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3 years ago

8. In a closed hydraulic brake system, the hydraulic pressure:

Engineering

2 answers:

beks73 [17]3 years ago

8 0

Answer:

Option A is correct.

The hydraulic pressure in a closed hydraulic brake system is constant.

Explanation:

The hydraulic brake system works on the principle of pressure transmission. It is termed a closed hydraulic brake system when there is no leakage of fluid all through the setup. The transmission works on the basis of Pascal's law for pressure and it explains that pressure applied to a fluid is transmitted undiminished to every part of the fluid.

The force applied on the brake pedals by the driver at the controlling end of the hydraulic system is transmitted by the pushrod on the piston(s) in the master cylinder.

This causes fluid from the brake fluid reservoir to flow into a pressure chamber. This in turn causes an increase in the pressure of the entire hydraulic system, forcing fluid through the hydraulic lines toward the caliper(s) where it then acts upon the caliper piston(s).

The force applied on the brakes pedal, F₁,

is applied on the cross sectional Area of the master piston, A₁ and this is transmitted in terms of pressure by the fluid (with no loss in pressure, that is, constant pressure, if the hydraulic braking system is especially a closed one) through the callipers to the calliper(s) pistons.

Hope this Helps!!!

DiKsa [7]3 years ago

3 0

Answer:

A. Is constant

Explanation:

This is in accordance with pascal's law of pressure that states the hydraulic pressure in a fluid is transmitted undiminished to every portion of the fluid.

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Air at 2.5 bar, 400 K is extracted from a main jet engine compressor for cabin cooling. The extracted air enters a heat exchange

Shkiper50 [21]

Answer:

a) Power developed by the turbine = 132.89 kW

b) magnitude of the rate of heat transfer from the air to the ambient, in kw = 251.25 kW

Explanation:

b) The process is a constant pressure process (Isobaric process)

The constant pressure specific heat of air, $c_{p} = 1.005 kJ/kg -K$

Specific heat ratio for air, $\gamma = 1.4$

The mass flow rate of air, $\dot{m} = 2.5 kg/s$

P₁ = 2.5 bar, T₁ = 400 K

P₂ = 2.5 bar, T₂ = 300 K

Using the steady flow energy equation:

$Q_{1-2} = \dot{m} c_{p} (T_{2} - T_{1} \\Q_{1-2} = 2.5 * 1.005 * (300 - 400)\\Q_{1-2} = -251.25 kW$

Therefore, the magnitude of the rate of heat transfer from the air to the ambient, in kw, $Q_{1-2} = 251.25 kW$

a) For the isentropic process:

Power developed by the turbine is given by the relation $\dot{W} = \dot{M} c_{p} (T_{2} - T_{3})$

Isentropic efficiency, $\eta_{t} = 80%$

P₂ = 2.5 bar, T₂ = 300 K

P₃ = 1 bar, $T_{3s}$ = ? where $T_{3s}$ is the isentropic temperature at 100% efficiency

The isentropic relation is given by:

$\frac{T_{3s} }{T_{2} } = (\frac{P_{3} }{P_{2} }) ^{\frac{\gamma - 1}{\gamma} } \\\frac{T_{3s} }{300 } = (\frac{1 }{2.5 }) ^{\frac{1.4 - 1}{1.4 }$

$T_{3s} = 230.9 K$

To get the temperature at 80% efficiency, we will use the relation:

$\eta_{t} = \frac{T_{2} - T_{3} }{T_{2} - T_{3s} } \\0.8= \frac{300 - T_{3} }{300 - 230.9 }$

T₃ = 244.72 K

Power developed by the turbine is given by the relation:

$\dot{W} = \dot{M} c_{p} (T_{2} - T_{3})\\ \dot{W} = 2.5 * 1.005* (300-244.72)\\ \dot{W} = 138.89 kW$

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3 years ago

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Which of the following is not a relationship set between elements in a sketch

Svet_ta [14]

Answer:

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2 years ago

The basic barometer can be used to measure the height of a building. If the barometric readings at the top and at the bottom of

bixtya [17]

Answer:

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Explanation:

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we have also

$(\rho gh)_{air} = p_b - p_t$

1.18*9.81*h = (100.4 -97.08)*10^3

h = 287.1 m

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Technician A is wrong.

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See the link below for more about automobile engineering:

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