Question

In an air standard diesel cycle, the compression ratio is 16 and at the beginning of isentropic compression, the temperature is 15c and the pressure is 0.1 j\.1pa. Heat is added until the t:mperature at constant process is 1480c. Calculate cut off ratio, heat supplied per kg of arr, cycle efficiency and mean effective pressure.

Answer 1

Answer:

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Explanation:

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Answer 2

The cut off ratio is 2.01 ,the heat supplied is 342.17 KJ /kg ,cycle efficiency is 61.3 and the mean effective pressure is 704.07 k pa

Step by Step Explanation:

Given:Compression Ratio r= 16

Temperature = 15 c

Pressure = 0.1

To find: Cut of ratio, heat supplied per kg and cycle efficiency and mean effective pressure for the diesel cycle.

r= 16 = $\frac{V_{1} }{V_{2} }$

$T_{1} = 15^{o} c$

$p_{1} = 0.1$ × $10^{3} kpa$

⇒ 1480 c (Constant Pressure)

= 1753 k

$r_{cut} =$ ?

$Q_{s} = ?$

n = ?

Mean effective pressure = ?

$r_{cut} = \frac{V_{3} }{V_{2} }$

$\frac{P_{2} V_{2} }{T_{2} } = \frac{P_{3} V_{3} }{T_{3} }$

$P_{2} = p_{3}$

$\frac{V_{3} }{V_{2} } = \frac{T_{3} }{T_{2} } = r_{cut}$

$\frac{T_{2} }{T_{1} } = [\frac{V_{1} }{V_{2} } ]^{R - 1}$

R = 1.4

$T_{2} = T_{1} [r]^{1.4 - 1}$

⇒ $288 (16) ^{1.4 - 1}$

$T_{2} = 873.05 k$

$r_{cut} = \frac{T_{3} }{T_{2} }$

$r_{cut} = \frac{1753}{873.05}$

⇒ 2.01

$r_{cut} = 2.01$

$Q_{s} = C_{p} (T_{3} - T_{2} )$

$C_{p} = 1.005 KJ/Kg$

$Q_{s} = 1.005(1753 - 873.05)$

$Q_{s} = 884. 4 KJ/kg$

$n = 1 - \frac{Q_{r} }{Q_{s} }$

$Q_{r} = C_{v} (T_{4} - T_{1} )$

$C_{v} = 0.718 KJ/Kg$

$T_{4} = 765 k$

$Q_{r} = C_{v} (T_{4} - T_{1} )$

$Q_{r} = 0.718 (765 - 288)$

⇒ 342.17 KJ / kg

$n = 1 - \frac{Q_{r} }{Q_{s} }$

$n = 1 - \frac{342.17}{884.4}$

n = 61.3

Mean effective pressure = $\frac{W_{net} }{V_{1} - V_{2} }$

$P_{1} V_{1} = R T_{1}$

$V_{1} = \frac{ R T_{1}}{P_{1} }$

R = 0.287

$V_{2} = \frac{0.82}{16}$

$V_{2} = 0.05$

$V_{1} = \frac{ R T_{1}}{P_{1} }$

= $\frac{(0.287)(288)}{(0.1)10^{3} }$

$V_{1} = 0.82 m^{3} / kg$

$n= \frac{w}{Q_{s} }$

w = n × $Q_{s}$

w = 542.13 × 884.4

= 542.13  kj/ kg

Mep = $\frac{542.13}{0.82 - 0.05}$

Mean effective pressure  =  704.07 kpa

Final answer:

In a standard diesel cycle the cut off ratio is 2.01 ,the heat supplied is 342.17 KJ /kg ,cycle efficiency is 61.3 and the mean effective pressure is 704.07 k pa

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