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2 years ago

The ratio of the volume of charge admitted at n. T. P. To the swept volume of the piston is called.

1 answer:

5 0

The ratio of the volume of charge admitted at n. T. P. To the swept volume of the piston is called volumetric efficiency.

<h3>What is volumetric efficiency?</h3>

Volumetric efficiency is the ratio of the volume of gas to that is filled to the displacement of the piston of the cylinder. It tells the efficiency of the compressor of a gas compressor.

The formula of the volumetric efficiency is, "ηV=Actual volumeSwept volume. ηV=VaVS." The meaning of NTP in volumetric efficiency is the ratio of volume.

Thus, the swept volume of the piston is called volumetric efficiency.

To learn more about volumetric efficiency, refer to the below link:

brainly.com/question/14783214

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3 years ago

In a creep test, increasing the temperature will (choose the best answer) A. increase the instantaneous initial deformation B. i

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2 years ago

As the impurity concentration in solid solution of a metal is increased, the tensile strength:________.a) decreasesb) increasesc

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Increases

Explanation:

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3 years ago

A series AC circuit contains a resistor, an inductor of 250 mH, a capacitor of 4.40 µF, and a source with ΔVmax = 240 V operatin

slega [8]

Answer:

Explanation:

Inductance = 250 mH = 250 / 1000 = 0.25 H

capacitance = 4.40 µF = 4.4 × 10⁻⁶ F ( µ = 10⁻⁶)

ΔVmax = 240, f frequency = 50Hz and I max = 110 mA = 110 /1000 = 0.11A

a) inductive reactance = 2πfl = 2 × 3.142 × 50 × 0.25 H =78.55 ohms

b) capacitive reactance = $\frac{1}{2\pi fC}$ = 1 / ( 2 × 3.142× 50 × 4.4 × 10⁻⁶ ) = 723.34 ohms

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3 years ago

Read 2 more answers

A person walks into a refrigerated warehouse with head uncovered. Model the head as a 25- cm diameter sphere at 35°C with a surf

galina1969 [7]

Answer:

Hello some parts of your question is missing below is the missing part

Convection coefficient = 11 w/m^2. °c

answer : 44.83 watts

Explanation:

Given data :

surface emissivity ( ε )= 0.95

head ( sphere) diameter( D ) = 0.25 m

Temperature of sphere( T ) = 35° C

Temperature of surrounding ( T∞ ) = 25°C

Temperature of surrounding surface ( Ts ) = 15°C

б = ( 5.67 * 10^-8 )

Determine the total rate of heat loss

First we calculate the surface area of the sphere

As = $\pi D^{2}$

= $\pi * 0.25^2$ = 0.2 m^2

next we calculate heat loss due to radiation

Qrad = ε * б * As( $T^{4} - T^{4} _{s}$ ) ---- ( 1 )

where ;

ε = 0.95

б = ( 5.67 * 10^-8 )

As = 0.2 m^2

T = 35 + 273 = 308 k

Ts = 15 + 273 = 288 k

input values into equation 1

Qrad = 0.95 * ( 5.67 * 10^-8 ) * 0.2 ( (308)^4 - ( 288)^4 )

= 22.83 watts

Qrad ( heat loss due to radiation ) = 22.83 watts

calculate the heat loss due to convection

Qconv = h* As ( ΔT )

= 11*0.2 ( 35 -25 ) = 22 watts

Hence total rate of heat loss

= 22 + 22.83

= 44.83 watts

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3 years ago