1. ME 200 L14: ME 200 L14:Conservation of Mass: Control Volume 4.1-4.3 HW 5 cancelled HW 6 assigned https://engineering.purdue.edu/ME200/ Spring 2014 MWF 1030-1120 AM J. P. Gore gore@purdue.edu Gatewood Wing 3166, 765 494 0061 Office Hours: MWF 1130-1230 TAs: Robert Kapaku rkapaku@purdue.edu Dong Han han193@purdue.edurkapaku@purdue.eduhan193@purdue.edu

2. Mass Rate Balance time rate of change of mass contained within the control volume at time t time rate of flow of mass in across inlet i at time t time rate of flow of mass out across exit e at time t (Eq. 4.1)

3. Mass Rate Balance (Eq. 4.2) Eq. 4.2 is the mass rate balance for control volumes with several inlets and exits. In practice there may be several locations on the boundary through which mass enters or exits. Multiple inlets and exits are accounted for by introducing summations:

4. Mass Flow Rate (One-Dimensional Flow) ► Flow is normal to the boundary at locations where mass enters or exits the control volume. ► All intensive properties are uniform with position over each inlet or exit area ( A ) through which matter flows. (Eq. 4.4b) where V is velocity v is specific volume

5. Example A.How long does a 50 gallon drum take to fill if H 2 O is added at 2 gallons/minute and density remains constant? B.If a 2 inch hose is currently feeding the water, find the average velocity of water in the hose. C.If the time to fill needs to be reduced to 25%, and only one hose can be used, what size hose would be needed? Control Volume Basic Equations Solution Water 1 2 5

6. Mass Rate Balance (Steady-State Form) ► Steady-state: all properties are unchanging in time. ► For steady-state control volume, dm cv / dt = 0. (Eq. 4.6) (mass rate in)(mass rate out)

7. 6 kg/min = π/4(0.025 m) 2 V/0.0247 m 3 /kg V 1 = 5.03 m/s d 2 = 2.07 cm Example R 134a enters the condenser of a refrigeration system operating at steady state and 9 bar, 50 ºC, through a 2.5 cm diameter pipe. At the exit, the pressure is 9 bar, the temperature is 30 ºC, and the velocity is 2.5 m/s. The mass flow rate of the entering refrigerant is 6 kg/min. Determine: the velocity at the inlet, in m/s, and the diameter of the exit pipe, in cm. Find –V 1 = ? in m/s –d 2 = ? in cm System Basic Equations Solution P 1 = 9 bar T 1 = 50 ºC d 1 = 2.5 cm m 1 = 6 kg/min R-134a P 2 = 9 bar T 2 = 30 ºC V 2 = 2.5 m/s 1 2 7

8. Example: Feed Water Heater A direct contact feed-water heater is shown in the sketch. Water at 7 bars and 40 C enters through a 25 cm2 port and is heated by steam entering at 7 bars and temperature of 200 C at a flow rate of 40 kg/s. The resulting saturated liquid leaves the FWH at 7 bars. Find: Mass flow rate at inlet 2 and exit 3. Assumptions: Steady state Solution P 2 = 7 bars T 2 = 40 ºC A 2 = 25 cm 2 FWH P 3 = 7 bars Sat. Liquid A 3 V 3 =0.06m3/s 2 3 8 P 1 = 7 bars T 1 = 200 ºC m 1 = 40 kg/s 1 Basic Equations

9. Example: Feed Water Heater A direct contact feed-water heater is shown in the sketch. Water at 7 bars and 40 C enters through a 25 cm2 port and is heated by steam entering at 7 bars and temperature of 200 C at a flow rate of 40 kg/s. The resulting saturated liquid leaves the FWH at 7 bars. Find: Mass flow rate at inlet 2 and exit 3. Assumptions: Steady state Solution P 2 = 7 bars T 2 = 40 ºC A 2 = 25 cm 2 FWH P 3 = 7 bars Sat. Liquid A 3 V 3 =0.06m3/s 2 3 9 P 1 = 7 bars T 1 = 200 ºC m 1 = 40 kg/s 1 State 2 is defined. Find v 2 and then design the volume flow rate and the area of the pipe.

10. Example: Aircraft Jet Engine Video September 17th, 2010ME 20010 http://www.youtube.com/watch?v=ON0sVe1yeOk