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sergeinik [125]
2 years ago
8

Consider the following grooves, each of width W, that have been machined from a solid block of material. (a) For each case obtai

n an expression for the view factor of the groove with respect to the surroundings outside the groove. (b) For the V groove, obtain an expression for the view factor F12, where A1 and A2 are opposite surfaces. (c) If H
Engineering
1 answer:
kogti [31]2 years ago
8 0

Answer:A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

Explanation:

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Refrigerant-134a enters the coils of the evaporator of a refrigeration system as a saturated liquid–vapor mixture at a pressure
dezoksy [38]

Answer:

(a). Entropy change of refrigerant  is  = 0.7077 \frac{KJ}{K}

(b). Entropy change of cooled space dS_{space} = - 0.6844 \frac{KJ}{K}

(c). Total entropy change is dS = 0.0232 \frac{KJ}{K}

Explanation:

Given data

Saturation pressure = 140 K pa

Saturation temperature from property table

T_{sat} = - 18.77 °c =  - 18.77 + 273 = 254.23 K

(a). Entropy change of refrigerant  is given by

dS_{ref} = \frac{Q}{T_{sat}}

Since heat absorbed by refrigerant Q = 180 KJ

dS = \frac{180}{254.23}

dS = 0.7077 \frac{KJ}{K}

(b). Entropy change of cooled space

dS_{space} = - \frac{Q}{T_{space}}

T_{space} = - 10 °c = 263 K

dS_{space} = - \frac{180}{263}

dS_{space} = - 0.6844 \frac{KJ}{K}

(c). Total entropy change is given by

dS = dS_{ref} + dS_{space}

dS = 0.7077 - 0.6844

dS = 0.0232 \frac{KJ}{K}

This is the value of total entropy change.

4 0
2 years ago
The hot water needs of an office are met by heating tab water by a heat pump from 16 C to 50 C at an average rate of 0.2 kg/min.
Alex777 [14]

Answer:

option B

Explanation:

given,

heating tap water from 16° C to 50° C

at the average rate of 0.2 kg/min

the COP of this heat pump is 2.8

power output = ?

COP = \dfrac{Q_H}{W_{in}}\\W_{in} = \dfrac{Q_H}{COP}\\W_{in} = \dfrac{\dfrac{0.2}{60}\times 4.18\times (50-16)}{2.8}\\W_{in} = 0.169

the required power input is 0.169 kW or 0.17 kW

hence, the correct answer is option B

7 0
2 years ago
Which statement concerning symbols used on plans is true?
Zielflug [23.3K]

Answer:

The symbols are to be noted on the title sheet or other introductory sheet of the plans.

Explanation:

in able to be understood for example a map key is always on a map so the reader can use it efficently without confusion

8 0
2 years ago
The pressure drop across a valve through which air flows is expected to be 10 kPa. If this differential were applied to the two
ozzi

Answer:hit that soulja boy

Explanation:

6 0
2 years ago
A shopaholic has to buy a pair of jeans , a pair of shoes l,a skirt and a top with budgeted dollar.Given the quantity of each pr
fredd [130]

Answer:

you might be facing some difficulty in observing this point of division between your question and me

0 0
2 years ago
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