A steam power plant operates on an ideal Carnot cycle. Dry saturated steam at 20 bar pressure is supplied to a turbine. It expands isentropically to a condenser pressure of 0.075 bar. Assuming that saturated water enters the boiler, calculate (a) the thermal efficiency, (b) the work ratio, and (c) the specific steam consumption.

Solution

For dry saturated steam at 20 bar, from the steam tables, we have

h₁ = h_g = 2799.5 kJ/kg, s₁ = s_g = 6.3408 kJ/kg.K

At 0.075 bar condenser pressure, from the steam tables, we have

h_f2 = 168.77 kJ/kg, h_fg2 = 2406.0 kJ/kg,

s_f2 = 0.5763 kJ/kg.K, s_fg2 = 7.6751 kJ/kg.K

Now s₁ = s₂ = s_f2 + x₂ s_fg2 for the isentropic expansion process 1−2. (Fig.4.2)

6.3408 = 0.5763 + x₂ × 7.6751

or x₂ = 0.751

∴ h₂ = h_f2 + x₂h_fg2 = 168.77 + 0.751 × 2406.0 = 1975.676 kJ/kg

At 20 bar, we have

h_f4 = 908.77 kJ/kg, s₄ = s_f4 = 2.4473 kJ/kg.K

For the isentropic compression process 3−4, we have

s₃ = s₄

or s_f3 + x₃ s_fg3 = s₄

Now s_f3 = s_fg3 and s_fg3 = s_fg2

0.5763 + x₃ × 7.6751 = 2.4473

or x₃ = 0.244

h_f3 = h_f2 and h_fg3 = h_fg2

h₃ = h_f3 + x₃ h_fg3 = 168.77 + 0.244 × 2406.0 = 755.834 kJ/kg

Pump work, w_p = h_f4 − h₃ = 908.77 − 755.834 = 152.936 kJ/kg

Turbine work, w_t = h₁ − h₂ = 2799.5 − 1975.676 = 823.824 kJ/kg

Net work, w_net = w_t − w_p = 823.824 − 152.936 = 670.888 kJ/kg

Heat supplied, q_s = h₁ − h_f4 = 2799.5 − 908.77 = 1890.73 kJ/kg

1. Thermal efficiency
2. Work ratio,
3. Specific steam consumption,

In an ideal Rankine cycle, saturated steam vapours enter the turbine at 160 bar and saturated liquid exits the condenser at 0.05 bar. The net power output of the cycle is 130 MW. Determine for the cycle (a) the thermal efficiency, (b) the steam mass flow rate, and (c) the specific steam consumption.

Solution

The Rankine cycle is shown in Fig. 4.4.

From steam tables,

For p₁ = 160 bar, h₁ = h_g = 2580.6 kJ/kg, s₁ = s_g1 = 5.2454 kJ/kg.K

p₂ = 0.05 bar, v_f3 = 0.001005 m³/kg, v_g3 = 28.193 m³/kg, h_f3 = 137.79 kJ/kg,

h_fg3 = 2423.7 kJ/kg, s_f3 = 0.4763 kJ/kg.K, s_fg3 = 7.9187 kJ/kg.K

For insentropic process 1−2,

s₁ = s₂ = s_f3 + x₂s_fg3

or 5.2454 = 0.4763 + x₂ × 7.9187

or x₂ = 0.602

h₂ = h_f3 + x₂h_fg3 = 137.79 + 0.602 × 2423.7 = 1596.86 kJ/kg

h₃ = h_f3 = 137.79 kJ/kg

h_M = h₃ + v_f3 (p₁ − p₂) × 10²

Turbine work, w_t = h₁ − h₂ = 2580.6 − 1596.86 = 983.74 kJ/kg

Pump work, w_p = h_M − h₃ = 153.86 − 137.79 = 16.07 kJ/kg

Heat added q_a = h₁ − h_M = 2580.6 − 153.86 = 2426.74 kJ/kg

1. Thermal efficiency =
2. Steam mass flow rate,
3. Specific steam consumption,

Figure 4.4 Ideal Rankine cycle

In a steam power plant working on ideal Rankine cycle, the steam turbine receives steam at 10 bar 250°C and discharges at 0.5 bar. Find the thermal efficiency.

Solution

At p₁ = 10 bar, t_s = 179.91°C. Therefore, steam is superheated. The T-s diagram is shown in Fig. 4.5.

Figure 4.5 Rankine cycle

From steam tables, at 10 bar, 250°C,

h′₁ = 2942.6 kJ/kg, s′₁ = 6.9246 kJ/kg.K

At p₂ = 0.5 bar, v_f3 = 0.001030 m³/kg, h_f3 = 340.47 kJ/kg, h_fg3 = 2305.4 kJ/kg, s_{f 3} = 1.091 kJ/kg.K,

s_fg3 = 6.5029 kJ/kg.K,

or 6.9246 = 1.091 + x′₂ × 6.5029

or x′₂ = 0.897

Pump work, w_p = v_f3 (p₁−p₂) × 10² = 0.001030 (10 − 0.5) × 10² = 0.9785 kJ/kg

Net work output

Heat added

Thermal efficiency

A high pressure boiler delivers steam at 90 bar and 480°C. The steam is expanded in the first stage of turbine to 12 bar and withdrawn and passed on the reheater. The expansion now takes place in the second stage of the steam turbine, down to the condenser pressure of 0.07 bar. Calculate (a) the efficiency of the ideal cycle and (b) the work output and efficiency of the reheat cycle operating through the same states neglecting pump work.

Figure 4.10 Reheat Rankine cycle

Solution

The cycle on T-s diagram is shown in Fig. 4.10.

At 90 bar, 480°C, from steam tables,

h₁ = 3336.5 kJ/kg, s₁ = 6.5942 kJ/kg.K

Now, s₁ = s₂

Since, s_g at 12 bar is less than 6.5942, the state 2 is in the superheated region. then, we get

h₂ = 2814.4 kJ/kg, t₂ = 201°C

At 12 bar, 480°C, h₃ = 3432.8 kJ/kg, s₃ = 7.6198 kJ/kg.K

At 0.07 bar, s_f4 = 0.5589 kJ/kg.K, s_fg4 = 7.7179 kJ/kg.K

or 7.6198 = 0.5589 + x₄ × 7.7179

or x₄ = 0.915

h₄ = h_f4 + x₄h_fg4 = 163.40 + 0.915 × 2409.1 = 2367.7 kJ/kg

h_fs = h_f4 = 163.4 kJ/kg

h₆ = h₅ + w_p

w_p = v_fs (p₁ − p₂) × 10²

= 0.001007 (90 − 0.07) × 10² = 9 kJ/kg

h₆ = 163.4 + 9 = 172.4 kJ/kg

1. q_a = (h₁ − h₆) + (h₃ − h₂) = (3336.5 − 172.4) + (3432.8 − 2814.4) = 3782.5 kJ/kg

   w_net = (h₁ − h₂) + (h₃ − h₄) − w_p

   = (3336.5 − 2814.4) + (3432.8 − 2367.7) − 9 = 1578.2 kJ/kg

   
2. Neglecting pump work

   q_a = 3782.5 + 9 = 3781.5 kJ/kg

   w_net = 1578.2 + 9 = 1587.2 kJ/kg

   

In Example 4.4, steam enters the reheater at 12 bar, 210°C, and leaves at 11.5 bar, 480°C. The combined steam rate is 3 kg/kWh, the generator efficiency is 94%, and the heat loss through the turbine casing is 1% of the throttle enthalpy. Determine the turbine thermal efficiency and condition of exhausted steam.

Solution

The processes in the T-s diagram is shown in Fig. 4.11.

Figure 4.11 Reheat Rankine cycle with turbine efficiency

Refer to Fig. 4.11

= 210°C, = 12 bar, t₃ = 480°C, p₃ = 11.5 bar

Now h₁ = 3336.5 kJ/kg, h₂ = 2814.4 kJ/kg

at 12 bar, 210°C from steam tables

= 2839.8 kJ/kg

h₃ at 11.5 bar, 480°C = 3433.4 kJ/kg

Now

For energy balance of turbine,

h₁ + h₃ = h′₂ + + w_net + q_loss

= h₁ − + h₃ − w_net − 0.01 × h₁

At p₄ = 0.07 bar, = 2620.1 kJ/kg, = 64°C

Since saturation temperature at 0.07 bar is 39.02°C, the exhaust steam is superheated and the degree of superheat is 24.98°C.

Efficiency of turbine upto extraction point,

Overall efficiency of turbine,

Steam at 100 bar and 500°C enters a prime mover that has one stage of reheat. The steam is exhausted from the prime mover at 0.07 bar and 85% dry. The net work developed by the prime mover is 1600 kJ/kg of steam. Calculate the thermal efficiency of the prime mover.

Solution

The cycle on T-s diagram is shown in Fig. 4.12.

Figure 4.12 Reheat Rankine cycle

Refer to Fig. 4.12.

From steam tables for p₁ = 100 bar, 500°C,

h₁ = 3373.6 kJ/kg

At 0.07 bar, h_f5 = 163.38 kJ/kg

h₄ = h_f4 + x₄ h_fg4

= 163.38 + 0.85 × 2409.2 = 2211.2 kJ/kg

w_net = 1600 kJ/kg

or h₃ − h₂ = w_net + h₄ − h₁ = 1600 + 2211.2 − 3373.6 = 435.6 kJ/hg

Thermal efficiency η_th =

A boiler feeds a turbine at 56 bar and 600°C. Before being passed on to the condenser at 30°C, the steam is bled for regenerative feed heating at 6.5 bar. For an ideal regenerative cycle and 1kg/s of throttle steam, determine (a) the amount of bled steam, (b) net work done, and (c) the ideal efficiency of the cycle.

Solution

The T-s diagram is shown in Fig. 4.19.

At p₁ = 56 bar, 600°C from steam tables,

h₁ = 3659.5 kJ/kg, s₁ = 7.2011 kJ/kg.K

At 6.5 bar, h₂ = 2987 kJ/kg

and t₂ = 265°C

s₃ = s₁ = s_f3 + x₃ s_gh3

At 30°C, s_f3 = 0.4369 kJ/kg.K, s_fg3 = 8.01674 kJ/kg.K

or 7.2011 = 0.436 + x₃ × 8.0164

or x₃ = 0.8438

h₃ = h_f3 + x₃h_gh3 = 125.79 + 0.8438 × 2430.5 = 2176.6 kJ/kg

At 6.5 bar, h_f6 = 684.26 kJ/kg, v_f6 = 0.001104 m³/kg

h₇ = h_f6 + v_f6 (p₁ − p₂) × 10² = 684.26 + 0.001104 (56 − 6.5) × 10²

= 684.26 + 5.47 = 689.72 kJ/kg

w_p2 = 5.47 kJ/kg

v_f4 = 0.001004 m³/kg

p₄ corresponding to 30°C = 0.042461 bar

w_p1 = v_{f 4} (p₂ − p₄) = 0.001004(6.5 − 0.042461) × 10²

= 0.648 kJ/kg

∑w_p = w_p1 + w_p2 = 0.648 + 5.47 = 6.118 kJ/kg

Figure 4.19 Regenerative Rankine cycle with a single heater

1. Neglecting pump work, for heat balance of heater,

   m₁ (h₂ − h_f6) = (1 − m₁) (h_f7 − h_f4)

   
2. w_net = w_t − ∑w_p = (h₁ − h₂) + (1 − m₁) (h₂ − h₃) −∑w_p

       = (3659.5 − 2987) + (1 − 0.195) (2987 − 2176.6) − 6.118

       = 1318.754 kJ/kg

3. Heat added, q_a = (h₁ − h₇) = 3659.5 − 689.72 = 2969.78 kJ/kg

   Cycle efficiency,

Determine the improvement of efficiency which would results if a single stage of regenerative feed heating were added to a steam cycle having terminal conditions of 16 bar, 300°C and 0.05 bar. The steam for feed heating is to be extracted at 1.16 bar. The drain from the heat exchanger at saturation temperature of bled steam pressure is returned to the system at a point downstream from the heat exchanger, the feed water getting heated to saturation temperature at pressure of bled steam. Neglect change in enthalpy due to liquid pump work.

Solution

The regenerative cycle and h-s diagram is shown in Fig. 4.20(a) and (b).

From the Mollier diagram,

From steam tables, h_f1 = 856.6 kJ/kg, h_f2 = 475.4 kJ/kg, h_f3 = 137.8 kJ/kg

For heat balance of heater,

m (h₂ − h_f3) = (h_f2 − h_f3)

Figure 4.20 Single stage regenerative heating Rankine cycle: (a) Schematic diagram, (b) Mollier diagram

or

w_net = (h₁ − h₂) + (1 − m) (h₂ − h₃)

= (3035 − 2575) + (1 − 0.138) (2575 − 2098) = 871 kJ/kg

q_a = h₁ − h_f2 = 3035 − 475.4 = 2559.6 kJ/kg

Thermal efficiency with regeneration,

Without feed water heating,

w_net = h₁ − h₃ = 3035 − 2098 = 937 kJ/kg

q_a = h₁ − h_f3 = 3035 − 137.8 = 2897.2 kJ/kg

Thermal efficiency without regeneration,

Improvement in thermal efficiency =

A regenerative cycle with three-stage bleed heating works between 30 bar, 450°C, and 0.04 bar. The bleed temperatures are chosen at equal temperature ranges. Determine the efficiency of the cycle. Neglect the pump work.

Solution

The schematic diagram of the cycle, T-s, and h-s diagrams are shown in Fig. 4.21(a) to (c).

At p = 0.04 bar, t_s = 30°C

Temperature range = 450 − 30 = 420°C

Equal temperature difference =

The temperatures of bled steam are: 450 − 105 = 345°C, 240°C, 135°C

From Mollier chart,

h₁ = 3340 kJ/kg, h₂ = 3150 kJ/kg, h₃ = 2935 kJ/kg,

h₄ = 2720 kJ/kg, h_s = 2120 kJ/kg

The pressure at 2, 3, 4 points are noted from the h-s chart. Then from steam tables,

h_f1 = 1008.35 kJ/kg, h_f2 = 837.45 kJ/kg, h_f3 = 684.0 kJ/kg,

h_f4 = 517.62 kJ/kg, h_f5 = 121.4 kJ/kg

For heat balance of heaters,

m₃ (h₄ − h_f4) = (1 − m₁ − m₂ − m₃) (h_f4 − h_f5)

m₂ (h₃ − h_f3) = (1− m₁ − m₂ − m₃) (h_f3 − h_f4)

m₁ (h₂ − h_f2) = (1 − m₁) (h_f2 − h_f3)

∴ m₃ (2720 − 517.62) = (1 − m₁ − m₂ − m₃) (517.62 − 121.4)

or 2202.38 m₃ = (1 − m₁ − m₂ − m₃) × 396.22

m₂ (2935 − 684) = (1 − m₁ − m₂) (684 − 517.62)

Figure 4.21 Regenerative Rankine cycle with three-stage bleeding: (a) Schematic diagram, (b) T-s diagram, (c) h-s diagram

or 2251 m₂ = (1 − m₁ − m₂) × 166.38

m₁ (3120 − 837.45) = (1 − m₁) (837.45 − 684)

or 2282.55 m₁ = (1 − m₁) × 153.45

Solving the above three equations, we get

m₁ = 0.063 kg, m₂ = 0.0645 kg, m₃ = 0.1332 kg

Net work done, w_net = 1 × (h₁ − h₂) + (1 − m₁) (h₂ − h₃) + (1 − m₁ − m₂) (h₃ − h₄) + (1 − m₁ − m₂ − m₃)(h₄ − h₅)

= (3340 − 3120) + (1 − 0.063) (3120 − 2935) + (1 − 0.063 − 0.0645) (2935 − 2720) + (1 − 0.063 − 0.0645 − 0.13327) (2720 − 2120) = 1024.6 kJ/kg

Heat added, q_a = h₁ − h_f2 = 3340 − 837.45 = 2502.55 kJ/kg

Thermal efficiency of the cycle,

A binary vapour cycle consists of two ideal Rankine cycles with mercury and steam as working substances. The following data is given:

Saturated mercury vapour pressure at entry to turbine = 4.5 bar

Mercury vapour exit pressure from turbine = 0.04 bar

Saturated steam generated in mercury condenser pressure = 15 bar

Steam superheated temperature in superheater in mercury boiler = 300°C

Condensate water pressure pumped through economiser located in exhaust flue of mercury boiler to saturation temperature = 15 bar

Isentropic efficiency of mercury turbine = 0.85

Isentropic efficiency of steam turbine = 0.88

Calculate (a) the overall thermal efficiency of the cycle and (b) the rate of flow of mercury turbine in kg/h.

Properties of saturated mercury

Figure 4.24 Hg − H₂O binary vapour cycle

Solution

Refer to Fig. 4.24.

Mercury cycle

h_A = 3559 kJ/kg, p_A = 4.5 bar, t_A = 450°C, s_A = 0.5397 kJ/kg.K

Now s_A = s_fB + x_B (s_gB − s_fB)

or 0.5397 = 0.0808 + x_B (0.6925 − 0.0808)

or x_B = 0.75

h_B = h_fB + x_B (h_gB − h_fB) = 29.98 + 0.75 (329.85 − 29.98) = 254.88 kJ/kg

h_B′ = h_A − (h_A − h_B) η_t1 = 355.98 − (355.98 − 254.88) × 0.85 = 270.04 kJ/kg

p_B′ = 0.04 bar, t_B′ = 216.9°C

h_c = h_fc = 29.98 kJ/kg, t_c = 216.9°C

h_D = h_c + v_fc (p_D − p_C) × 10² = 29.98 + 76.5 × 10⁻⁶ (4.5 − 0.04) × 10² = 30.0142 kJ/kg

w_p1 = v_fc (p_D − p_C) × 10² = 0.03412 kJ/kg

Steam cycle

p₁ = 15 bar, t₁ = 300°C, h₁ = 3038.9 kJ/kg, s₁ = 6.9207 kJ/kg. K

p₂ = 0.05 bar, h_f2 = 137.77 kJ/kg, h_g2 = 2561.6 kJ/kg, s_f2 = 0.4763 kJ/kg.K,

h₅ = 2789.9 kJ/kg, h₄′ = 844.66 kJ/kg

h_fg2 = 2423.8 kJ/kg, s_g2 = 8.3960 kJ/kg.K, v_f2 = 0.0010052 m³/kg

s₁ = s₂ = s_f2 + x₂ s_fg2

or 6.9207 = 0.4763 + x₂ (8.3980 − 0.4763)

or x₂ = 0.8137

h₂ = h_{f 2} + x₂h_fg2 = 137.77 + 0.8137 × 2423.8 = 2110 kJ/kg

h₂′ = h₁ − (h₁ − h₂) η_t2 = 3038.9 − (3038.9 − 2110) × 0.88 = 2221.47 kJ/kg

h₃ = h_f3 = 137.77 kJ/kg

w_p2 = v_{f 3} (p₄ − p₃) × 10² = 0.0010052 (15 − 0.05) × 10² = 1.503 kJ/kg

Total heat added, (q_a)_total = m_Hg (h_A − h_D) + 1 [(h₁ − h₅) + h_4′ − h₄)]

Work done, (w_t)_Hg = m_Hg (h_A − h_B′) = 8.103 (355.98 − 270.04) = 696.37 kJ/kg of steam

(w_t)_steam = 1 × (h₁ − ) = 1 × (3038.9 − 2221.47) = 817.43 kJ/kg

(w_t)_total = 696.37 + 817.43 = 1513.8 kJ/kg

Neglecting pump work being very small,

Considering pump work,

In a binary-vapour cycle, the steam cycle operates between 30 bar, 0.07 bar and uses a superheated temperature of 350°C. The mercury cycle works between 12.68 bar and 0.07 bar. The mercury vapour entering the turbine is in a dry saturated condition. Calculate the efficiency of the combined cycle assuming expansion in both cycles to be isentropic. Data for mercury is given below:

Solution

The binary vapour cycle is shown in Fig. 4.26.

Mercury cycle: s_a = s_b = s_fb + x_b (s_gb − s_fb)

0.50185 = 0.08548 + x_b (0.662906 − 0.08548)

or x_b = 0.721

Figure 4.26 Binary vapour cycle neglecting pump work

h_b = h_fb + x_b (h_gb − h_fb)

= 32.395 + 0.721 (326.667 − 32.395) = 244.565 kJ/kg of Hg

Isentropic work done, w_m = h_a − h_b = 360.734 − 244.565 = 116.169 kJ/kg Hg

Heat rejected, q_rm = h_b − h_fb = 244.565 − 32.395 = 212.17 kJ/kg Hg

Heat supplied, q_sm = h_a − h_fb = 360.734 − 32.395 = 328.339 kJ/kg Hg

Steam cycle: h₁ = 3115.3 kJ/kg, s₁ = 6.743 kJ/kg.K, h′₁ = 2804.2 kJ/kg

or 6.743 = 0.5589 + x₂ × 7.7179

or x₂ = 0.80127

Work done, w_s = h₁ − h₂ = 3115.3 − 2093.733 = 1021.567 kJ/kg

Heat supplied, q_ss = h₁ = h₁′ = 3115.3 − 2804.2 = 311.1 kJ/kg

Heat rejected by Hg/kg of steam = Heat received by water/kg steam

m_m (h_b − h_fb) = 1 (h₁′ − h_f3)

or

Total work done per kg of steam,

w₁ = w_s + m_m w_m = 1021.567 + 12.447 × 116.161 = 2467.379 kJ/kg steam

Heat supplied in the cycle,

q_s = m_m q_sm + q_ss = 12.447 × 328.339 + 311.1 = 4397.936 kJ/kg steam

Cycle efficiency =

In a single-heater regenerative cycle, the steam enters the turbine at 32 bar, 350°C and the exhaust pressure is 0.12 bar. The feed water heater is a direct contact type which operates at 4.5 bar. Find (a) the efficiency and the steam rate of the cycle and (b) the increase in efficiency and steam rate as compared to Rankine cycle. Neglect pump work.

Solution

From steam tables, h₁ = 3113.2 kJ/kg, s₁ = 6.712 kJ/kg.K = s₂ = s₃

s_g at 4.5 bar = 6.875 kJ/kg.K

The schematic diagram is shown in Fig. 4.27(a). The processes on the T-s and h-s diagrams are shown in Figs. 4.27(b) and (c) respectively. Since s₂ < s_g, state point 2 lies in the wet region.

s_f2 = 1.821 kJ/kg.K, s_fg2 = 5.036 kJ/kg.K

s₁ = s₂ = s_f2 + x₂ s_fg2

or 6.712 = 1.821 + x₂ × 5.036

or x₂ = 0.9712

h₂ = h_{f 2} + x₂ h_fg2 = 623.3 + 0.9712 × 2120.6 = 2682.83 kJ/kg

Figure 4.27 Single stage regenerative cycle: (a) Flow schematic diagram, (b) T-s diagram, (c) h-s diagram

s_f3 = 0.6963 kJ/kg.K s_fg3 = 7.3900 kJ/kg.K

s₁ = s₃ = s_f3 + x₃ s_fg3

or 6.712 = 0.6963 + x₂ × 7.3900

or x₃ = 0.814

h₃ = h_f3 + x₃ h_fg3 = 206.92 ÷ 0.814 × 2384.1 = 2147.58 kJ/kg

Since w_p = 0

∴ h₄ = h_f3 = 206.92 kJ/kg = h₅

h₆ = h_f2 = 623.3 kJ/kg = h₇

Energy balance for heater gives,

or m₁ (2682.83 − 623.3) = (1 − m₁) (623.3 − 206.92)

or 2059.53 m₁ = (1 − m₁) × 416.38

or m₁ = 0.168 kg

w_t = (h₁ − h₂) + (1 − m₁) (h₂ − h₃)

= (3113.2 − 2682.83) + (1 − 0.168) (2628.83 − 2147.58) = 875.698 kJ/kg

q_s = h₁ − h₆ = 3113.2 − 623.3 = 2489.9 kJ/kg

Steam rate =

Without regeneration, w_t = h₁ − h₃ = 3113.2 − 2147.58 = 965.62 kJ/kg

Increase in cycle efficiency due to regeneration = 35.17 − 33.22 = 1.95%

Steam rate =

Increase in steam rate due to regeneration = 4.11 − 3.728 = 0.382 kg/kWh

A steam power plant working on Rankine cycle is supplied with dry saturated steam at a pressure of 12 bar and exhausts into the condenser at 0.1 bar. Neglecting the pump work, calculate the cycle efficiency.

Solution

Corresponding to p = 12 bar, from steam tables we find that (see Fig. 4.3)

h_f1 = 798.4 kJ/kg; h_g1 = 1984.3 kJ/kg

s₁ = s_g1 = 6.519 kJ/kg°K

Corresponding to 0.1 bar, we find that

h_f2 = 191.8 kJ/kg; h_fg2 = 2393 kJ/kg

s_f2 = 0.649 kJ/kg K; s_fg2 = 7.502 kJ/kg·K

Process 1–2 is an isentropic expansion

s₁ = s₂

s₁ = s_g1 = 6.519 kJ/kg K

s₂ = s_{f 2} + x₂ s_fg2

s_{2 = 0.649 + x2} × _7.502

6.519 = 0.649 + 7.502 x₂

x₃ = 0.783

h₂ = h_f2 + h_fg2

=191.81 + 0.783 × 2392.8 = 2065.5 kJ/kg

h₁ = h_g2 = 2782.7 kJ/kg

Dry saturated steam at 10 bar is supplied to a prime mover working on Rankine cycle and the exhaust takes place at 0.2 bar. Determine the cycle efficiency, efficiency ratio, and specific steam consumption of the prime mover, if the indicated thermal efficiency is 20%. Determine the percentage change in the cycle efficiency if the steam is initially 90% dry.

Solution

The h-s diagram is shown in Fig. 4.28.

From Mollier diagram (Fig. 4.28), we have

h₁ = 2775 kJ/kg, h₂ = 2150 kJ/kg

From steam tables,

h_f2 = 251.5 kJ/kg corresponding to 0.2 bar

Thermal efficiency

Efficiency ratio =

Specific steam consumption =

Percentage change in cycle efficiency if steam is initially 90% dry

From Mollier diagram, h₁ = 2580 kJ/kg, h₂ = 2030 kJ/kg (see Fig. 4.29)

Thermal efficiency

∴ Percentage change in Rankine efficiency =

Figure 4.28 Mollier diagram for dry saturated steam

Figure 4.29 Mollier diagram for wet steam

The steam consumption of steam engine is 20 tonnes per shift of 8 hours when developing 220 kW. Dry and saturated steam enters the engine at 10 bar pressure and leaves at 0.1 bar pressure. Estimate the cycle efficiency and thermal efficiency of engine.

Solution

Given that m_s = 20/8 tonne/h = 2.5 tonne/h = 2500 kg/h, P = 220 kW

The h-s diagrams is shown in Fig. 4.30.

From steam tables corresponding to 10 bar

h₁ = h_g1 = 2778.1 kJ/kg; s₁ = s_g1 = 6.5864 kJ/kg.K

and corresponding to 0.1 bar, we find that

h_f2 = 191.81 kJ/kg; h_fg2 = 2392.8 kJ/kg;

s_f2 = 0.6492 kJ/kg.K; s_g2 = 7.5010 kJ/kg.K

Since 1−2 is an isentropic process

s₁ = s₃

or 6.5864 = 0.6492 + x₂ × 7.5010

or x₂ = 0.791

∴ h₃ = h_f2 + x₂ h_fg2 = 191.81 + 0.791 × 2392.8 = 2084.5 kJ/kg

Cycle Efficiency (η_R) =

Thermal efficiency of engine;

Figure 4.30 Mollier diagram for dry saturated steam

Steam at 50 bar, 400°C expands in a Rankine engine to 0.34 bar. For 150 kg/s of steam; determine (a) the power developed, (b) the thermal efficiency neglecting the pump work, and (c) the specific steam consumption.

Solution

The h-s diagrams is shown in Fig. 4.31.

Figure 4.31 Mollier diagram for superheated steam

1. Power developed = m(h₁ − h₂)

   From Mollier diagram (Fig. 4.31), we have

   h₁ = 3198.3 kJ/kg, h₂ = 2257.75 kJ/kg

   h_f2 = 301.5 kJ/kg (from steam tables)

   Power developed = 150 [3198.3 − 2257.75] = 141082 kW

2. Thermal efficiency
   
3. Specific steam consumption =

A steam engine is supplied with dry saturated steam at 15 bar. The pressure at release is 3 bar and the back pressure is 1 bar. Determine the efficiency of modified Rankine cycle.

Solution

Consider the Mollier diagram, as shown in Fig. 4.32.

From the Mollier diagram, we have

h₁ = 2790 kJ/kg, h₂ = 2510 kJ/kg, x₂ = 0.9

From steam tables, corresponding to 3 bar, we have

v_f2 =0.001073 m³/kg, v_g2 = 0.6058 m³/kg

∴ Volume of steam at point 3,

v₂ = v_f2 + x₂ (v_g2 − v_f2) = 0.001073 + 0.9 × (0.60585 − 0.001073) = 0.545 m³/kg

Figure 4.32 Mollier diagram for dry saturated steam

From steam tables corresponding to pressure of 1 bar, we find that sensible heat of water,

h_f4 = 417.44 kJ/kg

η_MR = Thermal efficiency of modified Rankine cycle

Steam at a pressure of 15 bar and 250°C is first expanded through a turbine to a pressure of 4 bar. It is then reheated at constant pressure to the initial temperature of 250°C and is finally expanded to 0.1 bar. Estimate the work done per kg of steam flowing through the turbine and the amount of heat supplied during the process of reheat.

Find the work output when there is direct expansion from 15 bar to 0.1 bar without any reheat. Assume all expansion processes to be isentropic.

Solution

Given that p₁ = 15 bar; T₁ = 250°C; p₂ = 4 bar; T₂ = 250°C; p₃ = 0.1 bar

The reheating of steam is represented on the Mollier chart as shown in Fig. 4.33. From the chart, we find that,

h₁ = 2930 kJ/kg; h₂ = 2660 kJ/kg; h₃ = 2965 kJ/kg; h₄ = 2345 kJ/kg; and h₅ = 2130 kJ/kg

From steam tables, corresponding to a pressure of 0.1 bar, we find that sensible heat of water at D,

h_f4 = h_f5 = 191.81 kJ/kg

Workdone per kg of steam:

We know that workdone per kg of steam,

w = (h₁ − h₂) + (h₃ − h₄)

= (2930 −2660) + (2965 − 2345) = 890 kJ/kg

Heat supplied during the process of reheat:

We know that the heat supplied during the process of reheat,

h = Heat supplied between 2 and 3

= (h₃ − h₂) − (h_f4) = (2965 − 2660) − 191.81 = 113.2 kJ/kg

Figure 4.33 Mollier diagram for superheated steam with reheating

Work output when the expansion is direct:

The direct expansion from 15 bar to 0.1 bar is shown by the line AE in Fig. 4.33. We know that work output

= Total heat drop = h₁ − h₅ = 2930 − 2130 = 800 kJ/kg

In a thermal plant, steam is supplied at a pressure of 30 bar and temperature of 300°C to the high pressure side of steam turbine where it is expanded to 5 bar. The steam is then removed and reheated to 300°C at a constant pressure. It is then expanded to the low pressure side of the turbine to 0.5 bar. Find the efficiency of the cycle with and without reheating.

Solution

Given that p₁ = 30 bar; T₁ = 300°C; p₂ = 5 bar; T₂ = 300°C; p₃ = 0.5 bar

Efficiency of the cycle with reheating:

The reheating of steam is represented on the Mollier chart as shown in Fig. 4.34.

From the chart, we find that

h₁ = 2990 kJ/kg; h₂ = 2625 kJ/kg; h₃ = 3075 kJ/kg; h₄ = 2595 kJ/kg; and h₅ = 2280 kJ/kg

From steam tables, corresponding to a pressure of 0.5 bar, we find that sensible heat of water at D,

h_f4 = h_f5 = 340.47 kJ/kg

Figure 4.34 Mollier diagram for superheated steam with reheating

We know that efficiency of the cycle with reheating,

Efficiency of the cycle without reheating:

We know that efficiency of the cycle without reheating

Steam supplied to a 10 MW turbo-alternator at 40 bar and 400°C. The auxiliaries consume 7% of the output. The condenser pressure is 0.05 bar and condensate is sub-cooled to 30°C. Assuming the boiler efficiency as 85%, the relative efficiency of turbine as 80%, and the mechanical efficiency of the alternator as 95%, determine (a) the steam consumption per hour, (b) the overall efficiency of the plant, and (c) the quality of steam at exit from turbine.

Solution

The h-s diagrams is shown in Fig. 4.35.

1. From Mollier chart,

   h₁ = 3215 kJ/kg, h₂ = 2064 kJ/kg

   Therefore, (h₁ − h₂) = 3215 − 2064 = 1151 kJ/kg

   Thus actual enthalpy drop,

   (h₁ − h₂) = η_relative × (h₁ − h₂)

       = 0.8 × 1151 = 920.8 kJ/kg

   Input to the alternator =

   

   Figure 4.35 Mollier diagram for superheated steam

   Therefore, steam consumption per hour,

   
2. Heat supplied to the boiler/hour =
   

   where h_f2 = enthalpy of liquid at saturation at 0.05 bar from steam tables

   The auxiliaries consume 7% of the output.

   Therefore, useful output = (10,000 × 0.93) = 9300 kW = 9300 × 60 × 60 kJ/h

   Thus overall efficiency =

3. The dryness fraction at exit, that is, B′ is read at h_2′ = 3215 − 920.8 = 2294.2 kJ on 0.05 bar line.

   Thus, x₂′ = 0.889

Two turbines, A and B, operate with steam at an initial pressure of 100 bar and temperature of 500°C. In each turbine, the steam is expanded in a high pressure turbine to 8.5 bar with efficiency ratio 80%. In turbine A, the expansion is further continued in low pressure turbine from 8.5 bar to 0.035 bar with efficiency ratio 0.75. In turbine B, steam is reheated after expansion in the high pressure turbine and is then fed to the low pressure unit at 7 bar and 500°C, after which it expands to 0.035 bar with efficiency ratio 0.85. Compare the two power cycle with respect to (a) thermal efficiency and (b) steam consumption for a full load output of 50,000 kW.

Solution

Turbine A: The expansion is shown on h-s chart in Fig. 4.36(a)

h_a = 3377 kJ/kg and h_b = 2753 kJ/kg

h_c = h_d = 3377 − 0.8 (3377−2753) = 2878 kJ/kg

h_e = 2060 kJ/kg

h_f = h_g = 2878 − (2878 − 2060) × 0.75 = 2264.5 kJ/kg

Total work done = (h_a − h_d) + (h_d − h_g) = (h_a − h_g)

Heat supplied = h_a − h_g

where h_g is the sensible heat of steam at 0.035 bar = 110 kJ

∴ Heat supplied = 3377 − 110 = 3267 kJ

Thermal efficiency =

Total work to be done = 53,000 kW = 50,000 kJ/s

Figure 4.36 Mollier diagrams: (a) h-s diagram for Turbine A, (b) h-s diagram for Turbine B

∴ Steam consumption =

Turbine B: The expansion is shown on h-s chart in Fig. 4.36(b)

h_a = 3377 kJ/kg: h_b = 2753 kJ/kg; h_c = h_d = 2878 kJ/kg

h_e = 3490 kJ/kg; h_f = 2385 kJ/kg.

h_h = 3490 − (3490 − 2315) 0.85 = 2551 kJ/kg

Total work done = (3377 − 2878) + (3490 − 2551) = 1438 kJ/kg

Total heat supplied = (3377 − 110) + (3490 − 2878) = 3879 kJ/kg.

∴ Thermal efficiency

Steam consumption =

In a regenerative cycle, having one feed water heater, the dry saturated steam is supplied from the boiler at a pressure of 30 bar and the condenser pressure is 1 bar. The steam is bled at a pressure of 5 bar. Determine the amount of bled steam per kg of steam supplied and the efficiency of the cycle. What would be the efficiency without regenerative feed heating? Determine the percentage increase in efficiency due to regeneration.

Solution

Given that p₁ = 30 bar; p₃ = 1 bar; p₂ = 5 bar

From Mollier diagram, as shown in Fig. 4.37, we find that

Enthalpy of steam at 30 bar, h₁ = 2800 kJ/kg

Enthalpy of steam at 5 bar, h₂ = 2460 kJ/kg

Enthalpy of steam at 1 bar, h₃ = 2220 kJ/kg

From steam tables, we also find the enthalpy or sensible heat of water at 5 bar.

Figure 4.37 Mollier diagram for dry saturated steam

h_f2 = 640.21 kJ/kg

and enthalpy or sensible heat of water at 1 bar,

h_f3 = 417.44 kJ/kg

Amount of bled steam per kg steam supplied

We know that amount of bled steam per kg of steam supplied,

Efficiency of the cycle: We know that efficiency of the cycle,

Efficiency of the cycle without regenerative feed heating: We know that efficiency of the cycle,

Percentage increase in efficiency due to regeneration: We know that percentage increase in efficiency due to regeneration

In a steam turbine plant, the steam is generated and supplied to the turbine at 50 bar and 370°C. The condenser pressure is 0.1 bar and the steam enters the condenser with dryness fraction of 0.9. Two feed water heaters are used, the steam in the heaters being bled at 5 bar and 0.5 bar. In each heater, the feed water is heated to saturation temperature of the bled steam. The condensate is also pumped at this temperature into the feed line immediately after the heater. Find the masses of the steam bled in the turbine per one kg of steam entering the turbine. Assuming the condition line for the turbine to be straight, calculate the thermal efficiency of the cycle.

Solution

Given that p₁ = 50 bar; T₁ = 370°C; p₄ = 0.1 bar; x₄ = 0.9; p₂ = 5 bar; p₃ = 0.5 bar

First, let us draw the Mollier diagram and condition line for the cycle, as shown in Fig. 4.38. From this diagram, we find that,

h₁ = 3110 kJ/kg, h₂ = 2780 kJ/kg, h₃ = 2510 kJ/kg, h₄ = 2350 kJ/kg

From steam tables, we also find that

h_f2 = 640.21 kJ/kg (at 5 bar)

h_f3 = 340.47 kJ/kg (at 0.5 bar)

h_{f 4} = 191.81 kJ/kg (at 0.1 bar)

Mass of steam bled in the turbine:

We know that mass of steam bled at 2.

and mass of steam bled at 3.

Thermal efficiency of the cycle:

We know that work done from 1 to 2 per kg of feed water

= h₁ − h₂ = 3110 − 2780 = 330 kJ/kg

Similarly, work done from 2 to 3 per kg of feed water

= (1 − m₁)(h₂ − h₃) = (1 − 0.123) (2780 − 2510) kJ/kg

= 236.8 kJ/kg

and work done from 3 to 4 per kg of feed water

= (1 − m₁ − m₂) (h₃ − h₄) = (1 − 0.123 − 0.056) (2510 − 2350) kJ/kg

= 131.4 kJ/kg

Figure 4.38 Mollier diagram

∴ Total work done = 330 + 236.8 + 131.4 = 698.2 kJ/kg

Heat supplied = h₁ − h_f2 = 3110 − 640.1 = 2469.9 kJ/kg

∴ Thermal efficiency of cycle,

In the Rankine cycle, steam leaves the boiler and enters the turbine at 4 MPa and 400°C. The condenser pressure is 10 kPa. Neglecting pump work, determine the cycle efficiency and the Carnot efficiency for the same temperature limits.

Solution

Refer to Fig. 4.39. Rankine cycle is represented by 1−2−3−4 ignoring pump work, Carnot cycle for the same temperature limits may be reproduced by 1−2′−3′−4′.

Carnot efficiency is given by

(It may be noted that Carnot efficiency is independent of the working substance.)

From steam tables, h₁ at 4MPa and 400°C = 3213.5 kJ/kg, s₁ = 6.7689 kJ/kg.K

At 10 kPa

h_f = 191.81 kJ/kg, h_fg = 2392.8 kJ/kg, s_f = 0.6492 kJ/kg.K, s_fg = 7.5010 kJ/kg.K

Also s₁ = s₂

or 6.7689 = 0.6492 + x₂ × 7.5010

or

∴ h₂ = h_f3 + x₂h_fg = 191.81 + 0.816 × 2392.8 = 2144.33 kJ/kg

Thermal efficiency η_Rankine =

Figure 4.39 T-s diagram for Rankine cycle

A steam power plant has the range of operation from 40 bar dry saturated to 0.05 bar. Determine (a) the cycle efficiency and (b) the work ratio and specific fuel consumption for (i) Carnot cycle and (ii) Rankine cycle.

Solution

From steam tables:

At 40 bar, t_g = 250.4°C, v_f = 0.001252 m³/kg, v_g = 0.049778 m³/kg, h_f = 1087.29 kJ/kg, h_g = 2801.40 kJ/kg, s_g = 6.07 kJ/kg.K

At 0.05 bar, t_g = 32.88°C, v_f = 0.0010005 m³/kg, v_g = 28.193 m³/kg, h_f = 137.79 kJ/kg

h_fg = 2423.7 kJ/kg, s_f = 0.4763 kJ/kg.K, s_fg = 7.9187 kJ/kg.K

1. Refer of Fig. 4.40. for Carnot cycle analysis on T-s plane.

   Cycle efficiency =

   

   (Note: Processes 1−2 and 3−4 are reversible adiabatic.)

   Also s₁ = s₂ = s_f + x₂s_fg

   or

   Also s₄ = s₃ = s_f + x₃s_fg

   or

   

   Figure 4.40 Carnot cycle on T-s plot

   

   Figure 4.41 Rankine cycle on T-s plot

   ∴ h₂ = 137.79 + 0.706 × 2423.7 = 1848.9 kJ/kg.

   and h₃ = 137.79 + 0.293 × 2423.7 = 847.9 kJ/kg.

   ∴ η_{Carnot cycle} =

   

   Alternatively,

   
2. Refer to Fig. 4.41 for Rankine cycle analysis

   Pump work = h_M − h_f3

   

   = 0.001005 × 39.95 × 10² = 4.015 kJ/kg

   Therefore, net work

   w_net = h₁ − h₂ − w_pump = 2801.4 − 1848.9 − 4.015 = 948.485 kJ/kg

   and η_{Rankine cycle} =

   = 0.3566 or 35.66%

1. b. Also work ratio =

   Specific steam consumption/kWh

Steam at 50 bar, 400°C expands in a steam tubine to 0.34 bar. For 150 kg/s of steam; determine (a) the power developed, (b) the thermal efficiency, and (c) the specific steam consumoption (i) for the Rankine cycle and (ii) for the Rankine engine. For an actual engine with same specifications, the brake steam rate is 4.75 kg/kWh and the driven electric generator has an electro mechanical efficiency of 94%.

Determine (a) the brake thermal efficiency, (b) the internal efficiency (expansion efficiency), (c) the power in kW, and (d) the exhaust dryness fraction of steam.

Solution

Refer to Fig. 4.42.

At 50 bar and 400°C

h₁ = 3195.6 kJ/kg, s₁ = 6.6458 kJ/kg.K, v₁ = 0.05781 m³/kg

At 0.34 bar, h_f = 301.48 kJ/kg, h_fg = 2328.9 kJ/kg, s_f = 0.9795 kJ/kg.K, s_fg =6.7471 kJ/kg.K

And s₁ = s₂ for ideal expansion

6.6458 = 0.9795 + x₂ ×6.7471

or x₂ = 0.84

Thus h₂ = h_f3 + x₂h_fg = 301.43 + 0.84 × 2328.9 = 2257.756 kJ/kg

1. For cycle w_pump = Pump work = h_M − h_f3
   
   1. Power developed = Steam per sec × (h₁ − h₂ − w_pump)

      = [150 (3195.6 − 2257.756 − 5.0857)] = 139913.7 kW

   2. η_Rankine =
      
      

      Figure 4.42 Rankine cycle on T-s plot

   3. Steam rate or specific consumption =
2. For engine
   1. Power developed = Steam per sec × (h₁ − h₂) = 150 × (3195.6 − 2257.756)

      = 140676.6 kW

   2. η_{Rankine engine} =
   3. Steam rate or specific steam consumption
      
3. Actual engine

   Brake specific steam consumption = 4.75 kg/kWh

   η_{brake thermal} =

   Expansion efficiency =

   Actual enthalpy drop = 0.808 × Isentropic enthalpy drop

   = 0.808 (3195.6 − 2257.756) = 757.8 kJ/kg

   Thus, h′₂ = h₁ − 757.8 = 3195.6 − 757.8 = 2437.8 kJ/kg

   Also, h′₂ = h_f2 + xh_fg2

   Thus,

In a steam power plant operating on Rankine cycle, the steam enters the turbine at 70 bar and 550°C with a velocity of 30 m/s. It discharges to the condenser at 0.20 bar with a velocity of 90 m/s. If the steam flow rate is 35 kg/s find the thermal efficiency and the net power produced. Neglect pump work.

[IES, 1992]

Solution

Given that p₁ = 70 bar, t₁ = 550°C, c₁ = 30 m/s, p₂ = 0.20 bar, c₂ = 90 m/s, ṁ_s = 35 kg/s

The simple Rankine cycle is shown in Fig. 4.43(a) and T-s diagram in Fig. 4.43(b).

Figure 4.43 Rankine cycle: (a) Schematic diagram, (b) T-s diagram

From steam tables, we have

At p₁ = 70 bar and t₁ = 550°C, h₁ = 3530.9 kJ/kg, s₁ = 6.9486 kJ/kg.K

At p₂ = 0.2 bar, h_f2 = 251384 kJ/kg, h_fg2 = 2358.3 kJ/kg, s_f2 = 0.8319 kJ/kg.K, s_fg2 = 7.0766 kJ/kg.K

Now s₁ = s₂ = s_f2 + x₂ s_fg2

or 6.9486 = 0.8319 + x₂ × 7.0766

or x₂ = 0.8644

∴ h₂ = h_f2 + x₂ h_fg2

= 251.38 + 0.8644 × 2358.3 = 2289.89 kJ/kg

Net power produced = ṁ_s (h₁ − h₂) = 35 (3530.9 − 2289.89) = 43435.35 kW

Thermal efficiency,

Now h_f3 = h_f2

The following data refer to a steam turbine power plant employing one stage of regenerative feed heating:

State of steam entering HP stage: 10 MPa, 600°C

State of steam entering LP stage: 2 MPa, 400°C

State of steam at condenser : 0.01 MPa, x = 0.9

The correct amount of steam is bled for feed heating at exit the HP stage. Calculate the mass of steam bled per kg of steam passing through the HP stage and the amount of heat supplied in the boiler per second for an output of 10 MW. Neglect pump work.

[IES, 1993]

Solution

The schematic diagram and T-s diagram are shown in Fig.4.44(a) and (b).

H.P. stage: p₁ = 10 MPa, t₁ = 600°C

L.P. stage: p₂ = 2 MPa, t₂ = 400°C

Condenser: p₃ = 0.01 MPa, x₃ = 0.9

W_out = 10 MW

From steam tables, we have (Fig. 4.44(a))

h₁ = 3625.3 kJ/kg

h₂ = 3247.6 kJ/kg

h₃ = h_f3 + x₃ h_fg3

= 191.81 + 0.9 × 2392.8 = 2345.35 kJ/kg

Heat balance for the heater:

(1 − m₁) (h₅ − h₄) = m₁(h₂ − h₅)

From steam tables,

h₅ = 908.77 kJ/kg at p₂ = 2 MPa

h₄ = 191.81 kJ/kg at p₃ = 0.01 MPa

∴ (1 − m₁)(908.77 − 191.81) = m₁(3247.6 − 908.77)

or (1 − m₁) × 716.96 = 2338.83 m₁

Figure 4.44 Regenerative feedwater heating cycle: (a) Schematic diagram, (b) T-s diagram

or

Heat supplied in boiler,

q_in = h₁ − h₅ = 3625.3 − 908.8 = 2716.5 kJ/kg

Work output from boiler,

W_out = (h₁ − h₂) + (1 − m₁) (h₂ − h₃)

= (3625.3 − 3247.6) + (1 − 0.235) (3247.6 − 2345.35) = 1067.92 kJ/kg

Cycle efficiency,

Heat supplied in the boiler for 10 MW output =

A small power plant produces 25 kg/s steam at 3 MPa, 600°C in the boiler. It cools the condenser with ocean water coming in at 12°C and returned at 15°C. Condenser exit is 45°C. Find (a) the net power output and (b) the required mass flow rate of ocean water.

[IAS 2012]

Solution

Given that ṁ_s= 25 kg/s, p₁ = 3 MPa, t₁ = 600°C, t_wi = 12°C, t_wo = 15°C, T₁ = 600 + 273 = 873 K,T₂ = 45 + 273 = 318 K

The T-s diagram is shown in Fig. 4.45.

1. h₁ = 3682 kJ/kg, h_f3 = 188.4 kJ/kg, p₁ = 9.5934 kPa, s₁ = 7.5084 kJ/kg.K

   v_f3 = 0.00101 m³/kg, h_fg3 = 2394.8 kJ/kg

   s_f3 = 0.6386 kJ/kg.K, s_fg3 = 7.5261 kJ/kg.K

   Now s₁ = s₂ = s_f3 + x₂ s_fg3

   

   Figure 4.45 T-s diagram

   or 7.5084 = 0.6386 + x₂ × 7.5261

   or x₂ = 0.9128

   h₂ = h_f3 + x₂ h_fg3 = 188.42 + 0.9128 × 2394.8 = 2374.39 kJ/kg

   Pump work, w_p = v_f3 (p₁ − p₂) = 0.00101 (300 − 9.59) = 0.29 kJ/kg

   Turbine work, w_t = h₁ − h₂ = 3682.3 − 2374.39 = 1307.91 kJ/kg

   Net work, w_net = w_t − w_p = 1307.91 − 0.29 = 1307.62 kJ/kg

   Net power output = ṁ_s × w_net = 25 × 1307.62 = 32.69 MW

2. Heat rejected constant pressure p₂, q_r = h₂ − h_f3 = 2374.39 − 188.42 = 2185.97 kJ/kg

   Let ṁ_w = mass of ocean water circulated

   ṁ_wc_pw × Δt_w = ṁ_s× q_r

   or ṁ_w × 4.187 × (15 − 12) = 25 × 2185.97

   or ṁ_w = 4350.7 kg/s