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3 years ago

What is the meaning of the measurement met?

1 answer:

6 0

Answer: MET(Metabolic equivalent of task) is the measurement technique that is based on the ratio of the energy that is being spend by an individual by any physical activity to the energy spent by individual sitting in quiet position.

This ration has some reference in which is ratio is measured.The energy being measured in the activity time of the person considers his/her relative mass at that time. This relative mass is compared in accordance with the oxygen amount of 3.5 ml per kg/min that is almost equal to energy of the person in the sitting position.

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The inverted U-tube is used to measure the pressure difference between two points A and B in an inclined pipeline through which

JulijaS [17]

Answer:

i) 0.610 m or 610 mm

ii) 0.4 m or 400 mm

Explanation:

The pressure difference between the pipes is

a) Air

Pa + πha +Ha = Pb + πhb +Hb

Pa - Pb = π(hb-ha) + Hb-Ha

Relative density of air = 1.2754 kg /m3

Pa - Pb = 1.2754 * 0.4 + (0.3-0.2) = 0.610 m or 610 mm

b) paraffin of relative density of 0.75

Pa - Pb = π(hb-ha) + Hb-Ha

Pa - Pb = 0.75 * 0.4 + (0.3-0.2) = 0.4 m or 400 mm

8 0

3 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

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