We have to find specific enthalpy steam at 0.024 pressure.We can determine this property as follows (Refer table-2 givenbelow)Atp = 0.02 bar pressure, h =2535.5 kJ/kgAtp = 0.03 bar pressure, h =2545.6 kJ/kg SaturationTemp.(p) bar0.010.020.03Table 7.15.2: (Refer Appendix A-2) Properties of saturated water and steam (Pressure) tablesAbsoluteSpecific entropySpecific volumeSpecific enthalpypressure t, (0° C)(kJ/kg K)(m³/kg)(kJ/kg)WaterEvaporationSteamWater EvaporationSteam Waterh₂begSigVISph₂82484.92514.20.106 8.8716.980.001000 129.21 29.32533.50.2618.46373.52460.00.001001 67.0017.032545.6101.1 2444.50.3558.22324.080.001003 45.67SteamS₂8.9778.7248.578

In this problem, we must calculate the specific enthalpy steam at 0.024 pressure.This attribute may…

We have to find specific enthalpy steam at 0.024 pressure. We can determine this property as follows (Refer table-2 given below) At p = 0.02 bar pressure, h =2535.5 kJ/kg p = 0.03 bar pressure, h =2545.6 kJ/kg At

We have to find specific enthalpy steam at 0.024 pressure. We can determine this property as follows (Refer table-2 given below) At p = 0.02 bar pressure, h =2535.5 kJ/kg p = 0.03 bar pressure, h =2545.6 kJ/kg At

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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Concept explainers

Properties of Pure Substances

A substance with fixed chemical composition throughout is called a pure substance like water, air, and nitrogen. A pure substance does not have to be a single element or compound. A mixture of two or more phases of a pure substance is also a pure substance as long as the chemical composition is the same for all phases. These substances mainly have a constant/ uniform composition. The substances also have fixed boiling and melting points. A pure substance takes part in a chemical reaction to form predictable products.

Question

We have to find specific enthalpy steam at 0.024 pressure.
We can determine this property as follows (Refer table-2 given
below)
At
p = 0.02 bar pressure, h =2535.5 kJ/kg
At
p = 0.03 bar pressure, h =2545.6 kJ/kg

Saturation
Temp.
(p) bar
0.01
0.02
0.03
Table 7.15.2: (Refer Appendix A-2) Properties of saturated water and steam (Pressure) tables
Absolute
Specific entropy
Specific volume
Specific enthalpy
pressure t, (0° C)
(kJ/kg K)
(m³/kg)
(kJ/kg)
Water
Evaporation
Steam
Water Evaporation
Steam Water
h₂
beg
Sig
VI
Sp
h₂
8
2484.9
2514.2
0.106 8.871
6.98
0.001000 129.21 29.3
2533.5
0.261
8.463
73.5
2460.0
0.001001 67.00
17.03
2545.6
101.1 2444.5
0.355
8.223
24.08
0.001003 45.67
Steam
S₂
8.977
8.724
8.578

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