Following data relate to a performance test of a single acting 14 cm 10 cm reciprocating compressor are given: suction pressure = 1 bar, suction temperature = 0 C, discharge pressure = 6 bar, discharge temperature = 180 C, speed of compressor = 100 rpm, shaft power = 6.5 kw, mass of air delivered = 1.7 kg/min. Calculate the following : (i) actual volumetric efficiency (ii) indicated power (iii) isothermal efficiency (iv) mechanical efficiency (v) overall efficiency (vi) motor power if motor transmission is 90% R= 0.87 kj/kgk 87 J/kgK L= 10 cm 0.1 m D= 14 cm 0.14 m suction pressure p1 = 1 bar 100000 Pa suction temperature T1= 0 C 93 K discharge pressure p= 6 bar discharge temperature T= 180 C 453 K speed of compressor N = 100 rpm shaft power P shaft = 6.5 kw or power input to compressor mass of air delivered m a = 1.7 kg/min 0.08333 kg/s Transmission eff ƞ trans = 90 % 0.9 Actual volumetric efficiency, ƞvol Displacement volume Vd=π/4*D L*N Vd= 1.847 m3/min p FAD=mRT1/p1 1.430 m3/min ƞvol=fad/vd 0.7740 77.4 % Indicated Power, IP T/T1=(p/P1)^(n-1/n) or (n-1/n)=ln(t/t1)/ln(p/p1) n= 1.3 IP=mRT1*(n/n-1)*((p/p1)^((n-1)/n)-1) IP= 5.3507 kj/s or kw 3 4 pv n = c v 1 Isothermal Efficiency, ƞ iso Isothermal Power, Piso = mrt1ln(p/p1) P iso = 4.69007 kj/s or kw ƞiso=p iso /IP 0.797911 79.8 % Mechanical Efficiency, ƞ mech ƞmech=ip/p shaft 0.856036 85.6 % Overall Thermal Efficiency, ƞ o ƞo=p iso /P shaft 0.683041 68.3 % Motor Power Motor power= P shaft /ƞ trans 6.94 kw specific heat cv= 0.718 kj/kgk Work done and heat transfer during compression per kg of air
work done, W=mRT*(n/n-1)*((p/p1)^((n-1)/n)-1) W= 8.7 kj/kg of air Heat transferred during compression, Q=W+ΔU = (p1v1-pv)/(n-1)+cv(t-t1) Q= (p1v1-pv)/(n-1)+cv(t-t1)=r(t1-t)/(n-1)+cv(t-t1)=(t-t1)*(cv-r/(n-1)) Q= -8.0316 kj/kg heat rejection
An air compressor takes in air at 1 bar and 0 C and compress it according to law pv 1. = constant. It is then delivered to a receiver at a constant pressure of 10 bar. R= 0.87 kj/kgk, cv=0.718kj/kgk. Determine: (i) Temperature at the end of compression (ii) work done and heat transferred during compression per kg of air T1= 0 C 93 K p1= 1 bar p= 10 bar n= (in pv n ) 1. R= 0.87 kj/kgk m= 1 kg (to find out per kg of air) specific heat cv= 0.718 kj/kgk Temperature at the end of compression, T compression process 1- T/T1=(p/P1)^((n-1)/n) p T/T1= 1.468 T= 430.1 K 157.1 C Work done and heat transfer during compression per kg of air work done, W=mRT*(n/n-1)*((p/p1)^((n-1)/n)-1) W= 36.03 kj/kg of air Heat transferred during compression, Q=W+ΔU = (p1v1-pv)/(n-1)+cv(t-t1) Q= (p1v1-pv)/(n-1)+cv(t-t1)=r(t1-t)/(n-1)+cv(t-t1)=(t-t1)*(cv-r/(n-1)) Q= -98.757 kj/kg heat rejection 3 pv 1. = c v 1 (p1v1-pv)*n/(n-1)= -36.06 * (p1v1-pv)/(n-1)= -196.689 * cv(t-t1)= 98.418 Q= (p1v1-pv)/(n-1)+cv(t-t1)= -98.757 kj/kg cp= 1.005 mcp(t-t1)= 137.7505
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A single stroke, double acting compressor has a free air delivery (FAD) of 14 m3/min measured at 1.013 bar at 15 C. The pressure and temperature in the cylinder during induction are 0.95 bar at 3 C. The delivery pressure is 7 bar and the speed of compressor is 300 rpm. Take the clearance volume as 5 % of the swept volume with compression and expansion index of n = 1.3. Calculate (i) swept volume of the cylinder (ii) volumetric efficiency (iii) the delivery temperature (iv) indicated power p= 1.013 bar 101300 Pa T= 15 C 88 K T1= 3 C 305 K p1= 0.95 bar p= 7 bar 3 n= (in pv n ) 1.3 N= 300 rpm pv 1.3 = constant Free air to be delivered, FAD= 14 m3/min p clerance volume Vc= 0.05 *Vs specific heat cv= 0.718 kj/kgk R= 0.87 kj/kgk 87 J/kgK swept volume of the cylinder mass delivered/min m=pv/rt 17.157859 kg/min V= FAD per minute at p, T Swept volume Vs = V1-V3 = V1-Vc V1= 1.05 *Vs Volume induced per cycle = V1-V4 m=pv/rt=p1(v1-v4)/rt1 or V1-V4 = (FAD/(*N))*(T1/T)*(p/p1), V1-V4= 0.063494 m3 V4/V3=(p/p1)^(1/n) V4/V3= 4.65 V4= 4.65 *V3 0.3371 *Vs V1-V4= 0.8 *Vs Vs= 0.03 m3 Volumetric efficiency, ƞ vol ƞ vol =V/Vs, V=FAD/cycle 0.740413 m=pv/rt=p1(v1-v4)/rt1 V=(V1-V4)*(T/T1)*(p1/p) V= 0.74 *Vs ƞ vol =V/Vs, V=FAD/cycle 0.74 7.40413 % ƞ vol =1+k-k(P/p1)^(1/n) 0.818 Temperature at the end of compression or delivery temparature, T compression process 1- T/T1=(p/P1)^((n-1)/n) T/T1= 1.585 T= 483.6 K 10.6 C Indicated power required IP=(n/n-1)*p1(V1-V4)*((p/p1)^((n-1)/n)-1) or (n/n-1)*mr(t-t1) IP= 3810.53 kj/min 63.50891 kj/s or kw vc vs 4 v 1
A single acting, two stage reciprocating air compressor having 4.5 kg of air per min are comprerssed from 1.013 bar and 15 C through a pressure ratio of 9 to 1. Both stages have the same pressure ratio, and the law of compression and expansion in both stages is pv 1.3 = constant. If the intercooling is complete, minimum work to be given to compressor, calculate (i) indicated power (ii) swept volume required. Assume that the clerance ratio is 5% and that the compressor runs at 300 rpm. Amount of air compressed 4.5 kg/min Ps = 1.013 bar 101300 Pa Ts = 15 C 88 K pd/ps= 9 n = 1.3 Vc = 0.5 Vs R= 0.87 kj/kgk 87 J/kgK N= 300 rpm clerance volume/swept volume (k)= 0.05 * clearance ratio Indicated Power, IP pi/ps=pd/pi pi =(ps*pd)= 9 *ps pi= 3 *ps = 3.039 bar 303900 Pa pi/ps= 3 Ti/Ts=(pi/ps)^((n-1)/n) Ti= 371.11 K Total work required per minute = *(n/(n-1))*mr(ti-ts) p 930.0 kj/min pd 7 6 IP 15.50 kw HP The cylinder swept volumes required mass induced per cycle m = Amount of air compressed/n 0.015 kg/cycle LP cylinder FAD for LP cylinder V1-V4=mRTs/ps 0.01 m 3 /cycle ɳ vol =(V1-V4)/Vs = 1+k-k(pi/ps)^(1/n) 0.934 Swept volume of LP cylinder Vs= 0.0131 m3/cycle pi 3 8 LP 5 HP cylinder Volume drawn in =mass induced per cycle m*r*ts/pi 0.00408 m3/cycle ɳ vol =(V5-V8)/Vs = 1+k-k(pd/pi)^(1/n) 0.934 Swept volume of HP cylinder Vs= 0.00437 m3 ps 4 v 1 It is also noted that clearance ratio (k) is same in each cylinder, and the suction temperature are same since intercooling is complete, therefore ratio of LP and HP swept volume is equal to the the ratio of the suction and deliver pressure of that stage. Vs (HP) = 0.00437
A single acting, two stage reciprocating air compressor with complete intercooling delivers 10.5 kg of air per min at 16 bar. the suction occurs at 1 bar and 7 C. The compression and expansion process are reversible, polytropic index n=1.3. Considering minimum work to be done on compressor, Calculate (i) power required to drive the compressor (ii) The isothermal efficiency (iii) the FAD (iv) the heat transferred in intercooler. Consider the compressor runs at 440 rpm. If the clearance ratio for L.P. and H.P. cylinders are 0.04 and 0.06 respectively, calculate swept and clearance volumes for each cylinder. Amount of air delivered 10.5 kg/min 0.175 kg/s P1 = 1 bar 100000 Pa P = 4 bar 400000 Pa P3 = 16 bar 1600000 Pa T1 = 7 C 300 K n = 1.3 R= 0.87 kj/kgk 87 J/kgK N= 440 rpm clerance ratio for L.P. (k)= 0.04 clerance ratio for H.P. (k)= 0.06 cp= 1.005 pressure ratio = (p1p3)= 4 Power required (for perfect intercooling) = n/(n-1)*mrt1*((p3/p1)^((n-1)/n)-1) 49.3 kw p Isothermal Work = mrt1*ln((p3/p1)) pd Piso= 41.77598 kw ƞiso= 0.849 7 HP 6 FAD: V=mRT1/p1 9.041 m3/min pi 3 8 5 Heat transfer in intercooler: T=T1*(p/p1)^((n-1)/n) 413.108 K Heat transferred in intercooler, Q= mc p (T-T5)=mc p (T-T1)= 19.89196 kw The cylinder swept volumes and clearance volume required ps LP 4 1
volumetric eficiency for L.P. Stage, ƞ LP = 1+k-k(p/p1)^(1/n), =(FAD/cycle)/Vs (LP) 0.93806 volumetric eficiency for H.P. Stage, ƞ HP = 1+k-k(p3/p)^(1/n), =(FAD/cycle)/Vs (HP) 0.885709 Swept Volume V (LP) =V1-V3=(FAD per cycle)/(ƞ LP )=(FAD /speed)/(ƞ LP ) 0.041 m3 clearance volume=k LP *V( LP ) 0.00089 m3 v Swept Volume V (HP) =V5-V7=(FAD per cycle)/(stage pr. Ratio*ƞ HP )=(FAD /speed)/(stage pr. Ratio*ƞ HP ) 0.005799 m3 clearance volume=k HP *V( HP ) 0.000348 m3
A centrifugal compressor used as a supercharger for aero-engines handles 150 kg/min of air. The suction pressure and temperature are 1 bar and 90 K. The suction velocity is 80 m/s. After compression in the impeller the conditions are 1.5 bar 345 K and 0 m/s. Calculate (i) Isentropic efficiency (ii) Power required to drive the compressor (iii) The overall efficiency of the unit. It may be assumed that KE of air gained in the impeller is entirely converted into pressure in the diffuser. ṁ= 150 kg/min.5 kg/s p1= 1 bar T1= 90 K V1= 80 m/s p= 1.5 bar T= 345 K T p V= 0 m/s T γ= 1.4 03 T 3 03' cp= 1.005 kj/kgk 3' p 1 T 0 T 0' ' p 3 T 01 1 (i) isentropic efficiency s T'/T1=(p/p1)^(γ-1/γ) T'= 35.619 K Isentropic work done =cp(t'-t1)+(v^-v1^)/(*1000) Isentropic work done = 56.80 kj/kg Work done in impeller=cp(t-t1)+(v^-v1^)/(*1000) Work done in impeller= 76.8 kj/kg ƞ isent = 0.745 (ii) Power required to drive the compressor P=ṁ Work done in impeller(kj/kg) P= 190.69 kw (iii) Overall efficiency ƞo As KE gained in the impeller is converted into the pressure hence, cp(t3-t)=(v^- V1^)/(*1000) T3= 365.90 K Pressure of air after leaving the diffuser, p3 T3/T=(p3/p)^(γ-1/γ) p3/p=(t3/t)^(γ/γ-1) p3= 1.843 bar After isentropic compression, the delivery temperature from diffuser, T3' T3'/T1=(p3/p1)^(γ-1/γ) T3'= 345.34 K ƞo=(t3'-t1)/(t3-t1) 0.79