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zzz [600]
2 years ago
10

A construction company distributes its products by trucks loaded at its loading station. A backacter in conjunction with trucks

are used for this purpose. If it was found out that on an average of 12 trucks per hour arrived and the average loading time was 3 minutes for each truck. A truck must queue until it is loaded. The backacter’s daily all-in rate is GH¢ 1000 and that of the truck is GH¢ 400.
a) Compute the operating characteristics: L, Lq, W, Wq, and P.

b) The company is considering replacing the backacter with a bigger one which will have an average service rate of 1.5 minutes to serve trucks waiting to have their schedules improved. As a manager, would you recommend the new backacter if the daily all-in rate is GH¢ 1300.

c) The site management is considering whether to deploy an extra backwater to assist the existing one. The daily all-in-rate and efficiency of the new backwater is assumed to be the same as that of the existing backwater. Should the additional backwater be deployed?
Engineering
1 answer:
blagie [28]2 years ago
6 0

Answer:

a) L = 1.5

L_q = 0.9

W =  \dfrac{1 }{8 } \, hour

W_q =  \dfrac{3}{40 } \, hour

P = \dfrac{3}{5 }

b) The new backacter should be recommended

c) The additional backacter should not be deployed

Explanation:

a) The required parameters are;

L = The number of customers available

L = \dfrac{\lambda }{\mu -\lambda }

μ = Service rate

L_q = The number of customers waiting in line

L_q = p\times L

W = The time spent waiting including being served

W = \dfrac{1 }{\mu -\lambda }

W_q = The time spent waiting in line

W_q = P \times W

P = The system utilization

P = \dfrac{\lambda }{\mu  }

From the information given;

λ = 12 trucks/hour

μ = 3 min/truck = 60/3 truck/hour = 20 truck/hour

Plugging in the above values, we have;

L = \dfrac{12 }{20 -12 } = \dfrac{12 }{8 } = 1.5

P = \dfrac{12 }{20 } = \dfrac{3}{5 }

L_q = \dfrac{3}{5 } \times \dfrac{3}{2 } = \dfrac{9}{10 } = 0.9

W = \dfrac{1 }{20 -12 } =  \dfrac{1 }{8 } \ hour

W_q = \dfrac{3}{5 } \times \dfrac{1}{8 } = \dfrac{3}{40 } \, hour

(b) The service rate with the new backacter = 1.5 minutes/truck which is thus;

μ = 60/1.5 trucks/hour = 40 trucks/hour

P = \dfrac{12 }{40  } = \dfrac{3}{10}

W = \dfrac{1 }{40 -12 } =  \dfrac{1 }{38 } \, hour

W_q = \dfrac{3}{10 } \times \dfrac{1}{38 } = \dfrac{3}{380 } \, hour

λ = 12 trucks/hour

Total cost = mC_s + \lambda WC_w

m = 1

C_s = GH¢ = 1300

C_w = 400

Total cost with the old backacter is given as follows;

1 \times 1000 + 12 \times \dfrac{1}{8}  \times 400 = \$ 1,600.00

Total cost with the new backacter is given as follows;

1 \times 1300 + 12 \times \dfrac{1}{38}  \times 400 = \$ 1,426.32

The new backacter will reduce the total costs, therefore, the new backacter is recommended.

c)

Here μ = 3 min/ 2 trucks = 2×60/3 truck/hour = 40 truck/hour

\therefore W = \dfrac{1 }{40 -12 } =  \dfrac{1 }{38 } \, hour

Total cost with the one backacter is given as follows;

1 \times 1000 + 12 \times \dfrac{1}{8}  \times 400 = \$ 1,600.00

Total cost with two backacters is given as follows;

2 \times 1000 + 12 \times \dfrac{1}{38}  \times 400 = \$ 2,126.32

The additional backacter will increase the total costs, therefore, it should not be deployed.

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2 years ago
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3 years ago
Liquid water is fed to a boiler at 24°C and 10 bar is converted at a constant pressure to saturated steam.
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Then the heat required in input is,

Q=\dot{m}\Delta h

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Q=57157.036kW

With the same value required of 15000m^3/h, we can calculate the velocity of the water, that is given by,

V= \frac{\dotV}{A}

V = \frac{\frac{15000}{3600}}{\pi /4 *(0.15)^2}

V=235.79m/s

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\dot{m}(h_1+\frac{V^2}{2000})+Q = \dot{m}h_2

Re-arrange for Q,

Q=\dot{m}(h_2-h_1-\frac{V^2}{2000})

Q=\dot{m}(\Delta h-\frac{V^2}{2000})

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Q= 56560.88kW

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4 0
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8 0
1 year ago
The boost converter of Fig. 6-8 has parameter Vs 20 V, D 0.6, R 12.5 , L 10 H, C 40 F, and the switching frequency is 200 kHz. (
mr Goodwill [35]

Answer:

a) the output voltage is 50 V

b)

- the average inductor current is 10 A

- the maximum inductor current is 13 A

- the maximum inductor current is 7 A

c) the output voltage ripple is 0.006 or 0.6%V₀

d) the average current in the diode under ideal components is 4 A

Explanation:

Given the data in the question;

a) the output voltage

V₀ = V_s/( 1 - D )

given that; V_s = 20 V, D = 0.6

we substitute

V₀ = 20 / ( 1 - 0.6 )

V₀ = 20 / 0.4

V₀ = 50 V

Therefore, the output voltage is 50 V

b)

- the average inductor current

I_L = V_s / ( 1 - D )²R

given that R = 12.5 Ω, V_s = 20 V, D = 0.6

we substitute

I_L = 20 / (( 1 - 0.6 )² × 12.5)

I_L = 20 / (( 0.4)² × 12.5)

I_L = 20 / ( 0.16 × 12.5 )

I_L = 20 / 2

I_L = 10 A

Therefore, the average inductor current is 10 A

- the maximum inductor current

I_{Lmax = [V_s / ( 1 - D )²R] + [ V

given that, R = 12.5 Ω, V_s = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz

we substitute

I_{Lmax = [20 / (( 1 - 0.6 )² × 12.5)] + [ (20 × 0.6 × 5) / (2 × 10) ]

I_{Lmax = [20 / 2 ] + [ 60 / 20 ]    

I_{Lmax = 10 + 3

I_{Lmax = 13 A

Therefore, the maximum inductor current is 13 A

- The minimum inductor current

I_{Lmax = [V_s / ( 1 - D )²R] - [ V

given that, R = 12.5 Ω, V_s = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz

we substitute

I_{Lmin = [20 / (( 1 - 0.6 )² × 12.5)] - [ (20 × 0.6 × 5) / (2 × 10) ]

I_{Lmin = [20 / 2 ] -[ 60 / 20 ]    

I_{Lmin = 10 - 3

I_{Lmin  = 7 A

Therefore, the maximum inductor current is 7 A

 

c)  the output voltage ripple

ΔV₀/V₀ = D/RCf

given that; R = 12.5 Ω, C = 40 μF = 40 × 10⁻⁶ F, D = 0.6, f = 200 Khz = 2 × 10⁵ Hz

we substitute

ΔV₀/V₀ = 0.6 / (12.5 × (40 × 10⁻⁶) × (2 × 10⁵) )

ΔV₀/V₀ = 0.6 / 100

ΔV₀/V₀ = 0.006 or 0.6%V₀

Therefore, the output voltage ripple is 0.006 or 0.6%V₀

d) the average current in the diode under ideal components;

under ideal components; diode current = output current

hence the diode current will be;

I_D = V₀/R

as V₀ = 50 V and R = 12.5 Ω

we substitute

I_D = 50 / 12.5

I_D = 4 A

Therefore, the average current in the diode under ideal components is 4 A

7 0
3 years ago
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