SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 4 SONNTAG BORGNAKKE VAN WYLEN. FUNDAMENTALS of. Thermodynamics. Sixth Edition

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1 SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 4 SONNTAG BORGNAKKE VAN WYLEN FUNDAMENTALS of Thermodynamics Sixth Edition

2 CHAPTER 4 SUBSECTION PROB NO. Concept-Study Guide Problems 7- Simple processes 3-8 Review Problems 9-30 Polytropic processes 3-34 Multi-step processes, other types of work Rates of work Heat Transfer Rates 4-43 Correspondence table The correspondence between the problem set in this sixth edition versus the problem set in the 5'th edition text. Problems that are new are marked new and the SI number refers to the corresponding SI unit problem. New 5 th Ed. SI New 5 th Ed. SI 7 new 3 new 50 8 new new new new new new New 4 40 new 9 7 new

3 Concept Problems 4.7E The electric company charges the customers per kw-hour. What is that in english units? The unit kw-hour is a rate multiplied with time. For the standard English Eng. units the rate of energy is in Btu/h and the time is in seconds. The integration in Eq.4. becomes kw-hour = 34.4 Btu/h h = 34.4 Btu Conversions are found in Table A. 4.8E Work as F x has units of lbf-ft, what is that in Btu? Conversions are found in Table A. lbf-ft = Btu = 778 Btu 4.9E A work of.5 Btu must be delivered on a rod from a pneumatic piston/cylinder where the air pressure is limited to 75 psia. What diameter cylinder should I have to restrict the rod motion to maximum ft? W = F dx = P dv = PA dx = PA x = P π 4 D x 4W D = πp x = = ft 4.5 Btu π 75 psia ft = lbf-ft π (lbf/ft ) ft

4 4.0E A force of 300 lbf moves a truck with 40 mi/h up a hill. What is the power? Ẇ = F V = 300 lbf 40 (mi/h) = = lbf-ft s =.6 Btu/s lbf-ft s

5 4.E A 00 hp dragster engine drives the car with a speed of 65 mi/h. How much force is between the tires and the road? Power is force times rate of displacement as in Eq.4. Power, rate of work Ẇ = F V = P V. = T ω We need the velocity in ft/s: V = 3600 = ft/s We need power in lbf-ft/s: hp = 550 lbf-ft/s F = Ẇ / V = lbf-ft/s ft/s = 693 lbf

6 4.E A 00 hp dragster engine has a drive shaft rotating at 000 RPM. How much torque is on the shaft? Power is force times rate of displacement as in Eq.4. Power, rate of work Ẇ = F V = P V. = T ω We need to convert the RPM to a value for angular velocity ω ω = RPM π π rad 60 s = s = s We need power in lbf-ft/s: hp = 550 lbf-ft/s 00 hp 550 lbf-ft/s-hp T = Ẇ / ω = rad/s = 35 lbf-ft

7 Simple Processes 4.3E A bulldozer pushes 000 lbm of dirt 300 ft with a force of 400 lbf. It then lifts the dirt 0 ft up to put it in a dump truck. How much work did it do in each situation? W = F dx = F x = 400 lbf 300 ft = lbf-ft = 54 Btu W = F dz = mg dz = mg Z = 000 lbm 3.74 ft/s 0 ft / (3.74 lbm-ft / s -lbf) = lbf-ft =.85 Btu

8 4.4E A steam radiator in a room at 75 F has saturated water vapor at 6 lbf/in. flowing through it, when the inlet and exit valves are closed. What is the pressure and the quality of the water, when it has cooled to 75F? How much work is done? After the valve is closed no flow, constant V and m. T 75 F P : x =, P = 6 lbf/in v = v g = ft 3 /lbm v : T = 75 F, v = v = ft 3 /lbm P = P g = 0.43 lbf/in v = = x ( ) x = W = PdV = 0

9 4.5E A linear spring, F = k s (x x o ), with spring constant k s = 35 lbf/ft, is stretched until it is.5 in. longer. Find the required force and work input. F = k s (x - x 0 ) = 35.5/ = 7.9 lbf W = Fdx = k s (x - x 0 )d(x - x 0 ) = k s (x - x 0 ) = 35 (.5/) = 0.76 lbf ft = Btu

10 4.6E Two hydraulic cylinders maintain a pressure of 75 psia. One has a cross sectional area of 0. ft the other 0.3 ft. To deliver a work of Btu to the piston how large a displacement (V) and piston motion H is needed for each cylinder? Neglect P atm W = F dx = P dv = PA dx = PA H = P V W = Btu = lbf-ft V = W P = lbf-ft lbf/ft = ft3 Both cases the height is H = V/A H = = ft H = = 0.09 ft F F cb

11 4.7E A piston/cylinder has 5 ft of liquid 70 F water on top of the piston (m = 0) with cross-sectional area of ft, see Fig. P.57. Air is let in under the piston that rises and pushes the water out over the top edge. Find the necessary work to push all the water out and plot the process in a P-V diagram. P = P o + ρgh = psia =.8 psia V = H A = 5 = 5 ft 3 lbm/ft 3 ft/s ft (lbm-ft/s -lbf) (in/ft) W = AREA = P dv = ½ (P + P o )(V max -V ) = ½ ( ) psia 5 ft 3 44 (in/ft) = lbf-ft = 49.8 Btu P o P HO P cb Air P0 V Vmax V

12 4.8E A cylinder fitted with a frictionless piston contains 0 lbm of superheated refrigerant R-34a vapor at 00 lbf/in., 300 F. The setup is cooled at constant pressure until the R-34a reaches a quality of 5%. Calculate the work done in the process. Constant pressure process boundary work. State properties from Table F.0 State : Table F.0. v = ft 3 /lbm; State : Table F.0. v = = ft 3 /lbm Interpolated to be at 00 psia, numbers at 0.5 psia could have been used. W = P dv = P (V -V ) = mp (v -v ) = ( ) = Btu P C.P. T C.P. 00 T P = 00 psia cb v v

13 Review Problems 4.9E The gas space above the water in a closed storage tank contains nitrogen at 80 F, 5 lbf/in.. Total tank volume is 50 ft 3 and there is 000 lbm of water at 80 F. An additional 000 lbm water is now forced into the tank. Assuming constant temperature throughout, find the final pressure of the nitrogen and the work done on the nitrogen in this process. Water is compressed liquid, so it is incompressible V H O = mv = = ft3 V N = V tank - V H O = = ft3 V N = V tank - V H O = = 7.85 ft3 N is an ideal gas so P N = P N V N /V N = = lbf/in W = PdV = P V ln V V = ln = Btu

14 4.30E A cylinder having an initial volume of 00 ft 3 contains 0. lbm of water at 00 F. The water is then compressed in an isothermal quasi-equilibrium process until it has a quality of 50%. Calculate the work done in the process assuming water vapor is an ideal gas. State : T, v = V/m = = 500 ft3 /lbm ( > v g ) since P g = 0.95 psia, very low so water is an ideal gas from to. P = P g v g v = V = mv = 0.*350 = 70 ft 3 = lbf/in v 3 = ( ) = 75.0 ft 3 /lbm W = PdV = P V ln V V = ln 00 = Btu W 3 = P =3 m(v 3 - v ) = (75-350) 44 / 778 = -6.6 Btu W 3 = = Btu P C.P. T C.P T 00 3 Psat P v v

15 Polytropic Processes 4.3E Helium gas expands from 0 psia, 600 R and 9 ft 3 to 5 psia in a polytropic process with n =.667. How much work does it give out? Process equation: PV n = constant = P V n = P Vn Solve for the volume at state V = V (P /P ) /n = = ft 3 Work from Eq.4.4 W = P V - P V n = lbf-ft = 43 lbf-ft = 5.43 Btu

16 4.3E Consider a mass going through a polytropic process where pressure is directly proportional to volume (n = ). The process start with P = 0, V = 0 and ends with P = 90 lbf/in., V = 0.4 ft 3.The physical setup could be as in Problem.. Find the boundary work done by the mass. 90 P W = PdV = AREA = (P + P )(V - V ) 0 0 W 0.4 V = (P + 0)( V - 0) = P V = = 59 ft lbf = 3.33 Btu

17 4.33E The piston/cylinder shown in Fig. P4.48 contains carbon dioxide at 50 lbf/in., 00 F with a volume of 5 ft 3. Mass is added at such a rate that the gas compresses according to the relation PV. = constant to a final temperature of 350 F. Determine the work done during the process. From Eq. 4.4 for PV n = const ( n =/ ) W = PdV = P V - P V - n W = mr(t - T ) - n W = ( ) -. Assuming ideal gas, PV = mrt, But mr = P V T = = Btu/R = Btu

18 4.34E Find the work for Problem 3.56E. State : Table F.9 P = 74.6 lbf/in, v = 0.94 ft 3 /lbm Process: Pv = C = P v = P v w = Pdv = C v - dv = C ln v v v State : P = 30 lbf/in P ; v = = / 30 =.76 ft 3 /lbm P w = P v ln v v = P v ln P P = ln = 6845 ft lbf/lbm =.65 Btu/lbm P T v v Notice T is not constant. It is not an ideal gas in this range.

19 Multi-step Processes, Other Types of Work 4.35E Consider a two-part process with an expansion from 3 to 6 ft 3 at a constant pressure of 0 lbf/in. followed by an expansion from 6 to ft 3 with a linearly rising pressure from 0 lbf/in. ending at 40 lbf/in.. Show the process in a P-V diagram and find the boundary work. By knowing the pressure versus volume variation the work is found P 3 W 3 = W + W 3 3 = PdV + PdV = P (V V ) 3 6 W = 0 44 (6-3) + (0 + 40)( - 6) 44 = = ft lbf. = ( / 778) = 44.4 Btu V + (P + P 3 )(V 3 -V )

20 4.36E A piston/cylinder has lbm of R-34a at state with 00 F, 90 lbf/in., and is then brought to saturated vapor, state, by cooling while the piston is locked with a pin. Now the piston is balanced with an additional constant force and the pin is removed. The cooling continues to a state 3 where the R-34a is saturated liquid. Show the processes in a P-V diagram and find the work in each of the two steps, to and to 3. Solution : C.V. R-34a This is a control mass. Properties from table F.0. and 0. State : (T,P) => v = ft 3 /lbm State given by fixed volume and x =.0 State : v = v = v g => W = T = = 55.3 F P = ( ) 0.59 = psia State 3 reached at constant P (F = constant) state 3: P 3 = P and v 3 = v f = ( ) 0.59 = 0.08 ft 3 /lbm W 3 = W + W 3 = 0 + W 3 = P dv = P(V 3 -V ) = mp(v 3 -v ) = ( ) = Btu P 3 cb V

21 4.37E A cylinder containing lbm of ammonia has an externally loaded piston. Initially the ammonia is at 80 lbf/in., 360 F and is now cooled to saturated vapor at 05 F, and then further cooled to 65 F, at which point the quality is 50%. Find the total work for the process, assuming a piecewise linear variation of P versus V. P State : (T, P) Table F.8. v =.767 ft 3 /lbm F State : (T, x) Table F.8. sat. vap. 9 P = 9 psia, v =.3 ft 3 /lbm 05 F State 3: (T, x) P 8 3 = 8 psia, 3 65 F v 3 = ( )/ = W 3 = PdV ( P + P v )m(v - v ) + ( P + P 3 )m(v 3 - v ) = [( )( ) + ( )( )] = -45. Btu

22 4.38E A -ft-long steel rod with a 0.5-in. diameter is stretched in a tensile test. What is the required work to obtain a relative strain of 0.%? The modulus of elasticity of steel is lbf/in.. - W = AEL 0 (e), A = π 4 (0.5) = π 6 in - W = ( π 6 ) (0-3 ) =.94 ft lbf

23 Rates of Work 4.39E An escalator raises a 00 lbm bucket of sand 30 ft in minute. Determine the total amount of work done and the instantaneous rate of work during the process. W = Fdx = F dx = F x = = 6000 ft lbf = (6000/778) Btu = 7.7 Btu. W = W / t = 7.7 / 60 = 0.9 Btu/s

24 4.40E A piston/cylinder of diameter 0 inches moves a piston with a velocity of 8 ft/s. The instantaneous pressure is 00 psia. What is the volume displacement rate, the force and the transmitted power? Rate of work is force times rate of displacement. The force is pressure times area. F = PA = P π D /4 = 00 lbf/in (π/4) 0 in = 7854 lbf. W = FV = 7854 lbf 8 ft s = 4 37 lbf-ft/s = 8.7 Btu/s V P

25 Heat Transfer Rates 4.4E The sun shines on a 500 ft road surface so it is at 5 F. Below the inch thick asphalt, average conductivity of Btu/h ft F, is a layer of compacted rubbles at a temperature of 60 F. Find the rate of heat transfer to the rubbles.. Q = k A T x = = 735 Βtu/h 5 60 /

26 4.4E A water-heater is covered up with insulation boards over a total surface area of 30 ft. The inside board surface is at 75 F and the outside surface is at 70 F and the board material has a conductivity of 0.05 Btu/h ft F. How thick a board should it be to limit the heat transfer loss to 70 Btu/h? Steady state conduction through a single layer board.. Q cond = k A T x = k Α Τ/ Q. x x = (75-70) / 70 = 0.9 ft =.6 in

27 4.43E The black grille on the back of a refrigerator has a surface temperature of 95 F with a total surface area of 0 ft. Heat transfer to the room air at 70 F takes place with an average convective heat transfer coefficient of 3 Btu/h ft R. How much energy can be removed during 5 minutes of operation?. Q = hα T; Q =. Q t = ha T t Q = 3 (Btu/h ft R) 0 ft (95 70) F (5/60) h = 87.5 Btu

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