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

Question

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

Answer 1

Solution;

Given;

$r=13.5$

k=1.35

$T_1=140+460=600°R$

$V_1=1cuft$

$C_v=\frac{R}{k-1}=\frac{53.34}{778(1.34)}=0.2016Btu/lb°R$

$C_p=kC_v=1.34×0.2016=0.3602Btu/lb°R$

$P_2=P_1(r)^{k-1}=14(13.5)^{0.34}=33.833psia$

$P_2=P_3$

$V_2=\frac{V_1}{13.5}=\frac{1}{13.5}=0.074ft^3$

T$T_2=T_1r^{k-1}=600×13.5^{0.34}=1454°R$

$V_3=V_2+0.06V_D$

$V_3=0.074+0.06(1-0.074)=0.1298ft^3$

$T_3=T_2(\frac{V_3}{V_2})^{k-1}$

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

$T_3=2445°R$

$T_4=T_3(\frac{V_3}{V_2})^{k-1}=2545(\frac{0.1297}{1})^{0.34}$

$T_4=1271°R$

$P_4=P_3(\frac{V_3}{V_4})^k$

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

$P_4=29.7psia$

$Q_r=mC_v(T_1-T_4)$

$Q_r=0.063×0.2016(600-1271)$

$Q_r=8.52Btu$

$W_{net}=Q_a-Q_r$

$Q_a=mC_p(T_3-T_2)$

$Q_a=0.063×0.2702(2545-1454)=18.57Btu$

$W_{net}=18.57-8.52=10.05Btu$

$\eta=\frac{W}{Q_a}=\frac{10.05}{18.57}=0.5412$$P_m=\frac{W}{V_1-V_2}=\frac{10.05×778}{(1-0.0741)×144}$

$P_m=58.64psi$

$W_r=\frac{10.05×1000}{42.4}=237hp$