Solved problems chapter 12 thermo dynamics, Exercises of Thermodynamics

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

SOLVED PROBLEMS OF CHAPTER # 12

TITLE: POSITIVE DISPLACEMENT MACHINES

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. m^3 /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 = Air Single Stage Compression Initial Pressure P 1 = 1.013 bar Delivery Pressure P 2 = 7 bar Initial Temperature T 1 = 15^0 C=288K Free Air Delivery = FAD = 0.3m^3 /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 m^3 /min (1.1) For Insentropic Process: Pi = (r/r-1) P 1 V {(P 2 /P 1 ) r-1/r^ -1} (1) Putting values, we get: Pi = 1.31 KW Now T 2 =? As (T 2 /T 1 ) = (P 2 /P 1 ) r-1/r Putting the values, we have: T 2 = 227.3 C (1.2) For Isothermal Process Pi = P 1 V 1 ln(P 2 /P 1 ) = 0.98 KW T 2 = T 1 = 15C (1.3) For Polytropic Process, Pi = (n/n-1) P 1 V 1 [(P 2 /P 1 ) n-1/n^ –1] = 1.2 KW (2) Delivery Temperature. T 2 = T 1 (P 2 /P 1 ) 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 # 1 Speed = N = 1000 rpm = 1000 / 60 rps. Stroke to Bore Ratio = l/d = 1. REQUIRED: Bore Size = d =? SOLUTION: As Cylinder Volume = Vol. Induced /cycle. (Πd^2 / 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 m^3 per second. So, V/cycle = Vo/ N = 0.3 /1000 = 0.0003 m^3 So Πd^2 L / 4 = 0. L /d = 1.2 => L = 1.2 d Πd^2 x1.2d / 4 = 0.0003 =>1.2 d^3 = 4 x 0.0003 / π So, Bore Size = d = 0.0683m = 68.3 mm

REQUIRED:

(i). FAD =? (ii) Pi =? (iii) ηv =? SOLUTION: (i) Mass delivered or mass of FAD = Mass of volume induced mo^ (FAD) = moVo^ = (P x FAD) / R.T. = P 1 V’^ / R T 1 FAD = 0.227 m^3 /min (ii) Pi = (n/n-1)m’RT 1 {(P 2 /P 1 )n-1/n^ –1} m’^ = P 1 V’^ / R T 1 = 4.63 x 10-3kg / sec. So Pi = 1.985 KW (iii) ηv = (FAD/cycle) / Vs ---------------------(A) FAD/cycle = (FAD/unit time) / (cycles/unit time) FAD/cycle = 0.227 x 10-3^ m^3 So equ: (A) ⇒ η 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. m^3 /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 = P 2 = 70 bar Induction Pressure = P 1 = 1 bar FAD = 2.4 m^3 /min at P = 1.013 bar T = 288 K Two Stage Compression (i) Ideal Intermediate Pressure, Pi/P 1 = P 2 /Pi (ii) Complete Intercooling ⇒ T 1 = T (iii) Clearance Volume is 3% of Swept Volume i.e. VcL^ = 0.03 VsL^ , VcH^ = 0.03 VsH As Intercooling is Complete & Clearance Ratio is same, we can say that; VsL^ / VsH^ = Pi / P 1 = P 2 / Pi ⇒ Pi^2 = P 1 P 2 => P 2 / P 1 = √(P 2 /P 1 ) Speed = N = 750 rpm, n = 1. Temp. at the end of Induction Stroke in each cylinder. T 1 = Ti = 32C = 305K, ηmech. = 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)moRT 1 {(P 2 /P 1 )n-1/2n^ –1} -----(A) But mo^ =? Mass induced = Mass of FAD mo^ = moFAD = P x FAD / R T = 2.94 kg/min (A)⇒ Pi = 22.7 KW

(ii) Saving Power = {(indicated power of single stage)- (indicated power of double stage)} For single stage, Pi = (n/n-1)moRT 1 {{P 2 /P 1 )n-1/n^ –1} = 28.7 KW Thus Saving Power = 28.7 – 22.7 = 6 KW. (iii) Swept Volume VsL^ & VsH^ for LP Stage. Va – Vd / cycle = (Va-Vd/unit time)/cycles /unit time = VoLp / N --------------------------------------------- (1) VoLp = m’RT 1 /P 1 = 2.57 m^3 /min (1)⇒Va – Vd / cycle = V’Lp / N Va – Vd = 3.43 x 10-3^ m^3 /cycles Va = VsL + Vc = VsL + 0.03 VsL = 1.03 VsL Vd =Vc(Pi/P 1 )1/n^ = 01641 VsL VsL = 0.00396 m^3 For HP Stage, VsH^ = Va/^ - Vc/ VsH^ = Va/^ - 0.03 VsH Va/^ = 1.03 VsH^ =>VsH^ = Va/^ / 1.03---------------------(2) As point a & a/^ touches the Isothermal line due to complete Intercooling, thus; Pi Va/^ = P 1 Va Va/^ = Va P 1 /Pi = Va / (Pi/P 1 ) ----------------------------(3) Va = 1.03 Vs = 1.03 x 0.003962 = 4.08 x 10-3m^3 /cycle Equ:(3)⇒ Va/^ = 0.0004877 m^3 / cycle Now VsH^ = Va/^ /1.03 = 0.00473 m^3 So Swept Volume of HP stage = 0.000473 m^3 (iv) Power output of motor = Shaft Power Shaft Power = Indicated Power / ηmech.= 22.7 / 6. Power 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 cm^3 ; 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 Compressor d = 200 mm = 0.2 m, L = 250 mm = 0.25 m N = 300 rpm, n = 1. Vc = 700 cm^3 = 700 x 10-6^ m^3 P 2 = 5 bar, P 1 = 1 bar, P = 1.013 bar T 1 = 32C = 305K, T = 288K REQUIRED: (i) FAD =? (ii) Power required, P =? When ηmech. = 80 % SOLUTION: As Vs = (π/4) d^2 L ⇒Vs = 7.85 x 10-3^ m^3 , Vc = 7 x 10-4^ m^3 Va = Vs + Vc = 7. Vo^ = FAD = (Va-Vd)x(T/T 1 )x(P 1 /P) Vd = Vc(P 2 /P 1 )1/n^ = 2.54 x 10-3^ m^3

Two Stages Air-Compressor L = d P 2 = 7 bar FAD Conditions = V’ 1 = 4.2 m^3 /min P 1 = 1.013 bar T 1 = 15C = 288K Intercooling T 2 = 311K, n = 1.3, clearance = 0 REQUIRED: (a) Intermediate pressure = Pi =? (b) Power reqd. to drive the Motor =? (c) Isothermal Efficiency = ηiso. =? DIAGRAM: SOLUTION: We know that PVo^ = mo^ RT mo^ = PVo^ / RT = 0.086 Kg / sec With Intercooler Pi = √ P 1 P 2 = √ 2.66 bar Now we know that Pi / P 1 = P 2 / Pi, are same for both stages. So I.P. = (n/n-1)moRT{(P 2 /P 1 )n-1/n^ – 1} = 15.3 KW ηiso. = Isothermal Work / Indicated Work ------------- (1) Isothermal work = moRT 1 ln(P 2 /P 1 ) = 137 KJ Indicated Work = (n/n-1)moRT 1 {(P 2 /P 1 )n-1/n^ –1} = 15.3 KJ η iso. = 89.8 %

PROBLEM 12.12:

Air at 1.013 bar and 15 C is to be compressed at the rate of 5.6 m^3 /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: P 1 = 1.013 bar P 2 = 1.75 bar T 1 = 15C = 288K Vo^ = 5.6 m^3 /min = 5.6/60 m^3 /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 = (P 2 – P 1 )xVo^ = 6.88 KW (2) For Vane Type Power = (r/r-1) P1Vo1 {(Pi/P 1 ) r-1/r^ –1}+(P 2 – P 1 )xVo 2 Here Pi = 0.75 (P 2 – P 1 ) + P 1 = 1.566 bar Vo 1 = 5.6 m^3 /min Voi = Vo 1 (Pi/P 1 )1/r^ = 4.1 bar Now Power = 5.64 KW

Positive Displacement Machines Designed by Sir Engr. Masood Khan