5. For an ideal Diesel cycle with the over-all value of k=1.33, rk=15, rc=2.1, P 1 = 14.2031 psia, find P 2 and
mean effective pressure.
P1=97.92kPaP_1 =97.92 kPaP1=97.92kPa
rp=2.1r_p = 2.1rp=2.1
P2=P1×(rk)γ−1=P1×(rk)γ−1P_2 = P_1\times (r_k)^{\gamma-1} = P_1\times (r_k)^{\gamma-1}P2=P1×(rk)γ−1=P1×(rk)γ−1
P2=97.92×(15)1.33−1=239.32kPaP_2 = 97.92 \times (15)^{1.33-1}=239.32kPaP2=97.92×(15)1.33−1=239.32kPa
Pm=P1rk(rp−1)(rk(γ−1)−1)(γ−1)(rk−1)P_m = \frac{P_1r_k(r_p-1)(r_k^{(\gamma-1) -1})}{(\gamma-1)(r_k-1)}Pm=(γ−1)(rk−1)P1rk(rp−1)(rk(γ−1)−1)
Pm=97.92×15(2.1−1)(15(1.33−1)−1)(1.33−1)(15−1)=505kPaP_m = \frac{97.92 \times 15 (2.1-1)(15^{(1.33-1) }-1)}{(1.33-1)(15-1)} = 505 kPaPm=(1.33−1)(15−1)97.92×15(2.1−1)(15(1.33−1)−1)=505kPa
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