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Mandarinka [93]
2 years ago
14

Bind hole, 38 diameter, .50 deep

Engineering
1 answer:
agasfer [191]2 years ago
4 0

Answer:

59.69021

Explanation:

38/.5 x 3.14159

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Cerrena [4.2K]
You are going to have to be more specific than that, perhaps add a picture next time ? Unfortunately, this question can’t be answered.
8 0
3 years ago
A steam power plant operating on a simple ideal Rankine cycle maintains the boiler at 6000 kPa, the turbine inlet at 600 °C, and
ohaa [14]

Answer:

ηa=0.349

ηb=0.345

Explanation:

The enthalpy and entropy at state 3 are determined from the given pressure and temperature with data from table:

h_{3}=3658.8kJ/kg\\ s_{3}=7.1693kJ/kg

The quality at state 4 is determined from the condition  s_{4} =s_{3} and the entropies of the components at the condenser pressure taken from table:

 q_{4} =\frac{s_{4}-s_{liq50}  }{s_{evap,50} } \\=\frac{7.1693-1.0912}{6.5019}=0.935

The enthalpy at state 4 then is:  

h_{4} =h_{liq50} +q_{4} h_{evap,50}\\ (340.54+0.935*2304.7)kJ/kg\\=2495.43kJ/kg\\

Part A

In the case when the water is in a saturated liquid state at the entrance of the pump the enthalpy and specific volume are determined from A-5 for the given pressure:  

h_{1}=340.54kJ/kg\\ \alpha _{1}=0.00103m^3/kg

The enthalpy at state 2 is determined from an energy balance on the pump:

h_{2} =h_{1} +\alpha _{1}( P_{2}-P_{1}  )

    =346.67 kJ/kg

The thermal efficiency is then determined from the heat input and output in the cycle:  

=1-\frac{q_{out} }{q_{in} } \\=1-\frac{h_{4} -h_{1} }{h_{3} -h_{2}} \\=0.349

Part B  

In the case when the water is at a lower temperature than the saturation temperature at the condenser pressure we look into table and see the water is in a compressed liquid state. Then we take the enthalpy and specific volume for that temperature with data from  and the saturated liquid values:  

h_{1}=293.7kJ/kg\\ \alpha _{1}=0.001023m^3/kg

The enthalpy at state 2 is then determined from an energy balance on the pump:  

h_{2} =h_{1} +\alpha _{1}( P_{2}-P_{1}  )

    =299.79 kJ/kg  

The thermal efficiency in this case then is:  

=1-\frac{q_{out} }{q_{in} } \\=1-\frac{h_{4} -h_{1} }{h_{3} -h_{2}} \\=0.345

8 0
3 years ago
How many numbers multiple of 3 are in the range [2,2000]?
Alekssandra [29.7K]
I don’t even know lol i’m just tryna bring my points up sorry.
6 0
3 years ago
A spherical tank for storing gas under pressure is 25 m in diameter and is made of steel 15 mm thick. The yield point of the mat
nalin [4]

Answer:

a) (option B) 230 kPa

b) (option A) 100 N/m

Explanation:

Given:

Diameter, d = 25 m

Thicknesses, t = 15 mm

Yield point = 240 MPa

Factor of safety = 2.5

a) To find the maximum internal pressure, let's use the formula:

\sigma l = \frac{\sigma y}{FOS} = \frac{PD}{4t}

\frac{\sigma y}{FOS} = \frac{PD}{4t}

Solving for P, we have:

P = \frac{\sigma y * 4t}{FOS * D}

P = \frac{240 * 4 * 15}{2.5 * 25}

P = 230.4 kPa

≈ 230 kPa

The maximum permissible internal pressure is nearly 230kPa

b) Given:

Thickness, t = 6.35 mm

L = 203 mm

Torque, T = 8 N m

Let's find the mean Area,

mA = (l - t)²

= (203 - 6.5)²

= 38671.22mm²

≈ 0.03867 m² (converted to meters)

To find the average shear flow, let's use the formula:

q = \frac{T}{2* mA}

= \frac{8}{2 * 0.03867}

q = 103.4 N/m approximately 100N/m

The average shear force flow is most nearly 100 N/m

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3 years ago
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Answer:

From the derivation in the attachment below it

Is clear that discrete time receives phase response

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