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2 years ago

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For the storage tank of Prob. 7.104, determine the maximum normal stress and the maximum shearing stress in the cylindrical wall

when the tank is filled to capacity (h

1 answer:

lara31 [8.8K]2 years ago

7 0

Answer:

The unpressurized cylindrical storage tank shown has a 5-mm wall thickness and is made of steel having a 400-MPa ultimate strength in tension. Determine the maximum height h to which it can be filled with water if a factor of safety of 4.0 is desired.

Explanation:

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In engineering, economic cost is a decision-making tangible factor. Group of answer choices True False

salantis [7]

Economic cost is a rescission making tangible factor true

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3 years ago

Read 2 more answers

A circuit contains a 40 ohm resistor and a 60 ohms resistor connected in parallel. If you test this circuit with a DMM you shoul

lorasvet [3.4K]

<h2>Answer:</h2>

24Ω

<h2>Explanation:</h2>

When resistors are connected in parallel, the reciprocal of their combined resistance, when read with a DMM (Digital Multimeter - for measuring various properties of a circuit or circuit element such as resistance...) is the sum of the reciprocals of their individual resistances.

For example if two resistors of resistances R₁ and R₂ are connected together in parallel, the reciprocal of their combined resistance Rₓ is given by;

$\frac{1}{R_x}$ = $\frac{1}{R_1}$ + $\frac{1}{R_2}$

Solving for Rₓ gives;

$R_{x}$ = $\frac{R_1 * R_2}{R_1 + R_2}$ ------------------(i)

From the question;

Let

R₁ = resistance of first resistor = 40Ω

R₂ = resistance of second resistor = 60Ω

Now,

To get their combined or total resistance, Rₓ, substitute these values into equation (i) as follows;

$R_{x}$ = $\frac{40 * 60}{40 + 60}$

$R_{x}$ = $\frac{2400}{100}$

$R_{x}$ = 24 Ω

Therefore, the total resistance is 24Ω

4 0

3 years ago

The idling engines of a landing turbojet produce forward thrust when operating in a normal manner, but they can produce reverse

sertanlavr [38]

Answer:

T = 5416.67 N

T = -2083.5 N

T = 0

Explanation:

Forward thrust has positive values and reverse thrust has negative values.

part a

Flight speed u = ( 150 km / h ) / 3.6 = 41.67 km / s

The thrust force represents the horizontal or x-component of momentum equation:

$T = flow(m_{exhaust})*(u_{exhaust} - u_{flight} )\\T = (50 kg/s ) * (150 - 41.67)\\\\T = 5416.67 N$

Answer: The thrust force T = 5416.67 N

part b

Now the exhaust velocity is now vertical due to reverse thrust application, then it has a zero horizontal component, thus thrust equation is:

$T = flow(m_{exhaust})*(u_{exhaust} - u_{flight} )\\T = (50 kg/s ) * (0 - 41.67)\\\\T = -2083.5 N$

Answer: The thrust force T = -2083.5 N reverse direction

part c

Now the exhaust velocity and flight velocity is zero, then it has a zero horizontal component, thus thrust is also zero as there is no difference in two velocities in x direction.

Answer: T = 0 N

5 0

3 years ago

During the pre-drive check, you'll want to observe the car from the _______.

DENIUS [597]

Answer:

Passenger seat

Explanation:

If im wrong correct me

5 0

3 years ago

The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me

masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

$\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0$ Equation 1

Similarly, the equation for outer node “o” will be

$\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0$ Equation 2

The conventive thermal resistance in i-node will be

$R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w$ Equation 3

The conventive hermal resistance per unit area is

$R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w$ Equation 4

The conductive thermal resistance per unit area is

$R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w$ Equation 5

Since $q_{rad}$ is given as 100, $T_{o}$ is 40 $T_ \infty$ is 300 $T_{\infty, o}$ is 25

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

$\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0$ Equation 6

$\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0$

$T_{i}-40= \frac {L}{0.05}*150$

$T_{i}-40=3000L$

$T_{i}=3000L+40$ Equation 7

From equation 6 we can substitute wherever there’s $T_{i}$ with 3000L+40 as seen in equation 7 hence we obtain

$\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0$

The above can be simplified to be

$\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0$

$\frac {260-3000L}{0.033}=50$

-3000L=1.665-260

$L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm$

Therefore, insulation thickness is 86mm

8 0

3 years ago