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Brilliant_brown [7]

3 years ago

15

A PMDC machine is measured to va120Vdc, ia 5.5A at the electrical terminals and the shaft is measured to have Tmech 5.5Nm, ns120

0RPM. State whether the machine acting as a motor or a generator and state the efficiency of the system. Q11. Motor, 89.4 %(a) Generator, 89.4%(b) Motor, 95.5%(c) Generator, 95.5 %(d) none of the abov

1 answer:

san4es73 [151]3 years ago

5 0

Answer:

(c) Generator, 95.5 %

Explanation:

given data

voltage va = 120V

current Ia = - 5.5 A

Tmech = 5.5Nm

ns = 1200 RPM

solution

first we get here electric input power that is express as

electric input power = va × Ia ......1

put here value and we get

electric input power = 120 × -5-5

electric input power -660 W

here negative mean it generate power

and here

Pin ( mech) will be

Pin ( mech) = Tl × ω .........2

Pin ( mech) = 5.5 × $\frac{2\pi N}{60}$

Pin ( mech) = 5.5 × $\frac{2\pi 1200}{60}$

Pin ( mech) = 691.150 W

and

efficiency will be here as

efficiency = $\frac{660}{691.150}$

efficiency = 95.5 %

so correct option is (c) Generator, 95.5 %

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Thermodynamics fill in the blanks The swimming pool at the local YMCA holds roughly 749511.5 L (749511.5 kg) of water and is kep

Talja [164]

Answer:

$95.914\ \text{GJ}$

$\$272.78$

Explanation:

m = Mass of water = 749511.5 kg

c = Specific heat of water = 4182 J/kg ⋅°C

$\Delta T$ = Change in temperature = $80.6-50=30.6^{\circ}\text{F}$

Cost of 1 GJ of energy = $2.844

Heat required is given by

$Q=mc\Delta T\\\Rightarrow Q=749511.5\times 4182\times 30.6\\\Rightarrow Q=95.914\times 10^9\ \text{J}=95.914\ \text{GJ}$

Amount of heat required to heat the water is $95.914\ \text{GJ}$ .

Cost of heating the water is

$95.914\times 2.844=\$272.78$

Cost of heating the water to the required temperature is $\$272.78$ .

7 0

3 years ago

Consider the circuit below where R1 = R4 = 5 Ohms, R2 = R3 = 10 Ohms, Vs1 = 9V, and Vs2 = 6V. Use superposition to solve for the

VladimirAG [237]

Answer:

The value of v2 in each case is:

A) V2=3v for only Vs1

B) V2=2v for only Vs2

C) V2=5v for both Vs1 and Vs2

Explanation:

In the attached graphic we draw the currents in the circuit. If we consider only one of the batteries, we can consider the other shorted.

Also, what the problem asks is the value V2 in each case, where:

$V_2=I_2R_2=V_{ab}$

If we use superposition, we passivate a battery and consider the circuit affected only by the other battery.

In the first case we can use an equivalent resistance between R2 and R3:

$V_{ab}'=I_1'R_{2||3}=I_1'\cdot(\frac{1}{R_2}+\frac{1}{R_3})^{-1}$

And

$V_{S1}-I_1'R_1-I_1'R_4-I_1'R_{2||3}=0 \rightarrow I_1'=0.6A$

$V_{ab}'=I_1'R_{2||3}=3V=V_{2}'$

In the second case we can use an equivalent resistance between R2 and (R1+R4):

$V_{ab}''=I_3'R_{2||1-4}=I_3'\cdot(\frac{1}{R_2}+\frac{1}{R_1+R_4})^{-1}$

And

$V_{S2}-I_3'R_3-I_3'R_{2||1-4}=0 \rightarrow I_3'=0.4A$

$V_{ab}''=I_3'R_{2||1-4}=2V$

If we consider both batteries:

$V_2=I_2R_2=V_{ab}=V_{ab}'+V_{ab}''=5V$

7 0

4 years ago

A heat engine that receives heat from a furnace at 1200°C and rejects waste heat to a river at 20°C has a thermal efficiency of

viktelen [127]

Answer:

second-law efficiency = 62.42 %

Explanation:

given data

temperature T1 = 1200°C = 1473 K

temperature T2 = 20°C = 293 K

thermal efficiency η = 50 percent

solution

as we know that thermal efficiency of reversible heat engine between same temp reservoir

so here

efficiency ( reversible ) η1 = 1 - $\frac{T2}{T1}$ ............1

efficiency ( reversible ) η1 = 1 - $\frac{293}{1473}$

so efficiency ( reversible ) η1 = 0.801

so here second-law efficiency of this power plant is

second-law efficiency = $\frac{thernal\ efficiency}{0.801}$

second-law efficiency = $\frac{50}{0.801}$

second-law efficiency = 62.42 %

3 0

3 years ago

A large particle composite consisting of tungsten particles within a copper matrix is to be prepared. If the volume fractions of

OverLord2011 [107]

Answer:

Upper bounds 22.07 GPa

Lower bounds 17.59 GPa

Explanation:

Calculation to estimate the upper and lower bounds of the modulus of this composite.

First step is to calculate the maximum modulus for the combined material using this formula

Modulus of Elasticity for mixture

E= EcuVcu+EwVw

Let pug in the formula

E =( 110 x 0.40)+ (407 x 0.60)

E=44+244.2 GPa

E=288.2GPa

Second step is to calculate the combined specific gravity using this formula

p= pcuVcu+pwTw

Let plug in the formula

p = (19.3 x 0.40) + (8.9 x 0.60)

p=7.72+5.34

p=13.06

Now let calculate the UPPER BOUNDS and the LOWER BOUNDS of the Specific stiffness

UPPER BOUNDS

Using this formula

Upper bounds=E/p

Let plug in the formula

Upper bounds=288.2/13.06

Upper bounds=22.07 GPa

LOWER BOUNDS

Using this formula

Lower bounds=EcuVcu/pcu+EwVw/pw

Let plug in the formula

Lower bounds =( 110 x 0.40)/8.9+ (407 x 0.60)/19.3

Lower bounds=(44/8.9)+(244.2/19.3)

Lower bounds=4.94+12.65

Lower bounds=17.59 GPa

Therefore the Estimated upper and lower bounds of the modulus of this composite will be:

Upper bounds 22.07 GPa

Lower bounds 17.59 GPa

7 0

3 years ago

When passing another vehicle, when is it acceptable to drive over the

miss Akunina [59]

Answer:

Under no circumstances

Explanation:

I'm not 100% sure why, but I remember hearing that you're not suposed to go over the speed limit no matter what

7 0

3 years ago

Read 2 more answers