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2 years ago

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the coil polarity in a waste spark system is determined by the direction in which the coil is wound (left hand rule for conventi

onal current flow) and cant be changed. for example if a V8 engine has a firing order of 18436527 and number one cylinder is on compression which cylinder would be on the exhaust stroke?

1 answer:

zaharov [31]2 years ago

3 0

The coil polarity in a waste-spark system is determined by the direction in which the coil is wound (left-hand rule for conventional current flow)and can’t be changed. For example, if a V-8 engine has a firing order of 18436572 and the number 1 cylinder is on compression, which cylinder will be on the exhaust stroke?

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a refrigerator uses refrigerant-134a as the working fluid and operates on the vapor-compression refrigeration cycle. the evapora

timurjin [86]

Answer:

I. 3.316 kW

II. 1.218 kW

III. 2.72

Explanation:

At state 1, the enthalpy and entropy are determined using the given data from A-13.

At P1 = 200kpa and T1 = 0,

h1 = 253.07 kJ/kg

s1 = 0.9699 kJ/kgK

At state 2, the isentropic enthalpy is determined at P2 = 1400kpa and s1 = s2 by interpolation. Thus

h2(s) = 295.95 kJ/kg

The actual enthalpy is then gotten by

h2 = h1 + [h2(s) - h1]/n

h2 = 253.07 + [295.95 - 253.07]/0.88

h2 = 253.07 + 48.73

h2 = 301.8 kJ/kg

h3 = h4 = 120.43 kJ/kg

Heating load is determined from energy balance, thus,

Q'l = m'(h1 - h4)

Q'l = 0.025(253.07 - 120.43)

Q'l = 0.025 * 132.64

Q'l = 3.316 kW

Power is determined by using

W' = m'(h2 - h1)

W'= 0.025(301.8 - 253.07)

W'= 0.025 * 48.73

W'= 1.218 kW

The Coefficient Of Performance is Q'l / W'

COP = 3.316/1.218

COP = 2.72

8 0

3 years ago

Read 2 more answers

A cable in a motor hoist must lift a 700-lb engine. The steel cable is 0.375in. in diameter. What is the stress in the cable?

UkoKoshka [18]

Answer:43.70 MPa

Explanation:

Given

mass of engine $700 lb \approx 317.515 kg$

diameter of cable $0.375 in.\approx 9.525 mm$

$A=\frac{\pi d^2}{4}=71.26 mm^2$

we know stress( $\sigma$ ) $=\frac{load\ applied}{area\ of\ cross-section}$

$\sigma =\frac{317.515\times 9.81}{71.26\times 10^{-6}}=43.70 MPa$

7 0

3 years ago

11. Technicians A and B are discussing

algol [13]

Answer:

C. Neither Technician A nor B

Explanation:

Just took the test

5 0

3 years ago

A reversible refrigeration cycle operates between cold and hot reservoirs at temperatures TC and TH, respectively. (a) If the co

podryga [215]

Answer:

a) $T_{H} = 1.967\,^{\circ}F$ , b) $COP_{R} = 9.105$ , c) $T_{H} = 115.934\,^{\circ}F$ , d) $COP_{R} = 6.995$ , e) $T_{H} = 25.129\,^{\circ}C$

Explanation:

a) The coefficient of performance of the reversible refrigeration cycle is:

$COP_{R} = \frac{T_{C}}{T_{H}-T_{C}}$

$10 = \frac{419.67\,R}{T_{H}-419.67\,R}$

The temperature of the hot reservoir is:

$10\cdot T_{H} - 4196.7 = 419.67$

$T_{H} = 461.637\,R$

$T_{H} = 1.967\,^{\circ}F$

b) The coefficient of performance is:

$COP_{R} = \frac{273.15\,K}{303.15\,K-273.15\,K}$

$COP_{R} = 9.105$

c) The temperature of the hot reservoir can be determined with the help of the following relation:

$\frac{Q_{C}}{Q_{H}-Q_{C}} = \frac{T_{C}}{T_{H}-T_{C}}$

$\frac{500\,BTU}{600\,BTU-500\,BTU} = \frac{479.67\,R}{T_{H}-479.67\,R}$

$5 = \frac{479.67\,R}{T_{H}-479.67\,R}$

$5\cdot T_{H} - 2398.35 = 479.67$

$T_{H} = 575.604\,R$

$T_{H} = 115.934\,^{\circ}F$

d) The coefficient of performance is:

$COP_{R} = \frac{489.67\,R}{559.67\,R-489.67\,R}$

$COP_{R} = 6.995$

e) The temperature of the cold reservoir is:

$8.9 = \frac{268.15\,K}{T_{H}-268.15\,K}$

$8.9\cdot T_{H} - 2386.535 = 268.15$

$T_{H} = 298.279\,K$

$T_{H} = 25.129\,^{\circ}C$

8 0

3 years ago

Pointssss 100 and brainliest :)

Delvig [45]

Answer:

thank you for the free point have a great rest of your day

7 0

3 years ago