Question #3:For the following data, a process takes place from State 1 to State 2. Find the properties h, v, u & s atboth sides. The initial state is P₁ = 6.5barAlso determine the change in properties h, v, u & sQ#State 1State 2Processuh sX1V₁ = 0.16m³/kgP2 = 2.5bar2x = 0.8T = 550ºCConstant volumeConstant pressure

To solve this problem, we need to follow these steps:Identify the initial and final states based on…

Answered: Question #3: For the following data, a…

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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Can you do part A and B
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An ideal gas expands isothermally along AB and does 630 J of work. The process is illustrated in figure below (not drawn to scale). Answer below questions. Hint How much heat does the gas exchange along AB? Net heat flow into gas is Correct J. The gas then expands adiabatically along BC and does 470 J of work. When the gas returns to A along CA, it exhausts 130 J of heat to its surroundings. How much work is done on the gas along this path? Work done on the gas is Incorrect J. What type of thermodynamic process does process CA represent? adiabatic process neither isothermal nor adiabatic not enough information given for determination isothermal process Correct Submit QuestionQuestion 9
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In a reciprocating air compressor, high pressure air in the cylinder pushes the piston against the external load and expand to a lower pressure. In a test the following measurements were noted:Internal energy of air before expansion = 27 kJInternal energy of air after expansion = 8 kJHeat entering the cylinder during the process = 2 kJSHOW THE DIAGRAM USING THE GIVEN PROVIDED
(9 blanks) Derive the work done of an ideal gas undergoing isothermal expansion at temperature T1 if the volume changes from V1 to V2. Express in terms of P1 and P2. V2 W = V1 Where: p= TV By substitution: V2 dv W = vị V Applying the rules of integration, the word done will become W= In| Using: P1V1 = nRT & P2V2 = nRT The work done in terms of pressure is expressed as W = In >
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You will start by getting used to the idea of the linear mass density by relating it to concepts you are already familiar with: the mass and the (volume mass) density.* First, you have a column of pure water that has a density PH₂0 = 1 g/cc. The water column is in a cylinder. The column has cross-sectional area 31 cm² and height 75 cm before salt is added. The linear mass density has two different definitions. One is going up in dimensionality, that is from the volume mass density p to the linear mass density λ = PA where A is the cross-sectional area of the object. The other is going down in dimensionality, that is from the mass down to the linear mass density λ = M H where H is the height of the column in our case. You will use both of these below: (a) What is the linear mass density of the water column? ÅH20 = (b) What is the mass of the water? M = Let's now try this after adding salt to the water. When we do, the column's height rises to a level 78 cm, and we measure its salinity…

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