Question:
Problems 1-3 refer to a power plant operating on an ideal Rankine steam cycle. The plant operates between the pressure limit of 1 MPa and 5.628 kPa. The temperature of the steam entering the turbines is 500 ⁰C the total gross power generated is 300 MW.
1 What is the quality of the steam after it has expanded in the turbines?
A. 78% B. 92% C. 97% D. 100%
2. What is the enthalpy of the steam as it enters the condenser
A. 2380 kJ/kg B. 2420 kJ/kg C. 2560 kJ/kg D. 2600 kJ/kg
3. What is the total steam flow-rate through all the turbines?
A. 10 kg/s B. 1 I 0 kg/s C. 200 kg/s D. 270 kg/s
Answer:
1. B
2. A
3. D
Explanation:
Given Parameters
Pressures
P₁ = 1 MPa, P₂ = 5.628kPa
Temperature
T₁ = 500 ⁰C
From Steam Tables
h₁ = 3479.1 kJ/Kg, S₁ = 7.7642 kJ/Kg
When P₂= 5.628kPa,
hf = 147 kJ/Kg hfg₂ = 2413 kJ/Kg
sf = 0.505 kJ/Kg sfg₂ = 7.845 kJ/KgK
(1).
s₁= s₂ (entropy)
s₂ = sf₂ + x₂ * sfg₂
7.7642 = 0.505 + x₂ * 7.845
solving for x₂, we have x₂ = 0.92
this implies, x₂ = 92%
hence, option B is the right answer
(2).
h₂ = hf₂ + x₂ * hfg₂
= 147 + 0.92 * 2413
= 2366.96 kJ/Kg
hence, Option A is correct
(3).
let the steam flow-rate be m
from the equation
[tex]m = \frac{p}{h1 - h2}[/tex]
where p is the power in Watt
m = 300000 / ( 3479.1 - 2366.96)
= 269.75Kg/sec
hence option D is the right answer
Answer:
Answer is 92%.
Explanation:
As per the question,
Pressure limits of 1 MPa and 5.628 kPa.
Temperature of the steam is 500⁰C.
(1).
s₁= s₂ (entropy)
s₂ = sf₂ + x₂ × sfg₂
7.7642 = 0.505 + x₂ × 7.845
=0.92 that is 92%.
