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4 years ago

What are the main causes of injuries when using forklifts?

1 answer:

5 0

The forklift overturning is a very common way of getting injured from a forklift. Overturning the forklift means it tips over onto it's side due to the operator turning it too fast.

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a cubical box 20-cm on a side is contructed from 1.2 cm thick concrete panels. A 100-W light bulb is sealed inside the box. What

Flura [38]

Answer:

Temperature on the inside ofthe box

Explanation:

The power of the light bulb is the rate of heat conduction of the bulb, $dq/dt = 100 W$

The thickness of the wall, L = 1.2 cm = 0.012m

Length of the cube's side, x = 20cm = 0.2 m

The area of the cubical box, A = 6x²

A = 6 * 0.2² = 6 * 0.04

A = 0.24 m²

Temperature of the surrounding, $T_0 = 20^0 C = 273 + 20 = 293 K$

Temperature of the inside of the box, $T_{in} = ?$

Coefficient of thermal conductivity, k = 0.8 W/m-K

The formula for the rate of heat conduction is given by:

$dq/dt = \frac{kA(T_{in} - T_0)}{L} \\\\100 = \frac{0.8*0.24(T_{in} - 293)}{0.012}\\\\T_{in} - 293 = \frac{100 * 0.012}{0.8*0.24} \\\\T_{in} - 293 = 6.25\\\\T_{in} = 293 + 6.25\\\\T_{in} = 299.25 K\\\\T_{in} = 299.25 - 273\\\\T_{in} = 26.25^0 C$

5 0

4 years ago

A square plate of titanium is 12cm along the top, 12cm on the right side, and 5mm thick. A normal tensile force of 15kN is appli

Scrat [10]

Answer:

$X_t=2.17391304*10^{-4}$

$X_r=2.89855072*10^{-4}$

$e_t=0.0026$

$e_r=0.0035$

Explanation:

From the question we are told that:

Dimension $12*12$

Thickness $l_t=5mm=5*10^-3$

Normal tensile force on top side $F_t= 15kN$

Normal tensile force on right side $F_r= 20kN$

Elastic modulus, $E=115Gpap=>115*10^9$

Generally the equation for Normal Strain X is mathematically given by

$X=\frac{Force}{Area*E}$

Therefore

For Top

$X_t=\frac{Force_t}{Area*E}$

Where

$Area=L*B*T$

$Area=12*10^{-2}*5*10^{-3}$

$Area=6*10^{-4}$

$X_t=\frac{15*10^3}{6*10^{-4}*115*10^9}$

$X_t=2.17391304*10^{-4}$

For Right side $X_r=\frac{Force_r}{Area*E}$

Where

Area=L*B*T

$Area=12*10^{-2}*5*10^{-3}$

$Area=6*10^{-4}$

$X_r=2.89855072*10^{-4}$

Generally the equation for elongation is mathematically given by

$e=strain *12$

For top

$e_t=2.17391304*10^{-4}*12$

$e_t=0.0026$

For Right

$e_r=2.89855072*10^{-4} *12$

$e_r=0.0035$

5 0

3 years ago

A converging-diverging nozzle with an exit to throat area ratio of 4.0 is designed to expand air isentropically to atmospheric p

34kurt

Answer

0.9, 1172.35kPa

Explanation:

<em>Question (in proper order) Attached below</em>

Air is flowing inside the throat has following inlet conditions

$P_{0}=1000 kPa$

$T_{0}=500 K$

$M=1.8$

$M=\frac{u}{c}=1.8$

'u' is the speed of sound in the air

$\Rightarrow u=1.8\times c$

$=1.8\times 340.29$

$=612.522\frac{m}{sec}$

Therefore volumetric flow rate entering,

$Q=612.522\times 0.0008$

$=0.4900176\frac{m^{3}}{sec}$

Using ideal gas equation

PV=nRT

$n=\frac{PV}{RT}$

$=\frac{1000\times 0.4900176}{8.314\times 500}$

=0.117878 gmoles/sec

Therefore , mass flow rate

$Mass = 0.117878\times 29$

$=3.4184 grams/sec$

Given

$\frac{A}{A_{0}}=2$

$\Rightarrow A=0.0016.m^{2}$

Using continuity equation

$A_{1}V_{1}=A_{2}V_{2}$

$\Rightarrow V_{2}=\frac{A_{1}V_{1}}{A_{2}}$

$=\frac{0.0008\times 612.522}{0.0016}$

$=306.261\frac{m}{sec}$

Hence exit velocity = 306.261 m/sec

Exit Mach number

$M=\frac{u}{c}=\frac{306.261}{340.29}=0.9$

Temperature will remain same as 500 K

Now

Using Bernoulli's equation

$\frac{P_{1}}{\rho g}+\frac{v_{1}^{2}}{2g}+z_{1}=\frac{P_{2}}{\rho g}+\frac{v_{2}^{2}}{2g}+z_{2}$

Here

$z_{1} = z_{2}$

$\frac{P_{1}}{\rho g}+\frac{v_{1}^{2}}{2g}-\frac{v_{2}^{2}}{2g}=\frac{P_{2}}{\rho g}$

$\Rightarrow \frac{1000000}{\rho g}+\frac{612.522^{2}}{2g}-\frac{306.261^{2}}{2g}=\frac{P_{2}}{\rho g}$

$\Rightarrow \frac{1000000}{1.225}+\frac{612.522^{2}}{2}-\frac{306.261^{2}}{2}=\frac{P_{2}}{1.225}$

$\Rightarrow P_{2}=1172.35kPa$

4 0

3 years ago

Engineering controls are the physical changes that employers make to the work environment or to equipment that make it safer to

sammy [17]

True, the Engineering controls are the physical changes that employers make to the work environment or to equipment that make it safer to use.

Explanation

Engineering controls are those techniques used to reduce or eliminate hazards of any condition, thereby protecting the workers.

These are mostly products that act as barriers between the worker and the hazard.

This may include machinery or equipment.

The common engineering controls used are glovebox, biosafety cabinet, fume hood, vented balance safety enclosure, HVAC system, lockout-tagout, sticky mat and rupture disc.

7 0

3 years ago

Air flows steadily through a turbine that produces 3.5x 105 ft-lbf/s of work. Using the data below at the inlet and outlet, wher

charle [14.2K]

The heat transfer rate through the turbine is $\bold{6.23 \times 10^{6} \mathrm{Btu} / \mathrm{hr}}$ .

<u>Explanation:</u>

Express the final form of the overall energy equation.

$\frac{\delta Q}{d t}-\frac{\delta W_{s}}{d t}=\iint_{c}\left(e+\frac{P}{\rho}\right) \rho(\mathbf{v} \cdot \mathbf{n}) d A+\frac{\partial}{\partial t} \iiint_{c, v} e \rho d V+\frac{\delta W_{\mu}}{d t} \ldots \ldots(1)$

Here, the rate of heat addition to the control volume is $\frac{\delta Q}{d t}$ , shaft work rate is $\frac{\delta W_{s}}{d t}$ , rate of accumulation of energy inside control volume is $\frac{\partial}{\partial t} \iiint_{c, v} e \rho d V$ , rate of work accomplished in viscous effects at the surface is $\frac{\delta W_{\mu}}{d t}$ , and the net energy efflux from the control volume is

$\iint_{c, s,}\left(e+\frac{P}{\rho}\right) \rho(\mathbf{v} \cdot \mathbf{n}) d A$

Write the following assumption.

There is no viscous work in the system, $\frac{\delta W_{\mu}}{d t}=0\\$

Steady state flow of the system, $\frac{\partial}{\partial t} \iiint_{c, v} e \rho d V=0$

Use the assumption condition in Equation (1).

$\frac{\delta Q}{d t}-\frac{\delta W_{s}}{d t}=\iint_{z}\left(e+\frac{P}{\rho}\right) \rho(\mathbf{v} \cdot \mathbf{n}) d A$

Integrate the above equation.

$\begin{array}{l}\frac{\delta Q}{d t}-\frac{\delta W_{s}}{d t}=\rho v A\left[e+\frac{P}{\rho}\right]_{2}-\rho v A\left[e+\frac{P}{\rho}\right] \\=\rho_{1} Q\left[u_{2}-u_{1}+\frac{v_{2}^{2}-v_{1}^{2}}{2}+\left(P_{2} / \rho_{2}-P_{1} / \rho_{1}\right)+g\left(z_{2}-z_{1}\right)\right] \\=\rho_{1} A_{1} v_{1}\left[u_{2}-u_{1}+\frac{v_{2}^{2}-v_{1}^{2}}{2}+\left(P_{2} / \rho_{2}-P_{1} / \rho_{1}\right)+g\left(z_{2}-z_{1}\right)\right]\end{array}$

Here, internal energy at entry section is $u_{1}$ internal energy at exit section is $u_{2}$ volumetric flow rate is Q, inlet fluid density is $\rho_{1}$ , outlet fluid density is $\rho_{2}$ , specific energy is e, exit velocity is $v_{2}$ , inlet velocity is $v_{1}$ , inlet pressure is $P_{1}$ , outlet pressure is $P_{2}$ , acceleration due to gravity is g and distance between inlet and outlet is z.

Calculate the cross section area at entry section.

$A_{1}=\frac{\pi}{4} d_{1}^{2}$

Here, inlet diameter of pipe is $d_{1}$

Substitute 0.962 ft for $d_{1}$

$\begin{aligned}&A_{1}=\frac{\pi}{4}(0.962 \mathrm{ft})^{2}\\&A_{1}=0.726 \mathrm{ft}^{2}\end{aligned}$

Calculate the heat transfer rate through the turbine.

Substitute $0.08101 \mathrm{b}_{m} / \mathrm{ft}^{3} \text { for } \rho_{2}, 0.05341 \mathrm{b}_{m} / \mathrm{ft}^{3} \text { for } \rho_{1}, 0.726 \mathrm{ft}^{2} \text { for } A_{1}, 244 \mathrm{ft} / \mathrm{s} \text { for } v_{2}$

$100 \mathrm{ft} / \mathrm{s} \text { for } v_{1}, 3.5 \times 10^{5} \mathrm{Ib}_{m} \mathrm{ft}^{2} / \mathrm{s}^{3} \text { for } \frac{\delta W_{s}}{d t},-16.16 \times 10^{5} \mathrm{ft}^{2} / \mathrm{s}^{2} \text { for } u_{2}-u_{1}, 32.2 \mathrm{ft} / \mathrm{s}^{2} \text { for }$

$g, 400 \mathrm{Ib}_{\mathrm{r}} / \mathrm{in}^{2} \text { for } P_{2}, 150 \mathrm{lb}_{\mathrm{r}} / \mathrm{in}^{2} \text { for } P_{1}, \text { and } 10 \mathrm{ft} \text { for } z_{2}-z_{1} \text { in Equation }(2)$

$\frac{\delta Q}{d t}-\frac{\delta W_{s}}{d t}=\rho_{1} A_{1} v_{1}\left[u_{2}-u_{1}+\frac{v_{2}^{2}-v_{1}^{2}}{2}+\left(P_{2} / \rho_{2}-P_{1} / \rho_{1}\right)+g\left(z_{2}-z_{1}\right)\right]$

$\begin{aligned}&\frac{\delta Q}{d t}=\left(9.98 \times 10^{5}+3.5 \times 10^{5}\right) \mathrm{ftlb}_{\mathrm{f}} / \mathrm{s}\\&=13.48 \times 10^{5} \mathrm{ftb}_{\mathrm{f}} / \mathrm{s}\left[\frac{1 \mathrm{Ba}}{778.17 \mathrm{ftbr}}\right]\left[\frac{3600 \mathrm{s}}{1 \mathrm{hr}}\right]\\&=6.23 \times 10^{6} \mathrm{Btu} / \mathrm{hr}\end{aligned}$

Hence, the heat transfer rate through the turbine is $6.23 \times 10^{6} \mathrm{Btu} / \mathrm{hr}$ .

3 0

3 years ago