Answer:
To solve this problem, we can use the first law of thermodynamics, which states that the change in internal energy of a closed system is equal to the heat added minus the work done:
ΔU = Q - W
where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system.
We can apply this equation to each stage of the power plant cycle:
a) Pump work:
Since water is pumped under saturated conditions, its specific volume can be assumed to be constant. Therefore, the work done by the pump is given by:
W_pump = m * v * (P_2 - P_1)
where m is the mass of water pumped, v is the specific volume of water, and P_1 and P_2 are the initial and final pressures, respectively. From the given data, we have:
P_1 = 0.7 bar
P_2 = 30 bar
v = v_f = 0.00106 m^3/kg (from saturated water table)
m = 1 kg (Assumed)
Plugging in these values, we get:
W_pump = 1 kg * 0.00106 m^3/kg * (30 bar - 0.7 bar) = 0.0307 kJ
Therefore, the work done by the pump is 0.0307 kJ.
b) Heat added to the boiler:
At constant pressure, the heat added to the water is given by:
Q_boiler = m * cp * (T_2 - T_1)
where m is the mass of water, cp is the specific heat of water, and T_1 and T_2 are the initial and final temperatures, respectively. From the given data, we have:
T_1 = T_sat = 100°C (from saturated water table)
T_2 = 500°C
cp = 4.18 kJ/kg·K
Plugging in these values, we get:
Q_boiler = 1 kg * 4.18 kJ/kg·K * (500°C - 100°C) = 1672 kJ
Therefore, the heat added to the boiler is 1672 kJ.
c) Turbine work:
Since the steam is expanded isentropically in the turbine, its specific entropy remains constant. Therefore, the work done by the turbine is given by:
W_turbine = m * (h_1 - h_2)
where m is the mass of steam, h_1 is the specific enthalpy of steam at the inlet to the turbine, and h_2 is the specific enthalpy of steam at the outlet of the turbine. From the given data, we have:
h_1 = h_sat + cp * (T_2 - T_sat) = 2882 kJ/kg (from steam tables)
h_2 = h_sat + cp * (T_3 - T_sat) = 1952 kJ/kg (from steam tables)
T_3 = T_sat = 100°C (from saturated water table)
m = 1 kg (Assumed)
Plugging in these values, we get:
W_turbine = 1 kg * (2882 kJ/kg - 1952 kJ/kg) = 930 kJ
Therefore, the work done by the turbine is 930 kJ.
d) Heat removed by the condenser:
The steam is condensed at constant pressure, and the heat removed by the condenser is given by:
Q_condenser = m * (h_2 - h_3)
where h_3 is the specific enthalpy of water at the outlet of the condenser, which is the same as the specific enthalpy of water at the inlet to the pump. From the given data, we have:
h_3 = h_f = 419 kJ/kg (from saturated water table)
Plugging in the values, we get:
Q_condenser = 1 kg * (1952 kJ/kg - 419 kJ/kg) = 1533 kJ
Therefore, the heat removed by the condenser is 1533 kJ.
e) Cycle thermal efficiency:
The cycle thermal efficiency is the ratio of the net work output to the heat input. The net work output is the difference between the turbine work and the pump work, i.e.,
W_net = W_turbine - W_pump = 930 kJ - 0.0307 kJ = 929.97 kJ
The heat input is the heat added to the boiler, i.e.,
Q_in = Q_boiler = 1672 kJ
Therefore, the cycle thermal efficiency is:
η = W_net / Q_in = 929.97 kJ / 1672 kJ = 0.555 or 55.5%
Therefore, the cycle thermal efficiency of the power plant is 55.5%.