Thermodynamics Assignment Questions

Thermodynamics Assignment Questions
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Question 1: Steam Nozzle
Absolute steam pressure = 275kPa
Steam quality = 0.95
Diameter of nozzle = 10mm
Mass flow rate = 11kg/hr
Steam pressure = 3Kpa
Velocity = 700m/s
Heat loss per unit volume = 50kJ/kg
The steam velocity at the inlet of nozzle

From first law, thermodynamics states that net energy change in any closed systems is zero (Dafermos, 2009).
W=0
Hin = Enthalpy at inlet of steam = 412kj/kg +0.95(521-412) = 515.55kJ/kg
Hout = Enthalpy of steam at outlet = 335+0.95(454-335) = 448.05Kj/kg
515+c22x9.81×11=448.05+11×7022×9.81+50
C = 20.37m/s
Quality of steam at exit
Hout= 448.05kJ/kg
448.05=312+x(467-312)
X= 0.9377
Question 2: Carnot Efficiency
Thermal temperature =90 degree Celsius
Rejection temperature =25 degree Celsius
(a) Thermal efficiency = TH-TCTH=90-2590=72.22%(b) Rate of delivering to environment = 2.2kW *0.7222= 1.59kW
Question 3: Air Compressor
Entry pressure = 100Kpa
Temperature at entry = 20 degree Celsius
Flow rate = 0.75m3/s
Exit pressure = 900kpa
Isentropic efficiency = 0.7
Exit temperature of air = x degree Celsius
hin = 514Kj/kg
Hin – Hout=0
0.7=514-378514-x
X= 319.71 kJ/kg
Temperature = temperature of air at 900kpa at 319.71kjkgfrom steam tables=47.8 degree celcius Power = Enthalpy*flow rate
Power = 514*0.075 = 38.55kW
Question 4: Turbine
Pressure at Entry = 1.2Mpa
Temperature at entry = 110 degree Celsius
Exit pressure =60Kpa
Exit temperature = 50 degree Celsius
Power output = 22Kw
Flow rate = power/enthalpy change
Enthalpy = 435 – 419.

52 = 15.48kJ/kg
Mass flow rate =22×100015.48×3600=0.418kg/s
Isentropic efficiency = 435-419.52435-416.67=84.44% Question 5: Piston Cylinder Cycle
Pressure 1 = 95kPa
Temperature 1 =170C
Pressure 2 = 480Kpa
Process 2-3 Isentropic expansion = 130Kpa
Process 3-4 = Constant volume heat rejection 95Kpa
Process 4-1 = Constant pressure heat rejection
Network cycle = (Qin- Qout) + (Wout-Win) =
0.718kjkg.k×17-130-950.718+130-951.00=3.026kJ
Thermal efficiency = Work output/work output
= 9517+130=64.6%
The Pressure-Volume Diagram

References
Dafermos, C. M. (2009). The second law of thermodynamics and stability. Archive for Rational Mechanics and Analysis, 70(2), 167-179.