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Investigating The Properties And Describe The Phase Of Each Mixture Essay

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Q3 T, oC P, kPa u, kJ/kg x Phase description
(a) 120.2 200 1720 0.6 Saturated liquid-vapour
(b) 125 232.1 1600 0.535 Saturated liquid-vapour
(c) 395.6 1000 2950 N/A Superheated vapour
(d) 75 500 313.9 N/A Compressed liquid
(e) 172.96 850 731.27 0.0 Saturated liquid

Outline: Determine the missing properties and describe the phase of each mixture. There are 2 properties given for every single mixture, this is sufficient to calculate the other properties.
(a) The quality of the mixture is 0.6 which means 60% of the mass of mixture is in vapour form while 40% of the ass is still in liquid phase. This means this is a saturated liquid-vapour mixture. We can now refer to the saturated water and steam table to find the temperature using the pressure given to us.
T=T_(sat@200kPa)=120.2^o C
To find the average internal energy we use the property table and equation; u=u_f+xu_fg =505 kJ⁄kg+(0.6)(2530 kJ⁄kg-505 kJ⁄kg)
=1720 kJ/kg
(b) We do not know this mixtures quality therefore we have to determine whether we have a compressed liquid, saturated vapour-liquid mixture or superheated water. To do this I will use the assumption that; uu_g superheated vapor
We can now look at the saturation table, u_f=524.74 kJ⁄kg and u_g=2534.6 kJ⁄kg and u=1600 kJ⁄kg. From this we can deduce we have a saturated liquid vapour mixture yet again so we refer back to the saturated liquid vapour table to find our pressure.
P=P_(saturated@125^0 C)=232.1 kPa
To acquire the quality we can rearrange the

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