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katovenus [111]
3 years ago
13

Find the percent change in cutting speed required to give an 80% reduction in tool life when the value of n is 0.12.

Engineering
1 answer:
vaieri [72.5K]3 years ago
6 0

Answer:21.3%

Explanation:

Given

80 % reduction in tool life

According to Taylor's tool life

VT^n=c

where V is cutting velocity

T=tool life of tool

80 % tool life reduction i.e. New tool Life is 0.2T

Thus

VT^{0.12}=V'\left ( 0.2T\right )^{0.12}

V'=\frac{V}{0.2^{0.12}}

V'=\frac{V}{0.824}=1.213V

Thus a change of 21.3 %(increment) is required to reduce tool life by 80%

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Technician A says that a magnetic field can be created by current flow. Technician B says that current can be induced by moving
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Answer:

Both

Explanation:

Magnetic fields are created by current flow (Biot-Savart Law) and current is induced by moving a wire through a magnetic field (Faraday’s Law).

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If there are 16 signal combinations (states) and a baud rate (number of signals/second) of 8000/second, how many bps could I sen
Mademuasel [1]

Answer:

32000 bits/seconds

Explanation:

Given that :

there are 16  signal combinations (states) = 2⁴

bits  n = 4

and a baud rate (number of signals/second) = 8000/second

Therefore; the number of bits per seconds can be calculated as follows:

Number of bits per seconds = bits  n × number of signal per seconds

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The water in an 8-m-diameter, 3-m-high above-ground swimming pool is to be emptied by unplugging a 3-cm-diameter, 25-m-long hori
Dima020 [189]

Answer:

The maximum discharge rate of water is 4.6 L/s

Explanation:

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h=height=3 m

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Here

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3 years ago
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A well-insulated tank in a vapor power plant operates at steady state. Saturated liquid water enters at inlet 1 at a rate of 125
Gekata [30.6K]

Answer:

a) \dot m_{3} = 135\,\frac{lbm}{s}, b) h_{3}=168.965\,\frac{BTU}{lbm}, c) T = 200.829\,^{\textdegree}F

Explanation:

a) The tank can be modelled by the Principle of Mass Conservation:

\dot m_{1} + \dot m_{2} - \dot m_{3} = 0

The mass flow rate exiting the tank is:

\dot m_{3} = \dot m_{1} + \dot m_{2}

\dot m_{3} = 125\,\frac{lbm}{s} + 10\,\frac{lbm}{s}

\dot m_{3} = 135\,\frac{lbm}{s}

b) An expression for the specific enthalpy at outlet is derived from the First Law of Thermodynamics:

\dot m_{1}\cdot h_{1} + \dot m_{2} \cdot h_{2} - \dot m_{3}\cdot h_{3} = 0

h_{3} = \frac{\dot m_{1}\cdot h_{1}+\dot m_{2}\cdot h_{2}}{\dot m_{3}}

Properties of water are obtained from tables:

h_{1}=180.16\,\frac{BTU}{lbm}

h_{2}=28.08\,\frac{BTU}{lbm} + \left(0.01604\,\frac{ft^{3}}{lbm}\right)\cdot (14.7\,psia-0.25638\,psia)

h_{2}=29.032\,\frac{BTU}{lbm}

The specific enthalpy at outlet is:

h_{3}=\frac{(125\,\frac{lbm}{s} )\cdot (180.16\,\frac{BTU}{lbm} )+(10\,\frac{lbm}{s} )\cdot (29.032\,\frac{BTU}{lbm} )}{135\,\frac{lbm}{s} }

h_{3}=168.965\,\frac{BTU}{lbm}

c) After a quick interpolation from data availables on water tables, the final temperature is:

T = 200.829\,^{\textdegree}F

8 0
3 years ago
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