Water at 70 kPa and 1008C is compressed isentropically in a closed system to 4 MPa. Determine the final temperature of the water
1 answer:
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Answer:
a) 5.2 kPa
b) 49.3%
Explanation:
Given data:
Thermal efficiency ( л ) = 56.9% = 0.569
minimum pressure ( P1 ) = 100 kpa
<u>a) Determine the pressure at inlet to expansion process</u>
P2 = ?
r = 1.4
efficiency = 1 - [ 1 / (rp) ]
0.569 = 1 - [ 1 / (rp)^0.4/1.4
1 - 0.569 = 1 / (rp)^0.285
∴ (rp)^0.285 = 0.431
rp = 0.0522
note : rp = P2 / P1
therefore P2 = rp * P1 = 0.0522 * 100 kpa
= 5.2 kPa
b) Thermal efficiency
Л = 1 - [ 1 / ( 10.9 )^0.285 ]
= 0.493 = 49.3%
Answer:
a) The ductility = -30.12%
the negative sign means reduction
Therefore, there is 30.12% reduction
b) the true stress at fracture is 658.26 Mpa
Explanation:
Given that;
Original diameter = 12.8 mm
Final diameter = 10.7
Engineering stress = 460 Mpa
a) determine The ductility in terms of percent reduction in area;
Ai = π/4( )² ; Ag = π/4(
)²
% = π/4 [ ( ( )² - (
)²) / ( π/4 (
)²) ]
= ( ( )² - (
)²) / (
)² × 100
we substitute
= [( (10.7)² - (12.8)²) / (12.8)² ] × 100
= [(114.49 - 163.84) / 163.84 ] × 100
= - 0.3012 × 100
= -30.12%
the negative sign means reduction
Therefore, there is 30.12% reduction
b) The true stress at fracture;
True stress =
( 1 +
)
is engineering strain
= dL / Lo
= (do² - df²) / df² = (12.8² - 10.7²) / 10.7² = (163.84 - 114.49) / 114.49
= 49.35 / 114.49
= 0.431
so we substitute the value of into our initial equation;
True stress = 460 ( 1 + 0.431)
True stress = 460 (1.431)
True stress = 658.26 Mpa
Therefore, the true stress at fracture is 658.26 Mpa