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Lina20 [59]
3 years ago
12

Diesel engines do not use spark plugs to ignite the fuel-air mixture. Instead, they rely on the compression stroke to raise the

temperature of the air in the cylinders before injecting the fuel.
Calculate the compression ratio (final/initial volume) required to raise the temperature of air, initially at 1 bar and 300 K, to 900 K. What is the final pressure?
Air can be considered an ideal gas with Cv = 5R/2 at all conditions of interest.
Engineering
1 answer:
Vlad [161]3 years ago
5 0

Answer:

1) The compression ratio required to raise the the temperature of air initially at 1 bar and 300 K to 900 K is 15.59

2) The final pressure is 46.765 bar

Explanation:

1) Here we have the relationship between the compression ratio and temperature given as follows;

\frac{T_{2}}{T_{1}}=\left (\frac{v_1}{v_2}   \right )^{\gamma -1}

Where:

T₁  = Initial temperature = 300 K

T₂  = Final temperature = 900 K

v₁  = Initial volume

v₂  = Final volume

γ = Ratio of specific heat capacities Cp/Cv

Cp - Cv = R

∴ Cp = R + Cv = R + 5·R/2 = 7·R/2

∴ γ = Cp/Cv = 7·R/2 ÷ 5·R/2 = 7/5 = 1.4

Plugging in the values, we have;

\frac{900}{300}=\left (\frac{v_1}{v_2}   \right )^{1.4 -1} \Rightarrow 3 = \left (\frac{v_1}{v_2}   \right )^{0.4}

log(3) ÷ 0.4 = log(v₁/v₂)

∴ The compression ratio is given as follows;

\left (\frac{v_1}{v_2}   \right ) = 10^{\frac{log(3)}{0.4} }= 15.59

2) The final pressure is found as follows;

\frac{P_{2}}{P_{1}}=\left (\frac{v_1}{v_2}   \right )^{\gamma}

Where:

P₁  = Initial pressure = 1 bar

P₂  = Final pressure = Required

\left (\frac{v_1}{v_2}   \right ) = Compression \ ratio= 15.59

γ = 1.4

Plugging in the values, we have;

\frac{P_{2}}{1}=15.59^{1.4} \Rightarrow P_{2} = 46.765 \ bar

Therefore the final pressure = 46.765 bar.

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A fatigue test was conducted in which the mean stress was 90 MPa (13050 psi), and the stress amplitude was 190 MPa (27560 psi).
Gwar [14]

Answer:

a) 280MPa

b) -100MPa

c) -0.35

d) 380 MPa

Explanation:

GIVEN DATA:

mean stress \sigma_m = 90MPa

stress amplitude \sigma_a = 190MPa

a) \sigma_m =\frac{\sigma_max+\sigma_min}{2}

    90 =\frac{\sigma_{max}+\sigma_{min}}{2} --------------1

\sigma_a =\frac{\sigma_{max}-\sigma_{min}}{2}

   190 = \frac{\sigma_{max}-\sigma_{min}}{2} -----------2

solving 1 and 2 equation we get

\sigma_{max} = 280MPa

b) \sigma_{min} = - 100MPa

c)

stress ratio=\frac{\sigma_{min}}{\sigma_{max}}

=\frac{-100}{280} = -0.35

d)magnitude of stress range

                      =(\sigma_{max} -\sigma_{min})

                       = 280 -(-100) = 380 MPa

3 0
3 years ago
Match each context to the type of the law that is most suitable for it.
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Answer:

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5 0
3 years ago
Assume a program requires the execution of 50 x 106 FP instructions, 110 x 106 INT instructions, 80 x 106 L/S instructions, and
Pavlova-9 [17]

Answer:

Part A:

1.3568*10^{-5}=\frac{5300* New\  CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8

CPI reduced by 1-\frac{0.8}{4} = 0.80=80%

Part C:

New Execution Time=\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s

Increase in speed=1-\frac{1.81472*10^{-5}}{2.7136*10^{-5}} =0.33125= 33.125\%

Explanation:

FP Instructions=50*106=5300

INT  Instructions=110*106=11660

L/S  Instructions=80*106=8480

Branch  Instructions=16*106=1696

Calculating Execution Time:

Execution Time=\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}

Execution Time=\frac{5300*1+11660*1+8480*4+1696*2}{2*10^9\ Hz}

Execution Time=2.7136*10^{-5}\ s

Part A:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time=\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s

1.3568*10^{-5}=\frac{5300* New\  CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time=\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s

1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8

CPI reduced by 1-\frac{0.8}{4} = 0.80=80%

Part C:

New\ CPI_1=0.6*Old\ CPI_1=0.6*1=0.6\\New\ CPI_2=0.6*Old\ CPI_2=0.6*1=0.6\\New\ CPI_3=0.7*Old\ CPI_3=0.7*4=2.8\\New\ CPI_4=0.7*Old\ CPI_4=0.7*2=1.4

New Execution Time=\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}

New Execution Time=\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s

Increase in speed=1-\frac{1.81472*10^{-5}}{2.7136*10^{-5}} =0.33125= 33.125\%

8 0
3 years ago
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