What causes the charging system warning lamp to go out when the engine starts up?

A room is kept at −5°C by a vapor-compression refrigeration cycle with R-134a as the refrigerant. Heat is rejected to cooling wa

Fed [463]

Answer:

note:

solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment

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4 0

3 years ago

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is released from under a vertical gate into a 2-mwide lined rectangular channel. The gate opening is 50 cm, and the flow rate in

SVEN [57.7K]

Answer:

hello your question is incomplete attached below is the complete question

answer: There is a hydraulic jump

Explanation:

First we have to calculate the depth of flow downstream of the gate

y1 = $C_{c} y_{g}$ ----------- ( 1 )

Cc ( concentration coefficient ) = 0.61 ( assumed )

Yg ( depth of gate opening ) = 0.5

hence equation 1 becomes

y1 = 0.61 * 0.5 = 0.305 m

calculate the flow per unit width q

q = Q / b ----------- ( 2 )

Q = 10 m^3 /s

b = 2 m

hence equation 2 becomes

q = 10 / 2 = 5 m^2/s

next calculate the depth before hydraulic jump y2 by using the hydraulic equation

answer : where y1 < y2 hence a hydraulic jump occurs in the lined channel

attached below is the remaining part of the solution

4 0

3 years ago

Ultimate tensile strength is: (a) The stress at 0.2% strain (b) The stress at the onset of plastic deformation (c) The stress at

MissTica

Answer:

By definition the ultimate tensile strength is the maximum stress in the stress-strain deformation. The stress at 0.2% strain, the stress at the onset of plastic deformation, the stress at the end of the elastic deformation and the stress at the fracture correspond, by definition, to other points of the stress-strain curve.

Explanation:

4 0

3 years ago

A 3-phase , 1MVA, 13.8kV/4160V, 60 Hz, transformer with Y-Delta winding connection is supplying a3-phase, 0.75 p.u. load on the

Tanya [424]

Answer:

a) 23.89 < -25.84 Ω

b) 31.38 < 25.84 A

c) 0.9323 leading

Explanation:

A) Calculate the load Impedance

current on load side = 0.75 p.u

power factor angle = 25.84

$I_{load}$ = 0.75 < 25.84°

attached below is the remaining part of the solution

B) Find the input current on the primary side in real units

load current in primary = 31.38 < 25.84 A

C) find the input power factor

power factor = 0.9323 leading

attached below is the detailed solution

8 0

3 years ago

QUESTION 3

lianna [129]

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6 0

3 years ago