Differential equation for pressure and heat release in combustion engine
$begingroup$
I have a differential equation on the following form, and I am interested in finding $p(theta)$
$frac{dp}{d theta}=frac{gamma-1}{V(theta)}frac{dQ_{HR}}{d theta} - gamma frac{p}{V(theta)} frac{dV}{d theta}$
and I know the following,
$frac{dQ_{HR}}{d theta} = frac{dm_{burnt}}{d theta}H_u$
$frac{dm_{burnt}}{d theta} = k_1 sinleft( pi frac{theta-theta_0}{Delta theta_c}right)$
and,
$frac{dV}{d theta} = k_2left(sin(theta)+k_3sin(2theta) right)$
I have tried integration by parts, but this only seems to dig my hole deeper. What are your suggestions to tackling a problem like this?
differential
$endgroup$
add a comment |
$begingroup$
I have a differential equation on the following form, and I am interested in finding $p(theta)$
$frac{dp}{d theta}=frac{gamma-1}{V(theta)}frac{dQ_{HR}}{d theta} - gamma frac{p}{V(theta)} frac{dV}{d theta}$
and I know the following,
$frac{dQ_{HR}}{d theta} = frac{dm_{burnt}}{d theta}H_u$
$frac{dm_{burnt}}{d theta} = k_1 sinleft( pi frac{theta-theta_0}{Delta theta_c}right)$
and,
$frac{dV}{d theta} = k_2left(sin(theta)+k_3sin(2theta) right)$
I have tried integration by parts, but this only seems to dig my hole deeper. What are your suggestions to tackling a problem like this?
differential
$endgroup$
add a comment |
$begingroup$
I have a differential equation on the following form, and I am interested in finding $p(theta)$
$frac{dp}{d theta}=frac{gamma-1}{V(theta)}frac{dQ_{HR}}{d theta} - gamma frac{p}{V(theta)} frac{dV}{d theta}$
and I know the following,
$frac{dQ_{HR}}{d theta} = frac{dm_{burnt}}{d theta}H_u$
$frac{dm_{burnt}}{d theta} = k_1 sinleft( pi frac{theta-theta_0}{Delta theta_c}right)$
and,
$frac{dV}{d theta} = k_2left(sin(theta)+k_3sin(2theta) right)$
I have tried integration by parts, but this only seems to dig my hole deeper. What are your suggestions to tackling a problem like this?
differential
$endgroup$
I have a differential equation on the following form, and I am interested in finding $p(theta)$
$frac{dp}{d theta}=frac{gamma-1}{V(theta)}frac{dQ_{HR}}{d theta} - gamma frac{p}{V(theta)} frac{dV}{d theta}$
and I know the following,
$frac{dQ_{HR}}{d theta} = frac{dm_{burnt}}{d theta}H_u$
$frac{dm_{burnt}}{d theta} = k_1 sinleft( pi frac{theta-theta_0}{Delta theta_c}right)$
and,
$frac{dV}{d theta} = k_2left(sin(theta)+k_3sin(2theta) right)$
I have tried integration by parts, but this only seems to dig my hole deeper. What are your suggestions to tackling a problem like this?
differential
differential
asked Dec 8 '18 at 16:23
Rasmus0909Rasmus0909
1
1
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2 Answers
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$begingroup$
The best I can get is the following:
Firstly, integrate the last equation gives
$$V(theta) = k_2 left(-cos theta - frac{k_3}{2} cos 2 theta + C_1 right)$$ where $C_1$ is some constant to be determined later.
Now substitute everything into the first equation, we get
$$frac{mathrm{d}p}{mathrm{d}theta} = frac{gamma - 1}{k_2 left(-cos theta - frac{k_3}{2} cos 2 theta + C_1 right)} cdot k_1 sin left(pi frac{theta - theta_0}{Delta theta_c} right) H_u - gamma cdot frac{p}{k_2 left(-cos theta - frac{k_3}{2} cos 2 theta + C_1 right)} cdot k_2 (sin theta + k_3 sin 2 theta)$$
Notice it is a $1^{text{st}}$ order ODE. Re-writing it into the standard form $y' + P(x) y = Q(x)$ gives
$$frac{mathrm{d}p}{mathrm{d}theta} + gamma frac{sin theta + k_3 sin 2 theta}{-cos theta - frac{k_3}{2} cos 2 theta + C_1} p = frac{k_1}{k_2}(gamma - 1)H_u frac{sin left(pi frac{theta - theta_0}{Delta theta_c} right)}{-cos theta - frac{k_3}{2} cos 2 theta + C_1}$$
Next, we compute
$$begin{align}
int gamma frac{sin theta + k_3 sin 2 theta}{-cos theta - frac{k_3}{2} cos 2 theta + C_1} mathrm{d}theta
&= gamma int frac{mathrm{d} (-cos theta - frac{k_3}{2} cos 2 theta + C_1)}{-cos theta - frac{k_3}{2} cos 2 theta + C_1} \
&= gamma log left|-cos theta - frac{k_3}{2} cos 2 theta + C_1 right| \
&= log left|-cos theta - frac{k_3}{2} cos 2 theta + C_1 right| ^ gamma
end{align}$$
So the integrating factor $mu(theta) = left|-cos theta - frac{k_3}{2} cos 2 theta + C_1 right| ^ gamma $
Finally, we need to compute
$$int frac{k_1}{k_2}(gamma - 1)H_u frac{sin left(pi frac{theta - theta_0}{Delta theta_c} right)}{-cos theta - frac{k_3}{2} cos 2 theta + C_1} cdot left|-cos theta - frac{k_3}{2} cos 2 theta + C_1 right| ^ gamma mathrm{d} theta \
= frac{k_1}{k_2}(gamma - 1)H_u int sin left(pi frac{theta - theta_0}{Delta theta_c} right) cdot operatorname{sgn} left(-cos theta - frac{k_3}{2} cos 2 theta + C_1 right) cdot left|-cos theta - frac{k_3}{2} cos 2 theta + C_1 right| ^ {gamma - 1} mathrm{d} theta$$
but this is how far I've got.
$endgroup$
$begingroup$
Thank you so much Alex. Looks like a huge effort! FYI, I used a forward Euler, which seemed to do the trick (although it's slightly cheating).
$endgroup$
– Rasmus0909
Dec 9 '18 at 15:20
add a comment |
$begingroup$
I ended up using a forward Euler numerical integration for this. I don't think the analytical solution is worth anybody's time for this.
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
The best I can get is the following:
Firstly, integrate the last equation gives
$$V(theta) = k_2 left(-cos theta - frac{k_3}{2} cos 2 theta + C_1 right)$$ where $C_1$ is some constant to be determined later.
Now substitute everything into the first equation, we get
$$frac{mathrm{d}p}{mathrm{d}theta} = frac{gamma - 1}{k_2 left(-cos theta - frac{k_3}{2} cos 2 theta + C_1 right)} cdot k_1 sin left(pi frac{theta - theta_0}{Delta theta_c} right) H_u - gamma cdot frac{p}{k_2 left(-cos theta - frac{k_3}{2} cos 2 theta + C_1 right)} cdot k_2 (sin theta + k_3 sin 2 theta)$$
Notice it is a $1^{text{st}}$ order ODE. Re-writing it into the standard form $y' + P(x) y = Q(x)$ gives
$$frac{mathrm{d}p}{mathrm{d}theta} + gamma frac{sin theta + k_3 sin 2 theta}{-cos theta - frac{k_3}{2} cos 2 theta + C_1} p = frac{k_1}{k_2}(gamma - 1)H_u frac{sin left(pi frac{theta - theta_0}{Delta theta_c} right)}{-cos theta - frac{k_3}{2} cos 2 theta + C_1}$$
Next, we compute
$$begin{align}
int gamma frac{sin theta + k_3 sin 2 theta}{-cos theta - frac{k_3}{2} cos 2 theta + C_1} mathrm{d}theta
&= gamma int frac{mathrm{d} (-cos theta - frac{k_3}{2} cos 2 theta + C_1)}{-cos theta - frac{k_3}{2} cos 2 theta + C_1} \
&= gamma log left|-cos theta - frac{k_3}{2} cos 2 theta + C_1 right| \
&= log left|-cos theta - frac{k_3}{2} cos 2 theta + C_1 right| ^ gamma
end{align}$$
So the integrating factor $mu(theta) = left|-cos theta - frac{k_3}{2} cos 2 theta + C_1 right| ^ gamma $
Finally, we need to compute
$$int frac{k_1}{k_2}(gamma - 1)H_u frac{sin left(pi frac{theta - theta_0}{Delta theta_c} right)}{-cos theta - frac{k_3}{2} cos 2 theta + C_1} cdot left|-cos theta - frac{k_3}{2} cos 2 theta + C_1 right| ^ gamma mathrm{d} theta \
= frac{k_1}{k_2}(gamma - 1)H_u int sin left(pi frac{theta - theta_0}{Delta theta_c} right) cdot operatorname{sgn} left(-cos theta - frac{k_3}{2} cos 2 theta + C_1 right) cdot left|-cos theta - frac{k_3}{2} cos 2 theta + C_1 right| ^ {gamma - 1} mathrm{d} theta$$
but this is how far I've got.
$endgroup$
$begingroup$
Thank you so much Alex. Looks like a huge effort! FYI, I used a forward Euler, which seemed to do the trick (although it's slightly cheating).
$endgroup$
– Rasmus0909
Dec 9 '18 at 15:20
add a comment |
$begingroup$
The best I can get is the following:
Firstly, integrate the last equation gives
$$V(theta) = k_2 left(-cos theta - frac{k_3}{2} cos 2 theta + C_1 right)$$ where $C_1$ is some constant to be determined later.
Now substitute everything into the first equation, we get
$$frac{mathrm{d}p}{mathrm{d}theta} = frac{gamma - 1}{k_2 left(-cos theta - frac{k_3}{2} cos 2 theta + C_1 right)} cdot k_1 sin left(pi frac{theta - theta_0}{Delta theta_c} right) H_u - gamma cdot frac{p}{k_2 left(-cos theta - frac{k_3}{2} cos 2 theta + C_1 right)} cdot k_2 (sin theta + k_3 sin 2 theta)$$
Notice it is a $1^{text{st}}$ order ODE. Re-writing it into the standard form $y' + P(x) y = Q(x)$ gives
$$frac{mathrm{d}p}{mathrm{d}theta} + gamma frac{sin theta + k_3 sin 2 theta}{-cos theta - frac{k_3}{2} cos 2 theta + C_1} p = frac{k_1}{k_2}(gamma - 1)H_u frac{sin left(pi frac{theta - theta_0}{Delta theta_c} right)}{-cos theta - frac{k_3}{2} cos 2 theta + C_1}$$
Next, we compute
$$begin{align}
int gamma frac{sin theta + k_3 sin 2 theta}{-cos theta - frac{k_3}{2} cos 2 theta + C_1} mathrm{d}theta
&= gamma int frac{mathrm{d} (-cos theta - frac{k_3}{2} cos 2 theta + C_1)}{-cos theta - frac{k_3}{2} cos 2 theta + C_1} \
&= gamma log left|-cos theta - frac{k_3}{2} cos 2 theta + C_1 right| \
&= log left|-cos theta - frac{k_3}{2} cos 2 theta + C_1 right| ^ gamma
end{align}$$
So the integrating factor $mu(theta) = left|-cos theta - frac{k_3}{2} cos 2 theta + C_1 right| ^ gamma $
Finally, we need to compute
$$int frac{k_1}{k_2}(gamma - 1)H_u frac{sin left(pi frac{theta - theta_0}{Delta theta_c} right)}{-cos theta - frac{k_3}{2} cos 2 theta + C_1} cdot left|-cos theta - frac{k_3}{2} cos 2 theta + C_1 right| ^ gamma mathrm{d} theta \
= frac{k_1}{k_2}(gamma - 1)H_u int sin left(pi frac{theta - theta_0}{Delta theta_c} right) cdot operatorname{sgn} left(-cos theta - frac{k_3}{2} cos 2 theta + C_1 right) cdot left|-cos theta - frac{k_3}{2} cos 2 theta + C_1 right| ^ {gamma - 1} mathrm{d} theta$$
but this is how far I've got.
$endgroup$
$begingroup$
Thank you so much Alex. Looks like a huge effort! FYI, I used a forward Euler, which seemed to do the trick (although it's slightly cheating).
$endgroup$
– Rasmus0909
Dec 9 '18 at 15:20
add a comment |
$begingroup$
The best I can get is the following:
Firstly, integrate the last equation gives
$$V(theta) = k_2 left(-cos theta - frac{k_3}{2} cos 2 theta + C_1 right)$$ where $C_1$ is some constant to be determined later.
Now substitute everything into the first equation, we get
$$frac{mathrm{d}p}{mathrm{d}theta} = frac{gamma - 1}{k_2 left(-cos theta - frac{k_3}{2} cos 2 theta + C_1 right)} cdot k_1 sin left(pi frac{theta - theta_0}{Delta theta_c} right) H_u - gamma cdot frac{p}{k_2 left(-cos theta - frac{k_3}{2} cos 2 theta + C_1 right)} cdot k_2 (sin theta + k_3 sin 2 theta)$$
Notice it is a $1^{text{st}}$ order ODE. Re-writing it into the standard form $y' + P(x) y = Q(x)$ gives
$$frac{mathrm{d}p}{mathrm{d}theta} + gamma frac{sin theta + k_3 sin 2 theta}{-cos theta - frac{k_3}{2} cos 2 theta + C_1} p = frac{k_1}{k_2}(gamma - 1)H_u frac{sin left(pi frac{theta - theta_0}{Delta theta_c} right)}{-cos theta - frac{k_3}{2} cos 2 theta + C_1}$$
Next, we compute
$$begin{align}
int gamma frac{sin theta + k_3 sin 2 theta}{-cos theta - frac{k_3}{2} cos 2 theta + C_1} mathrm{d}theta
&= gamma int frac{mathrm{d} (-cos theta - frac{k_3}{2} cos 2 theta + C_1)}{-cos theta - frac{k_3}{2} cos 2 theta + C_1} \
&= gamma log left|-cos theta - frac{k_3}{2} cos 2 theta + C_1 right| \
&= log left|-cos theta - frac{k_3}{2} cos 2 theta + C_1 right| ^ gamma
end{align}$$
So the integrating factor $mu(theta) = left|-cos theta - frac{k_3}{2} cos 2 theta + C_1 right| ^ gamma $
Finally, we need to compute
$$int frac{k_1}{k_2}(gamma - 1)H_u frac{sin left(pi frac{theta - theta_0}{Delta theta_c} right)}{-cos theta - frac{k_3}{2} cos 2 theta + C_1} cdot left|-cos theta - frac{k_3}{2} cos 2 theta + C_1 right| ^ gamma mathrm{d} theta \
= frac{k_1}{k_2}(gamma - 1)H_u int sin left(pi frac{theta - theta_0}{Delta theta_c} right) cdot operatorname{sgn} left(-cos theta - frac{k_3}{2} cos 2 theta + C_1 right) cdot left|-cos theta - frac{k_3}{2} cos 2 theta + C_1 right| ^ {gamma - 1} mathrm{d} theta$$
but this is how far I've got.
$endgroup$
The best I can get is the following:
Firstly, integrate the last equation gives
$$V(theta) = k_2 left(-cos theta - frac{k_3}{2} cos 2 theta + C_1 right)$$ where $C_1$ is some constant to be determined later.
Now substitute everything into the first equation, we get
$$frac{mathrm{d}p}{mathrm{d}theta} = frac{gamma - 1}{k_2 left(-cos theta - frac{k_3}{2} cos 2 theta + C_1 right)} cdot k_1 sin left(pi frac{theta - theta_0}{Delta theta_c} right) H_u - gamma cdot frac{p}{k_2 left(-cos theta - frac{k_3}{2} cos 2 theta + C_1 right)} cdot k_2 (sin theta + k_3 sin 2 theta)$$
Notice it is a $1^{text{st}}$ order ODE. Re-writing it into the standard form $y' + P(x) y = Q(x)$ gives
$$frac{mathrm{d}p}{mathrm{d}theta} + gamma frac{sin theta + k_3 sin 2 theta}{-cos theta - frac{k_3}{2} cos 2 theta + C_1} p = frac{k_1}{k_2}(gamma - 1)H_u frac{sin left(pi frac{theta - theta_0}{Delta theta_c} right)}{-cos theta - frac{k_3}{2} cos 2 theta + C_1}$$
Next, we compute
$$begin{align}
int gamma frac{sin theta + k_3 sin 2 theta}{-cos theta - frac{k_3}{2} cos 2 theta + C_1} mathrm{d}theta
&= gamma int frac{mathrm{d} (-cos theta - frac{k_3}{2} cos 2 theta + C_1)}{-cos theta - frac{k_3}{2} cos 2 theta + C_1} \
&= gamma log left|-cos theta - frac{k_3}{2} cos 2 theta + C_1 right| \
&= log left|-cos theta - frac{k_3}{2} cos 2 theta + C_1 right| ^ gamma
end{align}$$
So the integrating factor $mu(theta) = left|-cos theta - frac{k_3}{2} cos 2 theta + C_1 right| ^ gamma $
Finally, we need to compute
$$int frac{k_1}{k_2}(gamma - 1)H_u frac{sin left(pi frac{theta - theta_0}{Delta theta_c} right)}{-cos theta - frac{k_3}{2} cos 2 theta + C_1} cdot left|-cos theta - frac{k_3}{2} cos 2 theta + C_1 right| ^ gamma mathrm{d} theta \
= frac{k_1}{k_2}(gamma - 1)H_u int sin left(pi frac{theta - theta_0}{Delta theta_c} right) cdot operatorname{sgn} left(-cos theta - frac{k_3}{2} cos 2 theta + C_1 right) cdot left|-cos theta - frac{k_3}{2} cos 2 theta + C_1 right| ^ {gamma - 1} mathrm{d} theta$$
but this is how far I've got.
answered Dec 8 '18 at 20:34
Alex VongAlex Vong
1,309819
1,309819
$begingroup$
Thank you so much Alex. Looks like a huge effort! FYI, I used a forward Euler, which seemed to do the trick (although it's slightly cheating).
$endgroup$
– Rasmus0909
Dec 9 '18 at 15:20
add a comment |
$begingroup$
Thank you so much Alex. Looks like a huge effort! FYI, I used a forward Euler, which seemed to do the trick (although it's slightly cheating).
$endgroup$
– Rasmus0909
Dec 9 '18 at 15:20
$begingroup$
Thank you so much Alex. Looks like a huge effort! FYI, I used a forward Euler, which seemed to do the trick (although it's slightly cheating).
$endgroup$
– Rasmus0909
Dec 9 '18 at 15:20
$begingroup$
Thank you so much Alex. Looks like a huge effort! FYI, I used a forward Euler, which seemed to do the trick (although it's slightly cheating).
$endgroup$
– Rasmus0909
Dec 9 '18 at 15:20
add a comment |
$begingroup$
I ended up using a forward Euler numerical integration for this. I don't think the analytical solution is worth anybody's time for this.
$endgroup$
add a comment |
$begingroup$
I ended up using a forward Euler numerical integration for this. I don't think the analytical solution is worth anybody's time for this.
$endgroup$
add a comment |
$begingroup$
I ended up using a forward Euler numerical integration for this. I don't think the analytical solution is worth anybody's time for this.
$endgroup$
I ended up using a forward Euler numerical integration for this. I don't think the analytical solution is worth anybody's time for this.
answered Dec 9 '18 at 15:19
Rasmus0909Rasmus0909
1
1
add a comment |
add a comment |
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