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2 years ago

What are the main differences between pipefitters and plumbers? (Select all that apply.)

Engineering

1 answer:

romanna [79]2 years ago

8 0

Answer:

pipefitters design systems whereas plumbers maintain systems

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A slab-milling operation is performed on a 0.7 m long, 30 mm-wide cast-iron block with a feed of 0.25 mm/tooth and depth of cut

denis23 [38]

Answer:

a) $T_m=1.787min$

b) $MRR=35259.7mm^3/min$

Explanation:

From the question we are told that:

Cast-iron block Dimension:

Length $l=0.7m=>700mm$

Width $w=30mm$

Feed $F=0.25mm/tooth$

Depth $dp=3mm$

Diameter $d=75mm$

Number of cutting teeth $n=8$

Rotation speed $N=200rpm$

Generally the equation for Approach is mathematically given by

$x=\sqrt{Dd-d^2}$

$X=\sqrt{75*3-3^2}$

$X=14.69mm$

Therefore

Effective length is given as

$L_e=Approach +object Length$

$L_e=700+14.69$

$L_e=714.69mm$

Generally the equation for Machine Time is mathematically given by

$T_m=\frac{L_e}{F_m}$

Where

$F_m=F*n*N$

$F_m=0.25*8*200$

$F_m=400$

Therefore

$T_m=\frac{714.69}{400}$

$T_m=1.787min$

Generally the equation for Material Removal Rate. is mathematically given by

$MRR=\frac{L*B*d}{t_m}$

$MRR=\frac{700*30*3}{1.787}$

$MRR=35259.7mm^3/min$

3 0

2 years ago

How much to build a barber clipper?

goldenfox [79]

Like the price to manufacture?

8 0

3 years ago

The design specifications of a 1.2-m long solid circular transmission shaft require that the angle of twist of the shaft not exc

Verizon [17]

Answer:

c = 18.0569 mm

Explanation:

Strategy

We will find required diameter based on angle of twist and based on shearing stress. The larger value will govern.

Given Data

Applied Torque

T = 750 N.m

Length of shaft

L = 1.2 m

Modulus of Rigidity

G = 77.2 GPa

Allowable Stress

г = 90 MPa

Maximum Angle of twist

∅=4°

∅=4* $\pi$ /180

∅=69.813 *10^-3 rad

Required Diameter based on angle of twist

∅=TL/GJ

∅=TL/G* $\pi$ /2*c^4

∅=2TL/G* $\pi$ *c^4

c= $\sqrt[4]{2TL/\pi G }$ ∅

c=18.0869 *10^-3 rad

Required Diameter based on shearing stress

г = T/J*c

г = [T/(J* $\pi$ /2*c^4)]*c

г =[2T/(J* $\pi$ *c^4)]*c

c=17.441*10^-3 rad

Minimum Radius Required

We will use larger of the two values

c= 18.0569 x 10^-3 m

c = 18.0569 mm

3 0

3 years ago

Ammonia contained in a piston-cylinder assembly, initially saturated vapor at 0o F, undergoes an isothermal process during which

Rudik [331]

ANSWERS:

$-P_{2(a)} =15.6lbf/in^2\\-P_{2(b)} =30.146lbf/in^2\\ T_{2(a)} =0^oF\\T_{2(b)} =0^oF\\x_{2(b)} =49.87percent$

Explanation:

Given:

Piston cylinder assembly which mean that the process is constant pressure process P=C.

<u>AMMONIA </u>

state(1)

saturated vapor $x_{1} =1$

The temperature $T_{1} =0^0 F$

Isothermal process $T=C$

$-V_{2} =2V_{1}$ ( double)

$-V_{2} =.5V_{2}$ (reduced by half)

To find the final state by giving the quality in lbf/in we assume the friction is neglected and the system is in equilibrium.

state(1)

using PVT data for saturated ammonia

$-P_{1} =30.416 lbf/in^2\\-v_{1} =v_{g} =9.11ft^3/lb$

then the state exists in the supper heated region.

a) from standard data

$-v_{1(a)} =2v_{1} =18.22ft^3/lb\\-T_{1} =0^oF$

$at\\P_{x} =14lbf/in^2\\-v_{x} =20.289 ft^3/kg$

$at\\P_{y} =16 lbf/in^2\\-v_{y} =17.701ft^3/kg$

assume linear interpolation

$\frac{P_{x}-P_{2(b)} }{P_{x}- P_{y} } =\frac{v_{x}-v_{1(a)} }{v_{x}-v_{y} }$

$P_{1(b)}=P_{x} -(P_{x} -P_{y} )*\frac{v_{x}- v_{1(b)} }{v_{x}-v_{y} }\\ \\P_{1(b)} =14-(14-16)*\frac{20.289-18.22}{20.289-17.701} =15.6lbf/in^2$

$-v_{2(a)} =2v_{1} =4.555ft^3/lb\\v_{g}$

from standard data

$-v_{f} =0.02419ft^3/kg\\-v_{g} =9.11ft^3/kg\\v_{f}$

then the state exist in the wet zone

$-P_{s} =30.146lbf/in^2\\v_{2(a)} =v_{f} +x(v_{g} -v_{f} )$

$x=\frac{v_{2(a)-v_{f} } }{v_{g} -v_{f} } \\x=\frac{4.555-0.02419}{9.11-0.02419} =49.87%$

3 0

3 years ago

A gas stream flowing at 1000 cfm with a particulate loading of 400 gr/ft3 discharges from a certain industrial plant through an

Makovka662 [10]

<u>Solution and Explanation:</u>

Volume of gas stream = 1000 cfm (Cubic Feet per Minute)

Particulate loading = 400 gr/ft3 (Grain/cubic feet)

1 gr/ft3 = 0.00220462 lb/ft3

Total weight of particulate matter = $1000 \mathrm{cfm} \times 400 \mathrm{gr} / \mathrm{tt} 3 \times .000142857 \mathrm{lb} / \mathrm{ft} 3 \times 60=3428.568 \mathrm{lb} / \mathrm{hr}$

Cyclone is to 80 % efficient

So particulate remaining = $0.20 \times 3428.568 \mathrm{lb} / \mathrm{hr}=685.7136$

emissions from this stack be limited to = 10.0 lb/hr

Particles to be remaining after wet scrubber = 10.0 lb/hr

So particles to be removed = 685.7136- 10 = 675.7136

Efficiency = output multiply with 100/input = 98.542 %

4 0

3 years ago