such as superheating the steam, reducing condenser pressure, or reheating the steam between expansion stages are commonly employed.