Question 12.CSGP.129: Find the temperature after combustion and the specific energ......

Find the temperature after combustion and the specific energy release by combustion in Problem 12.92 using cold air properties. This is a difficult problem and it requires iterations.

The worlds largest diesel engine has displacement of 25 m³ running at 200 RPM in a two stroke cycle producing 100 000 hp. Assume an inlet state of 200 kPa, 300 K and a compression ratio of 20:1. What is the mean effective pressure?

Step-by-Step
The 'Blue Check Mark' means that this solution was answered by an expert.
Learn more on how do we answer questions.

We have 3 parameters for the cycle: T _1, P _1 and CR we need one more, so this comes from the total rate of work Eq.12.11.

\begin{aligned}& \dot{ W }= mP _{\text {meff }}\left( v _1- v _2\right) \times N _{ cy1 } \times RPM / 60= P _{\text {meff }} V _{\text {displ }} RPM / 60 \\& P _{\text {meff }}=\dot{ W } 60 / V _{\text {displ }} RPM =\frac{100000 \times 0.746 \,kW \times 60 s }{25\, m ^3 \times 200}=895 \,kPa\end{aligned}

\begin{gathered}v _1= RT / P =0.287 \times 300 / 200=0.4305 \,m ^3 / kg ; \\w _{\text {net }}= P _{\text {meff }}\left( v _1- v _2\right)=895 \times 0.4305\left(1-\frac{1}{20}\right)=366.03 \,kJ / kg \\T _2= T _1\left( v _1 / v _2\right)^{ k -1}=300(20)^{0.4}=994.3 \,K\end{gathered}

Combustion and expansion volumes

v _3= v _2 \times T _3 / T _2= v _1 T _3 /\left( T _2 \times CR \right) ; \quad v _4= v _1

Expansion process, isentropic from Eq.8.32

T _4= T _3\left( v _3 / v _4\right)^{ k -1}= T _3\left[ T _3 /\left( CR \times T _2\right)\right]^{ k -1}= T _3^{ k }\left( CR \times T _2\right)^{1- k }=0.01908 T _3^{ k }

The net work is also given by the heat transfers

w _{\text {net }}= q _{ H }- q _{ L }= C _{ p }\left( T _3- T _2\right)- C _{ v }\left( T _4- T _1\right)

Substitute w _{\text {net }} \text { and } T _4 into this equation and we get

366.03=1.004\left( T _3-994.3\right)-0.717\left(0.01908 T _3^{ k }-300\right)

divide by 1.004 and combine terms to get

1144.63= T _3-0.013626 T _3^{ k }

Trial and error, guess T _3 \Rightarrow RHS _{1400}=1054.1 ; RHS _{1600}=1183

Interpolate:

\begin{aligned}& T _3= 1 5 4 0 . 5 \,K \\& q _{ H }= C _{ P }\left( T _3- T _2\right)=1.004(1540.5-994.3)= 5 4 8 \,k J / k g\end{aligned}

Related Answered Questions