Answer to Question #309153 in Molecular Physics | Thermodynamics for Tobias Felix

Question #309153

A Diesel cycle operates with a compression ratio of 13.5 and with a cutoff

occurring at 6% of the stroke. State 1 is defined by 14 psia and 140 F. For the hot air

standard with k=1.34 and for an initial volume of 1 cu.ft. compute a) Qr, BTU, b)

Wnet, BTU, c) thermal efficiency, %, d) mean effective pressure, psi. e) For a rate of

circulation of 1000 cubic feet per minute, compute the horsepower


1
Expert's answer
2022-03-10T18:01:09-0500

Solution;

Given;

r=13.5r=13.5

k=1.35

T1=140+460=600°RT_1=140+460=600°R

V1=1cuftV_1=1cuft

Cv=Rk1=53.34778(1.34)=0.2016Btu/lb°RC_v=\frac{R}{k-1}=\frac{53.34}{778(1.34)}=0.2016Btu/lb°R

Cp=kCv=1.34×0.2016=0.3602Btu/lb°RC_p=kC_v=1.34×0.2016=0.3602Btu/lb°R

P2=P1(r)k1=14(13.5)0.34=33.833psiaP_2=P_1(r)^{k-1}=14(13.5)^{0.34}=33.833psia

P2=P3P_2=P_3

V2=V113.5=113.5=0.074ft3V_2=\frac{V_1}{13.5}=\frac{1}{13.5}=0.074ft^3

TT2=T1rk1=600×13.50.34=1454°RT_2=T_1r^{k-1}=600×13.5^{0.34}=1454°R

V3=V2+0.06VDV_3=V_2+0.06V_D

V3=0.074+0.06(10.074)=0.1298ft3V_3=0.074+0.06(1-0.074)=0.1298ft^3

T3=T2(V3V2)k1T_3=T_2(\frac{V_3}{V_2})^{k-1}

T3=1454(0.12970.0741)0.34T_3=1454(\frac{0.1297}{0.0741})^{0.34}

T3=2445°RT_3=2445°R

T4=T3(V3V2)k1=2545(0.12971)0.34T_4=T_3(\frac{V_3}{V_2})^{k-1}=2545(\frac{0.1297}{1})^{0.34}

T4=1271°RT_4=1271°R

P4=P3(V3V4)kP_4=P_3(\frac{V_3}{V_4})^k

P4=457.9(0.12971)1.34P_4=457.9(\frac{0.1297}{1})^{1.34}

P4=29.7psiaP_4=29.7psia

Qr=mCv(T1T4)Q_r=mC_v(T_1-T_4)

Qr=0.063×0.2016(6001271)Q_r=0.063×0.2016(600-1271)

Qr=8.52BtuQ_r=8.52Btu

Wnet=QaQrW_{net}=Q_a-Q_r

Qa=mCp(T3T2)Q_a=mC_p(T_3-T_2)

Qa=0.063×0.2702(25451454)=18.57BtuQ_a=0.063×0.2702(2545-1454)=18.57Btu

Wnet=18.578.52=10.05BtuW_{net}=18.57-8.52=10.05Btu

η=WQa=10.0518.57=0.5412\eta=\frac{W}{Q_a}=\frac{10.05}{18.57}=0.5412Pm=WV1V2=10.05×778(10.0741)×144P_m=\frac{W}{V_1-V_2}=\frac{10.05×778}{(1-0.0741)×144}

Pm=58.64psiP_m=58.64psi

Wr=10.05×100042.4=237hpW_r=\frac{10.05×1000}{42.4}=237hp


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