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4 years ago

6

The input shaft to a gearbox rotates at 2300 rpm and transmits a power of 42.6 kW. The output shaft power is 34.84 kW at a rotat

ional speed of 620 rpm. Determine the torque of the input shaft shaft, in N-m.

1 answer:

Marrrta [24]4 years ago

4 0

Answer:

Torque at input shaft will be 176.8695 N-m

Explanation:

We have given input power $P_{IN}=42.6KW=42.6\times 10^3W$

Angular speed = 2300 rpm

For converting rpm to rad/sec we have multiply with $\frac{2\pi }{60}$

So $2300rpm=\frac{2300\times 2\pi }{60}=240.855rad/sec$

We have to find torque

We know that power is given by $P=\tau \omega$ , here $\tau$ is torque and $\omega$ is angular speed

So $42.6\times 10^3=\tau \times 240.855$

$\tau =176.8695N-m$

So torque at input shaft will be 176.8695 N-m

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4 years ago

Air enters a cmpressor at 20 deg C and 80 kPa and exits at 800 kPa and 200 deg C. The power input is 400 kW. Find the heat trans

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Now,

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where

$h_{i}$ = specific enthalpy at inlet

$h_{o}$ = specific enthalpy at outlet

$v_{i}$ = air speed at inlet

$p_{s}$ = specific power input

H and H' = Elevation of inlet and outlet

Now, if

$v_{i} = 0$ and H = H'

Then the above eqn reduces to:

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$p_{s} = \frac{P_{i}}{ mass, m}$

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3 years ago

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statuscvo [17]

Answer: the mass flow rate of concentrated brine out of the process is 46,666.669 kg/hr

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solute

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8 0

4 years ago