Who is ur face actor

Differentiate between "Threshold and Resolution" with suitable examples.

9966 [12]

Answer:

to make the bace of a building more sturdy

Explanation:

example: the bace of the empire state building is stone very sturdy

6 0

2 years ago

When you first start a car after it has been sitting for more than an hour, it pollutes up to ......times more than when the eng

lesya [120]

20 time more then when it’s warm

6 0

3 years ago

Read 2 more answers

Imagine you are a process safety consultant and you have been tasked to make a metal refinery site DSEAR compliant. What are the

masya89 [10]

Complying with DSEAR involves:

Assessing risks. ...

Preventing or controlling risks. ...

Control measures. ...

Mitigation. ...

Preparing emergency plans and procedures. ...

Providing information, instruction and training for employees. ...

Places where explosive atmospheres may occur ('ATEX' requirements)

hse uk

4 0

2 years ago

A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu

Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

Negligible heat loss through the insulation
Steady state system
One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

$k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}$

During steady state

$\frac{dT}{dt}$ = 0 which gives $k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0$

From which we have $\frac{d^{2}T }{dx^{2} } = -\frac{q'_{G}}{k}$

Considering the boundary condition at x =0 where there is no heat loss

$\frac{dT}{dt}$ = 0 also at the other end of the plane wall we have

$-k\frac{dT }{dx } =$ hc (T - T∞) at point x = L

Integrating the equation we have

$\frac{dT }{dx } = \frac{q'_{G}}{k} x+ C_{1}$ from which C₁ is evaluated from the first boundary condition thus

0 = $\frac{q'_{G}}{k} (0)+ C_{1}$ from which C₁ = 0

From the second integration we have

$T = -\frac{q'_{G}}{2k} x^{2} + C_{2}$

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

$-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k} + C_{2}-T∞)$ → C₂ = $q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$

T(x) = $\frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$ and T(x) = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )$

∴ Tmax → when x = 0 = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))$

Substituting the values we get

T = 167 ° C

4 0

3 years ago

A pipe of 0.3 m outer diameter at a temperature of 160°C is insulated with a material having a thermal conductivity of k = 0.055

Alekssandra [29.7K]

Answer:

Q=0.95 W/m

Explanation:

Given that

Outer diameter = 0.3 m

Thermal conductivity of material

$K= 0.055(1+2.8\times 10^{-3}T)\frac{W}{mK}$

So the mean conductivity

$K_m=0.055\left ( 1+2.8\times 10^{-3}T_m \right )$

$T_m=\dfrac{160+273+40+273}{2}$

$T_m=373 K$

$K_m=0.055\left ( 1+2.8\times 10^{-3}\times 373 \right )$

$K_m=0.112 \frac{W}{mK}$

So heat conduction through cylinder

$Q=kA\dfrac{\Delta T}{L}$

$Q=0.112\times \pi \times 0.15^2\times 120$

Q=0.95 W/m

4 0

3 years ago