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EastWind [94]
3 years ago
12

8. What is the proper name of this starting system used on selected vertical crankshaft engines? Click here to enlarge image Sup

er-Start Easy-Spin Auto-Start Ready-Start​

Engineering
1 answer:
Kamila [148]3 years ago
4 0

Answer:

Ready-Start

Explanation:

The Ready-Start system is built to sense the temperature of the engine and to supply the proper fuel amount to the carburetor, such that the engine starts on or before the completion of two short attempts to crank the engine at weather temperatures of more than 32 °F

The two major components which the Ready-Start system incorporates into the engine are'

1) A thermostat

2) An air vane

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A spherical balloon with a diameter of 9 m is filled with helium at 20°C and 200 kPa. Determine the mole number and the mass of
SIZIF [17.4K]

Answer:

<em>number of mole is 31342.36 moles</em>

<em>mass is 125.369 kg</em>

<em></em>

Explanation:

Diameter of the spherical balloon d = 9 m

radius r = d/2 = 9/2 = 4.5 m

The volume pf the sphere balloon ca be calculated from

V = \frac{4}{3} \pi r^3

V = \frac{4}{3}* 3.142* 4.5^3 = 381.75 m^3

Temperature of the gas T = 20 °C = 20 + 273 = 293 K

Pressure of the helium gas = 200 kPa = 200 x 10^3 Pa

number of moles n = ?

Using

PV = nRT

where

P is the pressure of the gas

V is the volume of the gas

n is the mole number of the gas

R is the gas constant = 8.314 m^3⋅Pa⋅K^−1⋅mol^−1

T is the temperature of the gas (must be converted to kelvin K)

substituting values, we have

200 x 10^3 x 381.75 = n x 8.314 x 293

number of moles n  = 76350000/2436 = <em>31342.36 moles</em>

We recall that n = m/MM

or m = n x MM

where

n is the number of moles

m is the mass of the gas

MM is the molar mass of the gas

For helium, the molar mass = 4 g/mol

substituting values, we have

m = 31342.36 x 4

m = 125369.44 g

m =<em> 125.369 kg</em>

3 0
3 years ago
Assume you are an observer standing at a point along a three-lane roadway. All vehicles in lane 1 are traveling at 30 mi/h, all
garik1379 [7]

Answer:

Explanation:

Given data;

  • In line 1, v1 = 30mi/hr
  • in line 2, v2 = 45mi/hr
  • in line 3, v3 = 60mi/hr
  • therefore time mean speed = v1 + v2 + v3 /n
  • = VT = 45mi/hr

  • space mean speed ; Vs
  • harmonic mean = 1/V = 1/v1 + 1/v2 + 1/v3
  • V = 13.85mi/hr
  • Hence Vs = V x n = 3 x 13.85 = 41.55mi/hr
5 0
3 years ago
Rod of steel, 200 mm length reduces its diameter (50 mm) by turning by 2 mm with feed speed 25 mm/min. You are required to calcu
diamong [38]

Answer:

125 cm³/min

Explanation:

The material rate of removal is usually given by the formula

Material Rate of Removal = Radial Depth of Cut * Axial Depth of Cut * Feed Rate, where

Radial Depth of Cut = 25 mm

Axial depth of cut = 200 mm

Feed rate = 25 mm/min

On multiplying all together, we will then have

MRR = 25 mm * 200 mm * 25 mm/min

MRR = 125000 mm³/min

Or we convert it to cm³/min and have

MRR = 125000 mm³/min ÷ 1000

MRR = 125 cm³/min

4 0
3 years ago
A insulated vessel s has two compartments separated by a membreane. On one side is 1kg of steam at 400 degC and 200 bar. The oth
Lilit [14]

Answer:

See explaination

Explanation:

See attachment for the detailed step by step solution of the given problem.

5 0
3 years ago
Using the tables for water determine the specified property data at the indicated state. For H2O at T = 140 °C and v = 0.2 m3/kg
Dennis_Churaev [7]

Answer:

h = 1429.74\,\frac{kJ}{kg}

Explanation:

The determination of any further properties requires the knowledge of two independent properties. (Temperature and specific volume in this case). The specific volumes for saturated liquid and vapor at 140 °C are, respectively:

\nu_{f} = 0.001080\,\frac{m^{3}}{kg}

\nu_{g} = 0.50850\,\frac{m^{3}}{kg}

Since \nu_{f} < \nu < \nu_{g}, it is a liquid-vapor mixture. The quality of the mixture is:

x = \frac{\nu-\nu_{f}}{\nu_{g}-\nu_{f}}

x = \frac{0.2\,\frac{m^{3}}{kg} - 0.001080\,\frac{m^{3}}{kg} }{0.50850\,\frac{m^{3}}{kg} - 0.001080\,\frac{m^{3}}{kg} }

x = 0.392

The specific enthalpies for saturated liquid and vapor at 140 °C are, respectively:

h_{f} = 589.16\,\frac{kJ}{kg}

h_{g} = 2733.5\,\frac{kJ}{kg}

The specific enthalpy is:

h = h_{f}+x\cdot (h_{g}-h_{f})

h = 589.16\,\frac{kJ}{kg}+0.392\cdot \left( 2733.5\,\frac{kJ}{kg} - 589.16\,\frac{kJ}{kg} \right)

h = 1429.74\,\frac{kJ}{kg}

6 0
4 years ago
Read 2 more answers
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