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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.)

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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uysha [10]

Answer:

$\frac {p_2- p_1}{\rho g} = 31.06 m$

Explanation:

from bernoulli's theorem we have

$\frac{p_1}{\rho g} + \frac{v_1^{2}}{2g} +z_1 = \frac{p_2}{\rho g} + \frac{v_2^{2}}{2g} +z_2 + h_f$

we need to find pressure head difference i.e.

$\frac {p_2- p_1}{\rho g} = (z_1 - z_2) - h_f$

where h_f id head loss

$h_f = \frac{flv^{2}}{D 2g}$

velocity v = $\frac{1}{n} * R^{2/3} S^{2/3}$

$S = \frac{\delta h}{L} = \frac{40}{150} = 0.267$

hydraulic mean radius R = $\frac{A}{P} = \frac{hw}{2h+w}$

$R = \frac{40*1}{2*40+1} = 0.493 m$

so velocity is = $\frac{1}{0.013} * 0.493^{2/3} 0.267^{1/2}$

v = 24.80 m/s

head loss

$h_f = \frac{0.0019*150*24.80^{2}}{1* 2*9.81}$

$h_f =8.93 m$

pressure difference is

$\frac {p_2- p_1}{\rho g} = 40 - 8.93 = 31.06 m$

$\frac {p_2- p_1}{\rho g} = 31.06 m$

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3 years ago

PLS HELP! which statement is the best example of how society affects the useof technology? A. Farmers in hilly areas grow crops

Zielflug [23.3K]

Answer:

A

Explanation:

because Farmers in hilly areas grow crops on terraces built on the hillsides.

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2 years ago

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2 years ago

A pitfall cited in Section 1.10 is expecting to improve the overall performance of a computer by improving only one aspect of th

Oxana [17]

Answer:

a) For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time

$(1-0.2)*70 s =56s$

The reduction on this case is $70-56 s=14s$

And since the new total time would be given by $250-14=236 s$

b) For this case the total time is reduced 20% so that means that the new total time would be (1-0.2)=0.8 times the original total time $(1-0.2) *250s =200 s$

The original time for INT operations is calculated as:

$250 = 70+85+40 +t_{INT}$

$t_{INT}=55s$

For this part the only time that was changed is assumed the INT operations so then:

$200 = 70+85+40 \Delta t_{INT}$

And then: $\Delta t_{INT}= 200-70-85-40=5 s$

c) A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.

Explanation:

From the info given we know that a computer running a program that requires 250 s, with 70 s spent executing FP instructions, 85 s executed L/S instructions and 40 s spent executing branch instructions.

Part 1

For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time

$(1-0.2)*70 s =56s$

The reduction on this case is $70-56 s=14s$

And since the new total time would be given by $250-14=236 s$

Part 2

For this case the total time is reduced 20% so that means that the new total time would be (1-0.2)=0.8 times the original total time $(1-0.2) *250s =200 s$

The original time for INT operations is calculated as:

$250 = 70+85+40 +t_{INT}$

$t_{INT}=55s$

For this part the only time that was changed is assumed the INT operations so then:

$200 = 70+85+40 \Delta t_{INT}$

And then: $\Delta t_{INT}= 200-70-85-40=5 s$

And we can quantify the decrease using the relative change:

$\% Change = \frac{5s}{55 s} *100 = 9.09\%$ of reduction

Part 3

A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.

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3 years ago

Consider a system with two tasks, Task1 and Task2. Task1 has a period of 200 ms, and Task2 has a period of 300 ms. All tasks ini

Murrr4er [49]

<u>Explanation:</u>

Task 1 time period = 200ms, Task 2 time period = 300ms

Task ticked = $\frac{1000ms}{200ms}= 5$ → 5 times

Task 2 ticked = $\frac{1000ms}{300ms} = 3.33$ → 3 times

At 600 ms → 200ms 200ms 200ms

300ms → $\frac{30ms}{60ms}$

Largest time period = H.C.M of (200ms, 300ms)

= 600ms

4 0

3 years ago