What is refrigeration capacity and what is meant by a "ton" of refrigeration?

Answers

Answer 1

Answer:

Answer:

1 ton refrigeration =3.517 kJ/s = 3.517 kW

Explanation:

Refrigeration capacity is defined at the measure of the effective cooling capacity of a refrigerator which is expressed in Btu per hour or in tons.

1 ton capacity is a unit of air conditioning and refrigeration which measure the capacity of air conditioning and refrigeration unit.

One ton is equal to removal of 3025kcal heat per hour

1 ton refrigeration = 200 Btu/min = 3.517 kJ/s = 3.517 kW = 4.713 HP

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After the load impedance has been transformed through the ideal transformer, its impedance is: + . Enter the real part in the first blank and the imaginary part in the second blank. If a value is negative, include the negative sign. Provide up to four digits of precision. If the exact value can be provided with fewer digits, merely provide the exact value. These instructions pertain to the following blanks as well. What is the total impedance seen by the source? + . What is the current phasor Ig (expressed in rectangular form)?

Answers

Answer:

Ig =7.2 +j9.599

Explanation: Check the attachment

(a) For a given material, would you expect the surface energy to be greater than, the same as, or less than the grain boundary energy? Why? (b) The grain boundary energy of a small-angle grain boundary is less than for a high-angle one. Why is this so?

Answers

Answer:

(a) Surface energy is greater than grain boundary energy due to the fact that the bonds of the atoms on the surface are lower than those of the atoms at the grain boundary. The energy is also directly proportional to the number of bonds created.

(b) The energy of a high-angle grain boundary is higher than that of a small-angle grain boundary because the high-angle grain boundary has a higher misalignment and smaller number of bonds than a small-angle grain boundary.

Explanation:

At full load, a commercially available 100hp, three phase induction motor operates at an efficiency of 97% and a power factor of 0.88 lag. The motor is supplied from a three-phase outlet with a line voltage rating of 208V.a. What is the magnitude of the line current drawn from the 208 V outlet? (1 hp = 746 W.) b. Calculate the reactive power supplied to the motor.

Answers

Answer:

I = Line Current = 242.58 A

Q = Reactive Power = 41.5 kVAr

Explanation:

Firstly, converting 100 hp to kW.

Since, 1 hp = 0.746 kW,

100 hp = 0.746 kW x 100

100 hp = 74.6 kW

The power of a three phase induction motor can be given as:

$P_(in) = √(3) VI Cos\alpha\n$

where,

P in = Input Power required by the motor

V = Line Voltage

I = Line Current

Cosα = Power Factor

Now, calculating Pin:

$efficiency = \frac{{P_(out)} }{P_(in) }\n0.97 = (74.6)/(P_(in) ) \nP_(in) = (74.6)/(0.97)\n P_(in) = 76.9 kW$

a) Calculating the line current:

$P_(in) = √(3)VICos\alpha \n76.9 * 1000= √(3)*208*I*0.88\nI = (76.9*1000)/(√(3)*208*0.88 )\nI = 242.58 A$

b) Calculating Reactive Power:

The reactive power can be calculated as:

Q = P tanα

where,

Q = Reactive power

P = Active Power

α = power factor angle

Since,

$Cos\alpha =0.88\n\alpha =Cos^(-1)(0.88)\n\alpha=28.36$

Therefore,

$Q = 76.9 * tan (28.36)\nQ = 76.9 * (0.5397)\nQ = 41. 5 kVAr$

Indicate whether the following statements are true or false for an isothermal process: (A) Q=T(∆S). (B) ∆U=0.(C) The entropy change of the system is always zero. (D) The total entropy change of the system and the surroundings is always zero. (E) The entropy change of the surroundings is negative. (F) Q=W.

Answers

Answer:

A=False

B=False

C=False

D=False

E=False

F=False

Explanation:

A. In an isothermal process, only the reversibly heat transfer is 0, $Q_(rev)=T (\Delta S)$

B. Consider the phase change of boiling water. Here, the temperature remains constant but the internal energy of the system increases.

C. This is not true even in reversible process, as can be inferred from the equation in part A.

D. This is only true in reversible processes, but not in all isothermal processes.

E. Consider the phase change of freezing water. Here, the surroundings are increasing their entropy, as they are taking in heat from the system.

F. This is not true if $(\Delta U)\neq 0$ , like in answer B. One case where this is true is in the reversible isothermal expansion (or compression) of an ideal gas.

If you answer the whole question and show your work/coding I will rate 5 stars/brainliest!!! Walnut Orchard has two farms that grow wheat and corn. Because of different soil conditions, there are differences in the yields and costs of growing crops on the two farms. The yields and costs are shown in the following table. Each farm has 100 acres available for cultivation. 11,000 bushels of wheat and 7,000 bushels of corn must be grown. Please have an LP model to minimize the total cost while meeting the demand and solve it with Lindo or Excel. You need to have all parts of a model: notation, objective function, constraints, and sign restrictions.

Answers

Answer: 18,100

Explanation: two farms that grow wheat and corn.

The barrel of a bicycle tire pump becomes quite warm during use. Explain the mechanisms responsible for the temperature increase.

Answers

Answer:

The air heats up when being compressed and transefers heat to the barrel.

Explanation:

When a gas is compressed it raises in temperature. Assuming that the compression happens fast and is done before a significant amount of heat can be transferred to the barrel, we could say it is an adiabatic compression. This isn't exactly true, it is an approximation.

In an adiabatic transformation:

$P^(1-k) * T^k = constant$

For air k = 1.4

$P0^(-0.4) * T0^(1.4) = P1^(-0.4) * T1^(1.4)$

$T1^(1.4) = (P1^(0.4) * T0^(1.4))/(P0^(0.4))$

$T1^(1.4) = (P1)/(P0)^(0.4) * T0^(1.4)$

$T1 = T0 * (P1)/(P0)^(0.4/1.4)$

$T1 = T0 * (P1)/(P0)^(0.28)$

SInce it is compressing, the fraction P1/P0 will always be greater than one, and raised to a positive fraction it will always yield a number greater than one, so the final temperature will be greater than the initial temperature.

After it was compressed the hot air will exchange heat with the barrel heating it up.

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