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loris [4]
4 years ago
6

Disc brake rotors that are too thin cannot handle as much heat and will experience ___________.

Engineering
1 answer:
konstantin123 [22]4 years ago
6 0

Answer:

brake fade

Explanation:

Among the most frequent problems in the brake disc is their overheating. When there is overheating, the temperature of the disc rises to critical values, thus the brake pad starts to slide with respect to the disc, and the efficiency of the brake system decreases to a minimum.

Overheating of the brake disc is detected when inspecting the parts. Steel is the most common base material in the creation of brake discs. When heated, the material becomes its color. A disc of steel at critical temperature, turns bright orange and later on cooling becomes purple.

If a color change is seen on the brake discs when inspecting, a service station must be contacted without delay for an in-depth examination. After detecting the problem, the brake discs and pads will be changed in a mandatory way.

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Represent each of the following units as a combination of primitive
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Answer:

a. unit of length: [L]

b. unit of volume: [L^3]

c. unit of pressure:P=\frac{F}{A} \equiv\frac{[MLT^{-2}]}{[L^2]} [ML^{-1}T^{-2}]

d. unit of power: N.m.s^{-1}\equiv [ML^2T^{-3}]

e. unit of force: [kg.m/s^2]\equiv [MLT^{-2}]

f. unit of power: N.m.s^{-1}\equiv [ML^2T^{-3}]

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Power: P=\frac{W}{t}=\frac{F.x}{t}

where:

F = force

A = area

W = work

t = time

a = acceleration

v = velocity

x = displacement

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A power plant operates on a regenerative vapor power cycle with one open feedwater heater. Steam enters the first turbine stage
ivolga24 [154]

Answer:

a. 46.15%

b. 261.73 kg/s

c. 54.79 kW/K

Explanation:

a. State 1

The parameters given are;

T₁ = 560°C

P₁ = 12 MPa = 120 bar

Therefore;

h₁ = 3507.41 kJ/kg,  s₁ = 6.6864 kJ/(kg·K)

State 2

p₂ = 1 MPa = 10 bar

s₂ = s₁ = 6.6864 kJ/(kg·K)

h₂ = (6.6864 - 6.6426)÷(6.6955 - 6.6426)×(2828.27 - 2803.52) + 2803.52

= (0.0438 ÷ 0.0529) × 24.75 = 2824.01 kJ/kg

State 3

p₃ = 6 kPa = 0.06 bar

s₃ = s₁ = 6.6864 kJ/(kg·K)

sg = 8.3291 kJ/(kg·K)

sf = 0.52087 kJ/(kg·K)

x = s₃/sfg = (6.6864- 0.52087)/(8.3291  - 0.52087) = 0.7896

(h₃ - 151.494)/2415.17 = 0.7896

∴ h₃ = 2058.56 kJ/kg

State 4

Saturated liquid state

p₄ = 0.06 bar= 6000 Pa, h₄ = 151.494 kJ/kg, s₄ = 0.52087 kJ/(kg·K)

State 5

Open feed-water heater

p₅ = p₂ =  1 MPa = 10 bar = 1000000 Pa

s₄ = s₅ = 0.52087 kJ/(kg·K)

h₅ = h₄ + work done by the pump on the saturated liquid

∴ h₅ = h₄ + v₄ × (p₅ - p₄)

h₅ = 151.494 + 0.00100645 × (1000000 - 6000)/1000 = 152.4944113 kJ/kg

Step 6

Saturated liquid state

p₆ = 1 MPa = 10 bar

h₆ = 762.683 kJ/kg

s₆ = 2.1384 kJ/(kg·K)

v₆ = 0.00112723 m³/kg

Step 7

p₇ = p₁ = 12 MPa = 120 bar

s₇ = s₆ = 2.1384 kJ/(kg·K)

h₇ = h₆ + v₆ × (p₇ - p₆)

h₇ = 762.683  + 0.00112723 * (12 - 1) * 1000 = 775.08253 kJ/kg

The fraction of flow extracted at the second stage, y, is given as follows

y = \dfrac{762.683 - 152.4944113 }{2824.01 - 152.4944113 } = 0.2284

The turbine control volume is given as follows;

\dfrac{\dot{W_t}}{\dot{m_{1}}} = \left (h_{1} - h_{2}  \right ) + \left (1 - y  \right )\left (h_{2} - h_{3}  \right )

= (3507.41  - 2824.01) + (1 - 0.22840)*(2824.01 - 2058.56) = 1274.02122 kJ/kg

For the pumps, we have;

\dfrac{\dot{W_p}}{\dot{m_{1}}} = \left (h_{7} - h_{6}  \right ) + \left (1 - y  \right )\left (h_{5} - h_{4}  \right )

= (775.08253 - 762.683) + (1 - 0.22840)*(152.4944113 -  151.494)

= 13.17 kJ/kg

For the working fluid that flows through the steam generator, we have;

\dfrac{\dot{Q_{in}}}{\dot{m_{1}}} = \left (h_{1} - h_{7}  \right )

= 3507.41 - 775.08253 = 2732.32747 kJ/kg

The thermal efficiency, η, is given as follows;

\eta = \dfrac{\dfrac{\dot{W_t}}{\dot{m_{1}}} -\dfrac{\dot{W_p}}{\dot{m_{1}}}}{\dfrac{\dot{Q_{in}}}{\dot{m_{1}}}}

η = (1274.02122 - 13.17)/2732.32747 = 0.4615 which is 46.15%

(762.683 - 152.4944113)/(2824.01 - 152.4944113)

b. The mass flow rate, \dot{m_{1}}, into the first turbine stage is given as follows;

\dot{m_{1}} = \dfrac{\dot{W_{cycle}}}{\dfrac{\dot{W_t}}{\dot{m_{1}}} -\dfrac{\dot{W_p}}{\dot{m_{1}}}}

\dot{m_{1}} = 330 *1000/(1274.02122 - 13.17) = 261.73 kg/s

c. From the entropy rate balance of the steady state form, we have;

\dot{\sigma }_{cv} = \sum_{e}^{}\dot{m}_{e}s_{e} - \sum_{i}^{}\dot{m}_{i}s_{i} = \dot{m}_{6}s_{6} - \dot{m}_{2}s_{2} - \dot{m}_{5}s_{5}

\dot{\sigma }_{cv} = \dot{m}_{6} \left [s_{6} - ys_{2} - (1 - y)s_{5}  \right ]

= 261.73 * (2.1384 - 0.2284*6.6864 - (1 - 0.2284)*0.52087 = 54.79 kW/K

4 0
3 years ago
Ok there........................................................................
Juliette [100K]

Answer:

ok THERE

Explanation:

4 0
3 years ago
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