I will give rating to whom shows details and equations used show work please : )
ID: 1855548 • Letter: I
Question
I will give rating to whom shows details and equations used show work please :)
1.Determine the specific enthalpy (kJ/kg) of superheated water using the table below. The pressure and temperature are given as P = 400 kPa and T = 355.4oC, respectively.
2.Determine the specific volume (m3/kg) of superheated water using the table below. The pressure and temperature are given as P = 400 kPa and T = 148.57oC, respectively.
3.Determine the specific internal energy (kJ/kg) of superheated water using the table below. The pressure and temperature are given as P = 300 kPa and T = 401.7oC, respectively.
4.Determine the specific internal energy (kJ/kg) of superheated water using the table below. The pressure and temperature are given as P = 316.9 kPa and T = 400oC, respectively.
5.Determine the temperature (oC ) of superheated water using the table below. The pressure and specific internal energy are given as P = 400 kPa and u = 2,857.94 kJ/kg, respectively.
6.Determine the specific volume (m3/kg) of superheated water using the table below. The pressure and temperature are given as P = 375.3 kPa and T = 260oC, respectively.
7.Calculate the specific enthalpy (kJ/kg) of a substance given the following data. Use the equation forspecific enthalpy,h=u + Pv.
u= 2,821 kJ/kg
P = 2.67 bar
v= 0.851 m3/kg
I will give rating to whom shows details and equations used show work please :)
Explanation / Answer
1)at 4 bar see the 2nd table : at 340 : h1 = 3148.5 KJ/Kg and at 360 C it is : h2 = 3189.8 Kj/Kg
so at 355.4 C, h = 3148.5 + (355.4 - 340 )/(360-340) x (3189.8 - 3148.5) = 3180. 3 Kj/Kg
2) at 4 bar : at 143.63 , v1 = 0.4625 and at 150 C , v2 = 0.4708 m^3/Kg
so, at 148.57 c, v = 0.4625 + (148.57 - 143.63)/(150 - 143.63) x (0.4708 - 0.4625) = 0.46933 m^3/kg
3) at 3 bar : at 400 : u1 = 2965.1 Kj/kg and at 420 C , u2 = 2997.5
so, at 401.7 C : u = 2965.1 + (401.7 - 400)/(420-400) x (2997.5 - 2965.1) = 2967.854 Kj/Kg
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