Question 5 of 25

A vehicle of 1 200 kg is moving at a speed of 40 km/h on an incline of 1 in 50. The total constant rolling and wind resistance i

Readme [11.4K]

Answer:11.602 KW

Explanation:

mass of vehicle $\left ( m\right )=1200 kg$

speed=40Km/h

Resistance=600 N

$\eta =80%$

Gear ratio $\left ( G\right )=4:1$

$D_{effective}=500mm$

Net force to overcome by engine is

F=Resistance + sin component of weight

F=600+ $mgsin\theta$

Where $tan\theta =[tex]\frac{1}{50}$

$\theta =1.1457^{\circ}$

$F=600+1200\times 9.81\times sin\left ( 1.1457\right )$

F=600+235.38=835.38 N

power= $F.v=835.38\frac{100}{9}$

Engine Power= $\frac{835.\frac{100}{9}}{\eta }$ =11.602 KW

4 0

4 years ago

A data bus can be visualized as a multilane highway

iris [78.8K]

Answer:

B. with each component having an individual address

Explanation:

Data bus is a system within a computer or device, consisting of a connector or set of wires, that provides transportation for data. Data bus needs an address unique to each component in order to deliver the right data to the right place. Every memory location has a unique binary address. A microprocessor architecture is mainly composed of two main buses: The data bus and the address bus.

3 0

3 years ago

Where did fossil fuels come from

Levart [38]

Any fuel such as coal, petroleum (crude oil) or natural gas that has developed within the Earth over millions of years from the decayed remains of bacteria, plants or animals

6 0

3 years ago

Read 2 more answers

Technician A says that a lack of lubrication on the back of the disc brake pads can cause brake noise. Technician B says that pa

sweet-ann [11.9K]

Answer:

Technician A and B both are correct.

Explanation:

According to Technician A, we can stop noise by applying lubricant (Break friction).

According to Technician A, we can stop noise by set-up right position for break pad.

7 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) $pf_{C}=0.85$ $lagging$

(c) $I_{C} =32.37 A$

(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.

(c) Determine the magnitude of the line current from the source.

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

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