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2 years ago

Split point twist drills

Engineering

1 answer:

Ostrovityanka [42]2 years ago

4 0

Answer:

A split point twist drill is a self-centering speed bit with a cutting angle of 135 degrees. It is specifically designed to drill into metal and other hard materials without wandering around on the surface.

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A fluid at 300 K flows through a long, thin-walled pipe of 0.2-m diameter. The pipe is enclosed in a concrete casing that is of

andrew-mc [135]

Answer:

The correct answer is "1341.288 W/m".

Explanation:

Given that:

T₁ = 300 K

T₂ = 500 K

Diameter,

d = 0.2 m

Length,

l = 1 m

As we know,

The shape factor will be:

⇒ $SF=\frac{2 \pi l}{ln[\frac{1.08 b }{d} ]}$

By putting the value, we get

⇒ $=\frac{2 \pi l}{ln[\frac{1.08\times 1}{0.2} ]}$

⇒ $=3.7258 \ l$

hence,

The heat loss will be:

⇒ $Q=SF\times K(T_2-T_1)$

$=3.7258\times 1\times 1.8\times (500-300)$

$=3.7258\times 1.8\times (200)$

$=1341.288 \ W/m$

3 0

3 years ago

W10L1-Show It: Pythagorean Theorem<br> Calculate the total material in the picture.<br> 4<br> 3

Fantom [35]

Answer:

Explanation: I really dont even know, I just used up all my tries on it and got it wrong on every other thing i chose. So it's 35 i believe cause its the only answer i didnt choose.

7 0

3 years ago

Sometimes, steel studs may not be used on outside walls because they are?

Helen [10]

Answer:

We can describe 15×-10 as an expression. we would describe 6×-2< 35 as an...

Explanation:

We can describe 15×-10 as an expression. we would describe 6×-2< 35 as an...

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3 years ago

Car insurance incentives and discounts are available depending on _____. A. school attendance and driver skill B. vehicle type a

WARRIOR [948]

Answer:D. Location, vehicle type, and driving habits

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3 years ago

Read 2 more answers

A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu

Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

Negligible heat loss through the insulation
Steady state system
One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

$k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}$

During steady state

$\frac{dT}{dt}$ = 0 which gives $k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0$

From which we have $\frac{d^{2}T }{dx^{2} } = -\frac{q'_{G}}{k}$

Considering the boundary condition at x =0 where there is no heat loss

$\frac{dT}{dt}$ = 0 also at the other end of the plane wall we have

$-k\frac{dT }{dx } =$ hc (T - T∞) at point x = L

Integrating the equation we have

$\frac{dT }{dx } = \frac{q'_{G}}{k} x+ C_{1}$ from which C₁ is evaluated from the first boundary condition thus

0 = $\frac{q'_{G}}{k} (0)+ C_{1}$ from which C₁ = 0

From the second integration we have

$T = -\frac{q'_{G}}{2k} x^{2} + C_{2}$

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

$-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k} + C_{2}-T∞)$ → C₂ = $q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$

T(x) = $\frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$ and T(x) = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )$

∴ Tmax → when x = 0 = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))$

Substituting the values we get

T = 167 ° C

4 0

4 years ago