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kolezko [41]
3 years ago
8

An automobile engine consumes fuel at a rate of 22 L/h and delivers 85 kW of power to the wheels. If the fuel has a heating valu

e of 44,000 kJ/kg and a density of 0.8 g/cm3, determine the efficiency of this engine.
Engineering
2 answers:
attashe74 [19]3 years ago
7 0

Answer:

The efficiency of the engine is 39.53%

Explanation:

data given by the exercise

HV = heating value = 44000 kJ/kg

Vfuel = volume flow rate of fuel = 22 L/h = 0.0061 L/s

W = power output = 55 kW

p = density = 0.8 g/cm³

the mass flow rate of fuel is equal to

m_{fuel} =V*p=0.0061*0.8=4.89x10^{-3} kg/s

the rate of heat supplied is

Q_{in} =m_{fuel} *HV=4.89x10^{-3} *44000=215kW

the efficiency is equal to

n=\frac{W}{Q_{in} } *100=\frac{85}{215} *100=39.53%

zepelin [54]3 years ago
5 0

Answer:

Efficiency of the engine is 39.51 %

Explanation:

Mass of fuel consumed per hour = 22 L/h * 0.08 Kg/L = 17.6 Kg/h

Total Energy Consumed per hour = Mass of fuel consumed per hour * Heating  value of fuel

Total Energy Consumed per hour = 17.6 Kg/h * 44,000 KJ/Kg = 774,400 KJ/h

since, 1 KW = 3600 KJ/h

774,400KJ/h = 774,400/3600 kW = 215.1 kW

Efficiency of the engine = \frac{Power to the wheels}{Total Power}

Efficiency = 85/215.1 = 39.51 %

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Elanso [62]

Answer:

(b). T = 22.55 ⁰C

(c). q = 557.8 W

Explanation:

we take follow a step by step process to solving this problem.

from the question, we have that

The two glass pieces is separated by a 1.8 cm distance layer of air.

the thickness of glass piece is 1 cm

width = 4 m

the height = 3 m

(a). the sketch of the thermal circuit is uploaded in the picture below.

(b).  the thermal resistance due to the conduction in the first glass plane is given thus;

R₁ = Lg / Kg A ................(1)

given that Kg rep. the thermal conductivity of the glass plane

A = conduction surface area

Lg = Thickness of glass plane4

taking the thermal conductivity of glass plane as Kg = 0.78 w/mk

inputting values into equation (1) we have,

R₁ = [1 (cm) ˣ 1 (m)/100 (cm)] / [(0.78 w/mk)(4m ˣ 3m)]

R₁ = 1.068 ˣ 10 ⁻³ k/w

Being that we have same thermal resistance in the first and second plane,

therefore R₁ = R₃ = 1.068 ˣ 10 ⁻³ k/w

⇒ Also the thermal resistance between air and glass as a result of the conduction by the layer is given thus

R₂ = La/KaA .....................(2)

given Ka = thermal conductivity of air

A = surface area

La = thickness of air

substituting values into the equation we have

R₂ = [1.8 (cm) ˣ 1 (m)/100 (cm)] / [(0.0262 w/mk)(4m ˣ 3m)]

R₂ = 5.73 ˣ 10⁻² k/w

Given the thermal resistance on the outer surface due to convection, we have

R₄ = 1/hA

inputting value gives R₄ = 1 / (12 w/m² ˣ 12m) = 6.94 ˣ 10⁻³k/w

R₄ = 6.94 ˣ 10⁻³k/w

Finally the sum total of thermal resistance = R₁ + R₂ + R₃ + R₄

R-total = 0.0663 kw

From this we can calculate the rate of heat loss

using  q = Ti - To / R-total ..............(3)

given Ti and To is the inside and outside temperature i.e. 27⁰C and -10⁰C

from equation (3),

q = 27- (-10) / 0.0063 = 557.8 W

q = 557.8 W  

⇒ Applying the heat transfer formula for inside surface glass temperature gives;

q = Ti - T₂ / R₃ + R₄

T₂ = Ti - q (R₃ + R₄)

T₂ = 27 - 557.8 (1.068ˣ10⁻³ + 6.94ˣ10⁻³ ) = 22.55°C

T₂ = 22.55°C

cheers i hope this helps

8 0
3 years ago
Ti-6Al-4V has a fracture toughness of 74.6 MPa-m0.5. How much stress (in MPa) would it take to fail a plate loaded in tension th
Nikitich [7]

Answer:

critical stress  = 595 MPa

Explanation:

given data

fracture toughness =  74.6 MPa-\sqrt{m}

crack length = 10 mm

f = 1

solution

we know crack length = 10 mm  

and crack length = 2a as given in figure attach

so 2a = 10

a = 5 mm

and now we get here with the help of plane strain condition , critical stress is express as

critical stress  = \frac{k}{f\sqrt{\pi a}}    ......................1

put here value and we get

critical stress  = \frac{74.6}{1\sqrt{\pi 5\times 10^{-3}}}

critical stress  = 595 MPa

so here stress is change by plane strain condition because when plate become thinner than condition change by plane strain to plain stress.

plain stress condition occur in thin body where stress through thickness not vary by the thinner section.

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Reception of signals from a radio facility, located off the airway being flown, may be inadequate at the designated mea to ident
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The altitude ensures acceptable navigational signal coverage only within 22 NM of a VOR.

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Altitude or height exists as distance measurement, usually in the vertical or "up" approach, between a reference datum and a point or object. The exact meaning and reference datum change according to the context.

The MOCA exists in the lower published altitude in effect between fixes on VOR airways, off-airway routes, or route segments that satisfy obstacle support conditions for the whole route segment. This altitude also ensures acceptable navigational signal coverage only within 22 NM of a VOR.

The altitude ensures acceptable navigational signal coverage only within 22 NM of a VOR.

Therefore, the correct answer is 22 NM of a VOR.

To learn more about altitudes refer to:

brainly.com/question/1159693

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3 0
2 years ago
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Is this a question or a statement?

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