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erma4kov [3.2K]
3 years ago
13

The coupling is used to connect the two shafts together. Assuming that the shear stress in the bolts is uniform, determine the n

umber of bolts necessary to make the maximum shear stress in the shaft equal to the shear stress in the bolts. Each bolt has a diameter dd. Express your answer in terms of some or all of the variables rrr, RRR, ddd, and TTT. nn

Engineering
1 answer:
AleksAgata [21]3 years ago
8 0

Answer:

n = 2r³/Rd²

Explanation:

See the attached file for the derivation.

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Wet steam at 15 bar is throttled adiabatically in a steady-flow process to 2 bar. The resulting stream has a temperature of 130°
cricket20 [7]

Answer:

\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}

Explanation:

The adiabatic throttling process is modelled after the First Law of Thermodynamics:

m\cdot (h_{in} - h_{out}) = 0

h_{in} = h_{out}

Properties of water at inlet and outlet are obtained from steam tables:

State 1 - Inlet (Liquid-Vapor Mixture)

P = 1500\,kPa

T = 198.29\,^{\textdegree}C

h = 2726.9\,\frac{kJ}{kg}

s = 6.3068\,\frac{kJ}{kg\cdot K}

x = 0.967

State 2 - Outlet (Superheated Vapor)

P = 200\,kPa

T = 130\,^{\textdegree}C

h = 2726.9\,\frac{kJ}{kg}

s = 7.1776\,\frac{kJ}{kg\cdot K}

The change of entropy of the steam is derived of the Second Law of Thermodynamics:

\Delta s = 7.1776\,\frac{kJ}{kg\cdot K} - 6.3068\, \frac{kJ}{kg\cdot K}

\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}

6 0
3 years ago
Find the mathematical equation for SF distribution and BM diagram for the beam shown in figure 1.​
Novosadov [1.4K]

Answer:

i) SF: v(x) = \frac{(w_0* x )^2}{2L}

ii) BM : = \frac{(w_0*x)^3}{6L}

Explanation:

Let's take,

\frac{y}{w_0} = \frac{x}{L}

Making y the subject of formula, we have :

y = \frac{x}{L} * w_0

For shear force (SF), we have:

This is the area of the diagram.

v(x) = \frac{1}{2} * y = \frac{1}{2} * \frac{x}{L} * w_0

= \frac{(w_0* x )^2}{2L}

The shear force equation =

v(x) = \frac{(w_0* x )^2}{2L}

For bending moment (BM):

BM = v(x) * \frac{x}{3}

= \frac{(w_0* x )^2}{2L}  * \frac{x}{3}

= \frac{(w_0*x)^3}{6L}

The bending moment equation =

= \frac{(w_0*x)^3}{6L}

5 0
3 years ago
Strands of materials A and B are placed under a tensile force of 10 Newtons. Material A deforms more than Material B.
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Answer:

True

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Cite the phases that are present and the phase compositions for the following alloys: (a) 15 wt% Sn - 85 wt% Pb at 100 o C. (b)
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Answer:

a)  ∝  and β

   The phase compositions are :

    C_{\alpha } = 5wt% Sn - 95 wt% Pb

    C_{\beta } =  98 wt% Sn - 2wt% Pb

b)

The phase is; ∝  

The phase compositions is;   82 wt% Sn - 91.8 wt% Pb

Explanation:

a) 15 wt% Sn - 85 wt% Pb at 100⁰C.

The phases are ; ∝  and β

The phase compositions are :

C_{\alpha } = 5wt% Sn - 95 wt% Pb

C_{\beta } =  98 wt% Sn - 2wt% Pb

b) 1.25 kg of Sn and 14 kg Pb at 200⁰C

The phase is ; ∝  

The phase compositions is;  82 wt% Sn - 91.8 wt% Pb

Csn = 1.25 * 100 / 1.25 + 14 = 8.2 wt%

Cpb = 14 * 100 / 1.25 + 14 = 91.8 wt%

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