Question

Q1 (a) A U-tube manometer is used to measure the pressure differential between two locations in a building ventilation system. If this reads 80 mm H2O, which of the following correctly describes the pressure differential? (A) 0.080 Pa (B) 0.785 kPa (C) 0.785 Pa (D) 80 Pa

(b) The specific enthalpy of 0.20 kg of water contained in a volume of 0.06 m3 at 500 kPa is (A) 1686 kJ•kg-1 (B) 2326.8 kJ•kg-1 (C) 2108.5 kJ•kg-1 (D) 2748.7 kJ•kg-1  

(c) A tank contains liquid water at a pressure of 100 kPa and a temperature of 50.0oC. Referring to the data tables for H2O, the specific internal energy is best approximated by; (A) 2506.1 kJ.kg-1 (B) 2442.7 kJ.kg-1 (C) 417.4 kJ.kg-1 (D) 209.3 kJ.kg-1

(d) A gas expands in a closed container during a polytropic process where the index n = 1.35. The initial pressure is 0.8 MPa. If the volume increases by a factor of 4 during the process, what will be the final pressure? (A) 0.200 MPa (B) 0.270 MPa (C) 5.198 MPa (D) 0.123 MPa

Answer 1

Given pressure differential (Delta P) = 80 mm H_2 O 1 mm of H_2 O = 0.00980665 kPa Therefore 80 mm of H_2 O = 0.784532 kPa 0.785 kPa Therefore Delta P 0.785 kPa option (B) (b) Given m = 0.2 kg V = 0.06 m^3 P = 500 kPa Specific enthalpy (h) = U (kJ/kg) + PV/m (kJ/kg) From property tables of water: At P = 500 kPa U_f = 639.54 kJ: U_fg = 1921.2 kJ/kg But specific volume (V) = V/m = 0.06/0.2 = 0.3 m^3/kg At P = 500 kPa (V_f) = 0.001093 m^3/kg (V_g) = 0.37483 m^3/kg Therefore V = V_f + x (V_g - V_f) Therefore 0.3 = 0.001093 + x (0.37483 - 0.001093) Therefore x = 0.799 Now, U = U_f + x (U_f g) = 639.54 + (0.799) (1921.2) U = 2,174.58 k J/kg Therefore h = 2174.58 + (500) (0.06)/0.2 = 2325.211 k J/kg