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kenny6666 [7]
3 years ago
8

A jet impinges directly on to a plate that is oriented normal to the axis of the jet. The mass flow rate of the jet is 50 kg/min

and its velocity is 200 m/s. Perform a control volume analysis to calculate the force on the plate.
Engineering
1 answer:
PilotLPTM [1.2K]3 years ago
3 0

Answer:

166.67 N

Explanation:

Given:

Mass flow rate = 50 kg/min = 50 kg / 60 seconds = 0.833 kg/s

Initial velocity = 200 m/s

after striking the normal board the water will flow in the normal direction, thus the final velocity in the direction of the initial flow will be zero

therefore,

Force = (change in momentum)

or

Force = Initial momentum - final momentum

or

Force = 0.833 × 200 - 0.833 × 0

or

Force = 166.67 N

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A 03-series cylindrical roller bearing with inner ring rotating is required for an application in which the life requirement is
-BARSIC- [3]

Answer:

\mathbf{C_{10} = 137.611 \ kN}

Explanation:

From the information given:

Life requirement = 40 kh = 40 40 \times 10^{3} \ h

Speed (N) = 520 rev/min

Reliability goal (R_D) = 0.9

Radial load (F_D) = 2600 lbf

To find C10 value by using the formula:

C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}

where;

x_D = \text{bearing life in million revolution} \\  \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}

\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}

The Weibull parameters include:

x_o = 0.02

(\theta - x_o) = 4.439

b= 1.483

∴

Using the above formula:

C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}

C_{10}=3640 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{3}{10}}

C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}

C_{10} = 30962.449 \ lbf

Recall that:

1 kN = 225 lbf

∴

C_{10} = \dfrac{30962.449}{225}

\mathbf{C_{10} = 137.611 \ kN}

7 0
3 years ago
Currently, the lost time of each stage is 4 seconds, and intersection critical v/c ratio is set to be 0.75 to avoid cycle failur
KIM [24]

Find the solution in the attachments

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7 0
4 years ago
5. A pump operating at steady state receives 1.2 kg/s of liquid water at 50o C, 1.5 MPa. The pressure of the water at the pump e
Andrew [12]

Answer:

the work required by a reversible pump operating with the same conditions, in kW is 16.39 kW

the isentropic pump efficiency is 78%

Explanation:

Given that;

m = 1.2 kg/sec

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P1 = 1.5 Mpa

P2 = 15  Mpa

W-actual = 21 kw  

W reversible = m*Vf (p2 - p1)

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= 16.39 kW

the work required by a reversible pump operating with the same conditions, in kW is 16.39 kW

Isentropic Pump efficiency = W-reversible / W-actual

= 16.39 / 21 = 0.78

= 78%

the isentropic pump efficiency is 78%

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lesya [120]

Answer:

C

Explanation:

7 0
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