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2 years ago

15

A room is cooled by circulating chilled water through a heat exchanger located in the room. The air is circulated through the he

at exchanger by a 0.25-hp (shaft output) fan. Typical efficiency of small electric motors driving 0.25-hp equipment is 60 percent. Determine the rate of heat supply by the fan–motor assembly to the room.

1 answer:

Zanzabum2 years ago

6 0

Answer:

input power of the geothermal motor will be 0.4166 hP

Explanation:

We have given shaft output = 0.25 hP

hP is known as horse power which is unit of measuring the power of motor

Efficiency of motor is given as efficiency = 60 % = 0.6

We have to fond the heat supply by the fan motor Q

We know that efficiency of any motor is the ratio of shaft power and input power

So $0.6=\frac{0.25}{P_0}$

So $P_0=\frac{0.25}{0.6}=0.4166hP$

So input power of the geothermal motor will be 0.4166 hP

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<h3><u>CSMA/CD Protocol: </u></h3>

Carrier sensing can transmit the data at anytime only the condition is before sending the data sense carrier if the carrier is free then send the data.

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$T_{P}=1 H r$ , at time t = 10.00 AM, A starts, 10:59:59 AM B starts at time 11:00 AM collision starts.

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Pocket Size to detect the collision.

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CSMA/CD is widely used in Ethernet.

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In the previous example we have seen that in worst case $2 T_{P}$ time require to detect a collision.

There could be many collisions may happen before a successful completion of transmission of a packet.

We are given number of collisions (contentions slots)=4.

$\text { Propagation day }=\frac{\text {distance}}{\text {speed}}$

Distance = 1km = 1000m

$\begin{aligned}&\text { Speed }=2 \times 10^{8} \mathrm{m} / \mathrm{sec}\\ &T_{P}=\frac{1000}{2 \times 10^{8}}=(0.5) \times 10^{-5}=5 \times 10^{-6}\\ &T_{t}=5 \mu \mathrm{sec}\end{aligned}$

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2 years ago

Determine the dimensions for W if W = P L^3 / (M V^2) where P is a pressure, L is a length, M is a mass, and V is a velocity.

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Correct answer is option E. No dimensions

As we know formula Pressure (P) is $\frac{F}{A}$

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Dimensional formula of <em>Pressure is </em> $M^{1}L^{-1}T^{-2}$
Dimensional formula of <em>length is L </em>
Dimensional formula of <em>mass is M</em>
Dimensional formula of <em>velocity is </em> $L^{1} T^{-1}$

So, as given W= $\frac{P*L^{3} }{M*V^{2} }$

Dimensional formula of W = $\frac{M^{1}L^{-1}T^{-2} L^{3} }{M^{1} L^{2}T^{-2} }$

since all terms get cancelled

Work is dimensionless i.e no dimensions

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1 year ago

A specimen of commercially pure copper has a strength of 240 MPa. Estimate its average grain diameter using the Hall-Petch equat

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