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3 years ago

Can someone explain the answer for this question please? -metrology

Engineering

1 answer:

Mademuasel [1]3 years ago

5 0

Answer:

Metrology is “the science of measurement, embracing both experimental and theoretical determinations at any level of uncertainty in any field of science and technology,” as defined by the International Bureau of Weights and Measures

Explanation:

In simple terms, Metrology is the study or science of measurement. Metrology provides the quality assurance behind modern-day manufacturing processes. It is the science that enables production lines to produce thousands of identical pieces of sophisticated equipment.

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#5 Air undergoes an adiabatic compression in a piston-cylinder assembly from P1= 1 atm and Ti=70 oF to P2= 5 atm. Employing idea

otez555 [7]

Answer:

The work transfer per unit mass is approximately 149.89 kJ

The heat transfer for an adiabatic process = 0

Explanation:

The given information are;

P₁ = 1 atm

T₁ = 70°F = 294.2611 F

P₂ = 5 atm

γ = 1.5

Therefore, we have for adiabatic system under compression

$T_{2} = T_{1}\cdot \left (\dfrac{P_{2}}{P_{1}} \right )^{\dfrac{\gamma -1}{\gamma }}$

Therefore, we have;

$T_{2} = 294.2611 \times \left (\dfrac{5}{1} \right )^{\dfrac{1.5 -1}{1.5 }} \approx 503.179 \ K$

The p·dV work is given as follows;

$p \cdot dV = m \cdot c_v \cdot (T_2 - T_1)$

Therefore, we have;

Taking air as a diatomic gas, we have;

$C_v = \dfrac{5\times R}{2} = \dfrac{5\times 8.314}{2} = 20.785 \ J/(mol \cdot K)$

The molar mass of air = 28.97 g/mol

Therefore, we have

$c_v = \dfrac{C_v}{Molar \ mass} = \dfrac{20.785}{28.97} \approx 0.7175 \ kJ/(kg \cdot K)$

The work done per unit mass of gas is therefore;

$p \cdot dV =W = 1 \times 0.7175 \times (503.179 - 294.2611) \approx 149.89 \ kJ$

The work transfer per unit mass ≈ 149.89 kJ

The heat transfer for an adiabatic process = 0.

8 0

3 years ago

At time t the resultant force on a particle, of mass 250kg is (300ti-400tj)N. Initially, the particle is at the origin and is mo

Harlamova29_29 [7]

Answer:

Explanation:

4 0

2 years ago

Two balanced Y-connected loads in parallel, one drawing 15kW at 0.6 power factor lagging and the other drawing 10kVA at 0.8 powe

NemiM [27]

Answer:

(a) attached below

(b) $pf_{C}=0.85$ $lagging$

(d) $X_{C} =49.37$ Ω

(e) $I_{cap} =9.72 A$ and $I_{line} =27.66 A$

Explanation:

Given data:

$P_{1}=15 kW$

$S_{2} =10 kVA$

$pf_{1} =0.6$ $lagging$

$pf_{2}=0.8$ $leading$

$V=480$ $Volts$

(a) Draw the power triangle for each load and for the combined load.

$\alpha_{1}=cos^{-1} (0.6)=53.13$ °

$\alpha_{2}=cos^{-1} (0.8)=36.86$ °

$S_{1}=P_{1} /pf_{1} =15/0.6=25 kVA$

$Q_{1}=P_{1} tan(\alpha_{1} )=15*tan(53.13)=19.99$ ≅ $20kVAR$

$P_{2} =S_{2}*pf_{2} =10*0.8=8 kW$

$Q_{2} =P_{2} tan(\alpha_{2} )=8*tan(-36.86)=-5.99$ ≅ $-6 kVAR$

The negative sign means that the load 2 is providing reactive power rather than consuming

Then the combined load will be

$P_{c} =P_{1} +P_{2} =15+8=23 kW$

$Q_{c} =Q_{1} +Q_{2} =20-6=14 kVAR$

(b) Determine the power factor of the combined load and state whether lagging or leading.

$S_{c} =P_{c} +jQ_{c} =23+14j$

or in the polar form

$S_{c} =26.92$ °

$pf_{C}=cos(31.32) =0.85$ $lagging$

The relationship between Apparent power S and Current I is

$S=VI^{*}$

Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.

Current of the combined load can be found by

$I_{C} =S_{C}/\sqrt{3}*V$

$I_{C} =26.92*10^3/\sqrt{3}*480=32.37 A$

(d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity?Give your answer in Ω

$Q_{C} =3*V^2/X_{C}$