In the data set below what is the mean absolute deviation 41,56,38,45,55,51,52
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Answer:
Consider a regenerative vapor power cycle with two feedwater heaters, a closed one and an open one, and reheat. Steam enters the first turbine stage at 12 MPa, 480∘C, and expands to 2 MPa. Some steam is extracted at 2 MPa and fed to the closed feedwater heater. The remainder is reheated at 2 MPa to 440∘C and then expands through the second-stage turbine to 0.3 MPa, where an additional amount is extracted and fed into the open feedwater heater operating at 0.3 MPa. The steam expanding through the third-stage turbine exits at the condenser pressure of 6 kPa. Feedwater leaves the closed heater at 210∘C, 12 MPa, and condensate exiting as saturated liquid at 2 MPa is trapped into the open feedwater heater. Saturated liquid at 0.3 MPa leaves the open feedwater heater. Assume all pumps and turbine stages operate isentropically. Determine for the cycle
a. Draw the cycle on a T-S diagram using the same numbering in the schematic
b. Determine the thermal efficiency of the cycle.
c. Determine the mass flow rate of steam entering the first turbine of the cycle.
(i) Thermal efficiency of the cycle = 43.185 %
(ii) The mass flow rate of steam =93.66 kg/h
Step-by-step explanation:
So we have at
For Point 1 on the T-S diagram we have
p₁ = 80 bar,
t₁ = 480 °C,
From the super-heated steam tables we have
h₁ = 3349.6 kJ/kg, s₁ = 6.6613 kJ/kg·K
Point 2
p₂ = 20 bar
s₁ = s₂ =with x₂ = (6.6613 -6.6409)/(6.6849-6.6409) = 0.464
therefore h₂ =2953.1 + 0.464×(2977.1 - 2953.1) = 2964.22 kJ/kg
Point 3 on the T-S diagram we have
p₃ = 3 bar again s₁ = s₃ so we go to 3 bar on the steam tables and look up s = 6.6613 kJ/kg·K which is on the saturated steam tables
and x₃ is given as (6.6613 -1.6717)/(6.9916-1.6717) = 0.9379 and
h₃ = 561.43 + x₃×2163.5 = 2590.6 kJ/kg
Point 4
p₄ = 0.08 bar, s₁ = s₄, x₄ = 0.7949 and h₄ = 2083.45 kJ/kg
Point 5
p₅ = 0.08 bar, = 173.84 kJ/kg
Point 6
Here h₆ is given by plus the work done to move the water to the open heater therefore h₆ =
= 173.84 kJ/kg + 0.00100848×(3 - 0.08) × 100
= 173.84 kJ/kg + 0.29447616 kJ/kg = 174.13 kJ/kg
Point 7
p₇ = 3 bar, and = 561.43 kJ/kg
Point 8
Here again work is done to convey the fluid t constant pressure thus
h₈ = × (p₈ - p₇)
561.43 kJ/kg + 0.00107317×(80 - 3)×100 = 569.69 kJ/kg
Point 9
p₉ = 80 bar and T₉ = 205°C
By interpolating the values on the subcooled teperature tables we get
x₉ = 0.5 and h₉ = 854.94 + 0.5× (899.79 - 854.94) = 877.365 kJ/kg
Point 10
p₁₀ = 20 bar, h₁₀ = = 908.50 kJ/kg
point 11
Here h₁₁ = h₁₀ = 908.50 KJ/kg
For the closed feed water heater, energy and mass flow rate balance gives
m₁ × (h₂ - h₁₀) + (h₈ - h₉) = 0
Therefore m₁ = = 0.14967
while the open water heater we get
m₂×h₃+(1-m₁-m₂)×h₆+m₁×h₁₁ - h₇ = 0
from where m₂ = 0.11479
= (h₁-h₂) + (1 - m₁)(h₂ - h₃) +(1 - m₁ - m₂)(h₃ - h₄)
= 1076.11 kJ/kg
= (h₈ - h₇) + (1 - m₁ - m₂)×(h₆ - h₅)
= 8.4733 kJ/kg
Q = h₁ -h₉ = 2472.235 kJ/kg
Efficiency = η = = 43.185 %
(ii)
m'₁ = 100×10³/1066.63 = 93.66 kg/h
ΔACB and ΔMNB are similar. Therefore the corresponding sides are in proportion:
Substitute: