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Serggg [28]
3 years ago
7

An inventor claims to have devised a cyclical power engine that operates with a fuel whose temperature is 750 °C and radiates wa

ste heat to a sink at 0 °C. He also claims that this engine produces 3.3 kW while rejecting heat at a rate of 4.4 kW. Is this claim valid?
Engineering
1 answer:
Phantasy [73]3 years ago
5 0

Answer:

Yes

Explanation:

Given Data

Temprature of source=750°c=1023k

Temprature of sink =0°c=273k

Work produced=3.3KW

Heat Rejected=4.4KW

Efficiency of heat engine(η)=\frac{Work produced}{Heat supplied}

and

Heat Supplied {\left (Q_s\right)}=Work Produced(W)+Heat rejected\left ( Q_r \right )

{Q_s}=3.3+4.4=7.7KW

η=\frac{3.3}{7.7}

η=42.85%

Also the maximum efficiency of a heat engine operating between two different Tempratures i.e. Source & Sink

η=1-\frac{T_ {sink}}{T_ {source}}

η=1-\frac{273}{1023}

η=73.31%

Therefore our Engine Efficiency is less than the maximum efficiency hence the given claim is valid.

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2 years ago
Air at atmospheric pressure and at 300K flows with a velocity of 1.5m/s over a flat plate. The transition from laminar to turbul
Savatey [412]

Answer:3.47 m

Explanation:

Given

Temperature(T)=300 K

velocity(v)=1.5 m/s

At 300 K

\mu =1.846 \times 10^{-5} Pa-s

\rho =1.77 kg/m^3

And reynold's number is given by

Re.=\frac{\rho v\time x}{\mu }

5\times 10^5=\frac{1.77\times 1.5\times x}{1.846\times 10^{-5}}

x=\frac{5\times 10^5\times 1.846\times 10^{-5}}{1.77\times 1.5}

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5 0
3 years ago
An air-conditioning system operating on the reversed Carnot cycle is required to transfer heat from a house at a rate of 755 kJ/
Lyrx [107]

Answer:

There is 0.466 KW required to operate this air-conditioning system

Explanation:

<u>Step 1:</u> Data given

Heat transfer rate of the house = Ql = 755 kJ/min

House temperature = Th = 24°C = 24 +273 = 297 Kelvin

Outdoor temperature = To = 35 °C = 35 + 273 = 308 Kelvin

<u>Step 2: </u> Calculate the coefficient of performance o reversed carnot air-conditioner working between the specified temperature limits.

COPr,c = 1 / ((To/Th) - 1)

COPr,c = 1 /(( 308/297) - 1)

COPr,c = 1/ 0.037

COPr,c = 27

<u>Step 3:</u> The power input cna be given as followed:

Wnet,in = Ql / COPr,max

Wnet, in = 755  / 27

Wnet,in = 27.963 kJ/min

Win = 27.963 * 1 KW/60kJ/min  = 0.466 KW

There is 0.466 KW required to operate this air-conditioning system

3 0
3 years ago
Does a food market have any rooms in particular? Also whats units?
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Vivi is a drummer for a band. She burns 756756756 calories while drumming for 333 hours. She burns the same number of calories each hour.

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3 years ago
Read 2 more answers
Shows a closed tank holding air and oil to which is connected a U-tube mercury manometer and a pressure gage. Determine the read
damaskus [11]

Answer:

P_2-P_1=27209h

Explanation:

For pressure gage we can determine this by saying:

The closed tank with oil and air has a pressure of P₁ and the pressure of oil at a certain height in the U-tube on mercury is p₁gh₁. The pressure of mercury on the air in pressure gauge is p₂gh₂. The pressure of the gage is P₂.

P_1+p_1gh_1=p_2_gh_2+P_2

We want to work out P₁-P₂: Heights aren't given so we can solve it in terms of height: assuming h₁=h₂=h

P_1-P_2=p_1gh_1-p_2gh_2=(55)\cdot{32.2}h-845\cdot{32.2}h

P_2-P_1=27209h

3 0
3 years ago
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