Assignment of Power Plant-I Designed by Sir Engr. Masood Khan

SOLVED PROBLEMS OF CHAPTER # 12

TITLE: POSITIVE DISPLACEMENT MACHINES

Positive Displacement Machines Designed by Sir Engr. Masood Khan

PROBLEM: 12.1:Air is to be compressed in a single-stage-reciprocating compressor from 1.013 bar and 15 C to 7 bar. Calculate the indicated power required for free air delivery of 0.3 m3/min., when the compression is as follows:(i) isentropic;(ii) reversible isothermal;(iii) Polytropic, with n = 1.25.

What will be the delivery temperature in each case?

GIVEN:Working Fluid = AirSingle Stage CompressionInitial Pressure P1 = 1.013 barDelivery Pressure P2 = 7 barInitial Temperature T1 = 150C=288KFree Air Delivery = FAD = 0.3m3/min

REQUIRED:1. Indicated Power for Compression.1.1. Insentropic =1.2. Reversible Isothermal =1.3. Polytropic, n = 1.25.2. Delivery Temperature.

SOLUTION:As induction & FAD are same,So FAD = Volume Induced, V = 0.3 m3/min(1.1) For Insentropic Process:Pi = (r/r-1) P1V {(P2/P1) r-1/r -1} (1)Putting values, we get: Pi = 1.31 KWNow T2 =? As (T2/T1) = (P2/P1) r-1/r

Putting the values, we have: T2 = 227.3 C

(1.2) For Isothermal ProcessPi = P1V1 ln(P2/P1) = 0.98 KWT2 = T1 = 15C(1.3) For Polytropic Process,Pi = (n/n-1) P1V1 [(P2/P1) n-1/n –1] = 1.2 KW(2) Delivery Temperature.T2 = T1 (P2/P1) n-1/n = 423.9K = 150.9C

PROBLEM: 12.2:The compressor of problem 12.1 is to run at 1000 rpm. If the compressor is sigle acting and has a stroke / bore ratio of 1.2/1, calculate the bore size required.

GIVEN: As Problem # 1Speed = N = 1000 rpm = 1000 / 60 rps.Stroke to Bore Ratio = l/d = 1.2

REQUIRED:Bore Size = d =?

SOLUTION:As Cylinder Volume = Vol. Induced /cycle.(Πd2 / 4)x L = V / Cycle.---------------------------(A)As V/cycle = volume per unit time / Cycle per unit time = Vo / N.From Previous Problem, Vo = 0.3 / 60 m3 per second.So, V/cycle = Vo/ N = 0.3 /1000 = 0.0003 m3

So Πd2 L / 4 = 0.0003L /d = 1.2 => L = 1.2 dΠd2x1.2d / 4 = 0.0003 =>1.2 d3 = 4 x 0.0003 / πSo, Bore Size = d = 0.0683m = 68.3 mm

Positive Displacement Machines Designed by Sir Engr. Masood Khan

PROBLEM: 12.3:A single-stage, single-acting air compressor running at 1000 rpm delivers air at 25 bar. For this purpose the induction and free air conditions can be taken as 1.013 bar and 15 C, and the FAD as 0.25 m3/min. The clearance volume is 3% of the swept volume and the stroke/bore ratio is 1.2/1. Calculate:(i) the bore and stroke;(ii) the volumetric efficiency;(iii) the indicated power;(iv) The isothermal efficiency.Take the index of compression and re-expansion as 1.3.8

GIVEN:Speed = N = 1000 rpm.Delivery Pressure = P2 = 25 barInduction & FAD are:P = P1 = 1.013 bar, T = T1 = 15 CFAD = Vo = 0.25 m3/minVc = 0.03 Vs, L /d = 1.2

REQUIRED:Bore = d =? Stroke = L =?Volumetric Efficiency = v =?Isothermal Effeiency = iso =?Indicated Power = Pi =?

SOLUTION:We know that:Swept Volume = Volume of StrokeVs = π d2 L / 4, L = 1.2d So, Vs = π d2 1.2 d / 4d = (Vs x 4 / 1.2 π)1/3 ----------(1)Vo = Va – Vd = 0.25 m3 / min(Va – Vd) / cycle = V’/ N = 0.25/1000

Va – Vd = 0.25 x 10-3 m3/cycle.Va = Vs + Vc = Vs + 0.03 Vs = 1.03 Vs

Now Vd = Vc(P2/P1)1/n -----------------(2)So, Vd = 0.3533 VsFrom (2) Va – Vd = 0.25x10-3m3/cycleVs = 0.3694 x 10-3 m3 / cyclePutting in (1): d = 73.16 mmL/d = 1.2 L = 1.2 d = 87.8 mmVolumetric Efficiency:v = FAD / Vs ----------------------------(3)FAD / cycle = (FAD/min) / (cycles / min)= 0.25/1000 = 0.25 x 10-3 m3/cycleSo, v = 0.676 = 67.6 %Now Pi = (n/n-1) P1Vo[(P2/P1)n-1/n –1] = 2 KWIsothemal Efficiency:iso = Isothermal Power / Indicated Power = Piso / Pin.

Piso = P1Vo ln Pi / P1 = 1.353 KWPutting values:iso = 67.67 %

PROBLEM 12.4:The compressor of problem 12.3 has actual induction conditions of 1 bar and 40 C, and the delivery pressure is 25 bars. Taking the bore and stroke as calculated in problem 12.3, calculate the FAD Referred to 1.013 bar and 15 C and the indicated power required. Calculate also the volumetric efficiency and compare it with that of 12.3.

GIVEN:As in Problem 12.03Induction ConditionsP1 = 1 bar T1 = 40C = 313KP2 = 25 bar L/d = 1.2

Positive Displacement Machines Designed by Sir Engr. Masood Khan

REQUIRED:(i). FAD =? (ii) Pi =? (iii) v =?

SOLUTION:(i) Mass delivered or mass of FAD = Mass of volume inducedmo (FAD) = moVo = (P x FAD) / R.T. = P1 V’ / R T1

FAD = 0.227 m3 /min(ii) Pi = (n/n-1)m’RT1{(P2/P1)n-1/n –1}m’ = P1 V’ / R T1 = 4.63 x 10-3kg / sec.So Pi = 1.985 KW(iii) v = (FAD/cycle) / Vs ---------------------(A)FAD/cycle = (FAD/unit time) / (cycles/unit time)FAD/cycle = 0.227 x 10-3 m3

So equ: (A) v = 61.4 %

PROBLEM 12.5:A single-acting compressor is required to deliver air at 70 bar from an induction pressure of 1 bar, at the rate of 2.4 m3/min measured at free air conditions of 1.013 bar and 15 C. The compression is carried out in two stages with an ideal intermediate pressure and complete intercooling. The clearance volume is 3% of the swept volume in each cylinder and the compressor speed is 750 rpm. The index of compression and re-expansion is 1.25 for both cylinders and the temperature at the end of the induction stroke in each cylinder is 32 C. The mechanical efficiency of the compressor is 85%. Calculate:(i) the indicated power required;(ii) the saving in power over single-stage compression

between the same pressures;(iii) the swept volume of each cylinder;(iv) The required power output of the drive motor.

GIVEN:Delivery Pressure = P2 = 70 barInduction Pressure = P1 = 1 barFAD = 2.4 m3/min at P = 1.013 barT = 288 K Two Stage Compression(i) Ideal Intermediate Pressure, Pi/P1 = P2/Pi

(ii) Complete Intercooling T1 = T(iii) Clearance Volume is 3% of Swept Volumei.e. Vc

L = 0.03 VsL , Vc

H = 0.03 VsH

As Intercooling is Complete & Clearance Ratio is same, we can say that; Vs

L / VsH = Pi / P1 = P2 / Pi

Pi2 = P1 P2 => P2 / P1 = (P2/P1)

Speed = N = 750 rpm, n = 1.25Temp. at the end of Induction Stroke in each cylinder.T1 = Ti = 32C = 305K, mech. = 85 %

REQUIRED:(i) Indicated Power = Pi =?(ii) The savings in power over the single compressor b/w the same pressures.(iii) Swept volume of each cylinder = Vs=?(iv) Power output of the driver motor =?

DIAGRAM:

SOLUTION:(i) As the intermediate pressure is ideal & Intercooling is complete, thus minimum work conditions are,Hence; Pi = 2x(n/n-1)moRT1{(P2/P1)n-1/2n –1} -----(A)But mo =?Mass induced = Mass of FADmo = moFAD = P x FAD / R T = 2.94 kg/min(A) Pi = 22.7 KW

Positive Displacement Machines Designed by Sir Engr. Masood Khan

(ii) Saving Power = {(indicated power of single stage)-(indicated power of double stage)}For single stage,Pi = (n/n-1)moRT1{{P2/P1)n-1/n –1} = 28.7 KWThus Saving Power = 28.7 – 22.7 = 6 KW.(iii) Swept Volume Vs

L & VsH for LP Stage.

Va – Vd / cycle = (Va-Vd/unit time)/cycles /unit time= Vo

Lp / N --------------------------------------------- (1)Vo

Lp = m’RT1/P1 = 2.57 m3/min(1)Va – Vd / cycle = V’

Lp / NVa – Vd = 3.43 x 10-3 m3/cyclesVa = VsL + Vc = VsL + 0.03 VsL = 1.03 VsL

Vd =Vc(Pi/P1)1/n = 01641 VsL

VsL = 0.00396 m3

For HP Stage, VsH = Va

/ - Vc/

VsH = Va

/ - 0.03 VsH

Va/ = 1.03 Vs

H =>VsH = Va

/ / 1.03---------------------(2)As point a & a/ touches the Isothermal line due to complete Intercooling, thus; Pi Va

/ = P1 Va

Va/ = Va P1/Pi = Va / (Pi/P1) ----------------------------(3)

Va = 1.03 Vs = 1.03 x 0.003962 = 4.08 x 10-3m3/cycleEqu:(3) Va

/ = 0.0004877 m3 / cycleNow Vs

H = Va/ /1.03 = 0.00473 m3

So Swept Volume of HP stage = 0.000473 m3

(iv) Power output of motor = Shaft PowerShaft Power = Indicated Power / mech.= 22.7 / 6.85Power output of Motor = 26.71 KW

PROBLEM 12.7:A single-cylinder, single-acting air compressor of 200 mm bore by 250 mm stroke is constructed so that its clearance can be altered by moving the cylinder head, the stroke being unaffected.

(a) using the data below calculate:(i) the free air delivery;(ii) The power required from the drive motor.Data Clearance volume set at 700 cm3; rotational speed, 300rpm; delivery pressure, 5 bar; suction pressure and temperature, 1 bar and 32 C; free air conditions, 1.013 bar and 15 C; index of compression and re-expansion, 1.25; mechanical efficiency, 80%.To what minimum value can the clearance volume be reduced when the delivery pressure is 4.2 bar, assuming that the same driving power is available and that the suction conditions, speed, value of index, and mechanical efficiency, remain unaltered?

GIVEN:Single Stage Compressord = 200 mm = 0.2 m, L = 250 mm = 0.25 mN = 300 rpm, n = 1.25Vc = 700 cm3 = 700 x 10-6 m3

P2 = 5 bar, P1 = 1 bar, P = 1.013 barT1 = 32C = 305K, T = 288K

REQUIRED:(i) FAD =? (ii) Power required, P = ?When mech. = 80 %

SOLUTION:As Vs = (π/4) d2 LVs = 7.85 x 10-3 m3, Vc = 7 x 10-4 m3

Va = Vs + Vc = 7.85Vo = FAD = (Va-Vd)x(T/T1)x(P1/P)Vd = Vc(P2/P1)1/n = 2.54 x 10-3 m3

Positive Displacement Machines Designed by Sir Engr. Masood Khan

Va – Vd = 6.01 x 10-3m3

V = (Va – Vd)x(T/T1)x(P/P1) = 5.61 x 10-3 m3/cycleV/min = 300 x 5.61 x 10-3

FAD = 1.68 m3/min(ii) Power = (n/n-1)m’R(T2 – T1) --------------(1)Where mo = PVo/RT = 2.068 Kg/minSo, equ: (1) Indicated Power = Pi = 5.7 KNAs we know thatPower required to drive the motor= Indicated Power / mech.= 7.2 KW

PROBLEM 12.8:A single acting, single-cylinder air compressor running at 300 rpm is driven by an electric motor. Using the data given below, and assuming that the bore is equal to the stroke, calculate:(i) the free air delivery;(ii) the volumetric efficiency;(iii) The bore and stroke.Data Air inlet conditions, 1.013 bar and 15 C; delivery pressure, 8 bar; clearance volume, 7% of the swept volume; index of compression and re-expansion, 1.3; mechanical efficiency of the drive between motor and compressor, 87%; motor power output, 23 kW.

GIVEN:Single Acting Single Cylinder CompressorN = 300 rpm, Power of the motor = 23 KWEfficiency of Transmission = = 87 %P1 = 1.013 bar P2 = 8 barT1 = 288 K Vc = 0.07 Vs

L = d, n = 1.3

REQUIRED:(i) FAD =?(ii) Volumetric Efficiency = v =?(iii) L =? (iv) d = ?

SOLUTION:Motor power = Indicated Power / trans.

23 x 103 = Indicated power / 0.87Indicated Power = 20010 watts.Now Pi = (n/n-1)P1(Va – Vd){(P2/P1)n-1/n –1}Va – Vd = 4.47 m3 / min = FAD(ii) v = 1 – Vc/Vs{(P2/P1)1/n –1} = 72.6 %(iii)&(iv)As Va = Vs + Vc = 1.07 Vs

Vd = Vc(P2/P1)1/n = 0.343 Vs

Va – Vd = 0.727 Vs =>4.47/300 = 0.727 Vs

Vs = 0.0205Now Πd3/4 = Vs = 0.0205 d = 296 mm L = 296 mm.

PROBLEM 12.9:A two-stage air compressor consists of three cylinders having the same bore and stroke. The delivery pressure is 7 bar and the FAD is 4.2 m3/min. Air is drawn in at 1.013 bar, 15 C and an intercooler cools the air at 38 C. The index of compression is 1.3 for all three cylinders. Neglecting clearance, calculate:(i) the intermediate pressure;(ii) the power required to drive the compressor;(iii) The isothermal efficiency.

GIVEN:Two Stages Air-Compressor

Positive Displacement Machines Designed by Sir Engr. Masood Khan

L = d P2 = 7 barFAD Conditions = V’

1 = 4.2 m3/minP1 = 1.013 bar T1 = 15C = 288KIntercoolingT2 = 311K, n = 1.3, clearance = 0

REQUIRED:(a) Intermediate pressure = Pi =?(b) Power reqd. to drive the Motor =?(c) Isothermal Efficiency = iso. =?

DIAGRAM:

SOLUTION:We know that PVo = mo RTmo = PVo / RT = 0.086 Kg / secWith Intercooler Pi = P1P2 = 2.66 barNow we know thatPi / P1 = P2 / Pi, are same for both stages.So I.P. = (n/n-1)moRT{(P2/P1)n-1/n – 1} = 15.3 KWiso. = Isothermal Work / Indicated Work ------------- (1)Isothermal work = moRT1ln(P2/P1) = 137 KJIndicated Work = (n/n-1)moRT1{(P2/P1)n-1/n –1} = 15.3 KJiso. = 89.8 %

PROBLEM 12.12:Air at 1.013 bar and 15 C is to be compressed at the rate of 5.6 m3/min to 1.75 bar. Two machines are considered: (a) the Roots Blower; and (b) a sliding Vane Rotary

Compressor. Compare the powers required, assuming for the vane type that internal compression takes place through 75% of the pressure rise before delivery takes place, and that the compressor is an ideal uncooled machine.

GIVEN:P1 = 1.013 bar P2 = 1.75 barT1 = 15C = 288KVo = 5.6 m3/min = 5.6/60 m3/sec

REQUIRED:Power Required;(1) Root Blower (2) Vane Type

ASSUMPTION:(1) For Vane Type, Internal compression takes place through 75 % of pressure rise before delivery takes place.(2) Compressor is an ideal uncooled machine.

SOLUTION:(1) For Root Blower Power Required = (P2 – P1)xVo = 6.88 KW(2) For Vane Type Power = (r/r-1) P1Vo1 {(Pi/P1) r-1/r –1}+(P2 – P1)xVo

2 Here Pi = 0.75 (P2 – P1) + P1 = 1.566 barVo

1 = 5.6 m3/minVo

i = Vo1 (Pi/P1)1/r = 4.1 bar

Now Power = 5.64 KW

Positive Displacement Machines Designed by Sir Engr. Masood Khan