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4 years ago

What should the resistance value be on a size 5 motor starter coil

Engineering

1 answer:

Katen [24]4 years ago

4 0

Answer:

1000 Mega Ohms.

Explanation:

One mega Ohm is equal to 1,000,000 ohms, which is the resistance between two points of a conductor with one ampere of current at one volt. The mega Ohm is a multiple of the Ohm, which is the SI derived unit for electrical resistance. In the metric system, "mega" is the prefix for 10 raised to the power of 6.

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A pump of a water distribution system at 25°C is powered by a 15 kW electric motor whose efficiency is 90 percent. The water flo

IRISSAK [1]

The friction loss in the system is 3.480 kilowatts.

<h2>Procedure - Friction loss through a pump</h2><h2 /><h3>Pump model</h3><h3 />

Let suppose that the pump within a distribution system is an open system at steady state, whose mass and energy balances are shown below:

<h3>Mass balance</h3>

$\dot m_{in}-\dot m_{out} = 0$ (1)

$\dot m_{in} = \frac{\dot V_{in}}{\nu_{in}}$ (2)

$\dot m_{out} = \frac{\dot V_{out}}{\nu_{out}}$ (3)

<h3>Energy balance</h3>

$\eta \cdot \dot W_{el} + \dot m_{in}\cdot (h_{in}-h_{out}) - \dot W_{f} = 0$ (4)

Where:

$\dot m_{in}$ - Inlet mass flow, in kilograms per second.
$\dot m_{out}$ - Outlet mass flow, in kilograms per second.
$\dot V_{in}$ - Inlet volume flow, in cubic meters per second.
$\dot V_{out}$ - Outlet volume flow, in cubic meters per second.
$\nu_{in}$ - Inlet specific volume, in cubic meters per kilogram.
$\nu_{out}$ - Outlet specific volume, in cubic meters per kilogram.
$\eta$ - Pump efficiency, no unit.
$\dot W_{el}$ - Electric motor power, in kilowatts.
$h_{in}$ - Inlet specific enthalpy, in kilojoules per kilogram.
$h_{out}$ - Outlet specific enthalpy, in kilojoules per kilogram.
$\dot W$ - Work losses due to friction, in kilowatts.

<h3>Data from steam tables</h3>

From steam tables we get the following water properties at inlet and outlet:

Inlet

$p = 100\,kPa$ , $T = 25\,^{\circ}C$ , $\nu = 0.001003\,\frac{kJ}{kg}$ , $h = 104.927\,\frac{kJ}{kg}$ , Subcooled liquid

Outlet

$p = 300\,kPa$ , $T = 25\,^{\circ}C$ , $\nu = 0.001003\,\frac{kJ}{kg}$ , $h = 105.128\,\frac{kJ}{kg}$ , Subcooled liquid

<h3>Calculation of the friction loss in the system</h3>

If we know that $\dot V_{in} = 0.05\,\frac{m^{3}}{s}$ , $\nu_{in} = 0.001003\,\frac{m^{3}}{kg}$ , $h_{in} = 104.927\,\frac{kJ}{kg}$ , $h_{out} = 105.128\,\frac{kJ}{kg}$ , $\eta = 0.90$ and $\dot W_{el} = 15\,kW$ , then the friction loss in the system is:

$\dot W_{f} = \frac{\dot V_{in}}{\nu_{in}}\cdot (h_{in} - h_{out}) + \eta \cdot \dot W_{el}$

$\dot W_{f} = \left(\frac{0.05\,\frac{m^{3}}{s} }{0.001003\,\frac{m^{3}}{kg} } \right)\cdot \left(104.927\,\frac{kJ}{kg}-105.128\,\frac{kJ}{kg}\right) + (0.90)\cdot (15\,kW)$

$\dot W_{f} = 3.480\,kW$

The friction loss in the system is 3.480 kilowatts. $\blacksquare$

To learn more on pumps, we kindly invite to check this verified question: brainly.com/question/544887

6 0

2 years ago

Under normal operating conditions, the electric motor exerts a torque of 2.8 kN-m.on shaft AB. Knowing that each shaft is solid,

77julia77 [94]

Answer:

Explanation:

The image attached to the question is shown in the first diagram below.

From the diagram given ; we can deduce a free body diagram which will aid us in solving the question.

IF we take a look at the second diagram attached below ; we will have a clear understanding of what the free body diagram of the system looks like :

From the diagram; we can determine the length of BC by using pyhtagoras theorem;

SO;

$L_{BC}^2 = L_{AB}^2 + L_{AC}^2$

$L_{BC}^2 = (3.5+2.5)^2+ 4^2$

$L_{BC}= \sqrt{(6)^2+ 4^2}$

$L_{BC}= \sqrt{36+ 16}$

$L_{BC}= \sqrt{52}$

$L_{BC}= 7.2111 \ m$

The cross -sectional of the cable is calculated by the formula :

$A = \dfrac{\pi}{4}d^2$

where d = 4mm

$A = \dfrac{\pi}{4}(4 \ mm * \dfrac{1 \ m}{1000 \ mm})^2$

A = 1.26 × 10⁻⁵ m²

However, looking at the maximum deflection in length $\delta$ ; we can calculate for the force $F_{BC$ by using the formula:

$\delta = \dfrac{F_{BC}L_{BC}}{AE}$

$F_{BC} = \dfrac{ AE \ \delta}{L_{BC}}$

where ;

E = modulus elasticity

$L_{BC}$ = length of the cable

Replacing 1.26 × 10⁻⁵ m² for A; 200 × 10⁹ Pa for E ; 7.2111 m for $L_{BC}$ and 0.006 m for $\delta$ ; we have:

$F_{BC} = \dfrac{1.26*10^{-5}*200*10^9*0.006}{7.2111}$

$F_{BC} = 2096.76 \ N \\ \\ F_{BC} = 2.09676 \ kN$ ---- (1)

Similarly; we can determine the force $F_{BC}$ using the allowable maximum stress; we have the following relation,

$\sigma = \dfrac{F_{BC}}{A}$

${F_{BC}}= {A}*\sigma$

where;

$\sigma =$ maximum allowable stress

Replacing 190 × 10⁶ Pa for $\sigma$ ; we have :

${F_{BC}}= 1.26*10^{-5} * 190*10^{6} \\ \\ {F_{BC}}=2394 \ N \\ \\ {F_{BC}}= 2.394 \ kN$ ------ (2)

Comparing (1) and (2)

The magnitude of the force $F_{BC} = 2.09676 \ kN$ since the elongation of the cable should not exceed 6mm

Finally applying the moment equilibrium condition about point A

$\sum M_A = 0$

$3.5 P - (6) ( \dfrac{4}{7.2111}F_{BC}) = 0$

$3.5 P - 3.328 F_{BC} = 0$

$3.5 P = 3.328 F_{BC}$

$3.5 P = 3.328 *2.09676 \ kN$

$P =\dfrac{ 3.328 *2.09676 \ kN}{3.5 }$

P = 1.9937 kN

Hence; the maximum load P that can be applied is 1.9937 kN

4 0

3 years ago

Source 1 can supply energy at the rate of 11000 kJ/min at 310°C. A second Source 2 can supply energy at the rate of 110000 kJ/mi

VladimirAG [237]

Answer:

Source 2.

Explanation:

The efficiency of the ideal reversible heat engine is given by the Carnot's power cycle:

$\eta_{th} = 1 - \frac{T_{L}}{T_{H}}$

Where:

$T_{L}$ - Temperature of the cold reservoir, in K.

$T_{H}$ - Temperature of the hot reservoir, in K.

The thermal efficiencies are, respectively:

Source 1

$\eta_{th} = 1 - \frac{311.15\,K}{583.15\,K}$

$\eta_{th} = 0.466 \,(46.6\,\%)$

Source 2

$\eta_{th} = 1 - \frac{311.15\,K}{338.15\,K}$

$\eta_{th} = 0.0798 \,(7.98\,\%)$

The power produced by each device is presented below:

Source 1

$\dot W = (0.466)\cdot (11000\,\frac{kJ}{min})\cdot (\frac{1\,min}{60\,s} )$

$\dot W = 85.433\,kW$

Source 2

$\dot W = (0.0798)\cdot (110000\,\frac{kJ}{min})\cdot (\frac{1\,min}{60\,s} )$

$\dot W = 146.3\,kW$

The source 2 produces the largest amount of power.

8 0

3 years ago

The first thing you should do is develop a ____________________ to determine what vehicle you can afford.

swat32

The first thing you should do is develop a <u>budget</u> to determine what vehicle you can afford.

<h3>What is an automobile?</h3>

An automobile is also referred to as a vehicle, car or motorcar and it can be defined as a four-wheeled vehicle that is designed and developed to be propelled by an internal-combustion (gasoline) engine, especially for the purpose of transportation from one location to another.

<h3>What is a budget?</h3>

A budget can be defined as a financial plan that is typically used for the estimation of revenue and expenditures of an individual, business organization or government for a specified period of time, often one year.

In this context, we can reasonably infer and logically deduce that the first thing anyone should do is to develop a <u>budget</u> in order to determine what vehicle they can afford.

Read more on budget here: brainly.com/question/13964173

#SPJ1

7 0

1 year ago

I want a problems and there solutions of The inception of cavitation?

Ugo [173]

Answer:

The overview of the given scenario is explained in explanation segment below.

Explanation:

The inception of cavitation, that further sets the restriction for high-pressure and high-free operation, has always been the matter of substantial experimental study over the last few generations.
Cavitation inception would be expected to vary on the segment where the local "PL" pressure mostly on segment keeps falling to that are below the "Pv" vapor pressure of the fluid and therefore could be anticipated from either the apportionment of the pressure.

⇒ A cavitation number is denoted by "σ" .

4 0

3 years ago