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erik [133]
3 years ago
7

Water enters an ice machine at 55°F and leaves as ice at 25°F. If the COP of the ice machine is 2.45 during this operation, dete

rmine the required power input for an ice production rate of 28 lbm/h. (169 Btu of energy needs to be removed from each lbm of water at 55°F to turn it into ice at 25°F.)
Engineering
1 answer:
adell [148]3 years ago
8 0

Answer:

\dot W_{in} = 0.82363 hp

Explanation:

Given data:

T_1 = 55 degree F

t_2 - 25 degree F

COP = 2.45

Production rate = 28 lbm/h

we knwo that cooling load is given

\dot Q_L = \dot m q_L

\dot Q_L = 28 lbm/h \times 16 = 4732 Btu/h

Power is determined as

\dot W_{in} = \frac{\dot Q_L}COP}

\dot W_{in} = \frac{4732}{2.4} = 1931.42 Btu/h

\dot W_{in} =1931.42 Btu \times \frac{1 hp}{2345 Btu/h} = 0.82363 hp

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What is the metal removal rate when a 2 in-diameter hole 3.5 in deep is drilled in 1020 steel at cutting speed of 120 fpm with a
Studentka2010 [4]

Answer:

a) the metal removal rate is 14.4 in³/min

b) the cutting time is 0.98 min

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Ns = 12V/πD

V is the cutting speed(120 fpm) and D is the diameter of the hole( 2 in)

so we substitute

Ns = 12 × 120 / π2

Ns = 1440 / 6.2831

Ns = 229.18 rmp

Now we find the metal removal rate using the equation

MRR = (πD²/4) Fr × Ns

Fr is the feed rate( 0.02 ipr ),

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MRR = ((π × 2²)/4) × 0.02 × 229.18

MRR = 14.3998 ≈ 14.4 in³/min

Therefore the metal removal rate is 14.4 in³/min

Next we find the allowance for approach of the tip of the drill

A = D/2

A = 2/2

= 1 in

now find the time required to drill the hole

Tm = (L + A) / (Fr × Ns)

Lis the the depth of the hole( 3.5 in)

so we substitute our values

Tm = (3.5 + 1) / (0.02 × 229.18  )

Tm = 4.5 / 4.5836

Tm = 0.98 min

Therefore the cutting time is 0.98 min

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