Peratura ambiente 9 37e un motor de ignicion por chispa...

peratura ambiente. 9-37E Un motor de ignición por chispa tiene una relación de compresión de 10, una eficiencia isentrópica de compresión de 85 por ciento y una eficiencia isentrópica de expansión de 95 por ciento. Al principio de la compresión, el aire en el cilindro está a 13 psia y 60 °F. La temperatura máxima medida del gas es 2 300 °F. Determine el calor suministrado por unidad de masa, la eficiencia térmica y la presión media efectiva de este motor cuando se modela con el ciclo de Otto. Use calores específicos constantes a temperatura ambiente. Respuestas: 225 Btu/lbm, 49.7 por ciento, 45.3 psia

Transcript

00:01 We are given here spark ignition so here we are given a spark ignition spark ignition and the spark ignition engine a compression ratio so we are given here compression ratio which is equal to we can write 8 of 8 ascentrophic efficiency is 85 % so that is s ntropic compression efficiency 85 % and ascent efficiency expansion efficiency we have that is 95 % that is 85 % and 95 portion to beginning at compression that the ice cycle at cylinder that is 13 psia so pressure is 13 psia and there is 16 fahrenheit the temperature here is 60 fahrenheit and the maximum gas temperature found to be 23 here 230 fahrenheit by the measurement let's determine the heat supply by the unit mass and thermal efficiency see and the mean mean effective pressure of the engine so let's solve this question first find is the heat supply that is q that we want to find the first understand that the n compression eta you can weigh eta compression which is equal you can weigh t2s minus we can write t 2 s or we can see 25 divided by t1 minus here here you can wear t1 minus t 2 now here we know the ts which is equal to t and here we can write t is 2 k minus 1 and minus minus we can write t 1 divided by t 2 minus t 1 now we can write the t 2 which is simply find we can write here t 2 which is equal to t 1 1 plus the r here we can write r is 2 k minus 1 minus 1 divided but we can write data so this is the value of t 2 here and now we are substitute value here so value of t 1 here is to 520 into the bracket we can write 1 plus 10 is 2 1 .5 minus 1 1 1 1 4 minus 1 divided by we can write 0 .85 multiply 5 value of r the simple 5 we can write the answer is 1 445r so this is the value of t 2 now here we can write the heat here that is the heat supply power unit of mass so we can write q in which is equal to cv multiplied by t3 minus t2 and here we just substitute value here so we can receive the value is 171 1m 1u2d3 is 2760 minus 144 9 which is simply fine we can write the answer is 225 that is a 225 bt u for lbm so this is value of the heat supply now here we can write the heat exited here we can write the eta exerted which is equal to we can write t s minus t4 divided by t3 minus t 4 s and now here is a simplifying we can write 23 minus t4 divided by t 3 minus that is we can write t3 and r to 1 minus k now with a simplifying we can write t 4 which is equal to 3 3m whatever we can write 1 plus a external e xb and here we can invent r h 2 1 minus k minus 1 and here we just substitute of value into this formula so we can write 2760 and into the bracket we can write 1 plus 0 .95 here we can write 10 to 1 minus 1 .4 minus 1 multiply and we can write here answer is so answer here is 1 1 .9000 so this is the value of that t4.

04:17 Now let's find out q out.

04:20 So here, 8 out which is equal to we can write a cv.

04:24 You can write t4 minus t1.

04:27 We just simply substitute values so you can write 0.

04:32 0 .171.

04:33 In the bracket we can write 1182 minus we can wait 520.

04:38 We just simply find we can write here answer is 113 .2.

04:42 Vtu for lbm.

04:46 Vtu for lvm...