If work is postive when the system performs work on the surroundings, the energy balance on the system is written as

Transcription

1 Chapter.4 Energy Balance Q System W Q System Work is + when performed Work is + when received by the system by the system E - E 1 Q - W E - E 1 Q + W Figure.4-1 Sign convention for work If work is postive when the system performs work on the surroundings, the energy balance on the system is written as W E E E 1 KE + PE + U Q W (.4-1a) In this equation E and E 1 are the final and initial energy of the system, Q is the heat transferred to the system, and W is the work performed by the system. If work is postive when the system receives work from the surroundings, the energy balance on the system is written as E E E 1 KE + PE + U Q + W (.4-1b) In this equation W is the work received by the system. No matter what is the sign convention for work, the system energy will increase when work is performed on the system. As an example of the sign convention for work, let consider the expansion work when the system volume change by d. If work is considered to be positive when performed by the system then δw pd (.4-a) When the volume increases d > 0, the gas or system performs work on the surroundings and δw is positive. If If work is considered to be positive when performed by the surroundings on the system then δw pd (.4-b) When the volume decreases d < 0, the surroundings perform work on the gas and δw is positive. The differential form of the energy balance with positive work performed by the system is de δq δw δq pd (.4-3a) -17

2 The differential form of the energy balance with positive work performed by the surroundings is de δq + δw δq pd (.4-3b) The final result of the energy balance must be independent of the sign convention for work. The instantaneous time rate form of the energy balance is de dke + dpe + du Q + W (.4-4) Example Air is contained in a vertical-cylinder assembly fitted with an electrical resistor. The atmosphere exerts a pressure of of 14.7 lbf/in on top of the piston, which has a mass of 100 lb and a face area of. Electric current passes through the resistor, and the volume of the air slowly increases by 1.6 ft 3 while its pressure remains constant. The mass of the air is 0.6 lb, and its specific internal energy increases by 18 Btu/lb. The air and piston are at rest initially and finally. The piston-cylincer is a ceramic composite and thus a good insulator. Friction between the piston and cylinder wall can be ignored, and the local acceleration of gravity is g 3.0 ft/s. Determine the heat transfer from the resistor to the air, in Btu, for a system consisting of (a) the air alone, (b) the air and the piston. 9 p atm 14.7 lbf/in m piston 100 lb A piston Piston Air + _ m air 0.6 lb ft u 18 Btu/lb air Solution (a) Taking the air as the system, the energy balance becomes KE + PE + U Q + W Since KE 0, PE 0, solving for Q we have Q U W The work performed by the air is given by W pd p( 1 ) 1 The gas pressure on the piston can be determined from pa piston m piston g + p atm A piston 9 Moran, M. J. and Shapiro H. N., Fundamentals of Engineering Thermodynamics, Wiley, 008, p

3 p m A pistong piston + p atm p (100 lb)(3.0 ft/s ) 1 lbf 3. lb ft/s 144 in lbf in 15.4 lbf in Thus the work is lbf W p( 1 ) 15.4 in (1.6 ft3 ) 144 in 3,548. ft lbf W 3,548. ft lbf 1 Btu 778 ft lbf 4.56 Btu The heat transfer is Q U W m air u air W Q (0.6 lb)(18 Btu/lb) Btu 15.4 Btu (b) Taking the air and the piston as the system, the energy balance becomes ( KE + PE + U) air + ( KE + PE + U) piston Q + W Since KE air 0, PE air 0, and KE piston 0 solving for Q we have Q U air + PE piston W The work is done by the upper surface of the piston where p p atm : W pd p atm ( 1 ) 1 lbf W 14.7 in (1.6 ft3 ) 144 in 1 Btu 778 ft lbf 4.35 Btu The elevation change z can be evaluated from the volume change of the air. 1 z A piston 1.6 ft ft The potential energy change of the piston is PE piston m piston g z -19

4 ft PE piston (100 lb)(3.0 s )(1.6 ft) 1 lbf 3. lb ft/s 1 Btu 778 ft lbf 0. Btu Q m air u air + PE piston W Q (0.6 lb)(18 Btu/lb) + 0. Btu Btu 15.4 Btu Example A 10 small well insulated cylinder and piston assembly contains an ideal gas at bar and 94.3 K. A mechanical lock prevents the piston from moving. The length of the cylinder containing the gas is m and the piston cross sectional area is m. B A Piston moves from A to B Gas The piston, which weights 6 kg, is tightly fitted and when allowed to move, there are indications that considerable friction is present. When the mechanical lock is released, the pistion moves in the cylinder until it impacts and is engaged by another mechanical stop; at this point, the gas volume has just double. The heat capacity of the ideal gas is 0.93 J/mol K, independent of temperature and pressure. Consider the heat capacity of the piston and cylinder walls to be negligible. As an engineer, can you estimate the temperature and pressure of the gas after such an expansion? Clearly state any assumption. Repeat the calculations if the cylinder were rotated 90 o and 180 o before tripping the mechanical lock. Solution Let system gas in the cylinder Case 1: piston rises E surroundings Q + W Q 0. Work done on the surroundings is given by W P a ( f i ) + m p g z, where z m i initial gas volume (0.305)( ) m 3 10 Tester J. W., Modell M. Thermodynamics and Its Applications, 3 rd Edition, Prencetice Hall, 1997, pg

5 f final gas volume m m 3 W ( )( ) + (6)(9.81)(0.305) 1.50 kj E surroundings E system 1.50 kj U system N g C v (T T gi ) N g moles of gas P g RT gi gi 5 3 ( )( ) (8.314)(94.3).3461 mol T T gi 1,50 J/(N g C v ) ,50/( ) 68.8 K P N RT g (.3461)(8.314)(68.8) ,681 Pa Case : piston moves horizontally (cylinder was rotated 90 o ) E surroundings Q + W Q 0. Work done on the surroundings is given by W P a ( f i ) m p g z, where z m W ( )( ) + (6)(9.81)(0.305) 574 J E surroundings E system 574 J U system N g C v (T T gi ) T T gi 574 J/(N g C v ) ,50/( ) 8.6 K P N RT g (.3461)(8.314)(8.6) ,380 Pa Case 3: piston moves down (cylinder was rotated 180 o ) E surroundings Q + W Q 0. Work done on the surroundings is given by W P a ( f i ) m p g z, where z m W ( )( ) (6)(9.81)(0.305) 10 J -1

6 E surroundings E system 1.50 kj U system N g C v (T T gi ) T T gi + 10 J/(N g C v ) ,50/( ) 96.4 K P N RT g (.3461)(8.314)(96.4) ,000 Pa The following Matlab program determines the final gas temperature and pressure for three cases: (1) Piston rises, () Piston moves horizontally, and (3) Piston moves down. The program can be saved as exd4d.m mp6; % kg g9.81; % m/s dx.305 % m R8.314; Cv0.93 % J/mol K i5.6669e-3 % m3 f*i; Tgi94.3 % K Pgi10.13e5 % Pa NgPgi*i/(R*Tgi) % Moles Pa1.013e5; % Pa % Case 3: Piston rises disp('piston rises') WPa*(f-i)+mp*g*dx; TTgi-W/(Ng*Cv); PNg*R*T/f; fprintf('work on the surroundings (J) %g\n',w) fprintf('final gas temperature (K) %g\n',t) fprintf('final gas pressure (Pa) %g\n',p) % Case : Piston moves horizontally disp(' ') disp('piston moves horizontally') WPa*(f-i); TTgi-W/(Ng*Cv); PNg*R*T/f; fprintf('work on the surroundings (J) %g\n',w) fprintf('final gas temperature (K) %g\n',t) fprintf('final gas pressure (Pa) %g\n',p) % Case 3: Piston moves down disp(' ') disp('piston moves down') WPa*(f-i)-mp*g*dx; TTgi-W/(Ng*Cv); PNg*R*T/f; fprintf('work on the surroundings (J) %g\n',w) fprintf('final gas temperature (K) %g\n',t) fprintf('final gas pressure (Pa) %g\n',p) >> exd4d -

7 Piston rises Work on the surroundings (J) Final gas temperature (K) Final gas pressure (Pa) Piston moves horizontally Work on the surroundings (J) Final gas temperature (K) 8.61 Final gas pressure (Pa) Piston moves down Work on the surroundings (J) Final gas temperature (K) Final gas pressure (Pa) Example The rate of heat transfer between a certain electric motor and its surroundings varies with time as Q 0.[1 e (-0.05t) ] In this equation, t is in seconds and Q is in kw. The shaft of the motor rotates at a constant speed of ω 100 rad/s and applies a constant torque of I 18 N m to an external load. The motor draws a constant electric power input equal to.0 kw. For the motor, plot Q and W, each in kw, and the change in energy, E, in kj, as function of time from t 0 to t 100 s. 11 Solution Let the system be the motor, the instantaneous time rate form of the energy balance is de Q + W W is the net power on the system, therefore W W elec W shaft W elec.0 kw W shaft Iω (18 N m)(100 rad/s) 1,800 W 1.8 kw W W elec W shaft kw The accumulation of energy is de 0.[1 e (-0.05t) ] e (-0.05t) 11 Moran, M. J. and Shapiro H. N., Fundamentals of Engineering Thermodynamics, Wiley, 008, p. 6-3

8 The change in energy at any time is E t de 0 t 0. e 0 ( 0.05 t) E E E 1 0. ( 0.05) ( 0.05 t ) t e 4[1 e (-0.05t) ] The following Matlab program plots Q and W, and the change in energy, E, as a function of time. % Example.4-3 tw[0 100]; Power[0. 0.]; t0:100; Heat_rate-0.*(1-exp(-0.05*t)); Del_E4*(1-exp(-0.05*t)); % Plot Del_E subplot(,1,1);plot(t,del_e) xlabel('time, s');ylabel('e_-e_1, kj') % Plot Power and Heat_rate subplot(,1,),plot(tw,power,t,heat_rate,':') axis([ ]) xlabel('time, s');ylabel('power, Heat_rate, kw') legend('power','heat_rate') 0-4