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mrs_skeptik [129]
3 years ago
11

A workpiece of 2000 mm length and 300 mm width was machined by a planning operation with the feed set at 0.3 mm/stroke. If the m

achine tool executes 10 double strokes/min, the planning time for a single pass will be?
Engineering
1 answer:
tiny-mole [99]3 years ago
6 0

Answer:

The planning time of the planner machine is 100 minute

Explanation:

<u>Planning machine</u>

A planning machine is a metal working machine that gives a flat surface to the work piece. Here the work-piece reciprocates and the feed is given to the tool.  A planning machine is used for heavy duty work and are often large in size.

 The machining time or the planning time of a planning machine is given by,

t_{m} = \frac{L_{w}}{N_{s}\times f}

where, L_{w} is the total length of travel of job

                                    = width of the job

                                    = 300 mm

            N_{s} is number of strokes per min

                                     = 10 double strokes per min

            f is feed of the tool, mm per stroke

                                      = 0.3 mm per stroke

Therefore,  t_{m} = \frac{L_{w}}{N_{s}\times f}

                                            = \frac{300}{10\times 0.3}

                                           = 100 min

Therefore the planning time is 100 minute.

                                             

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Answer:

Option B  Lower than

Explanation:

Gauge pressure is a relative measurement based on atmospheric pressure. Gauge pressure can be positive if it is above atmospheric pressure or it can also be negative it is below.  On another hand, absolute pressure is an actual pressure in a space and its value has always to be zero or above. Basically absolute pressure is zero if it is in a perfect vacuum. So the measurement of absolute pressure is gauge pressure + atmospheric pressure.  This is the reason in normal condition the gauge pressure = absolute pressure - atmospheric pressure and therefore is lower than absolute pressure

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What are some possible reasons for the sudden development of the cell theory
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An insulated piston-cylinder device contains 0.15 of saturated refrigerant-134a vapor at 0.8 MPa pressure. The refrigerant is no
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Answer:

Assumption:

1. The kinetic and potential energy changes are negligible

2. The cylinder is well insulated and thus heat transfer is negligible.

3. The thermal energy stored in the cylinder itself is negligible.

4. The process is stated to be reversible

Analysis:

a. This is reversible adiabatic(i.e isentropic) process and thus s_{1} =s_{2}

From the refrigerant table A11-A13

P_{1} =0.8MPa   \left \{ {{ {{v_{1}=v_{g}  @0.8MPa =0.025645 m^{3/}/kg } } \atop { {{u_{1}=u_{g}  @0.8MPa =246.82 kJ/kg } -   also  {{s_{1}=s_{g}  @0.8MPa =0.91853 kJ/kgK } } \right.

sat vapor

m=\frac{V}{v_{1} } =\frac{0.15}{0.025645} =5.8491 kg\\and \\\\P_{2} =0.2MPa  \left \{ {{x_{2} =\frac{s_{2} -s_{f} }{s_{fg }}=\frac{0.91853-0.15449}{0.78339}   = 0.9753 \atop {u_{2} =u_{f} +x_{2} }(u_{fg}) =  38.26+0.9753(186.25)= 38.26+181.65 =219.9kJ/kg \right. \\s_{1} = s_{2}

T_{2} =T_{sat @ 0.2MPa} = -10.09^{o}  C

b.) We take the content of the cylinder as the sysytem.

This is a closed system since no mass leaves or enters.

Hence, the energy balance for adiabatic closed system can be expressed as:

E_{in} - E_{out}  =ΔE

w_{b, out}  =ΔU

w_{b, out} =m([tex]u_{1} -u_{2)

w_{b, out}  = workdone during the isentropic process

=5.8491(246.82-219.9)

=5.8491(26.91)

=157.3993

=157.4kJ

4 0
3 years ago
A storm sewer is carrying snow melt containing 1.2 g/L of sodium chloride into a small stream. The stream has a naturally occurr
galina1969 [7]

Answer:

Given Data:

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converting into mg/L = Csewer = 1.2 g/L x 1000 mg/g = 1200 mg/L

flow rate of sewer Qsewer = 2000 L/min

concentration of sewer Cstream = 20 mg/L

flow rate of sewer Qstream = 2m3/s

converting Q into L/min = 2m3/s x 1000 x 60 = 120000 L/min

mass diagram is

6 0
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