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3 years ago

Please Help It's a lot I'm sorry. =(

Engineering

2 answers:

kow [346]3 years ago

8 0

it the first one hope this helps

kap26 [50]3 years ago

8 0

The first one hope that this helps

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A heat engine receives 6 kW from a 250oC source and rejects heat at 30oC. Examine each of three cases with respect to the inequa

taurus [48]

Answer:

Explanation:

Given

$T_h=250^{\circ}C\approx 523\ K$

$T_L=30^{\circ}C\approx 303\ K$

$Q_1=6 kW$

From Clausius inequality

$\oint \frac{dQ}{T}=0$ =Reversible cycle

$\oint \frac{dQ}{T}$ =Irreversible cycle

$\oint \frac{dQ}{T}>0$ =Impossible

(a)For $P_{out}=3 kW$

Rejected heat $Q_2=6-3=3\ kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{3}{303}=1.57\times 10^{-3} kW/K$

thus it is Impossible cycle

(b) $P_{out}=2 kW$

$Q_2=6-2=4 kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{4}{303}=-1.73\times 10^{-3} kW/K$

Possible

(c)Carnot cycle

$\frac{Q_2}{Q_1}=\frac{T_1}{T_2}$

$Q_2=3.47\ kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{3.47}{303}=0$

and maximum Work is obtained for reversible cycle when operate between same temperature limits

$P_{out}=Q_1-Q_2=6-3.47=2.53\ kW$

Thus it is possible

6 0

4 years ago

Retaining<br>Function of<br>Wall

goblinko [34]

Answer:

A retaining of a wall is a protective structure, first and foremost.

Explanation:

Its main aim is to provide functional support for keeping soil in place. It acts as a wall to keep the soil on one side and the rest of the landscape area on the other, providing a platform for a garden to be created.

7 0

3 years ago

Using the tables for water determine the specified property data at the indicated state. For H2O at T = 140 °C and v = 0.2 m3/kg

Dennis_Churaev [7]

Answer:

$h = 1429.74\,\frac{kJ}{kg}$

Explanation:

The determination of any further properties requires the knowledge of two independent properties. (Temperature and specific volume in this case). The specific volumes for saturated liquid and vapor at 140 °C are, respectively:

$\nu_{f} = 0.001080\,\frac{m^{3}}{kg}$

$\nu_{g} = 0.50850\,\frac{m^{3}}{kg}$

Since $\nu_{f} < \nu < \nu_{g}$ , it is a liquid-vapor mixture. The quality of the mixture is:

$x = \frac{\nu-\nu_{f}}{\nu_{g}-\nu_{f}}$

$x = \frac{0.2\,\frac{m^{3}}{kg} - 0.001080\,\frac{m^{3}}{kg} }{0.50850\,\frac{m^{3}}{kg} - 0.001080\,\frac{m^{3}}{kg} }$

$x = 0.392$

The specific enthalpies for saturated liquid and vapor at 140 °C are, respectively:

$h_{f} = 589.16\,\frac{kJ}{kg}$

$h_{g} = 2733.5\,\frac{kJ}{kg}$

The specific enthalpy is:

$h = h_{f}+x\cdot (h_{g}-h_{f})$

$h = 589.16\,\frac{kJ}{kg}+0.392\cdot \left( 2733.5\,\frac{kJ}{kg} - 589.16\,\frac{kJ}{kg} \right)$

$h = 1429.74\,\frac{kJ}{kg}$

6 0

4 years ago

Read 2 more answers

Which of the following is lost when the computer is turned off? ILL MARK BRAINLEST

Pavel [41]

Answer:

cloud storage is the answer

6 0

3 years ago

Read 2 more answers

Explain difference(s) between strength, hardness and toughness.

harkovskaia [24]

Explanation:

Step1

Strength is the maximum stress induces in the material under applied load condition. More the strength more will be ability to sustain load. Strength can be measured from tensile test. It has same unit as stress. Generally ductile material has more strength as compare to brittle material.

Step2

Hardness is the resistance to scratch on the material under given load condition. More the hardness more will be the resistance towards scratching of material. Hardness can be measured by Rockwell or Birnell hardness test. This property of metal is opposite to the strength.

Step3

Toughness is the ability to absorb energy under given loading condition up to its fracture point. It is a type of strain energy that is stored in the metal. Generally ductile metal has more toughness as compare to brittle.

4 0

4 years ago