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serg [7]
3 years ago
9

A cooking pan whose inner diameter is 20 cm is filled with water and covered with a 4-kg lid. If the local atmospheric pressure

is 101 kPa, determine the temperature at which the water starts boiling when it is heated

Engineering
2 answers:
sashaice [31]3 years ago
8 0

Answer:

T=100.2^0C

Explanation:

The boiling point of the water depends only on the absolute pressure p. the atmospheric temperature is given by P=101000 pa and the pressure from the lid we calculate by dividing its weight with the surface area of lid. the weight is product of  mass and the gravitational acceleration. surface area is calculated by given diameter.

D=20cm=20cm.\frac{1m}{100cm} =0.2m\\A=D^2\pi /4\\p_{l} =W/A=\frac{m.g}{D^2\pi/4 }

The absolute pressure p is sum of atmospheric pressure and  the pressure from the weight of lid.

p=p_{a} +p_{l} \\\\p=p_{a} +\frac{m.g}{D^2\pi /4} \\\\p=101000Pa+\frac{4kg.9.81m/s^2}{(0.2m^2)^2.\pi /4} \\p=102249Pa

At this pressure the boiling temperature of water is 100.2.

Arisa [49]3 years ago
7 0

Explanation:

Below is an attachment containing the solution.

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Lady_Fox [76]

Answer:

The maximum possible volume flow of gasoline is 0.543 m^3/s

Explanation:

Power = pressure differential × volume flow rate

Power = 3.8 kW

Pressure differential = 7 kPa

Volume flow rate = power ÷ pressure differential = 3.8 ÷ 7 = 0.543 m^3/s

3 0
4 years ago
The shaft is made of A992 steel. It has a diameter of 1 in. and is supported by bearings at A and D, which allows free rotation.
zysi [14]

Answer:

the angle of twist of B with respect to D is -1.15°

the angle of twist of C with respect to D is 1.15°

Explanation:

The missing diagram that is supposed to be added to this image is attached in the file below.

From the given information:

The shaft is made of A992 steel. It has a diameter of 1 in and is supported by bearing at A and D.

For the Modulus of Rigidity  G = 11 × 10³ Ksi =  11 × 10⁶ lb/in²

The objective are :

1) To determine the angle of twist of B with respect to D

Considering the Polar moment of Inertia at the shaft J\tau

shaft J\tau = \dfrac{\pi}{2}r^4

where ;

r = 1 in /2

r = 0.5 in

shaft J \tau = \dfrac{\pi}{2} \times 0.5^4

shaft J\tau = 0.098218

Now; the angle of twist at  B with respect to D  is calculated by using the expression

\phi_{B/D} = \sum \dfrac{TL}{JG}

\phi_{B/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}

where;

T_{CD} \ \  and \ \  L_{CD} are the torques at segments CD and length at segments CD

{T_{BC} \  \ and  \ \ L_{BC}} are the torques at segments BC and length at segments BC

Also ; from the diagram; the following values where obtained:

L_{BC}} = 2.5  in

J\tau = 0.098218

G =  11 × 10⁶ lb/in²

T_{BC = -60 lb.ft

T_{CD = 0 lb.ft

L_{CD = 5.5 in

\phi_{B/D} = 0+ \dfrac{[(-60 \times 12 )] (2.5 \times  12 )}{ (0.9818)(11 \times 10^6)}

\phi_{B/D} = \dfrac{[(-720 )] (30 )}{1079980}

\phi_{B/D} = \dfrac{-21600}{1079980}

\phi_{B/D} = − 0.02 rad

To degree; we have

\phi_{B/D}  = -0.02 \times \dfrac{180}{\pi}

\mathbf{\phi_{B/D}  = -1.15^0}

Since we have a negative sign; that typically illustrates that the angle of twist is in an anti- clockwise direction

Thus; the angle of twist of B with respect to D is 1.15°

(2) Determine the angle of twist of C with respect to D.Answer unit: degree or radians, two decimal places

For  the angle of twist of C with respect to D; we have:

\phi_{C/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}

\phi_{C/D} = 0+\dfrac{T_{BC}L_{BC}}{JG}

\phi_{B/D} = 0+ \dfrac{[(60 \times 12 )] (2.5 \times  12 )}{ (0.9818)(11 \times 10^6)}

\phi_{C/D} = \dfrac{21600}{1079980}

\phi_{C/D} = 0.02 rad

To degree; we have

\phi_{C/D}  = 0.02 \times \dfrac{180}{\pi}

\mathbf{\phi_{C/D}  = 1.15^0}

3 0
3 years ago
In a planetary geartrain with a form factor of 8, the sun gear rotates clockwise at 5 rad⁄s and the ring gear rotates clockwise
lina2011 [118]

Answer:

D. N= 11. 22 rad/s (CW)

Explanation:

Given that

Form factor R = 8

Speed of sun gear = 5 rad/s (CW)

Speed of ring gear = 12 rad/s (CW)

Lets take speed of carrier gear is N

From Algebraic method ,the relationship between speed and form factor given as follows

\dfrac{N_{sun}-N}{N_{ring}-N}=-R

here negative sign means that ring and sun gear rotates in opposite direction

Lets take CW as positive and ACW as negative.

Now by putting the values

\dfrac{N_{sun}-N}{N_{ring}-N}=-R

\dfrac{5-N}{12-N}=-8

N= 11. 22 rad/s (CW)

So the speed of carrier gear is 11.22 rad/s clockwise.

8 0
3 years ago
18
a_sh-v [17]

Answer:

englishhhhhhhhh

Explanation:

we dont understand

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3 years ago
The purpose of pasteurizing milk is to
katen-ka-za [31]

Answer:

i think it c

Explanation:

6 0
3 years ago
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