Introduction to Engineering thermodynamics nd Edition, Sonntag and Borgnakke Solution manual Chapter 4 Claus Borgnakke The picture is a false color thermal image of the space shuttle s main engine. The sheet in the lower middle is after a normal shock across which you have changes in, T and density. Courtesy of NASA.
CONTENT SUBSECTION ROB NO. Concept problems -3 Force displacement work 4- Boundary work: simple one-step process 3-37 olytropic process 38-49 Boundary work: multistep process 50-6 Other types of work and general concepts 63-7 Rates of work 7-8 Heat transfer rates 83-9
Concept roblems
4. A car engine is rated at 60 hp. What is the power in SI units? The horsepower is an older unit for power usually used for car engines. The conversion to standard SI units is given in Table A. hp = 0.7355 kw = 735.5 W hp = 0.7457 kw for the UK horsepower 60 hp = 60 745.7 W = 9 3 W = 9.3 kw
4. Normally pistons have a flat head, but in diesel engines pistons can have bowls in them and protruding ridges. Does this geometry influence the work term? The shape of the surface does not influence the displacement dv = A n dx where A n is the area projected to the plane normal to the direction of motion. Work is A n = A cyl = π D /4 dw = F dx = dv = A n dx = A cyl dx and thus unaffected by the surface shape. Semi-spherical head is made to make room for larger valves. Ridge Bowl iston x normal plane
4.3 CV A is the mass inside a piston-cylinder, CV B is that plus the piston, outside which is the standard atmosphere. Write the process equation and the work term for the two CVs assuming we have a non-zero Q between a state and a state. o m p m A g C.V. A: rocess: = o + m p g/a cyl = C, C.V. B: rocess: = o = C, W = dv = m dv = m(v v ) W = o dv = o m dv = o m(v v )
4.4 Air at 90 K, 00 ka in a rigid box is heated to 35 K. How does each of the properties and v change (increase, about the same, or decrease) and what transfers do we have for Q and W (pos., neg., or zero)? Rigid box: V = C v = C, W = 0 Heat transfer in Q > 0 so positive As T goes up, air ideal gas V = mrt, so then increases
4.5 The sketch shows a physical setup. We now heat the cylinder. What happens to, T and v (up, down or constant) What transfers do we have for Q and W (pos., neg., or zero)? o m p rocess: = o + m p g/a cyl = C Heat in so T increases, v increases and Q is positive. As the volume increases the work is positive. W = dv
4.6 Two hydraulic piston/cylinders are connected through a hydraulic line so they have roughly the same pressure. If they have diameters of D and D = D respectively, what can you say about the piston forces F and F? For each cylinder we have the total force as: F = A cyl = π D /4 F = A cyl = π D /4 F = A cyl = π D /4 = π 4 D /4 = 4 F F F cb The forces are the total force acting up due to the cylinder pressure. There must be other forces on each piston to have a force balance so the pistons do not move.
4.7 For a buffer storage of natural gas (CH 4 ) a large bell in a container can move up and down keeping a pressure of 05 ka inside. The sun then heats the container and the gas from 80 K to 300 K during 4 hours. What happens to the volume and what is the sign of the work term? Solution The process has constant pressure Ideal gas: V = mrt as T increases then V increases W = dv > 0 so positive.
4.8 The sketch shows a physical situation, show the possible process in a -v diagram. a) b) c) o m p R- V V V V stop V V V stop
4.9 For the indicated physical set-up in a-b and c shown in Fig. 4.8 write a process equation and the expression for work. a) = and V V stop or V = V stop and W = (V V ) [ = float ] b) = A + BV; W = ( + )(V V ) c) = and V V stop or V = V stop and W = (V V ) [ = float ] a) b) c) o m p R- V V V V stop V V V stop
4.0 The sketch shows a physical situation; what is the work term a, b, c or d? a: w = (v v ) b: w = v ( ) c: w = ( + )(v v ) d: w = ( )(v + v ) work term is formula c
4. The sketch shows a physical situation; show the possible process in a -v diagram. a) b) c) o m p R- v v v v stop v v stop
4. Show how the polytropic exponent n can be evaluated if you know the end state properties, (, V ) and (, V ). olytropic process: V n = C Both states must be on the process line: V n = C = V n Take the ratio to get: = n V V and then take ln of the ratio ln = ln n V V now solve for the exponent n n = ln / ln V V = n ln V V
4.3 A drag force on an object moving through a medium (like a car through air or a submarine through water) is F d = 0.5 A ρv. Verify the unit becomes Newton. F d = 0.5 A ρv Units = m ( kg/m 3 ) ( m / s ) = kg m / s = N
Force displacement work 4.4 A piston of mass kg is lowered 0.5 m in the standard gravitational field. Find the required force and work involved in the process. F = ma = kg 9.80665 m/s = 9.6 N W = F dx = F dx = F x = 9.6 N 0.5 m = 9.805 J
4.5 An escalator raises a 00 kg bucket of sand 0 m in minute. Determine the total amount of work done during the process. The work is a force with a displacement and force is constant: F = mg W = F dx = F dx = F x = 00 kg 9.80665 m/s 0 m = 9807 J
4.6 A hydraulic cylinder of area 0.0 m must push a 000 kg arm and shovel 0.5 m straight up. What pressure is needed and how much work is done? F = mg = 000 kg 9.8 m/s = 980 N = A = F/A = 980 N/ 0.0 m = 98 000 a = 98 ka W = F dx = F x = 980 N 0.5 m = 4905 J
4.7 A hydraulic cylinder has a piston of cross sectional area 5 cm and a fluid pressure of Ma. If the piston is moved 0.5 m how much work is done? The work is a force with a displacement and force is constant: F = A Units: W = F dx = A dx = A x = 000 ka 5 0-4 m 0.5 m =.5 kj ka m m = kn m - m m = kn m = kj
4.8 Two hydraulic cylinders maintain a pressure of 00 ka. One has a cross sectional area of 0.0 m the other 0.03 m. To deliver a work of kj to the piston how large a displacement (V) and piston motion H is needed for each cylinder? Neglect atm. W = F dx = dv = A dx = A* H = V V = W = kj 00 ka = 0.000 833 m3 Both cases the height is H = V/A H = 0.000833 0.0 = 0.0833 m H = 0.000833 0.03 = 0.078 m F F cb
4.9 A linear spring, F = k s (x x 0 ), with spring constant k s = 500 N/m, is stretched until it is 00 mm longer. Find the required force and work input. F = k s (x - x 0 ) = 500 0. = 50 N W = F dx = k s (x - x 0 )d(x - x 0 ) = k s (x - x 0 ) / = 500 N m (0. /) m =.5 J
4.0 A work of.5 kj must be delivered on a rod from a pneumatic piston/cylinder where the air pressure is limited to 500 ka. What diameter cylinder should I have to restrict the rod motion to maximum 0.5 m? W = F dx = dv = A dx = A x = π 4 D x D = 4W π x = 4.5 kj π 500 ka 0.5 m = 0.3 m
4. The rolling resistance of a car depends on its weight as: F = 0.006 mg. How long will a car of 400 kg drive for a work input of 5 kj? Work is force times distance so assuming a constant force we get W = F dx = F x = 0.006 mgx Solve for x W x = 0.006 mg = 5 kj 0.006 400 kg 9.807 m/s = 303.5 m
4. The air drag force on a car is 0.5 A ρv. Assume air at 90 K, 00 ka and a car frontal area of 4 m driving at 90 km/h. How much energy is used to overcome the air drag driving for 30 minutes? ρ = v = RT = 00 kg =.05 0.87 90 m 3 V = 90 km h 000 m = 90 3600 s = 5 m/s x = V t = 5 m/s 30 min 60 s/min = 45 000 m F = 0.5 A ρ V = 0.5 4.05 5 = 675.8 m kg m 3 m s = 676 N W = F x = 676 N 45 000 m = 30 40 000 J = 30.4 MJ
Boundary work simple step process 4.3 A constant pressure piston cylinder contains 0. kg water as saturated vapor at 400 ka. It is now cooled so the water occupies half the original volume. Find the work in the process. Table B.. v = 0.465 m 3 /kg V = mv = 0.095 m 3 v = v / = 0.35 m 3 /kg V = V / = 0.0465 m 3 rocess: = C so the work term integral is W = dv = (V -V ) = 400 ka (0.0465 0.095) m 3 = -8.5 kj C.. T C.. = 400 ka 400 T 44 cb v v
4.4 A steam radiator in a room at 5 C has saturated water vapor at 0 ka 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 5 o C? How much work is done? Control volume radiator. After the valve is closed no more flow, constant volume and mass. : x =, = 0 ka v = v g =.566 m 3 /kg from Table B.. : T = 5 o C,? rocess: v = v =.566 m 3 /kg = [0.00003 + x 43.359] m 3 /kg x =.566 0.00003 43.359 State : T, x From Table B.. = 0.036 = sat = 3.69 ka W = dv = 0
4.5 Find the specific work in roblem 3.35. State from Table B.. at 00 ka v = v f + x v fg = 0.0006 + 0.5 0.88467 = 0.3 m 3 /kg State has same from Table B.. at 00 ka T = T sat + 0 = 0.3 + 0 = 40.3 o C so state is superheated vapor 0 v = 0.88573 + (0.95964 0.88573) 50-0.3 = 0.9354 m3 /kg rocess = C: w = dv = (v v ) = 00 (0.9354 0.3) = 4.6 kj/kg C.. T C.. 00 ka 00 40 C 0. C T 40 0 v v
4.6 A 400-L tank A, see figure 4.6, contains argon gas at 50 ka, 30 o C. Cylinder B, having a frictionless piston of such mass that a pressure of 50 ka will float it, is initially empty. The valve is opened and argon flows into B and eventually reaches a uniform state of 50 ka, 30 o C throughout. What is the work done by the argon? Take C.V. as all the argon in both A and B. Boundary movement work done in cylinder B against constant external pressure of 50 ka. Argon is an ideal gas, so write out that the mass and temperature at state and are the same A V A = m A RT A = m A RT = ( V A + V B ) => V B = 50 0.4 50-0.4 = 0.667 m 3 W = ext dv = ext (V B - V B ) = 50 ka (0.667-0) m 3 = 40 kj Argon A g o B B V B Notice there is a pressure loss in the valve so the pressure in B is always 50 ka while the piston floats.
4.7 A piston cylinder contains air at 600 ka, 90 K and a volume of 0.0 m 3. A constant pressure process gives 8 kj of work out. Find the final volume and temperature of the air. W = dv = V V = W/ = 8 600 = 0.03 m3 V = V + V = 0.0 + 0.03 = 0.04 m 3 Assuming ideal gas, V = mrt, then we have T = V mr = V V T = V V T = 0.04 0.0 90 = 60 K
4.8 Saturated water vapor at 00 ka is in a constant pressure piston cylinder. At this state the piston is 0. m from the cylinder bottom and cylinder area is 0.5 m. The temperature is then changed to 00 o C. Find the work in the process. State from B.. (, x): v = v g = 0.8857 m 3 /kg (also in B..3) State from B..3 (, T): v =.0803 m 3 /kg Since the mass and the cross sectional area is the same we get h = v v h =.0803 0.8857 0. = 0. m rocess: = C so the work integral is W = dv = (V - V ) = A (h - h ) W = 00 ka 0.5 m (0. 0.) m =. kj C.. T C.. 00 T 00 0 = 00 ka cb v v
4.9 A cylinder fitted with a frictionless piston contains 5 kg of superheated refrigerant R-34a vapor at 000 ka, 40 C. 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 B.5. State : v = 0.0350 m 3 /kg, State : v = 0.00087 + 0.5 0.0956 = 0.00576 m 3 /kg Interpolated to be at 000 ka, numbers at 07 ka could have been used in which case: v = 0.00566 m 3 /kg W = dv = (V -V ) = m (v -v ) = 5 000 (0.00576-0.0350) = -8.7 kj C.. T C.. 000 T 40 39 = 000 ka cb v v
4.30 A piston cylinder contains.5 kg water at 00 ka, 50 o C. It is now heated in a process where pressure is linearly related to volume to a state of 600 ka, 350 o C. Find the final volume and the work in the process. C.. Take as CV the.5 kg of water. m = m = m ; rocess Eq.: = A + BV (linearly in V) State : (, T) => v = 0.95964 m 3 /kg, 600 00 T v State : (, T) => v = 0.4744 m 3 /kg, V = mv = 0.74 m 3 From process eq.: W = dv = area = m ( + )(v v ) =.5 kg (00 + 600) ka (0.4744 0.95964) m3 /kg = 9.4 kj Notice volume is reduced so work is negative.
4.3 Find the specific work in roblem 3.39 for the case the volume is reduced. Saturated vapor R-34a at 50 o C changes volume at constant temperature. Find the new pressure, and quality if saturated, if the volume doubles. Repeat the question for the case the volume is reduced to half the original volume. R-34a 50 o C Table B.4.: v = v g = 0.05 m 3 /kg, v = v / = 0.00756 m 3 /kg W = dv = 38. ka (0.00756 0.05) m 3 /kg = -9.96 kj/kg C.. T C.. 38 T 50 = 38 ka cb v v
4.3 A piston/cylinder has 5 m of liquid 0 o C water on top of the piston (m = 0) with cross-sectional area of 0. m, see Fig..39. 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 -V diagram. = o + ρgh = 0.3 + 997 9.807 5 / 000 = 50. ka V = H A = 5 0. = 0.5 m 3 W = AREA = dv = ½ ( + o )(V max -V ) = ½ (50. + 0.3) ka 0.5 m 3 = 6.88 kj o HO cb Air 0 V V Vmax
4.33 A piston/cylinder contains kg water at 0 o C with volume 0. m 3. By mistake someone locks the piston preventing it from moving while we heat the water to saturated vapor. Find the final temperature, volume and the process work. Solution : v = V/m = 0. m 3 / kg = 0. m 3 /kg (two-phase state) : Constant volume: v = v g = v V = V = 0. m 3 W = dv = 0 T = T sat = 0 + 5 0. - 0.034 0.0936-0.034 =.7 C
4.34 Ammonia (0.5 kg) is in a piston cylinder at 00 ka, -0 o C is heated in a process where the pressure varies linear with the volume to a state of 0 o C, 300 ka. Find the work the ammonia gives out in the process. Take CV as the Ammonia, constant mass. Continuity Eq.: m = m = m ; rocess: = A + BV (linear in V) State : Superheated vapor v = 0.693 m 3 /kg State : Superheated vapor v = 0.6376 m 3 /kg Work is done while piston moves at increasing pressure, so we get W = dv = area = avg (V V ) = ( + )m(v v ) = ½(00 + 300) 0.5 (0.6376 0.693) =.683 kj v
4.35 A piston cylinder contains kg of liquid water at 0 o C and 300 ka. There is a linear spring mounted on the piston such that when the water is heated the pressure reaches 3 Ma with a volume of 0. m 3. a) Find the final temperature b) lot the process in a -v diagram. c) Find the work in the process. Take CV as the water. This is a constant mass: m = m = m ; State : Compressed liquid, take saturated liquid at same temperature. B..: v = v f (0) = 0.0000 m 3 /kg, State : v = V /m = 0./ = 0. m 3 /kg and = 3000 ka => Superheated vapor close to T = 400 o C Interpolate: T = 404 o C from B..3 Work is done while piston moves at linearly varying pressure, so we get: W = dv = area = avg (V V ) = ( + )(V - V ) = 0.5 (300 + 3000)(0. 0.00) = 63.35 kj C.. T C.. 300 T 0 300 ka v v
4.36 A piston cylinder contains 3 kg of air at 0 o C and 300 ka. It is now heated up in a constant pressure process to 600 K. a) Find the final volume b) lot the process path in a -v diagram c) Find the work in the process. Ideal gas V = mrt State : T, ideal gas so V = mrt V = mr T / = 3 0.87 93.5/300 = 0.843 m 3 State : T, = and ideal gas so V = mrt V = mr T / = 3 0.87 600/300 =.7 m 3 W = dv = (V - V ) = 300 (.7 0.843) = 64. kj T 300 T T v 600 93 300 ka v
4.37 A piston cylinder contains 0.5 kg air at 500 ka, 500 K. The air expands in a process so is linearly decreasing with volume to a final state of 00 ka, 300 K. Find the work in the process. rocess: = A + BV (linear in V, decreasing means B is negative) From the process: W = dv = AREA = ( + )(V - V ) V = mr T / = 0.5 0.87 (500/500) = 0.435 m 3 V = mr T / = 0.5 0.87 (300/00) = 0.4305 m 3 W = (500 + 00) ka (0.4305-0.435) m3 = 86. kj T 500 500 T T 00 300 cb v v
olytropic process 4.38 Consider a mass going through a polytropic process where pressure is directly proportional to volume (n = ). The process start with = 0, V = 0 and ends with = 600 ka, V = 0.0 m 3. The physical setup could be as in roblem 4.35. Find the boundary work done by the mass. The setup has a pressure that varies linear with volume going through the initial and the final state points. The work is the area below the process curve. 600 W = dv = AREA = ( + )(V - V ) 0 0 W 0.0 V = ( + 0)( V - 0) = V = 600 0.0 = 3 kj
4.39 Air at 500 K, 000 ka expands in a polytropic process, n =.5, to a pressure of 00 ka. How cold does the air become and what is the specific work out? rocess equation: V n = constant = V n = V n Solve for the temperature at state by using ideal gas (V = mrt) T T = V V = /n = (n-)/n T = T ( / ) (n-)/n = 500 00 000 (.5-)/.5 = 877. K Work from Eq.4.5 w = v - v -n = = R( - T ) -n 0.87(877. - 500) -.5 = 357.5 kj/kg The actual process is on a steeper curve than n =. W n = V
4.40 The piston/cylinder shown in Fig. 4.40 contains carbon dioxide at 300 ka, 00 C with a volume of 0. m 3. Mass is added at such a rate that the gas compresses according to the relation V. = constant to a final temperature of 00 C. Determine the work done during the process. From Eq. 4.4 for the polytopic process V n = const ( n =/ ) W = dv = V - V - n Assuming ideal gas, V = mrt W = mr(t - T ) - n, But mr = V 300 0. T = 373.5 W = 0.608(473. - 373.) -. ka m 3 K kj K K = 0.608 kj/k = -80.4 kj
4.4 A gas initially at Ma, 500 C is contained in a piston and cylinder arrangement with an initial volume of 0. m 3. The gas is then slowly expanded according to the relation V = constant until a final pressure of 00 ka is reached. Determine the work for this process. By knowing the process and the states and we can find the relation between the pressure and the volume so the work integral can be performed. rocess: V = C V = V / = 000 0./00 = m 3 For this process work is integrated to Eq.4.5 W = dv = CV - dv = C ln(v /V ) W = V ln V V = 000 0. ln (/0.) = 30.3 kj W V
4.4 A spring loaded piston/cylinder assembly contains kg water at 500 o C, 3 Ma. The setup is such that the pressure is proportional to volume, = CV. It is now cooled until the water becomes saturated vapor. Sketch the -v diagram and find the work in the process. Solution : State : Table B..3: v = 0.69 m 3 /kg rocess: m is constant and = C 0 V = C 0 m v = C v polytropic process with n = = Cv C = /v = 3000/0.69 = 580 ka kg/m 3 State : x = & = Cv (on process line) Trial & error on T sat or sat : Here from B..: at Ma v g = 0.09963 C = /v g = 0074 (low).5 Ma v g = 0.07998 C = /v g = 358 (high) C v.5 Ma v g = 0.08875 C = /v g = 535 (low) Now interpolate to match the right slope C: = 70 ka, v = /C = 70/580 = 0.0879 m 3 /kg is linear in V so the work becomes (area in -v diagram) W = dv = m ( + )(v - v ) = (3000 + 70)(0.0879-0.69) = - 74.5 kj/kg
4.43 Helium gas expands from 5 ka, 350 K and 0.5 m 3 to 00 ka in a polytropic process with n =.667. How much work does it give out? rocess equation: V n = constant = V n = V n Solve for the volume at state V = V ( / ) /n = 0.5 5 00 0.6 = 0.85 m 3 Work from Eq.4.4 W = V - V -n = 00 0.85-5 0.5 -.667 ka m 3 = 4.09 kj The actual process is on a steeper curve than n =. W n = V
4.44 A balloon behaves so the pressure is = C V /3, C = 00 ka/m. The balloon is blown up with air from a starting volume of m 3 to a volume of 3 m 3. Find the final mass of air assuming it is at 5 o C and the work done by the air. The process is polytropic with exponent n = -/3. = C V /3 = 00 /3 = 00 ka = C V /3 = 00 3 /3 = 44. ka W V W = dv = = m = V RT = V - V - n 44. 3-00 - (-/3) = 49.5 kj 44. 3 = 5.056 kg 0.87 98 (Equation 4.4)
4.45 A piston cylinder contains 0. kg nitrogen at 00 ka, 7 o C and it is now compressed in a polytropic process with n =.5 to a pressure of 50 ka. What is the work involved? Take CV as the nitrogen. m = m = m ; rocess Eq.: v n = Constant (polytropic) From the ideal gas law and the process equation we can get: State : T = T ( / ) n- n = 300.5 ( 50 00 )0.5.5 = 360.5 K From process eq.: W = dv = area = m n ( v v ) = mr n (T T ) = 0. 0.968 -.5 (360.5 300.5) = 7.65 kj = C T 5 = C v -.5 T T = C v -0.5 T T T v v
4.46 A balloon behaves such that the pressure inside is proportional to the diameter squared. It contains kg of ammonia at 0 C, 60% quality. The balloon and ammonia are now heated so that a final pressure of 600 ka is reached. Considering the ammonia as a control mass, find the amount of work done in the process. rocess : D, with V D 3 this implies D V /3 so V -/3 = constant, which is a polytropic process, n = /3 From table B..: V = mv = (0.00566 + 0.6 0.8783) = 0.3485 m 3 V = V 3/ = 0.3485 600 49.3 3/ = 0.5758 m 3 W = V - V dv = - n (Equation 4.4) = 600 0.5758-49.3 0.3485 - (-/3) = 7.5 kj W V
4.47 Consider a piston cylinder with 0.5 kg of R-34a as saturated vapor at -0 C. It is now compressed to a pressure of 500 ka in a polytropic process with n =.5. Find the final volume and temperature, and determine the work done during the process. Take CV as the R-34a which is a control mass. m = m = m rocess: v.5 = constant until = 500 ka : (T, x) v = 0.099 m 3 /kg, = sat = 0.7 ka from Table B.5. : (, process) v = v ( / ) (/.5) = 0.099 (0.7/500) /3 = 0.0546 Given (, v) at state from B.5. it is superheated vapor at T = 79 C rocess gives = C v -.5, which is integrated for the work term, Eq.(4.4) W = dv = = m -.5 ( v - v ) - 0.5 (500 0.0546-0.7 0.099) = -7.07 kj
4.48 Air goes through a polytropic process from 5 ka, 35 K to 300 ka and 500 K. Find the polytropic exponent n and the specific work in the process. rocess: v n = Const = v n = v n Ideal gas v = RT so v = RT v = RT From the process equation = 0.87 35 5 = 0.87 500 300 = 0.746 m 3 /kg = 0.47833 m 3 /kg ( / ) = (v / v ) n => ln( / ) = n ln(v / v ) n = ln( / ) / ln(v / v ) = The work is now from Eq.4.4 per unit mass w = v - v -n = R(T - T ) -n = ln.4 ln.56 =.969 0.87(500-35) -.969 = -5.8 kj/kg
4.49 A piston/cylinder contains water at 500 C, 3 Ma. It is cooled in a polytropic process to 00 C, Ma. Find the polytropic exponent and the specific work in the process. olytropic process: v n = C Both states must be on the process line: v n = C = v n Take the ratio to get: = n v v and then take ln of the ratio: ln = ln n v v now solve for the exponent n n = ln / ln v v =.0986 0.5746 =.99 w = dv = = v - v - n 000 0.0596-3000 0.69 -.99 (Equation 4.4) = 55. kj = n ln v v
4.50 Consider a two-part process with an expansion from 0. to 0. m 3 at a constant pressure of 50 ka followed by an expansion from 0. to 0.4 m 3 with a linearly rising pressure from 50 ka ending at 300 ka. Show the process in a -V diagram and find the boundary work. By knowing the pressure versus volume variation the work is found. If we plot the pressure versus the volume we see the work as the area below the process curve. 300 3 50 V 0. 0. 0.4 W 3 = W + W 3 = 3 dv + dv = (V V ) + ( + 3 )(V 3 -V ) = 50 (0.-.0) + (50 + 300) (0.4-0.) = 5 + 45 = 60 kj
4.5 A cylinder containing kg of ammonia has an externally loaded piston. Initially the ammonia is at Ma, 80 C and is now cooled to saturated vapor at 40 C, and then further cooled to 0 C, at which point the quality is 50%. Find the total work for the process, assuming a piecewise linear variation of versus V. 000 555 857 cb 3 40 0 80 o C o C v o C State : (T, ) Table B.. v = 0.057 m 3 /kg State : (T, x) Table B.. sat. vap. = 555 ka, v = 0.0833 m 3 /kg State 3: (T, x) 3 = 857 ka, v 3 = (0.00638 + 0.49)/ = 0.07543 m 3 /kg Sum the the work as two integrals each evaluated by the area in the -v diagram. 3 W 3 = = dv ( + 000 + 555 = -49.4 kj ) m(v - v ) + ( + 3 (0.0833-0.057) + 555 + 857 ) m(v 3 - v ) (0.07543-0.0833)
4.5 A helium gas is heated at constant volume from a state of 00 ka, 300 K to 500 K. A following process expands the gas at constant pressure to three times the initial volume. What is the specific work in the combined process? The two processes are: -> : Constant volume V = V 3 -> 3: Constant pressure 3 = V Use ideal gas approximation for helium. State : T, => v = RT / State : V = V => = (T /T ) State 3: 3 = => V 3 = 3V ; T 3 = T v 3 /v = 500 3 = 500 K We find the work by summing along the process path. w 3 = w + w 3 = w 3 = 3 (v 3 - v ) = R(T 3 - T ) =.077 kj/kgk (500-500) K = 077 kj/kg
4.53 A piston/cylinder arrangement shown in Fig. 4.53 initially contains air at 50 ka, 400 C. The setup is allowed to cool to the ambient temperature of 0 C. a. Is the piston resting on the stops in the final state? What is the final pressure in the cylinder? b. What is the specific work done by the air during this process? State : State : = 50 ka, T = 400 C = 673. K T = T 0 = 0 C = 93. K For all states air behave as an ideal gas. a) If piston at stops at, V = V / and pressure less than lift = = V V T T = 50 93. 673. = 30.7 ka < iston is resting on stops at state. b) Work done while piston is moving at constant ext =. W = ext dv = (V - V ) ; V = V = m RT / w = W /m = RT ( - ) = - 0.87 673. = -96.6 kj/kg T a V T T T a a V
4.54 A piston cylinder has.5 kg of air at 300 K and 50 ka. It is now heated up in a two step process. First constant volume to 000 K (state ) then followed by a constant pressure process to 500 K, state 3. Find the final volume and the work in the process. The two processes are: -> : Constant volume V = V 3 -> 3: Constant pressure 3 = V Use ideal gas approximation for air. State : T, => V = mrt / =.5 0.87 300/50 = 0.86 m 3 State : V = V => = (T /T ) = 50 000/300 = 500 ka State 3: 3 = => V 3 = V (T 3 /T ) = 0.86 500/000 =.95 m 3 We find the work by summing along the process path. W 3 = W + W 3 = W 3 = 3 (V 3 - V ) = 500(.95-0.86) = 5.3 kj
4.55 A piston cylinder contains air at 000 ka, 800 K with a volume of 0.05 m 3. The piston is pressed against the upper stops and it will float at a pressure of 750 ka. Then the air is cooled to 400 K. What is the process work? We need to find state. Let us see if we proceed past state a during the cooling. T a = T float / = 800 750 / 00 = 600 K so we do cool below T a. That means the piston is floating. Write the ideal gas law for state and to get V = mrt = V T 000 0.05 400 T = = 0.0333 m 750 800 3 W = a W = dv = (V - V ) = 750 (0.0333 0.05) =.5 kj o m p Air a V Vstop
4.56 The refrigerant R- is contained in a piston/cylinder as shown in Fig. 4.56, where the volume is L when the piston hits the stops. The initial state is 30 C, 50 ka with a volume of 0 L. This system is brought indoors and warms up to 5 C. a. Is the piston at the stops in the final state? b. Find the work done by the R- during this process. Initially piston floats, V < V stop so the piston moves at constant ext = until it reaches the stops or 5 C, whichever is first. a) From Table B.4.: v = 0.487 m 3 /kg, m = V/v = 0.00 0.487 = 0.0675 kg o m p R- a V Vstop Check the temperature at state a: a = 50 ka, v = V stop /m. v a = V/m = 0.0 0.0675 = 0.6357 m3 /kg => T a = -9 C & T = 5 C Since T > T a then it follows that > and the piston is against stop. b) Work done at constant ext =. W = ext dv = ext (V - V ) = 50(0.0-0.00) = 0.5 kj
4.57 A piston/cylinder contains 50 kg of water at 00 ka with a volume of 0. m 3. Stops in the cylinder restricts the enclosed volume to 0.5 m 3, similar to the setup in roblem 4.56. The water is now heated to 00 C. Find the final pressure, volume and the work done by the water. Initially the piston floats so the equilibrium lift pressure is 00 ka : 00 ka, v = 0./50 = 0.00 m 3 /kg, : 00 C, on line Check state a: a V v stop = 0.5/50 = 0.0 m 3 /kg V stop => Table B..: 00 ka, v f < v stop < v g State a is two phase at 00 ka and T stop 0. C so as T > T stop the state is higher up in the -V diagram with v = v stop < v g = 0.7 m 3 /kg (at 00 C) State two phase => = sat (T ) =.554 Ma, V = V stop = 0.5 m 3 W = W stop = 00 (0.5 0.) = 80 kj
4.58 A piston/cylinder (Fig. 4.58) contains kg of water at 0 C with a volume of 0. m 3. Initially the piston rests on some stops with the top surface open to the atmosphere, o and a mass so a water pressure of 400 ka will lift it. To what temperature should the water be heated to lift the piston? If it is heated to saturated vapor find the final temperature, volume and the work, W. (a) State to reach lift pressure of = 400 ka, Table B..: v f < v < v g = 0.465 m 3 /kg v = V/m = 0. m 3 /kg => T = T sat = 43.63 C (b) State is saturated vapor at 400 ka since state a is two-phase. o a H O V v = v g = 0.465 m 3 /kg, V = m v = 0.465 m 3, ressure is constant as volume increase beyond initial volume. W = dv = (V - V ) = lift (V V ) = 400 ka (0.465 0.) m 3 = 45 kj
4.59 A piston/cylinder contains kg of liquid water at 0 C and 300 ka. Initially the piston floats, similar to the setup in roblem 4.56, with a maximum enclosed volume of 0.00 m 3 if the piston touches the stops. Now heat is added so a final pressure of 600 ka is reached. Find the final volume and the work in the process. Take CV as the water which is a control mass: m = m = m ; Table B..: 0 C => sat =.34 ka State : Compressed liquid v = v f (0) = 0.0000 m 3 /kg State a: v stop = 0.00 m 3 /kg, 300 ka State : Since = 600 ka > lift then piston is pressed against the stops v = v stop = 0.00 m 3 /kg and V = 0.00 m 3 For the given : v f < v < vg so -phase T = T sat = 58.85 C Work is done while piston moves at lift = constant = 300 ka so we get W = dv = m lift (v -v ) = 300(0.00 0.0000) = 0.30 kj o cb H O V
4.60 0 kg of water in a piston cylinder arrangement exists as saturated liquid/vapor at 00 ka, with a quality of 50%. It is now heated so the volume triples. The mass of the piston is such that a cylinder pressure of 00 ka will float it, see Fig. 4.58. a) Find the final temperature and volume of the water. b) Find the work given out by the water. Take CV as the water m = m = m; rocess: State : v = constant until = lift then is constant. v = v f + x v fg = 0.00043 + 0.5.6996 = 0.8475 m 3 /kg State : v, lift => v = 3 0.8475 =.545 m 3 /kg; T = 89 C ; V = m v = 5.45 m 3 W = dv = lift (V - V ) = 00 ka 0 kg (.545 0.8475) m 3 /kg = 3390 kj o cb HO V
4.6 Find the work in roblem 3.47. Ammonia at 0 o C with a mass of 0 kg is in a piston cylinder arrangement with an initial volume of m 3. The piston initially resting on the stops has a mass such that a pressure of 900 ka will float it. The ammonia is now slowly heated to 50 o C. Find the work in the process. C.V. Ammonia, constant mass. rocess: V = constant unless = float State : T = 0 o C, v = V m = 0 = 0. m3 /kg a From Table B.. v f < v < v g x = (v - v f )/v fg = (0. - 0.006)/0.038 = 0.488 V cb State a: = 900 ka, v = v = 0. < v g at 900 ka This state is two-phase T a =.5 o C Since T > T a then v > v a State : 50 o C and on line(s) means 900 ka which is superheated vapor. From Table B.. linear interpolation between 800 and 000 ka: v = 0.648 m 3 /kg, V = mv =.648 m 3 W = dv = float (V - V ) = 900 (.648 -.0) = 583. kj
4.6 A piston cylinder setup similar to roblem 4.58 contains 0. kg saturated liquid and vapor water at 00 ka with quality 5%. The mass of the piston is such that a pressure of 500 ka will float it. The water is heated to 300 C. Find the final pressure, volume and the work, W. Take CV as the water: m = m = m rocess: v = constant until = lift To locate state : Table B.. v = 0.00043 + 0.5.6996 = 0.448 m 3 /kg a: v a = v = 0.448 m 3 /kg > v g at 500 ka lift a cb V so state a is Sup.Vapor T a = 00 C State is 300 C so heating continues after state a to at constant => : T, = lift => Table B..3 v = 0.556 m3/kg ; V = mv = 0.056 m3 W = lift (V - V ) = 500(0.056-0.0443) = 4.9 kj
Other types of work and general concepts 4.63 Electric power is volts times ampere ( = V i). When a car battery at V is charged with 6 amp for 3 hours how much energy is delivered? W = Ẇ dt = Ẇ t = V i t = V 6 Amp 3 3600 s = 777 600 J = 777.6 kj Remark: Volt times ampere is also watts, W = V Amp.
4.64 A 0.5-m-long steel rod with a -cm 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 0 8 ka. Solution : W = AEL 0 (e), A = π 4 (0.0) = 78.54 0-6 m W = 78.54 0-6 0 8 0.5 (0-3 ) = 3.93 J
4.65 A film of ethanol at 0 C has a surface tension of.3 mn/m and is maintained on a wire frame as shown in Fig. 4.65. Consider the film with two surfaces as a control mass and find the work done when the wire is moved 0 mm to make the film 0 40 mm. Solution : Assume a free surface on both sides of the frame, i.e., there are two surfaces 0 30 mm W = S da =.3 0-3 (800 600) 0-6 = 8.9 0-6 J = 8.9 µj
4.66 Assume a balloon material with a constant surface tension of S = N/m. What is the work required to stretch a spherical balloon up to a radius of r = 0.5 m? Neglect any effect from atmospheric pressure. Assume the initial area is small, and that we have surfaces inside and out W = S da = S (A A ) = S(A ) = S( π D ) = N/m π m = -.57 J W in = W =.57 J
4.67 A soap bubble has a surface tension of S = 3 0-4 N/cm as it sits flat on a rigid ring of diameter 5 cm. You now blow on the film to create a half sphere surface of diameter 5 cm. How much work was done? W = F dx = S da = S A = S ( π D - π 4 D ) = 3 0-4 N/cm 00 cm/m π 0.05 m ( - 0.5 ) =.8 0-4 J Notice the bubble has surfaces. A = π 4 D, A = ½ π D
4.68 A sheet of rubber is stretched out over a ring of radius 0.5 m. I pour liquid water at 0 o C on it so the rubber forms a half sphere (cup). Neglect the rubber mass and find the surface tension near the ring? F = F ; F = SL The length is the perimeter, πr, and there is two surfaces S πr = m Ho g = ρ Ho Vg = ρ Ho π (r) 3 g = ρ Ho π 3 r 3 S = ρ Ho 6 r g = 997 kg/m 3 6 0.5 m 9.8 m/s = 0.9 N/m
4.69 Consider a light bulb that is on. Explain where we have rates of work and heat transfer (include modes) that moves energy. Electrical power comes in, that is rate of work. In the wire filament this electrical work is converted to internal energy so the wire becomes hot and radiates energy out. There is also some heat transfer by conduction-convection to the gas inside the bulb so that become warm. The gas in turn heats the glass by conduction/convection and some of the radiation may be absorbed by the glass (often the glass is coated white) All the energy leaves the bulb as a combination of radiation over a range of wavelengths some of which you cannot see and conduction convection heat transfer to the air around the bulb.
4.70 Consider a window-mounted air conditioning unit used in the summer to cool incoming air. Examine the system boundaries for rates of work and heat transfer, including signs. Solution : Air-conditioner unit, steady operation with no change of temperature of AC unit. Cool side Inside Hot side Outside 5 C C 30 C 5 C 37 C - electrical work (power) input operates unit, +Q rate of heat transfer from the room, a larger -Q rate of heat transfer (sum of the other two energy rates) out to the outside air.
4.7 A room is heated with an electric space heater on a winter day. Examine the following control volumes, regarding heat transfer and work, including sign. a) The space heater. b) Room c) The space heater and the room together a) The space heater. Electrical work (power) input, and equal (after system warm up) Q out to the room. b) Room Q input from the heater balances Q loss to the outside, for steady (no temperature change) operation. c) The space heater and the room together Electrical work input balances Q loss to the outside, for steady operation.
Rates of work 4.7 A force of. kn moves a truck with 60 km/h up a hill. What is the power? Ẇ = F V =. kn 60 (km/h) =. 0 3 60 03 3600 = 0 000 W = 0 kw Nm s
4.73 An escalator raises a 00 kg bucket of sand 0 m in minute. Determine the rate of work done during the process. The work is a force with a displacement and force is constant: F = mg W = F dx = F dx = F x = 00 kg 9.80665 m/s 0 m = 9807 J The rate of work is work per unit time. W = W t = 9807 J 60 s = 63 W
4.74 A car uses 5 hp to drive at a horizontal level at constant 00 km/h. What is the traction force between the tires and the road? We need to relate the rate of work to the force and velocity dw = F dx => dw dt = Ẇ = F dx dt F = Ẇ / V = FV Ẇ = 5 hp = 5 0.7355 kw = 8.39 kw V = 00 000 3600 = 7.78 m/s F = Ẇ / V = (8.39 / 7.78) kn = 0.66 kn Units: kw / (ms ) = kw s m = kj s s m = kn m m = kn
4.75 A piston/cylinder of cross sectional area 0.0 m maintains constant pressure. It contains kg water with a quality of 5% at 50 o C. If we heat so g/s liquid turns into vapor what is the rate of work out? V vapor = m vapor v g, V liq = m liq v f m tot = constant = m vapor m liq V tot = V vapor + V liq ṁ tot = 0 = ṁ vapor + ṁ liq ṁ liq = -ṁ vapor V. tot = V. vapor + V. liq = ṁ vapor v g + ṁl iq v f = ṁ vapor (v g - v f ) = ṁ vapor v fg Ẇ = V. = ṁ vapor v fg = 475.9 ka 0.00 kg/s 0.3969 m 3 /kg = 0.864 kw = 86 W
4.76 Consider the car with the rolling resistance as in problem 4.7. How fast can it drive using 30 hp? F = 0.006 mg ower = F V = 30 hp = Ẇ V = Ẇ / F = Ẇ 30 0.7457 000 0.006 mg = = 7.5 m/s 0.006 00 9.8 Comment : This is a very high velocity, the rolling resistance is low relative to the air resistance.
4.77 Consider the car with the air drag force as in problem 4.9. How fast can it drive using 30 hp? ρ = v = RT = 00 0.87 90 Drag force: F drag = 0.5 A ρ V kg =.05 m 3 and A = 4 m ower for drag force: Ẇ drag = 30 hp 0.7457 =.37 kw Ẇ drag = F drag V = 0.5 4.05 V 3 V 3 = Ẇ drag /(0.5 4.05) = 0 688 V = 7.45 m/s = 7.45 3600 000 = 98.8 km/h
4.78 A battery is well insulated while being charged by.3 V at a current of 6 A. Take the battery as a control mass and find the instantaneous rate of work and the total work done over 4 hours. Solution : Battery thermally insulated Q = 0 For constant voltage E and current i, ower = E i =.3 6 = 73.8 W [Units V A = W] W = power dt = power t = 73.8 4 60 60 = 06 70 J = 06.7 kj
4.79 A current of 0 amp runs through a resistor with a resistance of 5 ohms. Find the rate of work that heats the resistor up.. W = power = E i = R i = 5 0 0 = 500 W R
4.80 A pressure of 650 ka pushes a piston of diameter 0.5 m with V = 5 m/s. What is the volume displacement rate, the force and the transmitted power? A = π 4 D = 0.049087 m. V = AV = 0049087 m 5 m/s = 0.454 m 3 /s F = A = 650 ka 0.049087 m = 3.9 kn. W = power = F V =. V = 650 ka 0.454 m 3 /s = 59.5 kw V
4.8 Assume the process in roblem 4.50 takes place with a constant rate of change in volume over minutes. Show the power (rate of work) as a function of time. W = dv since min = 0 secs. W = ( V / t) ( V / t) = 0.3 / 0 = 0.005 m 3 /s 300 3 kw 0.75 W 3 50 0.375 V V 0. 0. 0.4 0. 0. 0.4
4.8 Air at a constant pressure in a piston cylinder is at 300 ka, 300 K and a volume of 0. m 3. It is heated to 600 K over 30 seconds in a process with constant piston velocity. Find the power delivered to the piston. rocess: = constant :. Boundary work: dw = dv => W = V. V = V (T /T ) = 0. (600/300) = 0. m 3. W = V 0. - 0. = 300 t 30 ka m3 s = kw Q V Remark: Since we do not know the area we do not know the velocity
Heat Transfer rates 4.83 Find the rate of conduction heat transfer through a.5 cm thick hardwood board, k = 0.6 W/m K, with a temperature difference between the two sides of 0 o C. One dimensional heat transfer by conduction, we do not know the area so we can find the flux (heat transfer per unit area W/m ).. q =. Q/A = k T x = 0.6 W m K 0 K 0.05 m = 3 W/m
4.84 The sun shines on a 50 m road surface so it is at 45 C. Below the 5 cm thick asphalt, average conductivity of 0.06 W/m K, is a layer of compacted rubbles at a temperature of 5 C. Find the rate of heat transfer to the rubbles. Solution : This is steady one dimensional conduction through the asphalt layer.. Q = k A T x = 0.06 50 45-5 0.05 = 5400 W
4.85 A water-heater is covered up with insulation boards over a total surface area of 3 m. The inside board surface is at 75 C and the outside surface is at 0 C and the board material has a conductivity of 0.08 W/m K. How thick a board should it be to limit the heat transfer loss to 00 W? Solution : Steady state conduction through a single layer board.. Q cond = k A T x = k Α Τ/ Q. x x = 0.08 W m K 3 m = 0.066 m 75 0 00 K W
4.86 A large condenser (heat exchanger) in a power plant must transfer a total of 00 MW from steam running in a pipe to sea water being pumped through the heat exchanger. Assume the wall separating the steam and seawater is 4 mm of steel, conductivity 5 W/m K and that a maximum of 5 C difference between the two fluids is allowed in the design. Find the required minimum area for the heat transfer neglecting any convective heat transfer in the flows. Solution : Steady conduction through the 4 mm steel wall.. Q = k A T Α = Q. x / k Τ x A = 00 0 6 W 0.004 m / (5 W/mK 5 K) = 480 m Condensing water Sea water cb
4.87 The black grille on the back of a refrigerator has a surface temperature of 35 C with a total surface area of m. Heat transfer to the room air at 0 C takes place with an average convective heat transfer coefficient of 5 W/m K. How much energy can be removed during 5 minutes of operation? Solution :. Q = ha T; Q =. Q t = ha T t Q = 5 W/m K m (35-0) Κ 5 min 60 s/min = 0 500 J = 0.5 kj
4.88 A m window has a surface temperature of 5 o C and the outside wind is blowing air at o C across it with a convection heat transfer coefficient of h = 5 W/m K. What is the total heat transfer loss?. Q = h A T = 5 W/m K m (5 ) K = 350 W as a rate of heat transfer out. 5 o C o C
4.89 A wall surface on a house is at 30 C with an emissivity of ε = 0.7. The surrounding ambient to the house is at 5 C, average emissivity of 0.9. Find the rate of radiation energy from each of those surfaces per unit area. Solution : a) b). Q /A = εσat 4, σ = 5.67 0 8 W/m K 4. Q/A = 0.7 5.67 0-8 ( 73.5 + 30) 4 = 335 W/m. Q/A = 0.9 5.67 0-8 88.5 4 = 35 W/m
4.90 A log of burning wood in the fireplace has a surface temperature of 450 C. Assume the emissivity is (perfect black body) and find the radiant emission of energy per unit surface area. Solution :. Q /A = σ T 4 = 5.67 0 8 ( 73.5 + 450) 4 = 5 505 W/m = 5.5 kw/m
4.9 A radiant heating lamp has a surface temperature of 000 K with ε = 0.8. How large a surface area is needed to provide 50 W of radiation heat transfer? Radiation heat transfer. We do not know the ambient so let us find the area for an emitted radiation of 50 W from the surface. Q = εσat 4. Q A = εσt 4 = 50 0.8 5.67 0-8 000 4 = 0.0055 m