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4 years ago

A Carnot heat engine operates between 1000 deg F and 50 deg F, producing 120 BTU of work. What is the heat input to the engine?

Engineering

1 answer:

Mazyrski [523]4 years ago

4 0

Answer:

184.6 BTU

Explanation:

The thermal efficiency for a Carnot cycle follows this equation:

η = 1 - T2/T1

Where

η: thermal efficiency

T1: temperature of the heat source

T2: temperature of the heat sink

These temperatures must be in absolute scale:

1000 F = 1460 R

50 F = 510 R

Then

η = 1 - 510/1460 = 0.65

We also know that for any heat engine:

η = L / Q1

Where

L: useful work

Q1: heat taken from the source

Rearranging:

Q1 = L / η

Q1 = 120 / 0.65 = 184.6 BTU

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Answer:

Technician B

Explanation:

Resistance, $R=\frac {V}{I}$ where V is the voltage and I is the current in amps

Therefore, $R=\frac {12}{12}=1 ohm$

Power=VI=12*12=144 W

Therefore, the power is 144 W and resistance is 1 Ohm. This implies that technician A is wrong while technician B is correct

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4 years ago

The pressure and temperature at the beginning of compression of a cold air-standard Diesel cycle are 100 kPa and 300K, respectiv

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Answer:

Cut off ratio=2.38

Explanation:

Given that

$T_1=300K$

$P_1=100KPa$

$P_2=P_3=7200KPa$

$T_3=2250K$

Lets take $T_1$ is the temperature at the end of compression process

For air γ=1.4

$\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^{\dfrac{\gamma -1}{\gamma}}$

$\dfrac{T_2}{300}=\left(\dfrac{7200}{100}\right)^{\dfrac{1.4-1}{1.4}}$

$T_2=1070K$

At constant pressure

$\dfrac{T_3}{T_2}=\dfrac{V_3}{V_2}$

$\dfrac{2550}{1070}=\dfrac{V_3}{V_2}$

$\dfrac{V_3}{V_2}=2.83$

So cut off ratio

$cut\ off\ ratio =\dfrac{V_3}{V_2}$

Cut off ratio=2.38

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4 years ago

What are flip flops and what do they look like

Vinil7 [7]

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3 years ago

Read 2 more answers

A 1020 Cold-Drawn steel shaft is to transmit 20 hp while rotating at 1750 rpm. Calculate the transmitted torque in lbs. in. Igno

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Answer:

Question 1 A 1020 Cold-Drawn steel shaft is to transmit 20 hp while rotating at 1750 rpm. Calculate the transmitted torque in lbs. in. Ignore the effect of friction. Answer with three decimal points. 60.024 Question 2 Based on the maximum-shear-stress theory, determine the minimum diameter in inches for the shaft in Q1 to provide a safety factor of 3. Assume Sy = 57 Kpsi. Answer with three decimal points. 0.728 Question 3 If the shaft in Q2 was made of ASTM 30 cast iron, what would be the factor of safety? Assume Sut = 31 Kpsi, Suc = 109 Kpsi 0 2.1 O 2.0 O 2.5 0 2.4 2.3 O 2.2

Explanation:

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2 years ago

A centrifugal pump is used to extract water from a reservoir at 14,000 gal/min. The pipe connecting the pump inlet to the reserv

Dahasolnce [82]

This question is incomplete, the complete question is;

A centrifugal pump is used to extract water from a reservoir at 14,000 gal/min. The pipe connecting the pump inlet to the reservoir is 12 inches in diameter and is 65 ft long.

The average friction factor in this pipe is 0.018. The pump performance curves indicate that, at this flow rate, the head rise across the pump is 320 ft, the efficiency is 81% and the required NPSH is 25 ft.

Please estimate: The required brake horsepower.

Answer:

The required brake horsepower is 1400.08

Explanation:

Given the data in the question;

Power required to drive the pump can be determined using the formula;

P = r $_w$ QH / η₀(0.745)

given that; centrifugal pump is used to extract water from a reservoir at 14,000 gal/min.

Q = 14,000 gal/min = ( 14,000 × 0.00006309 )m³/sec = 0.883 m³/sec

the head rise across the pump is 320 ft,

H = 320 ft = ( 320 × 0.3048 )m = 97.536 m

the efficiency η₀ = 81% = 0.81

r $_w$ = 9.81 kN/m³

so we substitute our values into the formula

P = [ 9.81 × 0.883 × 97.536 ] / 0.81(0.745)

P = 844.87926528 / 0.60345

P = 1400.08 HP

Therefore, The required brake horsepower is 1400.08

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3 years ago