A video inspection snake is use

. Using the Newton Raphson method, determine the uniform flow depth in a trapezoidal channel with a bottom width of 3.0 m and si

Over [174]

Answer:

y ≈ 2.5

Explanation:

Given data:

bottom width is 3 m

side slope is 1:2

discharge is 10 m^3/s

slope is 0.004

manning roughness coefficient is 0.015

manning equation is written as

$v =1/n R^{2/3} s^{1/2}$

where R is hydraulic radius

S = bed slope

$Q = Av =A 1/n R^{2/3} s^{1/2}$

$A = 1/2 \times (B+B+4y) \times y =(B+2y) y$

$R =\frac{A}{P}$

P is perimeter $= (B+2\sqrt{5} y)$

$R =\frac{(3+2y) y}{(3+2\sqrt{5} y)}$

$Q = (2+2y) y) \times 1/0.015 [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} 0.004^{1/2}$

solving for y $100 =(2+2y) y) \times (1/0.015) [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} \times 0.004^{1/2}$

solving for y value by using iteration method ,we get

y ≈ 2.5

5 0

3 years ago

• Build upon the results of problem 3-85 to determine the minimum factor of safety for fatigue based on infinite life, using the

Rudik [331]

Answer:

minimum factor of safety for fatigue is = 1.5432

Explanation:

given data

AISI 1018 steel cold drawn as table

ultimate strength Sut = 63.800 kpsi

yield strength Syt = 53.700 kpsi

modulus of elasticity E = 29.700 kpsi

we get here

$\sigma a$ = $\sqrt{(\sigma a \times kb)^2+3\times (za\times kt)^2}$ ...........1

here kb and kt = 1 combined bending and torsion fatigue factor

put here value and we get

$\sigma a$ = $\sqrt{(12 \times 1)^2+3\times (0\times 1)^2}$

$\sigma a$ = 12 kpsi

and

$\sigma m$ = $\sqrt{(\sigma m \times kb)^2+3\times (zm\times kt)^2}$ ...........2

put here value and we get

$\sigma m$ = $\sqrt{(-0.9 \times 1)^2+3\times (10\times 1)^2}$

$\sigma m$ = 17.34 kpsi

now we apply here goodman line equation here that is

$\frac{\sigma m}{Sut} + \frac{\sigma a}{Se} = \frac{1}{FOS}$ ...................3

here Se = 0.5 × Sut

Se = 0.5 × 63.800 = 31.9 kspi

put value in equation 3 we get

$\frac{17.34}{63.800} + \frac{12}{31.9} = \frac{1}{FOS}$

solve it we get

FOS = 1.5432

6 0

3 years ago

A Carnot refrigeration cycle absorbs heat at -12 °C and rejects it at 40 °C. a)-Calculate the coefficient of performance of this

tresset_1 [31]

Answer:

a)COP=5.01

b) $W_{in}=2.998$ KW

c)COP=6.01

d) $Q_R=17.99 KW$

Explanation:

Given

$T_L$ = -12°C, $T_H$ =40°C

For refrigeration

We know that Carnot cycle is an ideal cycle that have all reversible process.

So COP of refrigeration is given as follows

$COP=\dfrac{T_L}{T_H-T_L}$ ,T in Kelvin.

$COP=\dfrac{261}{313-261}$

a)COP=5.01

Given that refrigeration effect= 15 KW

We know that $COP=\dfrac{RE}{W_{in}}$

RE is the refrigeration effect

So

5.01= $\dfrac{15}{W_{in}}$

b) $W_{in}=2.998$ KW

For heat pump

So COP of heat pump is given as follows

$COP=\dfrac{T_h}{T_H-T_L}$ ,T in Kelvin.

$COP=\dfrac{313}{313-261}$

c)COP=6.01

In heat pump

Heat rejection at high temperature=heat absorb at low temperature+work in put

$Q_R=Q_A+W_{in}$

Given that $Q_A=15$ KW

We know that $COP=\dfrac{Q_R}{W_{in}}$

$COP=\dfrac{Q_R}{Q_R-Q_A}$

$6.01=\dfrac{Q_R}{Q_R-15}$

d) $Q_R=17.99 KW$

5 0

3 years ago

Suppose you want to buy a new car. The car will be used mainly for going to work, shopping, running errands, and visiting friend

Sergeu [11.5K]

1. So you can complete the things you NEED to do faster. (I.e work, groceries)

2. So it’s easier for you to do leisure activities. (I.e visiting friends, shopping)

5 0

3 years ago

The function below takes a two parameters: a list called a_list and a value called a_value. Complete the function to first check

Alik [6]

Explanation:

def checklist(a_list, a_value):// def //used to name a function, the //arguments being passed into the //functions are a_list and a_value

if a_value in list://if statement to

//check if it's in the list or not

print("")//print statement to do

//nothing

else://else statement for if the

//a_value is in a_list

a_list.append(a_value)//this adds

//the value to the list

5 0

3 years ago