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8090 [49]
2 years ago
10

The pressure gauge on a tank of CO2 used to fill Dora-water bottles reads 51.0 psi.

Engineering
1 answer:
tensa zangetsu [6.8K]2 years ago
3 0

Known :

Pg = 51 psi

Patm = 28 inHg = 13.7 psi

Solution :

Absolute pressure is gauge pressure plus the atmospheric pressure :

Pabs = Pg + Patm

Pabs = 51 + 13.7

Pabs = 64.8 psi

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Semmy [17]
A boy eat a energy of a sandwich to run a race because when they eat a sandwich it helps them to help it mid workout and real nutritions of NYC and bring extra fuel and eating the right thing
I hope this help
4 0
3 years ago
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You find an unnamed fluid in the lab we will call Fluid A. Fluid A has a specific gravity of 1.65 and a dynamic viscosity of 210
Naily [24]

Answer:

1.2727 stokes

Explanation:

specific gravity of fluid A = 1.65

Dynamic viscosity = 210 centipoise

<u>Calculate the kinematic viscosity of Fluid A </u>

First step : determine the density of fluid A

Pa = Pw * Specific gravity =  1000 * 1.65 = 1650 kg/m^3

next : convert dynamic viscosity to kg/m-s

210 centipoise = 0.21 kg/m-s

Kinetic viscosity of Fluid A = dynamic viscosity / density of fluid A

                                            = 0.21 / 1650 = 1.2727 * 10^-4 m^2/sec

Convert to stokes = 1.2727 stokes

4 0
3 years ago
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
A 50-kN hydraulic press performs pressing and clamping actions. The clamping cylinder force is 4 kN. The pressing cylinder strok
galben [10]

Answer:

The attached figure shows the hydraulic circuit using one sequence valve to control two simultaneous operations performed in proper sequence in one direction only. In the other direction, both the operations are simultaneous.

When we keep the 4/2 DCV in crossed arrow position, oil under pressure is supplied to the inlet port of the sequence valve. It directly flows to Head end port-1. Hence Cylinder 'C1' extends first.

By the end of the extension of cylinder 'C1', pressure in the line increases and hence poppet of sequence valve is lifted off from its seat and allows oil to flow to port-2 and hence, Cylinder 'C2 extends completing the pressing operation.

In the straight-arrow position of 4/2 DCV the oil under pressure reaches the rod end of both the cylinders C1 and C2 simultaneously through port-3. This causes both the cylinders to retract simultaneously.

Also, a Flow control valve is provided tho control the velocity of clamping

Explanation:

find attached the figure

4 0
3 years ago
What is the metal removal rate when a 2 in-diameter hole 3.5 in deep is drilled in 1020 steel at cutting speed of 120 fpm with a
Studentka2010 [4]

Answer:

a) the metal removal rate is 14.4 in³/min

b) the cutting time is 0.98 min

Explanation:

Given the data from the question

first we find the rpm for the spindle of the drilling tool, using the equation

Ns = 12V/πD

V is the cutting speed(120 fpm) and D is the diameter of the hole( 2 in)

so we substitute

Ns = 12 × 120 / π2

Ns = 1440 / 6.2831

Ns = 229.18 rmp

Now we find the metal removal rate using the equation

MRR = (πD²/4) Fr × Ns

Fr is the feed rate( 0.02 ipr ),

so we substitute

MRR = ((π × 2²)/4) × 0.02 × 229.18

MRR = 14.3998 ≈ 14.4 in³/min

Therefore the metal removal rate is 14.4 in³/min

Next we find the allowance for approach of the tip of the drill

A = D/2

A = 2/2

= 1 in

now find the time required to drill the hole

Tm = (L + A) / (Fr × Ns)

Lis the the depth of the hole( 3.5 in)

so we substitute our values

Tm = (3.5 + 1) / (0.02 × 229.18  )

Tm = 4.5 / 4.5836

Tm = 0.98 min

Therefore the cutting time is 0.98 min

8 0
2 years ago
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