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djyliett [7]
3 years ago
8

An inventor claims to have developed a refrigerator that at steady state requires a net power input of 1.1 horsepower to remove

12,000 Btu/h of energy by heat transfer from the freezer compartment at 0°F and discharge energy by heat transfer to a kitchen at 70°F. Evaluate this claim.
Engineering
1 answer:
Lynna [10]3 years ago
6 0

Answer:

The inventor's claim is false in the sense that no thermal machine can violate the first thermodynamic law.

Explanation:

The inventor's claim could not be possible as no thermal machine can transfer more heat than the input work consumed. If we expose the thermal efficiency:

n=Q out / W in

Where Q and W both must be in the same power unit, so we will convert the remove heat from BTU/hr to hp:

12000 BTU/hr = 4.72 hp

Therefore by comparing, we notice that the removing heat of 4.75 hp is large than the delivered work of 1.11 hp. By evaluating the efficiency:

[tex]n=4.75 hp / 1.1 hp  = 4.3 > 1[/tex]

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A solid cylindrical workpiece made of 304 stainless steel is 150 mm in diameter and 100 mm is high. It is reduced in height by 5
goblinko [34]

Answer:

45.3 MN

Explanation:

The forging force at the end of the stroke is given by

F = Y.π.r².[1 + (2μr/3h)]

The final height, h is given as h = 100/2

h = 50 mm

Next, we find the final radius by applying the volume constancy law

volumes before deformation = volumes after deformation

π * 75² * 2 * 100 = π * r² * 2 * 50

75² * 2 = r²

r² = 11250

r = √11250

r = 106 mm

E = In(100/50)

E = 0.69

From the graph flow, we find that Y = 1000 MPa, and thus, we apply the formula

F = Y.π.r².[1 + (2μr/3h)]

F = 1000 * 3.142 * 0.106² * [1 + (2 * 0.2 * 0.106/ 3 * 0.05)]

F = 35.3 * [1 + 0.2826]

F = 35.3 * 1.2826

F = 45.3 MN

7 0
3 years ago
Can you guys please introduce yourself​
EleoNora [17]

Answer: why?

Explanation:

6 0
3 years ago
Read 2 more answers
A direct contact heat exchanger (where the fluid mixes completely) has three inlets and one outlet. The mass flow rates of the i
lara31 [8.8K]

Answer:

Enthalpy at outlet=284.44 KJ

Explanation:

m_1=1 Kg/s,m_2=1.5 Kg/s,m_3=22 Kg/s

h_1=100 KJ/Kg,h_2=120 KJ/Kg,h_3=500 KJ/Kg

We need to Find enthalpy of outlet.

Lets take the outlet mass m and outlet enthalpy h.

So from mass conservation

m_1+m_2+m_3=m

   m=1+1.5+2 Kg/s

  m=4.5 Kg/s

Now from energy conservation

m_1h_1+m_2h_2+m_3h_3=mh

By putting the values

1\times 100+1.5\times 120+2\times 500=4.5\times h

So h=284.44 KJ

4 0
3 years ago
Prefix version of 6600 volts​
GenaCL600 [577]

Answer:

6.6 kilo volts = 6.6 k volts

Explanation:

A prefix is a word, number or a letter that is added before another word. In physics we have different prefixes for the exponential powers of 10, that are placed before units in place of those powers. Some examples are:

deci (d)   ------  10⁻¹

centi (c)   ------  10⁻²

milli (m)   ------   10⁻³

kilo (k)     ------   10³

mega (M) -----   10⁶

giga (G)   ------   10⁹

We have:

6600 volts

converting to exponential form:

=> 6.6 x 10³ volts

Thus, we know that the prefix of kilo (k) is used for 10³.

Hence,

=> <u>6.6 kilo volts = 6.6 k volts</u>

7 0
3 years ago
An 1,840 W toaster, a 1,420 W electric frying pan, and a 70 W lamp are plugged into the same outlet in a 15 A, 120 V circuit. (T
Viktor [21]

Answer:

A)

Current drawn by toaster = 15.33 A

Current drawn by electric frying pan = 11.83 A

Current drawn by lamp = 0.58 A

B)

The fuse will definitely blow up since the current drawn by three devices (27.74 A) is way higher than 15 A fuse rating.

Explanation:

There are three devices plugged into the same outlet.

Toaster = 1840 W

Electric frying pan = 1420 W

Lamp = 70 W

Since the three devices are connected in parallel therefore, the voltage across them will be same but the current drawn by each will be different.

A) What current is drawn by each device?

The current flowing through the device is given by

I = P/V

Where P is the power and V is the voltage.

Current drawn by toaster:

I = 1840/120

I = 15.33 A

Current drawn by electric frying pan:

I = 1420/120

I = 11.83 A

Current drawn by lamp:

I = 70/120

I = 0.58 A

B) Will this combination blow the 15-A fuse?

The total current drawn by all three devices is

Total current = 15.33 + 11.83 + 0.58

Total current = 27.74 A

Therefore, the fuse will definitely blow up since the current drawn by three devices (27.74 A) is way higher than 15 A fuse rating.

5 0
4 years ago
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