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2 years ago

The alternator must be operated with the battery disconnected or with the terminals at the back of the alternator

Engineering

2 answers:

Llana [10]2 years ago

7 0

True

I think so

Hope it helps

QveST [7]2 years ago

4 0

Answer:

true

Explanation:

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How are sedimentary - gravity flows different from fluid - gravity flows

Nimfa-mama [501]

Answer:

Sediment gravity flows are mixtures of water and sediment particles where the gravity acting on the sediment particles moves the fluid, in contrast to rivers, where the fluid moves the particles.

Explanation:

3 0

2 years ago

Water is the working fluid in an ideal Rankine cycle. Superheatedvapor enters the turbine at 10MPa, 480°C, and the condenser pre

choli [55]

Answer:

Explanation:

Given that:

Superheated vapor enters the turbine at 10 MPa, 480°C,

From the tables of superheated steam tables; the following values are obtained

$h_1 = 3322.02 \ kJ/kg\\\\ s_1 = 6.52846 \ kJ/kg.K$

Also; from the system, the isentropic line is 1-2 in which s_2 is in wet state

$s_2 = s_{f \ 6 kpa} +xs_{fg \ 6 kpa}$

$s_2 =0.51624 + x(7.82)$

$s_2 =0.51624 + 7.82x$

From the values obtained;

$s_1 =s_2= 6.52846 \ kJ/kg.K$

Therefore;

6.52846 = 0.51624+7.82x

6.52846 - 0.51624 = 7.82 x

6.01222 = 7.82 x

x = 6.01222/7.82

x = 0.7688

The enthalpy for this process at state (s_2) can be determined as follows:

$h_2 = h _f +xh_{fg} \\ \\ h_2 = 150.15 +(0.77 \times 2415.92) \\ \\ h_2 =150.15 +( 1629.2584 ) \\ \\ h_2 =2010.4084 \ kJ/kg$

The actual enthalpy at s_2 by using the isentropic efficiency of the turbine can determined by using the expression:

$n_T = \dfrac{h_1-h_{2a}}{h_1-h_2}$

$0.8 = \dfrac{3322.02-h_{2a}}{3322.02-2010.4084}$

$0.8 = \dfrac{3322.02-h_{2a}}{1311.6116}$

$0.8 * {1311.6116}= {3322.02-h_{2a}$

$1049.28928= {3322.02-h_{2a}$

$h_{2a}= {3322.02- 1049.28928$

$h_{2a}= 2272.73072$ kJ/kg

The work pump is calculated by applying the formula:

$w_p = v_{f \ 6 kpa} (p_4-p_3)$

$w_p = 0.0010062 * (10000-6)$

$w_p = 0.0010062 *9994$

$w_p = 10.0559628 \ kJ/kg$

However;

$w_p = h_4 -h_3$

From the process;

$h_3 = h_{f(6 kpa)} = 150.15 \ kJ/kg$

$10.0559628 = h_4 - 150.15$

$10.0559628+ 150.15 = h_4$

$160.2059628= h_4$

$h_4= 160.2059628 \ kJ/kg$

The actual enthalpy at s_4 by using the isentropic efficiency of the turbine can determined by using the expression:

$n_P = \dfrac{h_4-h_{3}}{h_{4a}-h_3}$

6 0

2 years ago

I need help on the Coderz Challenge missions 3 part 3. PLEASE HELP!

allsm [11]

Answer:

the answer how you analyzs the problwm

6 0

2 years ago

Read 2 more answers

What is the mass of the same dragster body (volume of 150 cm3) if it is made of basswood instead?

dusya [7]

Answer:

the answer is 61.5

Explanation:

8 0

2 years ago

**Please Help. ASAP**

natima [27]

Answer:

The answer is below

Explanation:

$\frac{v-u}{a} =t\\\\Making \ v\ the \ subject\ of\ formula:\\\\First \ cross-multiply:\\\\v-u=at\\\\add\ u\ to \ both\ sides:\\\\v-u+u=at+u\\\\v=u+at$

$\frac{y-x^2}{x}=3z\\ \\Making\ y\ the\ subject\ of\ formula:\\\\First \ cross \ multiply:\\\\y-x^2=3xz\\\\y=3xz+x^2\\\\y=x(x+3z)$

$x+xy=y\\\\Making\ x\ the\ subject\ of\ formula:\\\\x(1+y)=y\\\\Divide\ through\ by\ 1+y\\\\\frac{x(1+y)}{1+y} =\frac{y}{1+y} \\\\x=\frac{y}{1+y}$

$x+y=xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ x\ from \ both\ sides:\\\\x+y-x=xy-x\\\\y=xy-x\\\\y=x(y-1)\\\\Divide\ through\ by \ y-1\\\\\frac{y}{y-1} =\frac{x(y-1)}{y-1}\\ \\x=\frac{y}{y-1}$

$x=y+xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ xy\ from \ both\ sides:\\\\x-xy=y+xy-xy\\\\x-xy=y\\\\x(1-y)=y\\\\Divide\ through\ by \ 1-y\\\\\frac{x(1-y)}{1-y} =\frac{y}{1-y}\\ \\x=\frac{y}{1-y}$

$E=\frac{1}{2}mv^2-\frac{1}{2}mu^2\\ \\Making\ u\ the\ subject \ of\ formula:\\\\Multiply \ through\ by \ 2\\\\2E=mv^2-mu^2\\\\mu^2=mv^2-2E\\\\Divide\ through\ by\ m:\\\\u^2=\frac{mv^2-2E}{m}\\ \\Take\ square\ root\ of \ both\ sides:\\\\u=\sqrt{\frac{mv^2-2E}{m}}$

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\ \\Making\ y\ the\ subject \ of\ formula:\\\\\frac{x^2}{a^2}-1=\frac{y^2}{b^2}\\\\Multiply\ through\ by\ b^2\\\\b^2(\frac{x^2}{a^2} -1)=y^2\\\\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{b^2(\frac{x^2}{a^2} -1)}$

$ay^2=x^3\\\\Make\ y\ the\ subject\ of\ formula:\\\\Divide\ through\ by\ a:\\\\y^2=\frac{x^3}{a}\\ \\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{\frac{x^3}{a}} \\$

4 0

3 years ago