Differential equation for pressure and heat release in combustion engine












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$begingroup$


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?










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    $begingroup$


    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?










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      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?










      share|cite|improve this question









      $endgroup$




      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






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      share|cite|improve this question











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      asked Dec 8 '18 at 16:23









      Rasmus0909Rasmus0909

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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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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



















          0












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
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            2 Answers
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            0












            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              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
















            0












            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              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














            0












            0








            0





            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$



            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.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 8 '18 at 20:34









            Alex VongAlex Vong

            1,284819




            1,284819












            • $begingroup$
              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




            $begingroup$
            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











            0












            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$



                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.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 9 '18 at 15:19









                Rasmus0909Rasmus0909

                1




                1






























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