All categories

3 years ago

4. Which type of duct undergoes more rigorous testing before it's labeled?

Engineering

1 answer:

Alenkasestr [34]3 years ago

7 0

Answer:

Flexible duct

Explanation:

it is flexible duct due to current flow

You might be interested in

6. Question

valkas [14]

Answer:

Check the 2nd, 3rd and 4th statements.

Explanation:

4 0

3 years ago

An engine operates on gasoline (LHV=44 MJ/kg) with a brake thermal efficiency of 37.9 % What is the brake specific fuel consumpt

scZoUnD [109]

Answer:

$s =0.21\ kg/Kw.hr$

Explanation:

Given that

Calorific value (CV) = 44 MJ/Kg

CV= 44,000 KJ/kg

Brake thermal efficiency(η) = 37.9 %

We know that

$\eta =\dfrac{BP}{\dot{m_f}\times CV}$

Where BP is the brake power

$\eta =\dfrac{BP}{\dot{m_f}\times CV}$

$0.379 =\dfrac{BP}{\dot{m_f}\times 44000}$

$\dfrac{BP}{\dot{m_f}}=16676$

Brake specific fuel consumption (s)

$s =\dfrac{\dot{m_f}}{BP}$

$s =\dfrac{3600\times \dot{m_f}}{BP}$

$s =\dfrac{3600}{16,676}\ kg/Kw.hr$

$s =0.21\ kg/Kw.hr$

7 0

3 years ago

If you are unsure about holding a piece of wood to be drilled, then you should always use a

alisha [4.7K]

C I took construction class

4 0

3 years ago

A DC servomotor has a torque constant of 0.075 N-m/A and a voltage constant of 0.12 V/(rad/sec). The armature resistance is 2.5

quester [9]

Answer:

The answer is below

Explanation:

Given that:

$k_t=torque\ constant=0.075\ Nm/A\\\\k_v=voltage\ constant=0.12\ V/(rad/sec)\\\\R_a=armature \ resistance=2.5 \Omega\\\\V_t=terminal\ voltage=24\ V\\\\a)The \ starting\ current\ I_a\ is\ given \ as:\\\\I_a=\frac{V_t}{R_a} =\frac{24}{2.5} =9.6\ A\\\\The \ starting\ torque(T)\ is:\\\\T=k_tI_a=0.075*9.6=0.72\ N.m$

b) The maximum speed occurs when the terminal voltage and back emf are equal to each other i.e.

$V_t=e_b=k_v\omega\\\\\omega=\frac{V_t}{k_v}=\frac{24}{0.12} =200\ rad/s$

c) The load torque is given as:

$T_L=0.0125\Omega\\\\The\ motor\ torque \ is:\\\\T=k_t(\frac{V_t-k_v\omega}{R_a} )\\\\but\ T = T_L,hence:\\\\0.0125\omega=0.075(\frac{24-0.12\omega}{2.5} )\\\\0.03125\omega=1.8-0.009\omega\\\\0.04025\omega=1.8\\\\\omega=44.72\ rad/sec\\\\N=\frac{60\omega}{2\pi} =\frac{60*44.72}{2\pi} =427\ rpm$

4 0

3 years ago

2. A fluid at 14.7 psi (lb-f per square inch) with kinematic viscosity (????????) 1.8 x10-4 ft2/sec and density(????????) 0.076

jenyasd209 [6]

Answer:

See explaination

Explanation:

We are going to define Pressure drop as the difference in total pressure between two points of a fluid carrying network. A pressure drop usually occurs when frictional forces, caused by the resistance to flow, act on a fluid as it flows through the tube.

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

4 0

4 years ago