2.

: 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:

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

The thermal resistance is 1 K/W

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 $\frac{750}{0.6}$ = 750 J/s or W

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

$R_{t}$ = 750/750 = 1 K/W

7 0

3 years ago

Under which of the following conditions is a Type B-1 Fire extinguisher required onboard a motorized vessel?

swat32

Answer:

The correct option is;

D. The vessel has closed living spaces onboard

Explanation:

Type B-1 Fire extinguishers

A fire extinguisher is required by the law to be installed in a boat that hs the following specifications

1) There are closed compartment in the boat that can be used for fuel storage

2) There exist double double bottom that is only partially filled with flotation materials

3) There are closed living spaces in the boat

4) The fuel tank is permanently installed in the boat

5) The engine is inboard.

7 0

3 years ago

Read 2 more answers

A rectangular open box, 25 ft by 10 ft in plan and 12 ft deep weighs 40 tons. Sufficient amount of stones is placed in the box a

dimulka [17.4K]

Answer:

44.95 tonnes

Explanation:

According to principle of buoyancy the object will just sink when it's weight is more than the weight of the liquid it displaces

It is given that empty weight of box = 40 tons

Let the mass of the stones to be placed be = M tonnes

Thus the combined mass of box and stones = (40+M) tonnes..........(i)

Since the box will displace water equal to it's volume V we have $volume of box = 25ft*10ft*12ft= 3000ft^{3}$

$Volume= 84.95m^{3}$

$Since 1ft^{3} =0.028m^{3}$

Now the weight of water displaced = $Weight =\rho \times Volumewhererho$ is density of water = 1000kg/ $m^{3}$

Thus weight of liquid displaced = $\frac{84.95X1000}{1000}tonnes=84.95 tonnes$ ..................(ii)

Equating i and ii we get

40 + M = 84.95

thus Mass of stones = 44.95 tonnes

3 0

3 years ago

List the main activities of exploration??

Trava [24]

Answer: Exploration includes plethora of activities and depend upon the kind of exploration a person is doing. But most include some of the basic activities like research , investigation, planning and execution.

Suppose we want to explore new petroleum sites then we would have to start with studying the geography of that area, then according to our research we will analyse the hot spots or the sector where probability of finding of oil field is highest, post that appropriate man power is skilled professionals, tools and machinery will be brought at the site so that execution can take place.

3 0

3 years ago

Water at a pressure of 3 bars enters a short horizontal convergent channel at 3.5 m/s. The upstream and downstream diameters of

earnstyle [38]

Answer:

The pressure reduces to 2.588 bars.

Explanation:

According to Bernoulli's theorem for ideal flow we have

$\frac{P}{\gamma _{w}}+\frac{V^{2}}{2g}+z=constant$

Since the losses are neglected thus applying this theorm between upper and lower porion we have

$\frac{P_{u}}{\gamma _{w}}+\frac{V-{u}^{2}}{2g}+z_{u}=\frac{P_{L}}{\gamma _{w}}+\frac{V{L}^{2}}{2g}+z_{L}$

Now by continuity equation we have

$A_{u}v_{u}=A_{L}v_{L}\\\\\therefore v_{L}=\frac{A_{u}}{A_{L}}\times v_{u}\\\\v_{L}=\frac{d^{2}_{u}}{d^{2}_{L}}\times v_{u}\\\\\therefore v_{L}=\frac{2500}{900}\times 3.5\\\\\therefore v_{L}=9.72m/s$

Applying the values in the Bernoulli's equation we get

$\frac{P_{L}}{\gamma _{w}}=\frac{300000}{\gamma _{w}}+\frac{3.5^{2}}{2g}-\frac{9.72^{2}}{2g}(\because z_{L}=z_{u})\\\\\frac{P_{L}}{\gamma _{w}}=26.38m\\\\\therefore P_{L}=258885.8Pa\\\\\therefore P_{L}=2.588bars$

6 0

3 years ago