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GrogVix [38]
3 years ago
12

Which of the following types of valves have a valve stem that moves up and down in a valve guide? Select one: a. Poppet valves b

. Reed valves c. Check valves d. All of the above.
Engineering
2 answers:
Alja [10]3 years ago
5 0

Answer:c

Explanation:

thats wht it is

puteri [66]3 years ago
4 0

Poppet valves are the one whose valve stem moves up and down in a valve guide.

Answer: A

Explanation

Poppet valve is used for controlling the flow of fluid in the engine.

The valves are present inside a cylindrical structure from where the fluids will get inserted or exhausted.

The timing of input and output as well the quantity of input and output is maintained by Poppet valve.

The valve consists of a large metal plate connected with a long metal rod.

The metal rod is termed as valve stem which is present in the valve guide.

The valve stem will be moving up and down to insert and exhaust the gases respectively in an internal combustion engine.

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Which sentence is correct about the exergy of an empty (pressure around zero Pascal) tank with a volume of V, located in an envi
sineoko [7]

Answer:

The correct option is;

c.  the exergy of the tank can be anything between zero to P₀·V

Explanation:

The given parameters are;

The volume of the tank = V

The pressure in the tank = 0 Pascal

The pressure of the surrounding = P₀

The temperature of the surrounding = T₀

Exergy is a measure of the amount of a given energy which a system posses that is extractable to provide useful work. It is possible work that brings about equilibrium. It is the potential the system has to bring about change

The exergy balance equation  is given as follows;

X_2 - X_1 = \int\limits^2_1 {} \, \delta Q \left (1 - \dfrac{T_0}{T} \right ) - [W - P_0 \cdot (V_2 - V_1)]- X_{destroyed}

Where;

X₂ - X₁ is the difference between the two exergies

Therefore, the exergy of the system with regards to the environment is the work received from the environment which at is equal to done on the system by the surrounding which by equilibrium for an empty tank with 0 pressure is equal to the product of the pressure of the surrounding and the volume of the empty tank or P₀ × V less the work, exergy destroyed, while taking into consideration the change in heat of the system

Therefore, the exergy of the tank can be anything between zero to P₀·V.

6 0
3 years ago
True or False: The differential lock in an AWD-equipped vehicle can be used at any time.
Bingel [31]

the answer would be false

7 0
3 years ago
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The substance is steam (H2O). NOTE: The purpose of this problem is to illustrate that there are conditions where water vapor is
Gennadij [26K]

Answer:

See the attached pictures for detailed answer.

Explanation:

See the attached pictures for step by step explanation.

7 0
3 years ago
An interior beam supports the floor of a classroom in a school building. The beam spans 26 ft. and the tributary width is 16 ft.
saul85 [17]

Answer:

a. L_o  = 40 psf

b. L ≈ 30.80 psf

c. The uniformly distributed total load for the beam = 812.8 ft./lb

d. The alternate concentrated load is more critical to bending , shear and deflection

Explanation:

The given parameters of the beam the beam are;

The span of the beam = 26 ft.

The width of the tributary, b = 16 ft.

The dead load, D = 20 psf.

a. The basic floor live load is given as follows;

The uniform floor live load, = 40 psf

The floor area, A = The span × The width = 26 ft. × 16 ft. = 416 ft.²

Therefore, the uniform live load, L_o  = 40 psf

b. The reduced floor live load, L in psf. is given as follows;

L = L_o \times \left ( 0.25 + \dfrac{15}{\sqrt{k_{LL} \cdot A_T} } \right)

For the school, K_{LL} = 2

Therefore, we have;

L = 40 \times \left ( 0.25 + \dfrac{15}{\sqrt{2 \times 416} } \right) = 30.80126 \ psf

The reduced floor live load, L ≈ 30.80 psf

c. The uniformly distributed total load for the beam, W_d = b × W_{D + L} =

∴  W_d =  = 16 × (20 + 30.80) ≈ 812.8 ft./lb

The uniformly distributed total load for the beam, W_d = 812.8 ft./lb

d. For the uniformly distributed load, we have;

V_{max} = 812.8 × 26/2 = 10566.4 lbs

M_{max} =  812.8 × 26²/8 = 68,681.6 ft-lbs

v_{max} = 5×812.8×26⁴/348/EI = 4,836,329.333/EI

For the alternate concentrated load, we have;

P_L = 1000 lb

W_{D} = 20 × 16 = 320 lb/ft.

V_{max} = 1,000 + 320 × 26/2 = 5,160 lbs

M_{max} =  1,000 × 26/4 + 320 × 26²/8 = 33,540 ft-lbs

v_{max} = 1,000 × 26³/(48·EI) + 5×320×26⁴/348/EI = 2,467,205.74713/EI

Therefore, the loading more critical to bending , shear and deflection, is the alternate concentrated load

7 0
3 years ago
A design that is found to have flaws and must be redesigned and retested is an example of what
Aleksandr [31]

Answer:

A. prototype

Explanation:

7 0
3 years ago
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