Letters G and H A. T- 310 K, v = 22.903 m'/kgRegion = ?P = ?B. T 310 K, v = 15 m'kgRegion = ?h = ?C. T = 310 K, P = 10 kPaRegion = ?v = ?%3DD. T = 310 K, P = 4 kPaRegion = ?u = ?E. T = 647.096 K, P = 22.064 MPaRegion = ?v = ?F. T = 800 K, P = 25 MPaRegion = ?p = ?G. T = 800 K, P = 5 MPaRegion = ?s = ?H. T 743.2 K, P = 4.61 MPaRegion = ?h = ? Given the following property data for H,0 designatethe region in T-v or P-v space (i.e., compressedliquid, liquid-vapor mixture, superheated vapor, etc.)and find the value(s) of the requested property orproperties. Use the tables in Appendix D or the NISTdatabase as necessary. Provide units with youranswers.

Answered: G. T 800 K, P = 5 MPa Region = ? s = ?…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Letters G and H

A. T- 310 K, v = 22.903 m'/kg
Region = ?
P = ?
B. T 310 K, v = 15 m'kg
Region = ?
h = ?
C. T = 310 K, P = 10 kPa
Region = ?
v = ?
%3D
D. T = 310 K, P = 4 kPa
Region = ?
u = ?
E. T = 647.096 K, P = 22.064 MPa
Region = ?
v = ?
F. T = 800 K, P = 25 MPa
Region = ?
p = ?
G. T = 800 K, P = 5 MPa
Region = ?
s = ?
H. T 743.2 K, P = 4.61 MPa
Region = ?
h = ?

Given the following property data for H,0 designate
the region in T-v or P-v space (i.e., compressed
liquid, liquid-vapor mixture, superheated vapor, etc.)
and find the value(s) of the requested property or
properties. Use the tables in Appendix D or the NIST
database as necessary. Provide units with your
answers.

Expert Solution

Step 1

G: Given Temperature T=800 K=527 $°$ C

Pressure P=5 MPa

Region=?, entropy s=?

From NIST steam tables we found above pressure 5 mpa and temperature 800 k are satisfies the SUPERHEATED REGION.In this region always liquid will exists in the form of steam.

and entropy we have values of entropy at 520 $°$ C and 540 $°$ C

temp( $° C$ ) entropy (KJ/Kg K)

520 7.0375

527 ?

540 7.0954

from interpolation

$\frac{520 - 540}{520 - 527} = \frac{7.0375 - 7.0954}{7.0375 - s}$

S=7.0577 KJ/kg k

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