THE UNIVERSITY OF MICHIGAN COLLEGE OF ENGINEERING Department of Mechanical Engineering Progress Report AIR-BOOSTED CYCLE ANALYSES E. T. Vincent ORA Project 04612 under contract with: DEPARTMENT OF THE ARMY ORDNANCE CORPS CONTRACT NO. DA-20-018-ORD-23664 DETROIT, MICHIGAN administered through: OFFICE OF RESEARCH ADMINISTRATION ANN ARBOR February 1962

TABLE OF CONTENTS Page LIST OF FIGURES v INTRODUCTION 1 ENGINE ARRANGEMENT 3 PERFORMANCE MANIFOLD CONDITIONS 5 INITIAL CONDITIONS IN CYLINDER5 END OF COMPRESSION6 CONSTANT-VOLUME COMBUSTION6 CONSTANT-PRESSURE COMBUSTION6 EXPANSION STROKE 12 EXHAUST MANIFOLD TEMPERATURE 13 WORK OF CYCLE 14 WORK SUMMARY 16 COMPOUNDING 17 HIGH-PRESSURE AIR COMPRESSOR 21 CONTINUOUS PERFORMANCE AT MAXIMUM POWER 22 AIR COMPRESSOR 24 CONCLUSIONS 26 iii

LIST OF FIGURES Figure Page 1. Engine indicator diagram. 2 2. Air-supply flow diagram 4 3. Engine performance per lb of basic airo 27 4. Compound I.HoP. corrected for compressor power. 28 5. Net B.HoP. performance compound engine plus compressor. 29 v

INTRODUCTION To increase the mean effective pressure of an engine to secure responsiveness, there is a method of approach in addition to that of increasing the degree of supercharge, viz., that of supplying the additional air and fuel necessary during the expansion stroke, maintaining the peak pressure approximatly constant. The engine indicator diagram will then vary as shown in Fig. 1, where 12345 represents the indicator diagram at what is called normal rated load and maximum rpm. As the vehicle resistance increases and the engine slows down, an increase in mean pressure is required to maintain maximum hp; this is secured by supplying additional air and fuel to the cylinder at such a rate that the maximum pressure is maintained constant and the diagram will change to 12567, 12589, etc., as the need arises. The advantages of such a procedure are: 1. The engine behaves and is rated as a normal engine of present-day design, without additional loads, etc., so long as normal operating conditions are applied. 2. The engine is capable of being operated at increased mean pressures for temporary overloads (period of overload can be long if suitably designed). 3. Increased load achieved without increase in maximum cylinder pressure and thus constant engine stress. 4. Increase of load achieved without increase (probably a decrease) in maximum combustion temperature. 5. Greater energy available for recovery by compounding. 6. Equipment to achieve this purpose only operates when required. 7. The increase in weight, space and costs involved can be kept small by designing the engine for its normal output requirements, say 300 hp, and employ the proposed principle to meet the emergency needs of, say, 500-600 hp only when needed; by this means little change in weight, space, etc., is seen. A typical design would need to be worked out to establish the requirements accurately. 1

46'6"6 8 3 T -\ \ \ \ \ \ \ VOLUME Fig. 1. Engine indicator diagram. 2 Fig. 1. Engine indicator diagram.

ENGINE ARRANGEMENT The proposed cycle of operation would be achieved by providing an air compressor capable of delivering air at, say, 1600 psi to a small receiver, to damp out oscillations and provide a small reserve. This receiver is connected to an engine supply manifold, and flow to the cylinders is provided by a mechanically controlled valve in the cylinder head capable of being varied as regards period or lift or both. When need arises for the use of responsiveness-as the engine stalls on high load-the compressor is coupled to the engine and the air supply is fed to the receiver and then to the cylinder as in Fig. 2. The appropriate increase in fuel is also added to the air supply. If high load is required at maximum speed for any reason the air supply is also available for this purpose. Thus a typical design could be as follows: Normal power at max rpm 400 hp Maximum emergency power at max rpm 600 hp Normal power at min rpm 200 hp Maximum power at min rpm 400 hp Engine responsiveness from rpmmax to rpmax/2 (constant 400 hp). Other combinations of hp and speed could also be achieved. In the above example, a 400-hp engine can produce 600 when required, and also be responsive at 400 hp over a 2:1 speed range. Alternatively, the full 600-hp normal engine could be designed and made responsive at 600 hp down to, for example, half maximum rpm; this unit would of course be larger than the former. A typical arrangement would be as shown in Fig. 2; the turbo-charger would be geared to the engine if compounding was to be employed. PERFORMANCE The performance estimate for this type of power plant was obtained in the manner given below. The method is admittedly an approximate one, but is believed to be within a reasonably small percentage of what can be expected, and sufficiently close to judge if the system has merit. First the basic engine cycle is analyzed, including heat losses by the method described in Progress Report No. 1. The details of the cycle employed are: Compression ratio 15:1 Supercharge ratio 2:1 F/A ratio O.05 5

EXHAUST TO ATMOSPHERE AIR RECEIVER AIR MANIFOD TURBO-CHARGER AND AIR MANIFOLD COMPOUND TURBINE ~~~~~~~~~~~~~HIGH-PRESSURE - |1 SUPERCHARGE INLET HIGH-PRESSURE AIR COMPRESSOR EXHAUST COMPRESSION TOTURB IGNITION AIR SUPPLY TO ENGINE ENGINE AIR INLET _________________ OUTPUT SHAFT CLUTCH Fig. 2. Air-supply flow diagram.

Calorific value of fuel 18500 Btu/lb Turbo-compressor efficiency Oo70 Turbine efficiency 0.86 Ambient air Pa = 14.7 psi and Ta = 125~F Total heat loss 21% Heat-loss compression 0.4% Heat-loss combustion 1.7% Incomplete combustion 2 0% Heat-loss expansion 5.0% Heat-loss exhaust 5.5% Heat-loss to oil 1,8% Miscellaneous 4.6% MANIFOLD CONDITIONS With the above efficiency values, the engine manifold conditions become Pm = 29.4 psia Tm = 7535 abs If an aftercooler of 60% effectiveness is employed, the manifold temperature is reduced to 612~ abs. INITIAL CONDITIONS IN CYLINDER For the above manifold conditions, the state in the cylinder at the beginning of compression would be approximately as follows for 1 lb of air trapped It is assumed that with turbo changing at the high pressures proposed and with valve Overlap, no exhaust gas is present in the cylinder and the volume of the 1 lb of air is the cylinder volume; this is considered a reasonable assumption for a first approach to the problem, P1 = 29.0 psia T, = 765~ abs, no aftercooling = 625~ abs with 60% aftercooling V1 = 9~75 cu ft/lb or 7.96 cu ft/lb with aftercooling. 5

END OF COMPRESSION Using the charts for cycle determination with the specified heat losses, the following states represent the end of compression. No 60% Effective Aftercooling Aftercooling Index of compression n 10343 1.353 Pressure P2 psia 1080 1110 Temperature T' abs 1912~ 1622~ Work of compression Btu 230,2 192.2 V2 cu ft/lb 0.65 0.531 CONSTANT-VOLUME COMBUSTION Assuming an arbitrary limit of 1500 psi for the peak cylinder pressure, a certain percentage of the fuel can be burnt approaching constant-volume combustion from point 2 to point 3 of the cycle. This will be the same for all the cycles; the only difference is that of temperature when using an aftercooler. Using the heat-loss factor for this process, the following gives the conditions at point 3, No Aftercooling 60% Aftercooling P3 (assumed) psia 1500 1500 Temperature increase AT 560 490 Btu's required 118 100 Btu's used including loss 135 117 T' abs 2472 2112 Va = v2 0o65 0.531 CONSTANT-PRESSURE COMBUSTION The balance of the heat available in the fuel is now added at constant pressure to the 1 Ilb of air asperated into the cylinder. Then 6

Total Btu of 0o05 lb of fuel = 0,05 x 18500 = 925 Btu Heat loss and incomplete combustion = 20 Btu Heat available = total Btu supplied - heat used at constant volume - heat loss Heat available = 925 - (135+20) = 770 Btu (no aftercooling) Heat available = 925 - (117+20) = 788 (60% aftercooling). The charts give the following for the ratio of V4/V3 for the above heat supply: V4/V3 = 2.06 no aftercooling = 2.25 with aftercooling. It follows that the temperatures can be calculated from V4 T4 = T3 14 *IV3 assuming that R, the gas constant, remains at a constant value. Thus T4 = 1500~ (no aftercooling) or 47600 (with aftercooling) and the contents of the cylinder at point 4 of the cycle becomes: No Aftercooling 60%o Aftercooling Air, lb 1.0 1.0 Fuel, lb 0o05 0.05 Gas constant for mixture 54,0 54.0 P4 = P3 psia 1500 1500 V4 cu ft 1.,34 o196 T4 5100o 4760 Work of expansion 144P3(V4-V3)/778 Btu 191.9 184. 3 The above represents the changes in state for the normal cycle combustion process. At point 4 when responsiveness is desired, additional fuel and air are supplied to the cylinder and, if the F/A is maintained at 0.05, then, for each lb of air supplied, 925 additional Btu will be added or, say, 910 Btu when heat loss is assumed, 7

If the additional supply is assumed to be at a temperature of 700OR when admitted and is at 1500 psia when in the cylinder, and if this supply is considered to enter as a separate entity, is supplied with the necessary fuel, then burnt at constant pressure, still in its separate container, followed by mixing with the charge that was normally asperated at constant pressure, then the approximate state at the end of this phase of the cycle can be obtained as given below. In actual conditions it is possible that the assumed 700~ for the compressed air is low, since the compressor will deliver the air at a high temperature. Such increase in inlet temperature will result in improved performance to that calculated and thus the figures below can be considered somewhat pessimistic in an over-all evaluation. The first operation, admitting air at 1500 psi and 7000F, can be determined as follows: Av = V6T - V4 (Fig. 1) Ve, = volume of normal change plus vol of Aw lb of air Aw RT Av =. P R = 53.34 for air T = 700~ P = 1500 psi Av = O.1728 Aw cu ft. Assume that the volume of the liquid fuel is negligible and determine the conditions when Aw is gradually increased from 0 to 1,0 lb of air, Then: Aw lb 0.0 0.2 0)4 0.6 0.8 1o0 Av cu ft 0,0 0.0345 o00691 0.1037 0.1381 0,1728 V6e cu ft 1 34 1.375 1.409 1,444 1o478 1o513 No aftercooling V6' cu ft 1.196 lo23 lo265 lo300 1o334 13569 60% aftercooling 8

On adding and burning the fuel at constant pressure, we have C QA - (Vs - V6')P4 () R where QA = Btu added Cp = mean specific heat for the temperature range R = J(Cp - Vc) for temperature range V6" = Volume of products after combustion, The process must be a trial-and-error one since the range of temperature is not known until the final temperature is available. Thus assume this temperature, proceed with the calculation, and see if the average calculated temperature is equal to the assumed average value used to determine Cp. By this means a correct assumed average temperature is obtained from which the values of Cp and Cv control the temperature: From graph for Tavg Cp = avg Cv = Cvavg R = J(Cp g- Cv ) -'avg avg and substitution in Eqo (1) will give QA = KAV C K = constant = -C x 1500 x 144 R Assume Tavg = 2200~ then Cp = 0.305 and Cv = 0.237 R = 778 (o305-,237) = 52.9 QA = 1245 AV or AV 910 0.731 cu ft/lb 1245 9

Aw lb =0 0.0 2 04 o06 0,8 1o0 AV = 0.0 0.146 0,292 0 438 0.585 0o731 V6" = 1 34 1.525 1o701 1.882 2.063 2 2244 No aftercooling V6" = 1.196 1.376 lo557 lo738 1.919 2.099 60% aftercooling The temperature at the end of combustion of the additional air and fuel is given by: QA = Aw(l+f)Cp(T6,,-700) but QA = Aw x 910 Thus T6, 910 + 700 Cpxl.05 910 + 700 0.305x1.05 2840 + 700 = 3540~ abs. Thus the average temperature of the process is Tavg = (53540+700) 21200 against the assumed average of 2200~ abs. This is close enough since Cp changes somewhat slowly at these temperatures. The cylinder contents now consist of (a) 1,05 lb of combustion products from the asperated air occupying V4 cu ft at P4 = 1500 psi and temp T4 abso 10

(b) Aw(l+f) lb of combustion products supplied at TDC occupying (V6"-V4) cu ft at P4 = 1500 psi and temp T6" = 3540~ abso These two quantities are now mixed at constant pressure with the result that: Heat lost by hot gas-heat gained by colder gas (l+f)Cp(T4-Tm) Aw(l+f)Cp(Tm-T6tt) T4 - Tm = AwTm - AwT6 Tm = T4+AwT6" l+Aw With no aftercooling, T4 = 5100~ With aftercooling, e = 0.60 T4 = 4760. Thus T = 100+Awx540 or 4760+Awx3540 which yields the following l+Aw l+Aw when Tg = 5100 for no aftercooling: Aw 0.0 0.2 0.4 0.6 0.8 1.0 Tm = T6 5100 4840 4650 4510 4405 4320 when Tg = 4760 and e =.60 aftercooling: Aw 0.0 0.2 0,4 o06 0,8 1.0 Tm = T6 4760 4560 4415 4300 4220 4160 The actual state at point 6 is now available for all values of Aw up to 1.0 lb of air. No Aftercooling Aw 0.0 0,2 0.4 0o6 0,8 1.0 P6 1500 1500 1500 1500 1500 1500 V6 1.34 1.525 1,701 1.882 2.063 2,244 Te 5100 4840 4650 4510 4405 4320 11

With 60_ Aftercooling Aw 0.0 00 2 0.4 0.6 0.8 lo0 P6 1500 1500 1500 1500 1500 1500 V6 1.196 1,576 1.557 1.738 1,919 2.099 T6 4760 4560 4415 4300 4220 4160 An additional advantage of the cycle being examined is a reduction of maximum gas temperature as additional air is added, EXPANSION STROKE The thermodynamic cycle can now be completed by expanding the gas, corrected for heat losses, back to the original volume of the 1 lb of air, This operation was carried out on the charts provided for the purpose with the following results. Aw = 0.0 0.2 0,4 0,6 0,8 1.0 Expansion ratio e = 0 7.26 6.4 5.72 5.18 4.72 4.34 P7 114.0 134 155 181 193 216 T7 2800 2660 2650 2730 2740 2750 Expansion ratio c = 0.60 6.65 5.78 5.09 4.56 4.13 3o77 P7 127 167 180 206 240 263 T7 2670 2680 2675 2755 2740 2770 The work of the expansion is given by We = P6V6-PVV7 (n-l)J 144 Qev P /P7 _ 0.3x778 P-/PV With no aftercooling V1 = 9.75 With 60% aftercooling V1 = 7.96 No Aftercooling Aw = 0.0 0.2 0.4 0,6 0.8 1,0 Work of expansion Btu 556 603 650 663 760 790 12

60% Aftercooling Work of expansion Btu 482 524 558 604 639 659 EXHAUST MANIFOLD TEMPERATURE The next step is to obtain the expected exhaust gas temperature. To do this we have the following equation Heat in gases at + Work done by piston jEnergy of end of expansion tin exhausting gas gas in manifold + Heat losses in process (l+Aw)(l+f)CvT7 + 144 Pe( V2) - (l+w) (l+f) CpTe + QL 778 where Pe = pressure in exhaust manifold Te = temperature in manifold QL = heat losses of exhaust = 50 Btu V7 and V2 = volume of gas, cu ft CvT 0.185 Pe(V7-V2)-50 Te = T7 + P (l+Aw)(l+f)Cp Since the variation of Te is of minor magnitude due to T7 having but a small variation, the values of Cv and Cp can be considered constant and equal at all values of Aw. The average values used are Cv = 0.23 and Cp = 0.30 for the gases and temperatures involved. Calculation gives the following values for the state in the exhaust pipe. 13

No Aftercooling V7 = 9.75, V2 = 0.65 cu ft Aw = 0.0 0.2 0.4 o06 0.8 1o0 T7 2800 2660 2650 2730 2740 2750 Gas lb 1.05 1.26 1.47 1,68 1.89 2.10 Pe (assumed) 25.5 25.5 25,5 25.5 25 255 T8 2127 2021 2013 2075 2087 2098 60% Aftercooling V7 = 7.96, V2 = 0.531 cu ft Aw = 0.0 0.2 0.4 0.6 0,8 1,0 T7 2676 2680 2675 2755 2740 2770 Gas lb 1 o05 1.26 1.47 1.68 1,89 2.10 Pe (assumed) 25~5 25.5 25,5 25~5 25o5 2505 T8 2002 2015 2016 2080 2073 2096 It is seen that the exhaust temperature does not vary over a wide range. This is reasonable since the same F/A ratio is being employed in all cases and the mass of gas varied. The values obtained vary from a low of 2000~ abs to a high of 2127~ abs or 1540~ to 1667~F, a not unreasonable value for an F/A ratio of 0.05, possibly 10% high at the most, but heat losses for a length of exhaust pipe to the point of temperature measurement can lower these values by 200~F or so quite easily. WORK OF CYCLE It is now possible to evaluate the net work of the cycle and the fuel quantity needed for this output, The work done by the gas during the constant pressure expansion process is as follows: W = 144 P3(V6-V3) Btu 778 Aw 0.0 0.2 0.4 0.6 0.8 1,0 With no Aftercooling W 191.8 24352 292.5 342,5 393.0 44305 60% Aftercooling W 184.8 234.8 284,8 335.5 386,0 435 5 14

There is a small pressure difference between inlet and exhaust pressures. Thus there is a net positive work performed here, which is as follows: (Pm-Pe)(V1-V2) Suction loop work (PmPe) (V1V2) Since Pm and Pe were maintained constant at 29,4 and 25.5 psia,respectively, this becomes Suction loop work = 0.722 (V1-V2) For the case with no aftercooling, V1-V2 = 9.75-0.65 = 9o10 cu ft, while with aftercooling, it becomes V1-V2 = 7o96-.531 = 7.429 cu fto Suction loop (no aftercooling) = 6.58 Btu Suction loop (with aftercooling) = 5.36 Btu. 15

WORK SUMMARY ~ _Item ~No Aftercooling Aftercooling, E =0.60 ~________________~0,0 0.2 0,4 0,6 0,8 1,0 0.0 0.2 0o.4 0,6 0,8 1,0 Suction loop 6.58 6,58 6.58 6.58 6.58 6._58 5.56 5.56 5.56 5.36 5.56 556 (positive Btu)____________________________________________ Compression work -230.2 -192Q2 (negative Btu)____________________________________________________ Const. press. ^^ ^ ^ ^ ^ ^ ^^ ^ g^^^ ^ g^ ^ Const, pressB 191.9 243.2 292.5 34525 39350 443.5 184.5 234.8 284,8 335.5 386.0 455 expo Btu ____________________________________ Expansion 556 605 650 665 760 790 482 524 558 604 659 659 B3tu _____________________________________________________________ Net work of cycle 524,5 622,6 718.9 7819 929)4 1009,9 479,5 572 656.0 752.7 858,2 907.7 Btu__________________ I.HP./lb of basic 742 881 1018 1107 1313 1442 679 809 928 1066 1187 128 air/sec_________________ Total air flow ^ _ ^ ^ ^g ^ ^ ^ ^ g ^ Total air flow, 1.0 1.20 1.l4o 1.60 1,80 2,00 1.0 1,20 lko 1,60 1.80 2.00 lb/sec_________________'FuIel flow, g ^ Q ^ ^ ^ ^ Q ^ ^ Ftiel flow,.05.06.07.08.09 0,10.05.06.07.08.09 0.10 lb/sec_________________ Total gas flow/lb ^ ^ ^ ^ g ^ ^ ^ ^ g ^ ^ Total gas flow/lb 1,0 1.26 1.47 1.68 1,89 2.10 105 1,26 147 168 1,89 210 of basic air/sec:516. iFuel per hr, lb/lb 180 216 252 288 324 560 180 216 252 288 524 0 of basic air___________________________________________ Fuel Ib per.24 245.248.260.247.250 265.267 271.270.27 0280 I, H.P./hr_________________________________________ IM.E.P3 512 570 426 465 552 599 344 416 477 547 610 660 psi___________________________________________

Considering these results and plotting the IoMoEPo on a diagram such as Fig. 1 of Progress Report No. 1, it is seen that if the engine with no aftercooler was rated at 312 I.M.E.P. at 2800 rpm, responsiveness would exist down to a speed of 1475 rpm for an additional air flow equal to that of the basic engine air flow: a not unattractive performance, particularly since the S.F.C. changes by about 10%, as a maximum, in the process on an indicated bases. Since the engine speed would decrease with responsiveness, the mechanical efficiency will generally improve under such conditions, with the result that the change in SoFCo can generally be less than the 10% calculated. COMPOUNDING The above calculations have been based upon turbo-charging, the assumption being that the exhaust gases are capable of driving the compressor onlyo In the cycle under consideration, it is apparent that as the auxiliary air flow (Aw) increases, the energy available at the end of the expansion stroke is also increased, e.g., the pressure P7 changes from 114 psi at Aw = 0 to 216 psi when Aw = 1.0; greater blow-down energy is thus available. To examine what improvement can be obtained from such conditions by means of compounding, the following cases were investigated, Case 1. Where it is assumed that conventional present-day turbocharger efficiencies are employed, Case 2. Where a reasonable increase in turbine and compressor efficiencies are employed, increases that can be easily justified with the best of modern design; however, such design will probably add something to their cost, unless produced in large numbers, CASE 1 Assumed isentropic compressor efficiency = 0.70 Assumed isentropic turbine efficiency = 0,86 Air manifold pressure = 29.4 psi Exhaust manifold pressure = 25,5 psi Atmospheric pressure = 14,7 Air flow (basic) = 1.0 lb/sec. Compressor —The compressor performance is given by the following: wCpTa k-1 W..... (R -W - 1) c17 17

where W = Btu of work for compression w = air flow, lb p = 0.241 Btu/lb/OF Ta = atmospheric temperature, ~R R = pressure ratio of compressor = Pm/Pa Pa = atmospheric pressure k = 164 nc = Isentropic compressor efficiency. For both cases, with andwithout aftercooling the air enters and leaves the charger under the same conditions; thus so long as Pm remains constant, the compressor work per lb of air is constant. If the inlet air to charger is at 115~F then W o1Ox. 241x565 (2.0 286_1) 0.70 = 194.5 (1.219-1) =42 7 Btu/lb of air. This will remain constant as Aw is varied since the auxiliary air does not flow through the charger. Turbine Output.-The gas flow through the turbo will be the basic plus auxiliary air and fuel or (l+Aw)(l+F/A), its pressure and temperature will be Pa and T8, and the relationship between these parameters and the work done is given to a first approximation by = (+Aw)(l+f)CpTs8 F1 k-l/k1 ~T P8 in which = 0,04322 (l+Aw)Ts Btu W = work output Btu TT = turbine efficiency = 0.86 Pa = atmospheric pressure, psia Cp = specific heat at an average temp of expansion of, say, 1800~R = 0.29 when F/A = 0o05 k = 131 18

No Aftercooling Aw 0.0 0.2 0.4 0.6 0.8 1.0 T8 2127 2021 2013 2075 2087 2098 W Btu 91.9 105.0 122.0 143.5 162.3 182.0 60% Aftercooling T8 2002 2015 2016 2080 2073 2096 W Btu 87.2 104.7 122.2 143.9 161.5 181.9 Output Compounded Engine No Aftercooling Aw 0.0 0.2 0.4 0.6 0.8 1.0 Net cycle work 524.5 622.6 718.9 781.9 929.4 1009.4 Turbine work 91.9 105.0 122.0 145.5 162.5 182.0 Total output 616.2 727.6 840.9 925.4 1091.7 1191.4 Input to comp. 42.7 42.7 42.7 42.7 42.7 42.7 Net output 575.5 684.9 798.2 882.7 1049.0 1148.7 I.H.P./lb of basic air/sec 811 969 1130 1250 1486 1627 Fuel flow lb/hr/lb basic air 180 216 252 288 524 360 S.F.C. lb/I.H.P./hr.222.225.225.250.218.222 I.M.E.P. psi 541.0 406.0 474.0 525.0 624.0 682.0 60% Aftercooling Net cycle work 479.5 572.0 656.0 752.7 858.2 907.7 Turbine work 87.2 104.7 122.2 143.9 161.5 181.9 Total output 566.7 676.7 778.2 896.6 999.7 1089.6 Input to comp. 42.7 42.7 42.7 42.7 42.7 42,7 Net output 524.0 654.0 755.5 855.9 957.0 1046.9 I.H.P./lb of basic air/sec 741.0 896 104l 1208 1558 1480,0 Fuel flows lb/hr/lb basic air 180 216 252 288 324 560 S.F.C. lb/I.H.P./hr.245.241.242.259.239.243 I.M.E.P. psi 576 462 556 620 696 761 It is seen that, despite a change in output of roughly 2:1, the S.F.C, remains practically unchanged in each case, a distinct advantage as far as engine operating temperatures are concerned. CASE 2 In view of the unit efficiencies obtained today in small gas-turbine power plants, it is believed that a substantial gain can be made in some of the units of a turbo charger, provided some additional cost can be justified in the process. It will be assumed that the characteristics of the turbo 19

charger can be changed from those of Case 1 to the following: Compressor efficiency = 0.75 Turbine efficiency = 0.88 The other items of the specification remain unchanged. The calculations, similar to Case 1, now give the following results: loxx02241x565 0.286 Work of compressor O075 (2:0-1) = 39,8 Btu/lb of air Turbine output = 0.0443 (l+Aw)T8 Btu. Thus the turbine output as Aw varies will be as below: No Aftercooling Aw 0.0 0,2 0o4 0.6 0,8 1.0 W Btu of turbine 9400 107.6 125,0 146,9 166.2 186,2 60% Aftercooling W Btu of turbine 89.3 107,1 125.2 147,1 166.2 186.0 It is assumed that the engine cycle remains unchanged. Actually, there will be a small change in the case with no aftercooling, since the engine manifold temperature will be reduced a few degrees. With the above figures, the output of a compounded unit then becomes: Output Compounded Engine No Aftercooling Aw 0,0 0,2 0,4 0.6 08 1,0 Net output Btu 578.5 690.4 804,1 889.0 0.8 1155.8 I.H.P./lb basic air/sec 819 0 978 0 1139.0 1259 1509 1658 IoMEP. psi 344 410 478 529 634 687.0 S.F.Co lb/IoHoP./hr 0.22 0,221 0.222 0.228 0.215 022 60% Aftercooling Aw 0.0 0.2 0.4 0o6 0.8 1.0 Net output Btu 529,0 639 6 741.4 860 964.6 105539 IoHoP,/lb basic air/sec 749.0 905 1049 0 1218.0 1364,0 1491o 0 IM.E.P. psi 380,0 466o0 541.0 625.0 699.0 768,0 SFC. lb/I,H.P./hr 0,241 0,239 0.240 0.237 0,237 0,241 20

Comparison of these results with those of Case 1 shows that little has been gained by a research effort to improve the turbo-charger efficiencies alone, other things remaining the same. However, it will be found that a reasonable gain in power is possible if the improved efficiencies are employed to increase the turbo-charger pressure ratio instead of holding it constant as here. HIGH-PRESSURE AIR COMPRESSOR So far no allowance has been made for the power required to supply the high-pressure air employed. This power must be subtracted from the engine output and is calculated on the following basis. It is assumed that a three-stage air compressor is employed which is to deliver at a pressure of 1600 psi for injection of the air into the cylinder where the pressure is to be maintained at 1500 psi. The previous calculations have assumed a maximum of 1 lb of compressed air to be supplied by the compressor per lb asperated into the engine. On this basis the power for the compression of 1 lb of air in a three-stage compressor can be obtained. The conventional relationship for maximum efficiency of a multi-staged compressor will be employed. P2 P3 - P4 Pn P1 P2 P3 Pn-1 where P1 = initial pressure P2 = pressure in 2nd stage P3 = pressure in 3rd stage Pn = pressure in nth stage. Then with no losses between stages, but with intercooling to the original temperature, it can be shown, for a three-stage machine, that n-l RT1 3n -(P43n - Btu/lb J n-1 Pj 0.38 53.34x585 3x1038 0160&\3 3T 778 0 38 14. 7/ = 234.5 Btu/lb of air. 21

This assumes that the inlet temperature T, is 125~F and that n in PV1n = C for the compressor is 1.38, In this case the efficiency of the process is given by loger n lr3n - n-l -X where r = P4/P1 the pressure ratio, Then = 80%o These values are close enough for present purpose (they may be a little optimistic), It follows that the net performance of an engine fitted with such a compressor will be as follows. CONTINUOUS PERFORMANCE AT MAXIMUM POWER In this case, where the engine output can be maintained at all times equal to the maximum desired, Aw will vary from 0 to 1.0 for the complete range covered previously. It follows that: Net Output Compounded No Aftercooling Aw 0,0 0,2 0.4 0.6 0.8 1,0 Engine output Btu 573.5 684,9 798.2 882.7 1049o0 1148.7 Compressor input Btu 0.0 46.9 93~9 140.8 18706 234,5 Net output engine and comp. Btu 57305 638,0 704.3 741,9 861,4 914,2 Fuel lb/hr 180 216 252 288 324 360 S.F,C, 0.222 0,237 0,253 0,274 0,266 0,279 Over-all net I.MoE.P, 341,0 379~0 418 441 512 543 22

60% Aftercooling Aw 0 0 0,2 04 0,6 0,8 1.0 Engine output Btu 524.0 634.0 755.5 853.9 957O0 1046.9 Compressor input Btu 0.0 46.9 93.9 140.8 18706 234,5 Net output engine and comp. Btu 524.0 587o1 641.6 71351 770.4 812.4 Fuel lb/hr 180 216 252 288 324 360 S.F.C. 0.243 0.260 0.277 0.285 0.297 0,513 Over-all net I.M.E.P. 376 428 467 518 560 691 The above ideal results would be reduced to a brake horse power performance approximately as given below when corrected for normal losses. B.H.P. Performar e Aw 0.0 0.2 0.4 0.6 0,8 1,0 No cooling 700 788 877 926 1036 1158 B. H.P. Aftercooling 639 725 800 895 974 1028 S.F.C. No cooling.257.274.287 0.31.322.347 lb/B.H.P./hr Aftercooling o282.298.315 0.322.333 o357 INTERMITTENT PERFORMANCE AT MAXIMUM POWER The results given above represent the engine fitted with a compressor capable of maintaining the air supply indefinitely. At the opposite end of the scale, if the maximum output was maintained for a limited time only, then the responsive effect could be improved since the input to the compressor would be reduced proportionally. Assume that the peak power is required for a period of 5 minutes per hour. There would be no change in the maximum power available except that resulting from the small change in compressor power, but the average compressor input would be reduced by 1:12 and the performance would be as shown below. Net Output Compounded Aw for 5 min 000 0,2 0.4 0.6 0.8 1.0 Max. net output No cooling 700 831 963 1055 1208 1373 Bo HP. With cooling 639 768 886 924 1046 1243 S.FoC. No cooling 0.257 0.26 0.262 0.272 0.276 0.293 lb/B.H.P./hr With cooling 0,282 0.282 0.284 0.512 0.o510 0.295 25

The above figures are based upon a storage vessel capable of holding sufficient air for the five-minute burst of power, plus a compressor capable of re-charging the vessel in the remaining fifty-five minutes. Two such bursts of power per hour could be obtained with little change in the above values. The longer the time required, the closer the results approach the continuous valueso AIR COMPRESSOR To visualize the final power plant, some idea of the compressor displacement necessary for the purpose is required. Taking the case of continuous operation at the maximum value of Aw, ioe., 1.0 lb per lb of basic air, then for a 500-hp output engine with aftercooler: Basic air supply = 1 x 500 x 3600 1028 = 1752 lb/hr with an ambient temperature of 115~F at 14.7 psia. Volume of air = wRT 1752x53.34x575 P 14.7x144 = 25500 cu ft/hr. This must be the capacity of the compressor since Aw = lo,0 Assuming a volumetric efficiency of 0.80 with a compressor (reciprocating type) operating at 4000 rpm and employing a double-acting L.P. compressor cylinder, then IrD2 Cylinder displacement/rev. = x L x 2 cu ino/rev. ~cD2 Air displacement = 4172 x L x 2 x 0.80 x 4000 x 60 cu ft/hro where D = diameter of L.P, cylinder in in. L = stroke of LP cylinder in in, Thus -fTD2 25500 = x L x 2 x 0,80 x 4000 x 60 4x1728 D2L = 25500x4x1728 2xxrxOo 80x4000x60 = 146o2 cu in, 24

If D/L = 1,0, then D3 = 146.2 D = 5.26 in. and L = 5,26 in. This is a fairly large compressor, but not significantly out of proportion to a 500 BH.P. Diesel engine when it is considered that due to its attachment to the engine a B.oHP. of 805 can be developed continuously. Looking at the case for a 5-minute burst of power, then the maximum output of a basic 500-hp engine will become 974 B.H.P.; for this case, a compressor of the same characteristics as above is calculated as follows. Compressor delivery = 25500 x 5/60 cu ft/hr = 2130 cu ft/hr or 2 2130x4x1728 2xjxxO 80x4000x60 12.2 cu in./rev. If D = L, then D = 23, and L = 235. This is a quite small compressor, compared with a 500-hp Diesel, to permit a maximum power of 974 B.H.P. for five minutes. To the small compressor above, a receiver must be added to provide air storage. Compressed air/hr = 1752 x 5/60 = 146.1 lb/hr. Assume delivery air temperature of 700~R at a pressure of 1600 psi. w RT Volume of air = -RT P 146. 1x5 35 34x700 1600x144 = 2357 cu fto Thus an air receiver of 24 cu ft volume will contain all the air for one boost without any allowance for the air delivered by the compressor during the period. This means a vessel of about 2 ft diam by 7.5 ft long, perhaps a not impossible size since the bulk of the engine, if the full 975 hp was 25

required, will be reduced by a volume of about 50 cu ft if it is figured at an output of 10 hp per cut ft of engine volume. In other words, a 500-hp engine plus receiver and compressor will occupy 50+2+24 = 76.0 cu ft of space and develop a maximum of 975 hp. A normal engine of 975 hp would occupy 97.5 cu ft, or a net saving of 20 cu ft of engine space. If the comparison is confined to the unit giving a continuous maximum output, when aftercooled, then for a normal output of 500 hp when Aw = 00, a maximum of 805 hp will be obtained for Aw = 1.0. In this case no receiver will be necessary and the engine bulk will still be 50 cu ft at 10 hp/cu ft; the compressor of 5-1/2 x 5-1/2 size could be expected to occupy not more than 3-5 cu ft, giving a total bulk of, say, 55 cu ft for an engine with a maximum capacity of 805 B.HoP. against about 80 cu ft for a normal turbocharged unit of the same power. It must be keptin mind that the arbitrary rating of a normal 500 BHP. was picked at random for purely comparative purposes. It is recognized that the 800 to 900 maximum horse power resulting is more than is necessary for the purpose in mind at the present time. The engine size, bulk, etc., for a moderate change, say to 300 hp normal and 480 to 550 hp maximum, would roughly result in volumes of 355 to 45 cu ft for continuous or intermittent operation, respectively. Figures 3, 4, and 5 show the performance at maximum output under the various conditions, on an engine I.H.P. and B.H.P. basis, The specific fuel consumptions given represent the minimum attainable on the assumption of perfect combustion of the fuel with the air. In practice, particularly at the high F/A ratio of 0.05, some departure from these values is to be expected, since at such a ratio as the above some smoke is usual and a realistic value of S.F.C. would be some 15% greater than that shown..In addition, the B.H.P. given is that of the bare engine and must be reduced by the accessories driven, such as cooling fan, generator, etc. An allowance for these items cannot be made at the present time since the cooling fan power will depend not only on the engine but the efficiency of the transmission attached to ito At a later date in the analysis, when the transmission is finalized and the required generator size is established, actual performance curves can be developed for the above conditions. CONCLUSIONS The method of additional air supply at the TDC of the compression stroke has advantages for the purpose of military vehicle propulsion where high 26

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power is needed for a small fraction of the timeo In addition, a degree of responsiveness can also be obtained at the same time by the use of this principle (roughly a 2:1 speed change at constant horse-power)~ If the engine and compressor are properly proportioned, it appears that a normal 3550-hp compression ignition engine could take the place of a 600-hp one, resulting in a reduction of engine bulk and possibly weight There would be an increase in the fuel used at the maximum load condition, but a decrease at the 350 hp and less. Since the engine operates most of the time at part load, with 40% of the time at idle, it is believed that an actual reduction in the fuel rate in lb/hr will be achieved when this engine is coupled to the transmission and the operational performance finally establishedo The penalty to be paid would be the additional complication of the airsupply system. However, the increased power is obtained without increase of stress levels. 50

UNIVERSITY OF MICHIGAN 3 9015 03527 2486