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valentinak56 [21]

1 year ago

14

Derive the expression ε=ln(1+e), where ε is the true strain and e is the engineering strain. Note that this expression is not va

lid after the onset of necking

1 answer:

Rudiy271 year ago

6 0

The formula for true strain after derivation from basic terms is; ε_t = In(1 + ε_e)

<h3>How to derive the expression for True Strain?</h3>

Formula for Engineering Stress is;

σ_e = Load/Area

Formula for true stress is;

σ_t = Force/Instantaneous Area

Formula for Engineering Strain is;

ε_e = ΔL/L₀

Formula for true strain is;

dε_t = dL/L

Total true strain is gotten from;

ε_t = ∫(dL/L) between boundaries of L_f and L_o

When we integrate between those boundaries, we have;

ε_t = In[(L₀ + ΔL)/L₀

⇒ ε_t = In[(1+ ΔL/L₀)

⇒ ε_t = In(1 + ε_e)

Read more about True Strain at; brainly.com/question/20717759

#SPJ1

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(35-39) A student travels on a school bus in the middle of winter from home to school. The school bus temperature is 68.0° F. Th

arlik [135]

Answer:

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

Explanation:

From Heat Transfer we determine that heat transfer rate due to electromagnetic radiation ( $\dot Q$ ), measured in BTU per hour, is represented by this formula:

$\dot Q = \epsilon\cdot A\cdot \sigma \cdot (T_{s}^{4}-T_{b}^{4})$ (1)

Where:

$\epsilon$ - Emissivity, dimensionless.

$A$ - Surface area of the student, measured in square feet.

$\sigma$ - Stefan-Boltzmann constant, measured in BTU per hour-square feet-quartic Rankine.

$T_{s}$ - Temperature of the student, measured in Rankine.

$T_{b}$ - Temperature of the bus, measured in Rankine.

If we know that $\epsilon = 0.90$ , $A = 16.188\,ft^{2}$ , $\sigma = 1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}}$ , $T_{s} = 554.07\,R$ and $T_{b} = 527.67\,R$ , then the heat transfer rate due to electromagnetic radiation is:

$\dot Q = (0.90)\cdot (16.188\,ft^{2})\cdot \left(1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}} \right)\cdot [(554.07\,R)^{4}-(527.67\,R)^{4}]$

$\dot Q = 417.492\,\frac{BTU}{h}$

Under the consideration of steady heat transfer we find that the net energy transfer from the student's body during the 20 min-ride to school is:

$Q = \dot Q \cdot \Delta t$ (2)

Where $\Delta t$ is the heat transfer time, measured in hours.

If we know that $\dot Q = 417.492\,\frac{BTU}{h}$ and $\Delta t = \frac{1}{3}\,h$ , then the net energy transfer is:

$Q = \left(417.492\,\frac{BTU}{h} \right)\cdot \left(\frac{1}{3}\,h \right)$

$Q = 139.164\,BTU$

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

7 0

2 years ago

Which bulb has the lowest total cost of operation? (a) Incandescent (b) Fluorescent (c) LED

Finger [1]

Answer: LED have the lowest cost of operation.

Explanation:

If we ignore the initial procurement cost of the items the operational cost of any device consuming electricity is given by

$Cost=Energy\times cost/unit$

Among the three item's LED consumes the lowest power to give the same level of brightness as compared to the other 2 item's thus LED's shall have the lowest operational cost.

6 0

3 years ago

Name some technical skills that are suitable for school leavers .

Natalka [10]

Answer:

Welding, carpentry, masonry, construction worker, barber

Explanation:

7 0

2 years ago

A general contractor has received plans for a new high-rise hotel in an urban area. The hotel will be 12 stories tall and will h

liberstina [14]

Answer:

Ano klassing tanong yn?

Explanation:

Ang taas namn yn? Paki linaw po para matulungan po kita.!!

8 0

2 years ago

An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500

max2010maxim [7]

Answer: The exit temperature of the gas in deg C is $32^{o}C$ .

Explanation:

The given data is as follows.

$C_{p}$ = 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

$P_{1}$ = 100 kPa, $V_{1} = 15 m^{3}/s$

$T_{1} = 27^{o}C = (27 + 273) K = 300 K$

We know that for an ideal gas the mass flow rate will be calculated as follows.

$P_{1}V_{1} = mRT_{1}$

or, m = $\frac{P_{1}V_{1}}{RT_{1}}$

= $\frac{100 \times 15}{0.5 \times 300}$

= 10 kg/s

Now, according to the steady flow energy equation:

$mh_{1} + Q = mh_{2} + W$

$h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}$

$C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}$

$(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}$

$(T_{2} - T_{1})$ = 5 K

$T_{2}$ = 5 K + 300 K

$T_{2}$ = 305 K

= (305 K - 273 K)

= $32^{o}C$

Therefore, we can conclude that the exit temperature of the gas in deg C is $32^{o}C$ .

7 0

2 years ago