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Gre4nikov [31]
3 years ago
11

A heat engine is coupled with a dynamometer. The length of the load arm is 900 mm. The spring balance reading is 16. Applied wei

ght is 500 N. Rotational speed is 1774. How many kW of power will be developed?
Engineering
1 answer:
miss Akunina [59]3 years ago
4 0

Answer:

P = 80.922 KW

Explanation:

Given data;

Length of load arm is 900 mm = 0.9 m

Spring balanced  read 16 N

Applied weight is 500 N

Rotational speed is 1774 rpm

we know that power is given as

P = T\times \omega

T Torque = (w -s) L = (500 - 16)0.9 = 435.6 Nm

\omega angular speed =\frac{2 \pi N}{60}

Therefore Power is

P =\frac{435.6 \time 2 \pi \times 1774}{60} = 80922.65  watt

P = 80.922 KW

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That is a false statement.
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Thin film deposition is a process where: a)-elemental, alloy, or compound thin films are deposited onto a bulk substrate! b)-Pho
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Answer:

(A) elemental, alloy, or compound thin films are deposited on to a bulk substrate

Explanation:

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Answer:

Tmax=14.5MPa

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Explanation:

T = 600 * 0.15 = 90N.m

T_max =\frac{T_c}{j}  = \frac{x}{y}  = \frac{90 \times 0.0175}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}

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T_{min} =\frac{T_c}{j}  = \frac{x}{y}  = \frac{90 \times 0.0125}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}

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Cloud [144]

Answer:

The answer is "-121\  \frac{KJ}{Kg}".

Explanation:

Please find the correct question in the attachment file.

using formula:

\to W=-P_1V_1+P_2V_2 \\\\When \\\\\to W= \frac{P_1V_1-P_2V_2}{n-1}\ \   or \ \  \frac{RT_1 -RT_2}{n-1}\\\\

W =\frac{R(T_1 -T_2)}{n-1}\\\\

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Answer:

What that means please explain

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