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3 years ago

PLEASE HELP!! Its easy!!!

Engineering

2 answers:

tankabanditka [31]3 years ago

5 0

Answer: give me brainliest, fellow hamster friend

Explanation:

Rina8888 [55]3 years ago

4 0

Answer:

C is tire

F is cassette

D is hub

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A fully braced structural member in a building is subjected to several different loads, including roof loads of D = 5 k and L_r

allochka39001 [22]

Answer: 37.4K

Explanation:

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7 0

3 years ago

Air enters a cmpressor at 20 deg C and 80 kPa and exits at 800 kPa and 200 deg C. The power input is 400 kW. Find the heat trans

aksik [14]

Answer:

The heat is transferred is at the rate of 752.33 kW

Solution:

As per the question:

Temperature at inlet, $T_{i} = 20^{\circ}C$ = 273 + 20 = 293 K

Temperature at the outlet, $T_{o} = 200{\circ}C$ = 273 + 200 = 473 K

Pressure at inlet, $P_{i} = 80 kPa = 80\times 10^{3} Pa$

Pressure at outlet, $P_{o} = 800 kPa = 800\times 10^{3} Pa$

Speed at the outlet, $v_{o} = 20 m/s$

Diameter of the tube, $D = 10 cm = 10\times 10^{- 2} m = 0.1 m$

Input power, $P_{i} = 400 kW = 400\times 10^{3} W$

Now,

To calculate the heat transfer, $Q$ , we make use of the steady flow eqn:

$h_{i} + \frac{v_{i}^{2}}{2} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH' + p_{s}$

where

$h_{i}$ = specific enthalpy at inlet

$h_{o}$ = specific enthalpy at outlet

$v_{i}$ = air speed at inlet

$p_{s}$ = specific power input

H and H' = Elevation of inlet and outlet

Now, if

$v_{i} = 0$ and H = H'

Then the above eqn reduces to:

$h_{i} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH + p_{s}$

$Q = h_{o} - h_{i} + \frac{v_{o}^{2}}{2} + p_{s}$ (1)

Also,

$p_{s} = \frac{P_{i}}{ mass, m}$

Area of cross-section, A = $\frac{\pi D^{2}}{4} =\frac{\pi 0.1^{2}}{4} = 7.85\times 10^{- 3} m^{2}$

Specific Volume at outlet, $V_{o} = A\times v_{o} = 7.85\times 10^{- 3}\times 20 = 0.157 m^{3}/s$

From the eqn:

$P_{o}V_{o} = mRT_{o}$

$m = \frac{800\times 10^{3}\times 0.157}{287\times 473} = 0.925 kg/s$

Now,

$p_{s} = \frac{400\times 10^{3}}{0.925} = 432.432 kJ/kg$

Also,

$\Delta h = h_{o} - h_{i} = c_{p}\Delta T =c_{p}(T_{o} - T_{i}) = 1.005(200 - 20) = 180.9 kJ/kg$

Now, using these values in eqn (1):

$Q = 180.9 + \frac{20^{2}}{2} + 432.432 = 813.33 kW$

Now, rate of heat transfer, q:

q = mQ = $0.925\times 813.33 = 752.33 kW$

4 0

3 years ago

Statement true about fats in food

Elodia [21]

Answer:

What that means please explain

7 0

3 years ago

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Discuss the differences between conduction and convective heat transfer.

FrozenT [24]

Answer:

Basically there are two principal differences between the convection and conduction heat transfer

Explanation:

The conduction heat transfer is referred to the transfer between two solids due a temperature difference, while for, the convective heat transfer is referred to the transfer between a fluid (liquid or gas) and a solid. Also, they used different coefficients for its calculation.

We can include on the explanation that conduction thermal transfer is due to temperature difference, while convection thermal transfer is due to density difference.

8 0

3 years ago

Anna makes arrangements to reuse waste water that has been used in sinks and showers. Which term refers to the waste water that

victus00 [196]

Answer:

Greywater.

Explanation:

Greywater is also known as sullage and it can be defined as any form of gently used wastewater derived from sources within a residential or office building such as showers, washing machines, bathroom sinks, bathroom tub, etc.

Generally, greywater or sullage is completely free of fecal materials (faeces) because it is independent from all toilet activities. However, greywater is not clean for direct use because it usually contains food particles, dirt, oil from dishes, hair, etc.

In this scenario, Anna makes arrangements to reuse waste water that has been used in sinks and showers. Greywater is a term which refers to the waste water that Anna reuses to conserve resources.

Therefore, Anna reuses greywater to conserve resources.

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3 years ago