4 years ago

Which term defines the amount of mechanical work an engine can do per unit of heat energy it uses?

Engineering

1 answer:

skad [1K]4 years ago

7 0

Answer:

Explanation:

is the because that's the amount of work in making machine can do producing heat

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Vapor lock occurs when the gasoline is cooled and forms a gel, preventing fuel flow and

Sergeu [11.5K]

Answer:

True

Explanation:

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3 years ago

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HELP ASAP PLEASE What are the 7 types of transportation? Give me an example of each of them.

viktelen [127]

Answer:they are information, materials, tools and machines, captial, time, energy, and people

Explanation:

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3 years ago

Which of the following is the typical carbon concentration range for cast irons?

AveGali [126]

Answer:

typical carbon concentration range for cast irons is

F. 3.0 wt% - 4.5wt% C

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3 years ago

Water is flowing steadily through a 3-m long, 20 mm innerdiameter cast iron pipe. The water enters at a uniform velocity U and e

devlian [24]

Answer:

Hello your question is incomplete attached below is the complete question

answer :

U/r = R = $U_{max}$ [ 1 - $(\frac{R}{R} )^2$ ] = 0

∴ No slip condition is satisfied

Explanation:

Given that

Outlet velocity U(r) = $U_{max}$ [ 1 - $(\frac{r}{R} )^2$ ]

<u>prove that the output velocity profile satisfies the no slip condition</u>

at no slip u = 0 ( i.e. at the pipe's inner surface )

at , r = R ( inference is at center )

hence U/r = R = $U_{max}$ [ 1 - $(\frac{R}{R} )^2$ ] = 0

∴ No slip condition is satisfied

8 0

3 years ago

A steam turbine has isentropic efficiency of 0.8. Isentropically, it is supposed to deliver work of 100 kW. What is the actual w

trapecia [35]

Answer:

80 kW; 11 kW; 8 kW; 0.6

Explanation:

Part 1

Isentropic turbine efficiency:

$\eta_t = \frac{\text{Real turbine work}}{\text{isentropic turbine work}} = \frac{W_{real}}{W_s}$

$W_{real} = \eta_t*W_s$

$W_{real} = 0.8*100 kW$

$W_{real} = 80 kW$

Part 2

Coefficient of performance COP is defined by:

$COP = \frac{Q_{out}}{W}$

$Q_{out} = W*COP$

$Q_{out} = 5 kW*2.2$

$Q_{out} = 11 kW$

Part 3

(a)

Energy balance for a refrigeration cycle gives:

$Q_{in} + W = Q_{out}$

$3 kW + 5 kW = Q_{out}$

$8 kW = Q_{out}$

(b)

$COP = \frac{Q_{in}}{W}$

$COP = \frac{3 kW}{5 kW}$

$COP = 0.6$

3 0

3 years ago