I need solution fast plesss

A square power screw has a mean diameter of 30 mm and a pitch of 4 mm with single thread. The collar diameter can be assumed to

poizon [28]

Answer:

i) The torque required to raise the load is 15.85 N*m

ii) The torque required to lower the load is 6.91 N*m

iii) The minimum coefficient of friction is -0.016

Explanation:

Given:

dm = mean diameter = 0.03 m

p = pitch = 0.004 m

n = number of starts = 1

The lead is:

L = n * p = 1 * 0.004 = 0.004 m

F = load = 7000 N

dc = collar diameter = 0.035 m

u = 0.05

i) The helix angle is:

$tan\alpha =\frac{L}{\pi *d_{m} } =\frac{0.004}{\pi *0.03} \\\alpha =2.43$

The torque is:

$T=F\frac{d_{m} }{2} (\frac{\pi *u*d_{m}+L }{\pi *d_{m}-uL } )+(u_{c} F+\frac{d_{2} }{2} )=7000*\frac{0.03}{2} (\frac{\pi *0.05*0.03+0.004}{\pi *0.03-0.05*0.004} )+(0.05*7000*\frac{0.035}{2} )=15.85Nm$

ii) The torque to lowering the load is:

$T=7000*\frac{0.03}{2} (\frac{\pi *0.05*0.03-0.004}{\pi *0.03+0.05*0.004} )+(0.05*7000*\frac{0.035}{2} )=6.91Nm$

iii)

$T=F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )+u_{c} *F*\frac{d_{c}}{2}\\ 0=F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )+u_{c} *F*\frac{d_{c}}{2}\\F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )=-u_{c} *F*\frac{d_{c}}{2}\\\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )=-u_{c} *\frac{d_{c}}{2}\\\\\frac{0.03}{2} (\frac{u*\pi *0.03-0.004}{\pi *0.03+u*0.004} )=-0.05*\frac{0.035}{2}$

Clearing u:

u = -0.016

5 0

3 years ago

In a simple ideal Rankine cycle, water is used as the working fluid. The cycle operates with pressures of 2000 psi in the boiler

weqwewe [10]

Answer:

Explanation:

The pressures given are relative

p1 = 2000 psi

P1 = 2014 psi = 13.9 MPa

p2 = 4 psi

P2 = 18.6 psi = 128 kPa

Values are taken from the steam pressure-enthalpy diagram

h2 = 2500 kJ/kg

If the output of the turbine has a quality of 85%:

t2 = 106 C

I consider the expansion in the turbine to adiabatic and reversible, therefore, isentropic

s1 = s2 = 6.4 kJ/(kg K)

h1 = 3500 kJ/kg

t2 = 550 C

The work in the turbine is of

w = h1 - h2 = 3500 - 2500 = 1000 kJ/kg

The thermal efficiency of the cycle depends on the input heat.

η = w/q1

q1 is not a given, so it cannot be calculated.

3 0

3 years ago

Engine vacuum is being discussed. Technician A says that when the engine is operating under light loads, engine vacuum is low. T

sdas [7]

Neither of the two technicians (Technician A and Technician B) is correct.

<h3>What is an engine vacuum?</h3>

An engine vacuum can be defined as a type of engine which is designed and developed to derive its force from air pressure that's being pushed against one side of the piston of an automobile, while having a partial vacuum on the other side.

In this scenario, we can infer and logically conclude that neither of the two technicians (Technician A and Technician B) is correct because engine vacuum is high when the engine is operating under light loads and vice-versa.

Read more on engine vacuum here: brainly.com/question/14602340

#SPJ12

6 0

2 years ago

If the same type of thermoplastic polymer is being tensile tested and the strain rate is increased, it will: g

Serggg [28]

Answer:

It would break I think need to try it out

Explanation:

3 0

2 years ago

What are the inputs and outputs of a sailboat?

bearhunter [10]

Answer:

sailing ships were the primary means of maritime trade and transportation; exploration across the seas and oceans was reliant on sail for anything other than the shortest distances. Naval power in this period used sail to varying degrees depending on the current technology, culminating in the gun-armed sailing warships of the Age of Sail.

7 0

2 years ago

I need solution fast plesss​

I need solution fast plesss