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Elan Coil [88]
3 years ago
8

An inventor claims to have developed a food freezer that at steady state requires a power input of 0.25 kW to extract energy by

heat transfer at a rate of 3000 J/s from freezer contents at 270 K. Investigate this claim for an ambient temperature of 293 K.
Engineering
1 answer:
Serga [27]3 years ago
6 0

Answer:

investor claim is acceptable

Explanation:

given data

Win = 0.25 kW

Qc = 3000 J/s = 3kW

Th = 293 K

Tc = 270 K

solution

we get here coefficient of performance of cycle is

coefficient of performance = \frac{Qc}{Win}    ..................1

put here value and we get

coefficient of performance = \frac{3}{0.25}  

coefficient of performance = 1.2

and

coefficient of performance of ideal refrigeration is

coefficient of performance = \frac{Tc}{Th-Tc}    ..................2

coefficient of performance = \frac{270}{293-270}  

coefficient of performance = 11.74

and

we can see here that coefficient of performance of ideal refrigeration  is is more than real cycle coefficient of performance

so investor claim is acceptable

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katovenus [111]

The false statement about onStep is: B. The default number of steps per second is 30.

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In conclusion, the default number of steps per second for onStep isn't 30.

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7 0
2 years ago
Suppose there are n chairs in a row. We want to compute the number of ways to put 2 students into seats so that they are not nex
icang [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) f_{(n)} = f_{(n-1)} + n-2

b) g_{(n)} = g_(n+1) + (n-2)

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The explanation is shown on the second and third uploaded image

8 0
3 years ago
What is the velocity of flow in an asphalt channel that has a hydraulic radius of 3.404 m, length of 200 m and bed slope of 0.00
yKpoI14uk [10]

Answer:

The velocity of flow is 10.0 m/s.

Explanation:

We shall use Manning's equation to calculate the velocity of flow

Velocity of flow by manning's equation is given by

V=\frac{1}{n}R^{2/3}S^{1/2}

where

n = manning's roughness coefficient

R = hydraulic radius

S = bed slope of the channel

We know that for an asphalt channel value of manning's roughness coefficient = 0.016

Applying values in the above equation we obtain velocity of flow as

V=\frac{1}{0.016}\times 3.404^{2/3}\times 0.005^{1/2}\\\\\therefore V=10.000m/s

7 0
3 years ago
Refrigerant-134a at 400 psia has a specific volume of 0.1144 ft3/lbm. Determine the temperature of the refrigerant based on (a)
vekshin1

Answer:

a) Using Ideal gas Equation, T = 434.98°R = 435°R

b) Using Van Der Waal's Equation, T = 637.32°R = 637°R

c) T obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is T = 559.67°R = 560°R

Explanation:

a) Ideal gas Equation

PV = mRT

T = PV/mR

P = pressure = 400 psia

V/m = specific volume = 0.1144 ft³/lbm

R = gas constant = 0.1052 psia.ft³/lbm.°R

T = 400 × 0.1144/0.1052 = 434.98 °R

b) Van Der Waal's Equation

T = (1/R) (P + (a/v²)) (v - b)

a = Van Der Waal's constant = (27R²(T꜀ᵣ)²)/(64P꜀ᵣ)

R = 0.1052 psia.ft³/lbm.°R

T꜀ᵣ = critical temperature for refrigerant-134a (from the refrigerant tables) = 673.6°R

P꜀ᵣ = critical pressure for refrigerant-134a (from the refrigerant tables) = 588.7 psia

a = (27 × 0.1052² × 673.6²)/(64 × 588.7)

a = 3.596 ft⁶.psia/lbm²

b = (RT꜀ᵣ)/8P꜀ᵣ

b = (0.1052 × 673.6)/(8 × 588.7) = 0.01504 ft³/lbm

T = (1/0.1052) (400 + (3.596/0.1144²) (0.1144 - 0.01504) = 637.32°R

c) The temperature for the refrigerant-134a as obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is

T = 100°F = 559.67°R

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An amplifier with 40 dB of small-signal, open-circuit voltage gain, an input resistance of 1 MO, and an output resistance of 100
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