Thermodynamics 1 (MEP 261)
Quiz 2(B) ( 20 Marks)
Model Answer
Class ZA, Date: October 26, 2011
Name:
University ID:
Time: 2 hs
(12 Marks)
1) Complete the table for H2O:
T, °C
50
250
110
p, kPa
h, kJ/kg
200
400
600
v, m3/kg Phase description
4.16
Saturated vapor
x
2) Consider a river flowing toward a lake at an average velocity of 3 m/s at a rate of 500 m3/s at a
location 90 m above the lake surface. Determine the total mechanical energy of the river water per unit
mass and the power generation potential of the entire river at that location.
(5 Marks)
Solution: (Slides 11&12)
Kinetic energy per unit mass (ke) is
Ke = (3 m/s)2/2= 4.5 J/kg = 0.00454 kJ/kg
Potential energy per unit mass (pe) is
pe = 9.81 m/s2 * 90= 882.9 J/kg= 0.8829 kJ/kg
1
The total mechanical energy of the river water per unit mass (e) = ke + pe = 0.00454 kJ/kg +0.8829
kJ/kg = 0.88744 kJ/kg
= 5×105 kg/s
power generation potential of the entire river at that location;
3) Determine the mass of the air in a room whose dimensions are 6 m× 5 m× 4 m at 105 kPa and 30°C.
(3 Marks)
2