Why did they chose 1.40MPA and 1.60MPA and not 1.45MPA and 1.50MPa since it is closer to 0.15436? 4G. 46 17:27 O A0.00* 36 4KB/sStepla)"Temperature (T)= 250°csptei fie volume (v) = 0.15436 m³/rgP, u, h, s.Givenfind.Step2b)fromsteamP= 1•40 M1aP = 1.60 MPaO·141842692.3 291g.2 6.6732O. 163502698-3 2927.2 6.7467use linear faterpolation.Na - NI P- Pi)Pa -PiV= V+0.14184 -D.16350'xLP -1-40)O.154 36 =O.163 50 +1.60 - 1-4o1.484 MPaspecițic intennal Euvrgy2692.3 - 2698-32698. 3 +x(t-484 -1-40)1.60 -1-402695• 78 KT/KgEnthalpy.2919.2 -2027-22927.2 +x (1.484 - 1-40)1.60 - 1-4029a3. 84 KT/rgEutrophyS= 6.7467 +6.6732 - 6.74671.b0 -l.40x (1.484 -l140)6.71 683VX

GivenT=250oCspeciic volume v=0.15436 m3/kg

Answered: Why did they chose 1.40MPA and 1.60MPA…

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Why did they chose 1.40MPA and 1.60MPA and not 1.45MPA and 1.50MPa since it is closer to 0.15436?

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)

8th Edition

ISBN:9781305387102

Author:Kreith, Frank; Manglik, Raj M.

Publisher:Kreith, Frank; Manglik, Raj M.

Chapter6: Forced Convection Over Exterior Surfaces

Section: Chapter Questions

Problem 6.12P

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Concept explainers

Stress transformation

The term stress indicates the measure of internal forces that develop due to an external force's application on an object. When a specific external force is applied to an object, internal stress generates in the material. The value of stress can obtain by taking the ratio of the externally applied force and cross-sectional area of the object. There are different types of stress depending on the type of external force.

Question

Why did they chose 1.40MPA and 1.60MPA and not 1.45MPA and 1.50MPa since it is closer to 0.15436?

4G. 46 17:27 O A
0.00
* 36 4
KB/s
Stepl
a)
"Temperature (T)= 250°c
sptei fie volume (v) = 0.15436 m³/rg
P, u, h, s.
Given
find.
Step2
b)
from
steam
P= 1•40 M1a
P = 1.60 MPa
O·14184
2692.3 291g.2 6.6732
O. 16350
2698-3 2927.2 6.7467
use linear faterpolation.
Na - NI P- Pi)
Pa -Pi
V= V+
0.14184 -D.16350
'xLP -1-40)
O.154 36 =
O.163 50 +
1.60 - 1-4o
1.484 MPa
specițic intennal Euvrgy
2692.3 - 2698-3
2698. 3 +
x(t-484 -1-40)
1.60 -1-40
2695• 78 KT/Kg
Enthalpy.
2919.2 -2027-2
2927.2 +
x (1.484 - 1-40)
1.60 - 1-40
29a3. 84 KT/rg
Eutrophy
S= 6.7467 +
6.6732 - 6.7467
1.b0 -l.40
x (1.484 -l140)
6.71 683
VX

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