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3 years ago

14

1. Design a circuit, utilizing set/reset coils where PB 1 starts Motor 1 and PB2 stops Motor 1. Pressing and releasing either pu

sh button should latch the motor either on or off depending on the function

1 answer:

lianna [129]3 years ago

5 0

Answer:

Circuit attached with explanation

Explanation:

Hi Dear,

A circuit is attached for your reference.

When you press "start" PB, the supply reaches the motor starter relay coil "M" that is also in parallel with the "start" PB which allows the motor to remain ON even when you release "start" PB as supply to relay coil is directly from supply "L" through "M".

To stop motor just press "stop" PB and the circuit breaks which de-energize the relay coil and the motor stops.

Hope this finds easy to you.

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Opposition to current flow, restricts or resists current flow

BlackZzzverrR [31]

Answer:

The answer is c-resistance

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3 years ago

A pool of contaminated water is lined with a 40 cm thick containment barrier. The contaminant in the pit has a concentration of

konstantin123 [22]

This question is incomplete, the complete question is;

A pool of contaminated water is lined with a 40 cm thick containment barrier. The contaminant in the pit has a concentration of 1.5 mol/L, while the groundwater circulating around the pit flows fast enough that the contaminate concentration remains 0. There is initially no contaminant in the barrier material at the time of installation. The governing second order, partial differential equation for diffusion of the contaminant through the barrier is:

dC/dt = D( d²C / dz²)

where c(z,t) represent the concentration of containment of any depth into the barrier at anytime and D is the diffusion coefficient (a constant) for the containment in the barrier material.

a) write all boundary and initial conditions needed to solve this equation for C(z, t)

b) Find the steady state solution (infinite time) for C(z)

Answer:

a) At t = 0, z= 0, c = 1.5 mol/L

at t =0, z = 0.4m, c = 0 mol/L

b) C(z) = z² - 4.15z + 1.5

Explanation:

a)

The boundary and initial conditions are as follows

At t = 0, z= 0, c = 1.5 mol/L

at t =0, z = 0.4m, c = 0 mol/L

b)

The governing second order, partial differential equation for diffusion of the contaminant through the barrier is :

(dC/dt) = D*(d²C/dz²) ..............equ(1)

For steady state, above equation becomes,

(d²C/dz²) =0

Integrating above equation,

(dC/dz) = Z + C1 { where C1 is integration constant) }

again integrating above equation,

C = z² + C1*z + C2 ...................equ(2)

applying boundary condition : at t =0, z= 0, c = 1.5 mol/L, to above equation

C = z² + C1*z + C2

1.5 = 0 + 0*0 + c2

C2 = 1.5

applying boundary condition : at t =0, z= 0.4m, c = 0 mol/L, to equation (2) ,

0 = 0.4² + C1*0.4 + 1.5

0 = 0.16 + 0.4C1 + 1.5

0.4C1 = - 1.66

C1 = -1.66/0.4

C1 = -4.15

So, the steady state solution for C(z) is:

C(z) = z² - 4.15z + 1.5

6 0

3 years ago

A simple non-ideal Rankine cycle with water as the working fluid operates between the pressure limits of 15 MPa in the boiler an

konstantin123 [22]

Answer:

The right solution is "28.45%".

Explanation:

The given values are:

$P_4=50\ kPa$

$h_4=0.7(2304.7)+340.5$

$=1953.83 \ KJ/Kg$

and,

$P_3=15 \ mPa$

$h_3=hg$

$=2610.8 \ KJ/Kg$

$s_3=sg$

$=5.3108 \ KJ/Kgh$

At 45,

⇒ $x_{45} = \frac{5.3108-1.0912}{6.5019}$

$=0.66$

At $P_4=50 \ Kpa$ ,

$h_f=340.54$

or,

$V_f=0.001030 \ m^3/Kg$

then,

⇒ $h_2=340.54+0.001030(15\times 10^{3}-50)$

$=355.94 \ kJ/kg$

hence,

The isentropic efficiency of turbine will be:

⇒ $n_T=\frac{h_3-h_4}{h_3-h_{45}}$

$=\frac{2610.8-1953.83}{2610.8-1836.26}$

$=84.818$ (%)

The thermal efficiency of cycle will be:

⇒ $n_C=\frac{W_T-W_P}{2_{in}}$

$=\frac{(2610.8-1953-83)-(355.93-340.54)}{2610.8-355.93}$

$=28.45$ (%)

4 0

3 years ago

Yuliya22 [10]

No ground to connect to show me attach to any terminal or lead size reverse designated polarity

Answer B.False

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4 years ago

2. If the voltage is doubled and the resistance is not changed, what will happen to

Valentin [98]

Answer:

Remember Ohms law (the holy grail of electronics): V=IR

Explanation:

V=IR, where:

V=Voltage

I=Current

R=Resistance.

Looking at the equation (V=IR),

if R remains constant, but V is double, that means that the current has to also double, in order to keep the equation true.

Answer: the Current will double if voltage is doubled and resistance is kept constant.

8 0

3 years ago