1. ME 200 L15: ME 200 L15:Conservation of Energy: Control Volumes 2.5, 4.4-4.5 HW 5 cancelled; HW 6 assigned: Due 02/26/14 https://engineering.purdue.edu/ME200/ ThermoMentor © Program Launched 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. 2 In this Lecture … Allowing flow of mass (with associated energy) into and out of a system to make the “closed system,” an “open system.” Heat and work transfer are common to both closed and open systems. In flow and out flow of mass occurs with associated energy transport. In flow and out flow of mass requires energy expenditure called “Flow Work.” Keeping track of energy that comes in and the energy that goes out to ensure what stays in.

3. Energy Rate Balance/Energy Conservation/1 st Law of Thermo time rate of change of the energy contained within the control volume at time t net rate at which energy is being transferred in by heat transfer at time t net rate at which energy is being transferred out by work at time t net rate of energy transfer into the control volume accompanying mass flow Exit Flow work Inlet Flow work

4. Energy Rate Balance/Energy Conservation/1 st Law of Thermo time rate of change of the energy contained within the control volume at time t net rate at which energy is being transferred in by heat transfer at time t net rate at which energy is being transferred out by work at time t net rate of energy transfer into the control volume accompanying mass flow Exit Flow work=0 Inlet Flow work=0 =0 Control Mass Mass =0 CM Does not matter =0 Does not matter

5. Energy Rate Balance/Energy Conservation/1 st Law of Thermo time rate of change of the energy contained within the control volume at time t net rate at which energy is being transferred in by heat transfer at time t net rate at which energy is being transferred out by work at time t net rate of energy transfer into the control volume accompanying mass flow Exit Flow work Inlet Flow work

6. Evaluating Work for a Control Volume (Eq. 4.12) The expression for work is accounts for boundary work (associated with rotating shafts, displacement of the boundary, and electrical effects) where ► is the flow work at exit e. ► is the flow work at inlet i. ►

7. 7 Total Energy For a non-flowing fluid: For a flowing fluid, we must add the fluid’s potential for doing flow work: The energy contained in a flowing fluid is: Now you see the utility of h! 7

8. 8 First Law for Open Systems 8 dE/dt = net rate of Q net rate of W -+ Rate of energy addition due to inflow Rate of energy loss due to outflow -

9. 9 First Law for Open Systems What is our convention for Q and W? What assumptions are inherent in this form of the 1 st Law? 9

10. 10 First Law for Open Systems We also have steady-flow forms of mass and energy conservation. We use this form most often in ME 200. 10

11. 11 Common Assumptions In addition, there are common assumptions for each term: –Insulated (adiabatic) –Negligible ΔKE –Negligible ΔPE –No work –1-inlet, 1-outlet 11

12. 12 Common Assumptions We can still use previous assumptions such as –Incompressible (for liquids) –ideal gas –absence of Q and/or W types 12 (if applicable)

13. 13 Example Steam with a specific enthalpy of 3000 kJ/kg and a mass flow rate of 0.5 kg/s enters a horizontal pipe. At the exit, the specific enthalpy is 1700 kJ/kg. If there is no significant change in kinetic energy, determine the rate of heat transfer between the pipe and its surroundings, in kW. Assume steady state. Find –Q = ? in kW Sketch Assumptions –The control volume is at steady state. –ΔW cv = Δke = Δpe = 0 Basic Equations 12 h 2 = 1700 kJ/kgh 1 = 3000 kJ/kg m 1 = 0.5 kg/s steam 13 ;

14. 14 Example Steam with a specific enthalpy of 3000 kJ/kg and a mass flow rate of 0.5 kg/s enters a horizontal pipe. At the exit, the specific enthalpy is 1700 kJ/kg. If there is no significant change in kinetic energy, determine the rate of heat transfer between the pipe and its surroundings, in kW. Assume steady state. Find –Q = ? in kW System (mass flowing through pipe) 12 h 2 = 1700 kJ/kgh 1 = 3000 kJ/kg m 1 = 0.5 kg/s steam 14

15. 15 Example Assumptions W cv = 0 ΔKE = 0 ΔPE = 0 Steady State, Steady Flow Operation Basic Equations Solution 15