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juin [17]
3 years ago
14

Technician A says that the most effective way to diagnose wheel speed sensor circuit faults is to utilize the live data function

of your scan tool. Technician B says that once you have identified the affected circuit, the next step is to test at the sensor with a graphing multi-meter or an oscilloscope. Who is correct
Engineering
1 answer:
Marianna [84]3 years ago
7 0

Answer:

Both Technicians A & B are correct

Explanation:

In diagnosing wheel systems, both steps said by technicians A & B are correct because it's the normal procedure to diagnose wheel speed sensor circuit faults.

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Linus is using a calculator to multiply 5,426 and 30. He enters 5,426 x 300 by mistake. What can Linus do to correct his mistake
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3 0
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Read 2 more answers
1 kg of saturated steam at 1000 kPa is in a piston-cylinder and the massless cylinder is held in place by pins. The pins are rem
BARSIC [14]

Answer:

The final specific internal energy of the system is 1509.91 kJ/kg

Explanation:

The parameters given are;

Mass of steam = 1 kg

Initial pressure of saturated steam p₁ = 1000 kPa

Initial volume of steam, = V₁

Final volume of steam = 5 × V₁

Where condition of steam = saturated at 1000 kPa

Initial temperature, T₁  = 179.866 °C = 453.016 K

External pressure = Atmospheric = 60 kPa

Thermodynamic process = Adiabatic expansion

The specific heat ratio for steam = 1.33

Therefore, we have;

\dfrac{p_1}{p_2} = \left (\dfrac{V_2}{V_1} \right )^k = \left [\dfrac{T_1}{T_2}   \right ]^{\dfrac{k}{k-1}}

Adding the effect of the atmospheric pressure, we have;

p = 1000 + 60 = 1060

We therefore have;

\dfrac{1060}{p_2} = \left (\dfrac{5\cdot V_1}{V_1} \right )^{1.33}

P_2= \dfrac{1060}{5^{1.33}}  = 124.65 \ kPa

\left [\dfrac{V_2}{V_1} \right ]^k = \left [\dfrac{T_1}{T_2}   \right ]^{\dfrac{k}{k-1}}

\left [\dfrac{V_2}{V_1} \right ]^{k-1} = \left \dfrac{T_1}{T_2}   \right

5^{0.33} = \left \dfrac{T_1}{T_2}   \right

T₁/T₂ = 1.70083

T₁ = 1.70083·T₂

T₂ - T₁ = T₂ - 1.70083·T₂

Whereby the temperature of saturation T₁ = 179.866 °C = 453.016 K, we have;

T₂ = 453.016/1.70083 = 266.35 K

ΔU = 3×c_v×(T₂ - T₁)

c_v = cv for steam at 453.016 K = 1.926 + (453.016 -450)/(500-450)*(1.954-1.926) = 1.93 kJ/(kg·K)

cv for steam at 266.35 K = 1.86  kJ/(kg·K)

We use cv given by  (1.93 + 1.86)/2 = 1.895 kJ/(kg·K)

ΔU = 3×c_v×(T₂ - T₁) = 3*1.895 *(266.35 -453.016) = -1061.2 kJ/kg

The internal energy for steam = U_g = h_g -pV_g

h_g = 2777.12 kJ/kg

V_g = 0.194349 m³/kg

p = 1000 kPa

U_{g1} = 2777.12 - 0.194349 * 1060 = 2571.11 kJ/kg

The final specific internal energy of the system is therefore, U_{g1} + ΔU = 2571.11 - 1061.2 = 1509.91 kJ/kg.

3 0
3 years ago
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