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2 years ago

Sinks must be used for the correct intended purpose to prevent

Engineering

1 answer:

irakobra [83]2 years ago

8 0

Answer:

... spilling water or getting anything cascading onto the floor

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Air is compressed isothermally from 13 psia and 55°F to 80 psia in a reversible steady-flow device. Calculate the work required,

solong [7]

Answer:64.10 Btu/lbm

Explanation:

Work done in an isothermally compressed steady flow device is expressed as

Work done = P₁V₁ In { P₁/ P₂}

Work done=RT In { P₁/ P₂}

where P₁=13 psia

P₂= 80 psia

Temperature =°F Temperature is convert to °R

T(°R) = T(°F) + 459.67

T(°R) = 55°F+ 459.67

=514.67T(°R)

According to the properties of molar gas, gas constant and critical properties table, R which s the gas constant of air is given as 0.06855 Btu/lbm

Work = RT In { P₁/ P₂}

0.06855 x 514.67 In { 13/ 80}

=0.06855 x 514.67 In {0.1625}

= 0.06855 x 514.67 x -1.817

=- 64.10Btu/lbm

The required work therefore for this isothermal compression is 64.10 Btu/lbm

8 0

2 years ago

A 179 ‑turn circular coil of radius 3.95 cm and negligible resistance is immersed in a uniform magnetic field that is perpendicu

suter [353]

Answer:

The energy, that is dissipated in the resistor during this time interval is 153.6 mJ

Explanation:

Given;

number of turns, N = 179

radius of the circular coil, r = 3.95 cm = 0.0395 m

resistance, R = 10.1 Ω

time, t = 0.163 s

magnetic field strength, B = 0.573 T

Induced emf is given as;

$emf= N\frac{d \phi}{dt}$

where;

ΔФ is change in magnetic flux

ΔФ = BA = B x πr²

ΔФ = 0.573 x π(0.0395)² = 0.002809 T.m²

$emf = N\frac{d \phi}{dt} = 179(\frac{0.002809}{0.163} ) = 3.0848 \ V$

According to ohm's law;

V = IR

I = V / R

I = 3.0848 / 10.1

I = 0.3054 A

Energy = I²Rt

Energy = (0.3054)² x 10.1 x 0.163

Energy = 0.1536 J

Energy = 153.6 mJ

Therefore, the energy, that is dissipated in the resistor during this time interval is 153.6 mJ

6 0

3 years ago

Find the value of L

KonstantinChe [14]

Answer:

the value is 356732 Volt

6 0

2 years ago

Five kilograms of air at 427°C and 600 kPa are contained in a piston–cylinder device. The air expands adiabatically until the pr

son4ous [18]

Answer:

The entropy change of the air is $0.240kJ/kgK$

Explanation:

$T_{1} =427+273K,T_{1} =700K\\P_{1} =600kPa\\P_{2} =100kPa$

$T_{2}$ is unknown

we can apply the following expression to find $T_{2}$

$-w_{out} =mc_{v} (T_{2} -T_{1} )$

$T_{2} =T_{1} -\frac{w_{out } }{mc_{v} }$

now substitute

$T_{2} =700K-\frac{600kJ}{5kg*0.718kJ/kgK} \\T_{2}=533K$

To find entropy change of the air we can apply the ideal gas relationship

Δ $s_{air}$ $=c_{p} ln\frac{T_{2} }{T_{1} } -Rln\frac{P_{2} }{P_{1} }$

Δ $s_{air} =$ $1.005*ln(\frac{533}{700})-0.287* in(\frac{100}{600} )$

Δ $s_{air} =0.240kJ/kgK$

4 0

3 years ago

A frying pan is connected to a 150-volt circuit. If the resistance of the frying pan is 25 ohms, how many amperes does the fryin

mario62 [17]

Answer:

Explanation:

Ohms Law I=E/R (resistive requires no power factor correction)

150/25= 6 amps

5 0

3 years ago