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2 years ago

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The conditions at the beginning of compression in an Otto engine operating on hot-air standard with k=1.35 and 101.325 kPa, 0.05

m3 and 32C. The clearance is 8% and 15kJ are added per cycle. Determine the mean effective pressure.

1 answer:

Katyanochek1 [597]2 years ago

6 0

Answer:

$P_m=181.42 KPa$

Explanation:

$P_1=101.325 KPa,V_1=0.05m^3,T_1=32C$ ,K=1.35

Clearance is 8%.

Heat added=15 KJ

We know that compression ratio $r=1+\dfrac{1}{C}$

$r=1+\dfrac{1}{0.08}$

r=13.5

$r=\dfrac{V_1}{V_2}$

$13.5=\dfrac{0.05}{V_2}$

$V_2=3.7\times 10^{-3}$

We know that efficiency of otto cycle

$\eta =1-\dfrac{1}{r^{k-1}}$

$\eta =1-\dfrac{1}{13.5^{1.35-1}}$

$\eta =0.56$

$\eta =\dfrac{W}{Q}$

W is the work out put and Q is the heat addition.

$0.56 =\dfrac{W}{15}$

W=8.4 KJ

We know that Work =Mean effective pressure x swept volume.

Here swept volume $V_s=V_1-V_2$

$V_s=0.05-3.7\times 10^{-3}$

$V_s=0.0463 m^3$

Noe by putting the values

Work =Mean effective pressure x swept volume.

$8.4=P_m\times 0.0463$

$P_m=181.42 KPa$

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if the mine winch drum diameter is 6 m, calculate how far the cage will drop for each single rotation of the drum

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Answer:

The depth to which the cage will drop for each single rotation of the drum is approximately 18.85 m.

Explanation:

The mine winch drum consists of a drum around around which the haulage rope is attached

Therefore, given that the diameter of the winch drum = 6 m

The length of rope unwound by each rotation = How far the cage will drop for each single rotation of the drum

The length of the rope unwound by each rotation = The rope that goes round the circumference of the winch drum, once

∴ Since the rope that goes round the circumference of the winch drum, once = The circumference of the winch drum, we have;

The length of the rope unwound by each rotation = The circumference of the winch drum = π × The diameter of the winch drum

The length of the rope unwound by each rotation = π × 6 m = 6·π m

The length of rope unwound by each rotation = How far the cage will drop for each single rotation of the drum = 6·π m

How far the cage will drop for each single rotation of the drum = 6·π m ≈ 18.85 m

The depth to which the cage will drop for each single rotation of the drum ≈ 18.85 m.

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2 years ago

air at 600 kPa, 330 K enters a well-insulated, horizontal pipe having a diameter of 1.2 cm and exits at 120 kPa, 300 K. Applying

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Answer:

a) 251.31 m/s and 55.29 m/s

b) The mass flow rate is 0.0396 kg/s

c) The rate of entropy production is 0.0144 kW/K

Explanation:

a) The steady state is:

mi = mo

$\frac{A_{i}V_{i} }{v_{i} } =\frac{A_{o}V_{o} }{v_{o} } \\V_{i} =V_{o}(\frac{v_{i}}{v_{o}} )=V_{o}(\frac{T_{i}P_{o} }{T_{o}P_{i} } )=V_{o}(\frac{330*120}{300*600}) =0.22V_{o}$

The energy balance is:

$h_{i}+\frac{V_{i}^{2} }{2} =h_{2}+\frac{V_{o}^{2} }{2} \\h_{i}-h_{o}+(\frac{V_{i}^{2}-V_{o}^{2} }{2} )=0\\V_{o}=\sqrt{\frac{2(h_{i}-h_{o})}{0.9516} } =\sqrt{\frac{2*(330.24-300.19)x10^{3} }{0.9516} } =251.31m/s$

Vi = 0.22 * 251.31 = 55.29 m/s

b) The mass flow rate is:

$m=\frac{A_{o}V_{o}}{v_{o}} =\frac{\pi d^{2}V_{o}P_{o} }{4RT_{o}} =\frac{\pi *(0.012^{2})*251.31*120x10^{3} }{4*287*300} =0.0396kg/s$

c) The entropy produced is equal to:

$\frac{Q}{T} +S_{gen} =m(s_{2} -s_{1} )\\0+S_{gen} =m(s_{2} -s_{1} )\\S_{gen} =m(s_{2} -s_{1} )\\S_{gen}=0.0396*(c_{p} ln\frac{T_{o}}{T_{i}} -Rln\frac{P_{o}}{P_{i}} )=0.0396*(1.004ln\frac{300}{330} -0.287ln\frac{120}{600} )=0.0144kW/K$

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The fatigue data for a brass alloy are given as follows: Stress Amplitude (MPa) Cycles to Failure 170 3.7 × 104 148 1.0 × 105 13

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Answer:

i) S–N plot is attached

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i) I have attached the S–N plot (stress amplitude versus logarithm of cycles to failure)

ii) The question says we should find the fatigue strength at 4 × 10^(6) cycles.

So let's find the log of this and trace it on the graph attached.

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Thus, from the graph, at stress amplitude of 120 MPa, the log of cycles is approximately 5.75.

Thus,the fatigue life will be the inverse log of 5.75.

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