Mathcad - Example 2_2 Finite Heat Release
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Heat Release Example
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Heat Release Example
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Example 2.2 Finite Heat Release: A single cylinder Otto cycle engine is operated at full throttle. The engine has a bore of 0.1m, stroke of 0.1m and connecting rod length of 0.15m, with a compression ratio of 10. The initial cylinder pressure, P 1, and temperature T1 , at bottom dead center are 1 bar and 300K. The engine speed is 3000rpm. The heat addition, Qin, is 1800J. Compare the effect of start of heat release at θs=-20 and θs=0o atdc on (a) the net work, the thermal efficiency of the cycle, and the mean effective pressure, and (b) the pressure, temperature and work profiles versus crank angle. Assume that the ideal gas specific heat ratio is 1.4, the gas molecular weight is 29, the combustion duration is constant at 40 o, and that the Weibe parameters are a=5 and n=3.
Given: Nc := 1
L := .15m
T1 := 300K
θs1 := −20deg
γ := 1.4
kJ := 1000J
B := .1m
rc := 10
N := 3000rpm
θs2 := 0deg
a := 5
kmol := 1000mol
S := .1m
P1 := 1bar
Qin := 1800J
θd := 40deg
n := 3
R ideal := .287
kJ kg⋅ K
Asked : (a) the net work, the thermal efficiency of the cycle, and the mean effective pressure, and (b) the pressure, temperature and work profiles versus crank angle. Assume that the ideal gas specific heat ratio is 1.4, the gas molecular weight is 29, the combustion duration is constant at 40o , and that the Weibe parameters are a=5 and n=3.
Solution: a)
Solved for using the "Simple Heat Release Applet" http://www.engr.colostate.edu/~allan/thermo/page6/EngineParm1/engine.html
Heat Release Example
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Numerical Solution: ηt := 1 −
1 r c
γ− 1
= 0.602
R := 2
L S
m Up := 2 ⋅ S⋅ N = 62.832 s π 2 Vd := ⋅ B ⋅ S = 0.785 L 4 Volume as a function of Crank Angle
V( θ) :=
Vd rc − 1
8× 10
6× 10
V( θ) L
4× 10
2× 10
+
Vd 2
⋅ R + 1 − cos( θ) − R − 1 − 2
cos( 2 ⋅ θ) 2
−3
−3
−3
−3
0
− 100
0
100
θ deg
Q as a function of Crank Angle
θ− θ n s1 − a⋅ θ d xb ( θ) := 1 − e Q( θ) :=
Qin⋅ xb ( θ) if θs1 ≤ θ ≤ θs1 + θd 0 otherwise
θ− θ n s2 − a⋅ θ d xb2( θ) := 1 − e Q2 ( θ) :=
Qin⋅ xb2( θ) if θs2 ≤ θ ≤ θs2 + θd 0 otherwise
Heat Release Example
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2× 10
3
2
1.5× 10
3
1.5
1× 10
3
xb( θ) 1
500
0.5
Q( θ) J
0
0
− 100
0
− 100
100
0
θ
θ
deg
deg
100
Boundary Conditions: 5
θ0 := −180 deg θ :=
P0 := P1 = 1 × 10 Pa
θf := 180deg
n := 128
for i ∈ 0 .. n M ← θ0 + i ⋅ i
(θf − θ0) n
M
f ( θ , P) := −γ⋅
P :=
h←
γ − 1 d d V( θ) + ⋅ Q( θ) V( θ) dθ V( θ) dθ P
P d γ − 1 d f2 ( θ , P) := −γ⋅ ⋅ V( θ) + ⋅ Q2 ( θ) V( θ) dθ V( θ) dθ
⋅
(θf − θ0)
P2 :=
n
(θf − θ0)
h←
n
M ← P0 0
N ← P0 0
for i ∈ 0 .. n − 1
for i ∈ 0 .. n − 1
( ) k2 ← f ( θ0 + i ⋅ h + 0.5⋅ h , M + 0.5⋅ k1⋅ h ) i k3 ← f ( θ0 + i ⋅ h + 0.5⋅ h , M + 0.5⋅ k2⋅ h ) i k4 ← f ( θ0 + i ⋅ h + h , M + k3⋅ h ) i
( ) k2 ← f2 ( θ0 + i ⋅ h + 0.5⋅ h , N + 0.5⋅ k1⋅ h ) i k3 ← f2 ( θ0 + i ⋅ h + 0.5⋅ h , N + 0.5⋅ k2⋅ h ) i k4 ← f2 ( θ0 + i ⋅ h + h , N + k3⋅ h ) i
k1 ← f θ0 + i ⋅ h , M i
M M
i+ 1
←M + i
1 6
k1 ← f2 θ0 + i ⋅ h , N i
( k1 + 2 ⋅ k2 + 2 ⋅ k3 + k4) ⋅ h
N
i+ 1
N
←N + i
1 6
( k1 + 2 ⋅ k2 + 2 ⋅ k3 + k4) ⋅ h
Heat Release Example
P kPa
8× 10
3
6× 10
3
4× 10
3
2× 10
3
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Engine 1
0 − 200
− 100
0
100
200
θ deg
8× 10
3
6× 10
3
Engine 1 Engine 2
P kPa P2
4× 10
3
2× 10
3
kPa
0 − 200
− 100
0
100
200
θ deg
The solution of just the Pressure vs. Crank Angle is shown. The question ask for other parameters which can be solved for given the necessary equations.
Heat Release Example
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Heat Release Example
h
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