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3 years ago

7

For a bronze alloy, the stress at which plastic deformation begins is 297 MPa and the modulus of elasticity is 113 GPa. (a) What

is the maximum load that can be applied to a specimen having a cross-sectional area of 316 mm2 without plastic deformation? (b) If the original specimen length is 128 mm, what is the maximum length to which it may be stretched without causing plastic deformation?

1 answer:

Alenkinab [10]3 years ago

5 0

Answer:

a) 93.852 kN

b) 128.043 mm

Explanation:

Stress is load over section:

σ = P / A

If plastic deformation begins with a stress of 297 MPa, the maximum load before plastic deformation will be:

P = σ * A

316 mm^2 = 3.16*10^-4

P = 297*10^6 * 3.16*10^-4 = 93852 N = 93.852 kN

The stiffness of the specimen is:

k = E * A / l

k = 113*10^9 * 3.16*10^-4 / 0.128 = 279 MN/m

Hooke's law:

x' = x0 * (1 + P/k)

x' = 0.128 * (1 + 93.852*10^3 / 279*10^6) = 0.128043 m = 128.043 mm

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3 years ago

: The interior wall of a furnace is maintained at a temperature of 900 0C. The wall is 60 cm thick, 1 m wide, 1.5 m broad of mat

Snowcat [4.5K]

Answer:

<em>Heat is lost at the rate of 750 J/s or W</em>

<em>The thermal resistance is 1 K/W</em>

Explanation:

interior temperature $T_{2}$ = 900 °C

wall thickness t = 60 cm = 0.6 m

width = 1 m

breadth = 1.5 m

thermal conductivity k = 0.4 W/m-K

outside temperature $T_{1}$ = 150 °C

heat through the wall = ?

The area of the wall A = w x b = 1 x 1.5 = 1.5 m^2

Temperature difference $dt$ = $T_{2}$ - $T_{1}$ = 900 - 150 = 750 °C

note that $dt$ is also equal to 750 K since to convert from °C to K we'll have to add 273 to both temperature, which will still cancel out when we subtract the two temperatures.

To get the heat that escapes through the wall, we use the equation

Q = Ak $\frac{dt}{t}$

substituting values, we have

Q = 1.5 x 0.4 x $\frac{750}{0.6}$ = <em>750 J/s or W</em>

Thermal resistance $R_{t}$ = $\frac{dt}{Q}$

$R_{t}$ = 750/750 =<em> 1 K/W</em>

7 0

3 years ago

A rigid 10-L vessel initially contains a mixture of liquid and vapor water at 100 °C, with a quality factor of 0.123. The mixtur

masya89 [10]

Answer:

$Q_{in} = 46.454\,kJ$

Explanation:

The vessel is modelled after the First Law of Thermodynamics. Let suppose the inexistence of mass interaction at boundary between vessel and surroundings, changes in potential and kinectic energy are negligible and vessel is a rigid recipient.

$Q_{in} = U_{2} - U_{1}$

Properties of water at initial and final state are:

State 1 - (Liquid-Vapor Mixture)

$P = 101.42\,kPa$

$T = 100\,^{\textdegree}C$

$\nu = 0.2066\,\frac{m^{3}}{kg}$

$u = 675.761\,\frac{kJ}{kg}$

$x = 0.123$

State 2 - (Liquid-Vapor Mixture)

$P = 476.16\,kPa$

$T = 150\,^{\textdegree}C$

$\nu = 0.2066\,\frac{m^{3}}{kg}$

$u = 1643.545\,\frac{kJ}{kg}$

$x = 0.525$

The mass stored in the vessel is:

$m = \frac{V}{\nu}$

$m = \frac{10\times 10^{-3}\,m^{3}}{0.2066\,\frac{m^{3}}{kg} }$

$m = 0.048\,kg$

The heat transfer require to the process is:

$Q_{in} = m\cdot (u_{2}-u_{1})$

$Q_{in} = (0.048\,kg)\cdot (1643.545\,\frac{kJ}{kg} - 675.761\,\frac{kJ}{kg} )$

$Q_{in} = 46.454\,kJ$

3 0

3 years ago

An Otto cycle engine is analyzed using the air standard method. Given the conditions at state 1, compression ratio (r), and pres

My name is Ann [436]

Answer:

A) 222.58 kJ / kg

B) 0.8897 M^3/ kg

c) 0.7737 m^3/kg

D) 746.542 k

E) 536.017 kj/kg

efficiency = 58% ( approximately )

Explanation:

Given Data :

Gas constant (R) = 0.287 kJ/ kg.K

T1 = 310 k

P1 ( Kpa ) = 100

r = 11.5 ( compression ratio )

rp = 1.95 ( pressure ratio )

A ) specific internal energy at state 1

= Cv*T1 = 0.718 * 310 = 222.58 kJ / kg

B) Relative specific volume at state 1

= P1*V1 = R*T1 ( ideal gas equation )

V1 = R*T1 / P1 = (0.287* 10^3*310 ) / 100 * 10^3

V1 = 88.97 / 100 = 0.8897 M^3/ kg

C ) relative specific volume at state 2

Applying r ( compression ratio) = V1 / V2

11.5 = 0.8897 / V2

V2 = 0.8897 / 11.5 = 0.7737 m^3/kg

D) The temperature (k) at state 2

since the process is an Isentropic process we will apply the p-v-t relation

$\frac{T1}{T2} = (\frac{V1}{V2}^{n-1} ) = (\frac{P2}{P1} )^{\frac{n-1}{n} }$

hence T2 = $9^{1.4-1} * 310$ = 2.4082 * 310 = 746.542 k

e) specific internal energy at state 2

= Cv*T2 = 0.718 * 746.542 = 536.017 kj/kg

efficiency = output /input = 390.3511 / 667.5448 ≈ 58%

attached is a free hand diagram of an Otto cycle is attached below

3 0

3 years ago

A two-bus power system is interconnected by one transmission line. Bus 1 is a generator bus with specified terminal voltage magn

mixas84 [53]

Answer:

a) | v2 | , β2 ( load, bus voltage at bus 2 )

p1 , q1 ( slack, bus power at bus 1 )

b) q2 , β2

p1 and q1 ( slack, bus power at bus 1 )

Explanation:

Attached below is a schematic representation of the solution

<u>a) Identify the variables in the solution vector assume Bus 2 is a load bus</u>

The specified parameters are ; P2 and q2

while | v2 | and β2 are not specified

given that bus 2 is a load bus, bus 1 is a slack bus with ; | v1 | and β1 been specified while p1 and q1 are not specified

<em>Hence the variables in the solution </em>

<em>= | v2 | , β2 ( load, bus voltage at bus 2 )</em>

<em> p1 , q1 ( slack, bus power at bus 1 ) </em>

<u>b) Identify the variables in the solution vector ( assume Bus 2 is a PV bus )</u>

specified at Bus 2 are ; | p2 | , | v2 |

unspecified : q2 , β2

Bus 1 ( still a slack bus )

specified parameter : | v1 | and β1

unspecified : p1 and q1

<em>Hence the variables in the solution </em>

<em>= q2 , β2 </em>

<em> p1 and q1 ( slack, bus power at bus 1 ) </em>

7 0

3 years ago