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vova2212 [387]
3 years ago
14

Link AB is to be made of a steel for which the ultimate normal stress is 65 ksi. Determine the cross-sectional area of AB for wh

ich the factor of safety will be 3.20. Assume that the link will be adequately reinforced around the pins at A and B.
Engineering
1 answer:
Jlenok [28]3 years ago
7 0

Explanation:

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shutvik [7]

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yes it's C

Explanation:

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Saturated liquid-vapor mixture of water, called wet steam, in a steam line at 1500 kPa is throttled to 50 kPa and 100°C. What is
frozen [14]

Answer:

x = 0.944

Explanation:

Steam at outlet is an superheated steam, since T > T_{sat}. From steam tables, the specific enthalpy is:

h_{out}=2682.4\,\frac{kJ}{kg}

The throttle valve is modelled after the First Law of Thermodynamics:

h_{in} = h_{out}

Hence, specific enthalpy at inlet is:

h_{in}=2682.4\,\frac{kJ}{kg}

The quality in the steam line is:

x = \frac{2682.4\,\frac{kJ}{kg}-844.55\,\frac{kJ}{kg}}{2791.0\,\frac{kJ}{kg} - 844.55\,\frac{kJ}{kg} }

x = 0.944

6 0
2 years ago
A rectangular channel 3-m-wide carries 12 m^3/s at a depth of 90cm. Is the flow subcritical or supercritical? For the same flowr
hram777 [196]

Answer:

Super critical

1.2 m

Explanation:

Q = Flow rate = 12\ \text{m}^3/\text{s}

w = Width = 3 m

d = Depth = 90 cm = 0.9 m

A = Area = wd

v = Velocity

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

Q=Av\\\Rightarrow v=\dfrac{Q}{wd}\\\Rightarrow v=\dfrac{12}{3\times 0.9}\\\Rightarrow v=4.44\ \text{m/s}

Froude number is given by

Fr=\dfrac{v}{\sqrt{gd}}\\\Rightarrow Fr=\dfrac{4.44}{\sqrt{9.81\times 0.9}}\\\Rightarrow F_r=1.5

Since F_r>1 the flow is super critical.

Flow is critical when Fr=1

Depth is given by

d=(\dfrac{Q^2}{gw^2})^{\dfrac{1}{3}}\\\Rightarrow d=(\dfrac{12^2}{9.81\times 3^2})^{\dfrac{1}{3}}\\\Rightarrow d=1.2\ \text{m}

The depth of the channel will be 1.2 m for critical flow.

4 0
2 years ago
The wall shear stress in a fully developed flow portion of a 12-in.-diameter pipe carrying water is 2.00 lb/ft^2. Determine the
soldier1979 [14.2K]

Answer:

a) -8 lb / ft^3

b) -70.4 lb / ft^3

c) 54.4 lb / ft^3

Explanation:

Given:

- Diameter of pipe D = 12 in

- Shear stress t = 2.0 lb/ft^2

- y = 62.4 lb / ft^3

Find pressure gradient dP / dx when:

a) x is in horizontal flow direction

b) Vertical flow up

c) vertical flow down

Solution:

- dP / dx as function of shear stress and radial distance r:

                      (dP - y*L*sin(Q))/ L = 2*t / r

                      dP / L - y*sin(Q) = 2*t / r

Where            dP / L = - dP/dx,

                      dP / dx = -2*t / r - y*sin(Q)

Where            r = D /2 ,

                      dP / dx = -4*t / D - y*sin(Q)

a) Horizontal Pipe Q = 0

Hence,           dP / dx = -4*2 / 1 - 62.4*sin(0)

                      dP / dx = -8 + 0

                      dP/dx   = -8 lb / ft^3

b) Vertical pipe flow up Q = pi/2

Hence,           dP / dx = -4*2 / 1 - 62.4*sin(pi/2)

                      dP / dx = 8 - 62.4

                      dP/dx   = -70.4 lb / ft^3

c) Vertical flow down Q = -pi/2

Hence,           dP / dx = -4*2 / 1 - 62.4*sin(-pi/2)

                      dP / dx = -8 + 62.4

                      dP/dx   = 54.4 lb / ft^3                      

7 0
2 years ago
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