Question

Steam at 20 bar, 360oC is expanded in a steam turbine to 0.08 bar. It then enters a

condenser, where it is condensed to saturated liquid water. The pump feeds back the

water into the boiler. (a) Assuming ideal processes, find per kg of steam the net work

and the cycle efficiency. (b) If the turbine and the pump have each 80% efficiency, find

the percentage reduction in the net work and cycle efficiency​

Answer 1

Given :

P₁ = 20bar

P₂ = 0.08bar

To find :

(a) Assuming ideal processes, find per kg of steam the net work  and the cycle efficiency.

(b) If the turbine and the pump have each 80% efficiency, find  the percentage reduction in the net work and cycle efficiency​.

Solution :

All the property values at different state points in a T-S diagram are taken from the steam table are given below.

h₁ = 3159.3KJ/kg

h₃ = $h_f_p_2$ = 173.88KJ/kg

$h_f_g_p_2$ = 2403.1KJ/kg

$v_f_p$ = 0.001008m³/kg

S₁ = 6.9917 KJ/kgK

S₃ = $S_f_p_2$ = 0.5926 KJ/kgK

$S_g_p_2$ = 8.2287 kJ/kgK

$S_f_g_p_2$ = 7.6361 kJ/kgK

Now,

S₁ = S₂s = 6.9917

= It is also written as or in the form of

S₁ = S₂s

S₁ = $S_f_p_2$ + $x_2_sS_f_g_p_2$ = 0.5926 + x₂ × 7.6361

from this we have to find the value of

$x_2_s$ = 6.3991 / 7.6361 = 0.838

∴ $h_2_s - h_f_p_2 + x_2_s h_f_g_p_2$ = 173.88 + 0.838 × 2403.1 = 2187.68KJ/kg

a).  $W_{P}$ = $h_{4s}$ - $h_{3}$ = $U_{fp2}$ ($p_{1}-p_{2}$)

           = 0.001008 m³/kg x 19.92 xl00 KN/m²

           = 2. 008 K/Kg

     $h_{4s}$ = 175.89 KJ/kg

    $W_{T}$ = $h_{1}-h_{2s$

           = 3159.3 - 2187 68

           = 971.62 KJ/kg

    $W_{net}$ = = $W_T-W_P}$ =  969.61 KJ/kg

    $Q_1=h_1-h_{4s}$ = 3159.3- 175.89

           = 2983.41 KJ/kg

$\eta_c_y_c_l_e = \frac{W_n_e_t }{Q_1}$ = 969.61 81 / 2983.41 = 0.325 or 32.5%

b) If $\eta_p$ = 80% and $\eta_T$ = 80%

$W_p$ = 2.008 / 0.8 = 2.51 kJ/kg

$W_T$ = 0.8 X 971.62 = 777.3 kJ/kg

$W_n_e_t$ = $W_T - W_p$ = 774.8 KJ/kg

∴ % Reduction in work output = $\frac{969.61 - 774.8}{969.61} *100$ = 20.1%

$h_4_s$ = 173.88 + 2.51 = 176.39 KJ/kg

Q₁ = 3159.3 - 176.39 = 2982.91 KJ/kg

$\eta_c_y_c_l_e$ = 774.8 / 2982.91 = 0.2597 or 25.97%

∴ % Reduction in Cycle efficiency = $\frac{0.325 - 0.2597}{0.325} *100$ = 20.1%