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Natasha2012 [34]

4 years ago

14

The following information is available for an aircraft four-stroke reciprocating engine:indicated mean effective pressure (IMEP)

= 160 psistroke = 5 inbore = 5 inengine rpm = 3000number of cylinders = 9cylinder clearance volume = 15 in3engine mechanical efficiency, ηmech = 90%propeller efficiency, ηprop = 80%Calculate:brake horsepower (BHP)indicated horsepower (IHP)thrust horsepower (THP)friction horsepower (FHP)brake mean effective pressure (BMEP)Compression Ratio (is your answer reasonable for a spark ignition engine?)

1 answer:

Strike441 [17]4 years ago

5 0

Answer:

See explaination

Explanation:

Please kindly check attachment for the detailed step by step solution of the given problem.

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Thermodynamics fill in the blanks The swimming pool at the local YMCA holds roughly 749511.5 L (749511.5 kg) of water and is kep

Talja [164]

Answer:

$95.914\ \text{GJ}$

$\$272.78$

Explanation:

m = Mass of water = 749511.5 kg

c = Specific heat of water = 4182 J/kg ⋅°C

$\Delta T$ = Change in temperature = $80.6-50=30.6^{\circ}\text{F}$

Cost of 1 GJ of energy = $2.844

Heat required is given by

$Q=mc\Delta T\\\Rightarrow Q=749511.5\times 4182\times 30.6\\\Rightarrow Q=95.914\times 10^9\ \text{J}=95.914\ \text{GJ}$

Amount of heat required to heat the water is $95.914\ \text{GJ}$ .

Cost of heating the water is

$95.914\times 2.844=\$272.78$

Cost of heating the water to the required temperature is $\$272.78$ .

7 0

3 years ago

A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 30.00 a

IrinaVladis [17]

Answer:

The original length of the specimen $l_{o} = 104.7 mm$

Explanation:

Original diameter $d_{o}$ = 30 mm

Final diameter $d_{1}$ = 30.04 mm

Change in diameter Δd = 0.04 mm

Final length $l_{1}$ = 105.20 mm

Elastic modulus E = 65.5 G pa = 65.5 × $10^{3}$ M pa

Shear modulus G = 25.4 G pa = 25.4 × $10^{3}$ M pa

We know that the relation between the shear modulus & elastic modulus is given by

$G = \frac{E}{2(1 + \mu)}$

$25.5 = \frac{65.5}{2 (1 + \mu)}$

$\mu = 0.28$

This is the value of possion's ratio.

We know that the possion's ratio is given by

$\mu = \frac{\frac{0.04}{30} }{\frac{change \ in \ length}{l_{o} } }$

${\frac{change \ in \ length}{l_{o} } = \frac{\frac{0.04}{30} }{0.28}$

${\frac{change \ in \ length}{l_{o} } =$ 0.00476

$\frac{l_{1} - l_{o} }{l_{o} } = 0.00476$

$\frac{l_{1} }{l_{o} } = 1.00476$

Final length $l_{o}$ = 105.2 m

Original length

$l_{o} = \frac{105.2}{1.00476}$

$l_{o} = 104.7 mm$

This is the original length of the specimen.

5 0

3 years ago

In the construction of a timber-framed house, select and justify three timber based

kvasek [131]

Timber frame construction uses timber studs and rails, together with a structural sheathing board, to form a structural frame that transmits all vertical and horizontal loads to the foundations.

3 0

3 years ago

When handling chemicals and solvents, technicians are recommended to

Luda [366]

Answer:

To wear PPE
Have prior knowledge of explosive levels and elemental properties
Know procedure to eliminate any threat

7 0

3 years ago

Read 2 more answers

A 500-km, 500-kV, 60-Hz, uncompensated three-phase line has a positivesequence series impedance. z = 5 0.03 1 + j 0.35 V/km and

Anni [7]

Answer:

A) 282.34 - j 12.08 Ω

B) 0.0266 + j 0.621 / unit

C)

A = 0.812 < 1.09° per unit

B = 164.6 < 85.42°Ω

C = 2.061 * 10^-3 < 90.32° s

D = 0.812 < 1.09° per unit

Explanation:

Given data :

Z ( impedance ) = 0.03 i + j 0.35 Ω/km

positive sequence shunt admittance ( Y ) = j4.4*10^-6 S/km

A) calculate Zc

Zc = $\sqrt{\frac{z}{y} }$ = $\sqrt{\frac{0.03 i + j 0.35}{j4.4*10^-6 } }$

= $\sqrt{79837.128< 4.899^o}$ = 282.6 < -2.45°

hence Zc = 282.34 - j 12.08 Ω

B) Calculate gl

gl = $\sqrt{zy} * d$

d = 500

z = 0.03 i + j 0.35

y = j4.4*10^-6 S/km

gl = $\sqrt{0.03 i + j 0.35* j4.4*10^-6} * 500$

= $\sqrt{1.5456*10^{-6} < 175.1^0} * 500$

= 0.622 < 87.55 °

gl = 0.0266 + j 0.621 / unit

C) exact ABCD parameters for this line

A = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )

B = Zc sin h (gl) Ω = 164.6 < 85.42°Ω ( as calculated )

C = 1/Zc sin h (gl) s = 2.061 * 10^-3 < 90.32° s ( as calculated )

D = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )

where : cos h (gl) = $\frac{e^{gl} + e^{-gl} }{2}$

sin h (gl) = $\frac{e^{gl}-e^{-gl} }{2}$

7 0

3 years ago