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# Answer to Question #99940 in Electrical Engineering for Hardi Amiru Sulemana

Question #99940
The pressure, volume and temperature at the beginning of the compression of a constant cycle are 101 kilo Newton per metre square, 0.003m cube and 18 degrees Celsius respectively. The maximum pressure of the cycle is 4.5Mega Newton per metre square. The volume ratio of the circle is 9.1. the cycle is repeated 3000 times a minute. Determine for the cycle;
a) the pressure volume and temperature at each of the cycle process change points
b) the thermal efficiency
c) theoretical out put in kilowatts
d)the mean effective pressure

Take Cp=1.006Kj/KgK, Cv=0.716Kj/KgK
1
2019-12-10T10:53:30-0500

a) Assume it's an Otto cycle process:

The compression 1-2 from the beginning is adiabatic:

"pV^\\gamma=pV^{c_p\/c_V}=\\text{ const}."

Calculate the volume at point 2 (end of 1-2 adiabatic process):

"p_2=p_1\\bigg(\\frac{V_1}{V_2}\\bigg)^{c_p\/c_V}=\\\\\n\\space\\\\=101\\cdot9.1^{1.006\/0.716}=2248\\text{ kN\/m}^2."

Apply the ideal gas law to find the temperature at point 2:

"\\frac{p_1V_1}{T_1}=\\frac{p_2V_2}{T_2},"

"T_2=T_1\\frac{p_2V_2}{p_1V_1}=\\\\\n\\space\\\\=(18+273)\\cdot\\frac{2248}{101}\\cdot\\frac{1}{9.1}=712\\text{ K}."

The process 2-3 is at constant volume, so the temperature at point 3 is:

"T_3=T_2\\frac{p_3}{p_2}=712\\cdot\\frac{4500}{2248}=1452\\text{ K}."

The pressure at point 4 (3-4 is adiabatic expansion):

"p_4=p_3\\bigg(\\frac{V_3}{V_4}\\bigg)^{c_p\/c_V}=\\\\\n\\space\\\\=4500\\cdot\\frac{1}{9.1^{1.006\/0.716}}=202\\text{ kN\/m}^2."

Temperature at 4:

"T_4=T_1\\frac{p_4}{p_1}=(18+273)\\cdot\\frac{202}{101}=582\\text{ K}."

b) The thermal efficiency:

"\\eta=1-\\frac{1}{n^{\\gamma-1}}=1-\\frac{1}{9.1^{1.006\/0.716-1}}=0.59."

c) The theoretical output in kW can be calculated from the following reasoning: the cycle is repeated 3000 times a minute, or 3000/60=50 times a second. The work done per one cycle is

"W_1=c_V[(T_3-T_2)-(T_4-T_1)]=\\\\\n=0.716[(1452-712)-(582-(18+273))]=\\\\=321\\text{ J}."

50 cycles per second will produce power (which is work per time, i.e. watt=joule/second):

"P=NW_1=50\\cdot321=16074\\text{ W}."

d) The mean effective pressure can be calculated through the displacement volume "V_d", number of revolutions per power stroke "n_c" (2 for a 4-stroke engine), revolutions per second "N" and power output "P" in watts:

"p_{\\text{me}}=\\frac{n_cP}{V_dN}=\\frac{2\\cdot16074}{V_1(1-\\frac{1}{9.1})\\cdot50}=240\\space779\\text{ Pa}."

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