Burn in hell i watched your stupid video and i still could not get the answer

An engine has been diagnosed with blowby.

12345 [234]

B, tech b only, Blow by is describe or defined as air going by the compression rings. Hope it helps

8 0

3 years ago

The air velocity in the duct of a heating system is to be measured by a Pitot-static probe inserted into the duct parallel to th

Ilia_Sergeevich [38]

Answer:

Flow velocity

50.48m/s

Pressure change at probe tip

1236.06Pa

Explanation:

Question is incomplete

The air velocity in the duct of a heating system is to be measured by a Pitot-static probe inserted into the duct parallel to the flow. If the differential height between the water columns connected to the two outlets of the probe is 0.126m, determine (a) the flow velocity and (b) the pressure rise at the tip of the probe. The air temperature and pressure in the duct are 352k and 98 kPa, respectively

solution

In this question, we are asked to calculate the flow velocity and the pressure rise at the tip of probe

please check attachment for complete solution and step by step explanation

8 0

3 years ago

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)

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