E2231 Quiz2docx Word Nicolas Makhoul File Home Insert Desig

E2231 Quiz2.docx - Word Nicolas Makhoul File Home Insert Design Layout References Mailings Review View Grammarly Tell me what you want to do Share Cut ind Copy abc Replace Enable Grammarly Grammarly Paste -v-.3 m |1Normal TNo Spac Heading 1 Heading 2 Title Subtitle Subtle Em...- Select Format Painter Clipboard Font Paragraph Styles Editing 4. You are told you have a closed system containing water. p- 80 bar, T- 90°C. What are the procedures (tables, calculations, etc.) you would use to find the system\'s specific internal energy? + 18096 10:02 AM 9/9/2016 Page 2 of 2 198 words Ask me anything

Solution

Given Data:

P = 80 bar and T = 90 °C = 363 K

To find:

Specific Internal Energy (u)= ?

Solution:

            h = u + pv

            h = u + RT       (pv = RT From ideal gas equation)

            R = Charateristics gas constant = R_U / M

            M = Molecular Weight

            M for H₂O = 18

            R_U = Universal Gas Constant = 8.314 kJ/kgmol K

            Therefore R = 8.314 / 18 = 0.4619 kJ/kgK

            From Steam table (saturated water and temperature table) at T = 90 °C

            h = h_f = 376.9 kJ/kg (For pure liquid h = h_f )

            (Hint: You should not take h_f value corresponding to P = 80 bar, because at this pressure there is a saturation temperature in that table, at this condition it may be wet steam or dry steam. If you go for Temperature table, there is a saturation pressure which is lesser than the given pressure. So its condition at that point is a liquid)

            h = u + RT:                u = h – RT = 376.9 – (0.4619 * 363) = 209.23 kJ / kg

Therefore, the specific internal energy u = 209.23 kJ/Kg

Since u is positive, specific internal energy increases.