7–2 Cooling of a Hot Block by Forced Air at High

What is the composition, in atom percent, of an alloy that consists of 4.5 wt% Pb and 95.5 wt% Sn?

jeka57 [31]

Answer: Option A is correct -- 2.6 at% Pb and 97.4 at% Sn.

Explanation:

Option A is the only correct option -- 2.6 at% Pb and 97.4 at% Sn. While option B, which is 7.6 at% Pb and 92.4 at% Sn. and option C, which is 97.4 at% Pb and 2.6 at% Sn. and option D, which is 92.4 at% Pb and 7.6 at% Sn. are wrong.

6 0

3 years ago

3. Suppose you work for this company that evaluates Boolean circuits in exponential time. Since you are very smart, your manager

hammer [34]

Answer:

Ergr5

Explanation:

3 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

In a black box experiment, when the amount of material exiting a closed system is less than the amount of material entering the

Oksi-84 [34.3K]

When the material that exits is lesser in amount than that of the entering material in a black box experiment, the parts of the system need to be changed.

<h3>What happens in a black box experiment?</h3>

In a black box experiment, the experimenters need to make assumptions regarding the drawing of conclusions. One such conclusion is the amount of material that exits.

If such amount is lesser than the one that enters the system, such experiment concludes that it is the time to change the parts of the system.

Hence, option D holds true regarding the black box experiment.

Learn more about black box experiment here:

brainly.com/question/13403296

#SPJ1

4 0

2 years ago

Radio Frequency IDentification (RFID) tags and readers are a category of low-end wireless devices that people may not recognize

Fittoniya [83]

Answer:

See explaination.

Explanation:

Radio Frequency Identification (RFID) tags and readers uses the electromagnetic waves to identify and track the attached objects.

A tag is attached to the object which is to be identified or tracked, and reader is used to read the response and send the acknowledgement. Therefore, RFID tags and readers are used in many industries, passports, transportations and pet identification etc.

i.

RFID technology is used in smartcards, implants for pets, passports and library books to identify and track the persons, objects and pets etc.

Hence, we can say that option (i) is true.

ii.

Electronic Product Code (EPC) is a small code stored in the RFID tag. The code stored in the memory is 96 bits which are used to identify the organization which manages the data, unique number to identify the product and a number to identify the particular tag and etc.

EPC is a unique identification number, it can read and be written by the RFID reader. It is used in supply chains instead of a bar code even though expansive.

Hence, option (ii) is true.

iii.

Tags are used to identify and track the objects. It doesn’t belong to base stations and access points as a Wi-Fi networks.

Therefore, option (iii) is false.

iv.

The EPC Generation 2 RFID tag is used to improve the security by enabling the authentication features. It is not the similar way of data transmission in the other wireless situations.

Therefore, option (iv) is false.

Finally, the options (i) and (ii) are TRUE, while (iii) and (iv) are FALSE.

8 0

3 years ago