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Angelina_Jolie [31]
3 years ago
12

11) (10 points) A large valve is to be used to control water supply in large conduits. Model tests are to be done to determine h

ow the valve will operate. Both the model and prototype will use water as the fluid. The model will be 1/6 scale (the modeled valve will be 1/6 the size of the prototype valve). If the prototype flow rate is to be 700 ft3 /s, determine the model flow rate. Use Reynolds scaling for the velocity.
Engineering
1 answer:
IrinaVladis [17]3 years ago
7 0

Answer:

7.94 ft^3/ s.

Explanation:

So, we are given that the '''model will be 1/6 scale (the modeled valve will be 1/6 the size of the prototype valve)'' and the prototype flow rate is to be 700 ft3 /s. Then, we are asked to look for or calculate or determine the value for the model flow rate.

Note that we are to use Reynolds scaling for the velocity as par the instruction from the question above.

Therefore; kp/ks = 1/6.

Hs= 700 ft3 /s and the formula for the Reynolds scaling => Hp/Hs = (kp/ks)^2.5.

Reynolds scaling==> Hp/ 700 = (1/6)^2.5.

= 7.94 ft^3/ s

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The ABC Corporation manufactures and sells two products: T1 and T2. 20XX budget for the company is given below:
Sliva [168]

Answer:

The ABC Corporation

a) Total Expected Revenue (in dollars) for 20XX:

Revenue from T1 = 60,000 x $165 = $26,400,000

Revenue from T2 = 40,000 x $250 = $10,000,000

Total Revenue from T1 and T2 = $36,400,000

b) Production Level (in units) for T1 and T2

                                           T1                       T2

Total Units sold             160,000           40,000

Add Closing Inventory   25,000             9,000

Units Available for sale 185,000           49,000

less opening inventory  20,000             8,000

Production Level          165,000 units 41,000 units

c) Total Direct Material Purchases (in dollars):

Cost of direct materials used    T1                T2

A:       (165,000 x 4 x $12)   $7,920,000   $2,460,000 (41,000 x 5 x $12)

B:       (165,000 x 2 x $5)       1,650,000          615,000 (41,000 x 3 x $5)

C:                                                           0          123,000 (41,000 x 1 x$3)

Total cost                            $9,570,000     $3,198,000 Total = $12,768,000

Cost of direct per unit = $58 ($9,570,000/165,000) for T1 and $78 ($3,198,000/41,000) for T2

Cost of direct materials used for production $12,768,000

Cost of closing direct materials:

                 A  (36,000 x $12)  $432,000

                 B (32,000 x $5)        160,000

                 C (7,000 x $3)            21,000             $613,000

Cost of direct materials available for prodn   $13,381,000

Less cost of beginning direct materials:

                 A  (32,000 x $12)        $384,000

                 B  (29,000 x $5)            145,000

                 C  (6,000 x $3)                18,000        $547,000

Cost of direct materials purchases               $12,834,000

d) The Total Direct Manufacturing Labor Cost (in dollars):

                                             T1                         T2

Direct labor per unit              2 hours                  3 hours

Direct labor rate per hour    $12                        $16

Direct labor cost per unit   $24                          $48

Production level              165,000 units        41,000 units

Labor Cost ($)                $3,960,000        $1,968,000

Total labor cost  $5,928,000 ($3,960,000 + $1,968,000)

e) Total Overhead cost (in dollars):

Overhead rate  = $20 per labor hour

Overhead cost per unit: T1 = $40 ($20 x 2) and T2 = $60 ($20 x 3)

T1 overhead = $20 x 2  x 165,000) = $6,600,000

T2 overhead = $20 x 3 x 41,000) =    $2,460,000

Total Overhead cost =                        $9,060,000

Cost of goods produced:

Cost of opening inventory of materials  = $547,000

Purchases of directials materials             12,834,000

less closing inventory of materials     =      $613,000

Cost of materials used for production    12,768,000

add Labor cost                                           5,928,000

add Overhead cost                                    9,060,000

Total production cost                            $27,756,000

f) Total cost of goods sold (in dollars):

Cost of opening inventory =          $3,928,000

Total Production cost             =    $27,756,000

Cost of goods available for sale  $31,684,000

Less cost of closing inventory       $4,724,000

Total cost of goods sold            $26,960,000

g) Total expected operating income (in dollars)

Sales Revenue:  T1 and T2  $36,400,000

Cost of goods sold                 26,960,000

Gross profit                             $9,440,000

less marketing & distribution      400,000

Total Expected Operating Income = $9,040,000

Explanation:

a) Cost of beginning inventory of finished goods:

T1, (Direct materials + Labor + Overhead) X inventory units =

T1 = 20,000 x ($58 + 24 + 40) = $2,440,000

T2 = 8,000 ($78 + 48 + 60) = $1,488,000

Total cost of beginning inventory = $3,928,000

b) Cost of closing Inventory of finished goods:

T1 = 25,000 x ($58 + 24 + 40) = $3,050,000

T2 = 9,000 ($78 + 48 + 60) = $1,674,000

Total cost of closing inventory = $4,724,000

5 0
3 years ago
Consider a thermal energy reservoir at 1500 K that can supply heat at a rate of 150,000 kJ/h. Determine the exergy of this suppl
anzhelika [568]

Answer:

33.4KW

Explanation:

Firstly, we calculate the power index in a reversible heat engine which is given as;

n = 1 - T1/T2

T1 = atmospheric temperature 298k

T2 = reservoir temperature 1500k

n = 1 - 298/1500

n = 0.8013

output energy = n × energy input

0.8013×150000 = 120195KJ/hr

Power output = 120195KJ/hr/3600 = 33.4KW

8 0
3 years ago
A product whose total work content time = 50 min is assembled on a manual production line at a rate of 24 units per hour. From p
valentinak56 [21]

Answer:

a)Cycle time = 2.37 min

b)Numbers of workers =21

c)Stations on the line =24

Explanation:

Given that

Total work content time(TWC) = 50 min

Production rate Rp= 24 units/hr

manning level will be close =1.5

Line balancing efficiency =0.94

a)

Cycle time

T_c=\dfrac{60E}{R_P}

T_c=\dfrac{60\times 0.95}{24}

Cycle time = 2.37 min

b)

Numbers of workers ,W

W=\dfrac{TWC}{T_c}

W=\dfrac{50}{2.37}

W= 21

Numbers of workers =21

c)

Stations on the line(n)

Lets find service time Ts

Ts = Cycle time -  Time for repositioning

Ts = Tc- Tr

Ts= 2.37  - 9/ 60 min

Ts= 2.22 min

We  know that efficiency

\eta=\dfrac{TWC}{n.T_s}

0.94=\dfrac{50}{n\times 2.22}

n=23.94  ⇒n=24

n=24

Stations on the line =24

7 0
3 years ago
A rigid canister with a radius of 5 in and a height of 10 in is filled with air. The initial pressure and temperature of air in
Elza [17]

Answer:1.458 Btu

Explanation:

Given

radius of canister\left ( r\right )=5in

Height of canister\left ( h\right )=10 in.

Initial pressure\left ( P_i\right )=14.7 Psi

Initia ltemprature\left ( T_i\right )=70^{\circ}F

Final pressure\left ( P_f\right )=30Psi

as canister is rigid therefore change in volume is zero

therefore

\frac{P_i}{T_i}=\frac{P_f}{T_f}

\frac{14.7}{70}=\frac{30}{T_f}

T_f=142.85^{\circ}F

volume of canister=\pi \times r^{2}\times h

                               =\pi \times 5^{2}\times 10=250\pi

volume of canister=12872,038.8 mm^3

now calculating mass of air

PV=mRT

substituting values

\left ( 14.7Psi\right )\left ( 12872,038.8 mm^3\right )=m\left ( 0.287\right )\left ( 70^{\circ}F\right )

m=21.0192 gm

Therefore heat transferred =mc_p\left ( T_f-T_i\right )

                                             =21.0192\times 10^{-3}\times \left ( 142.857-70\right )

                                             =1539.052J=1.458Btu

3 0
4 years ago
Burn rate can be affected by: A. Variations in chamber pressure B. Variations in initial grain temperature C. Gas flow velocity
Digiron [165]

Answer: D) All of the above

Explanation:

Burn rate can be affected by all of the above reasons as, variation in chamber pressure because the pressure are dependence on the burn rate and temperature variation in initial gain can affect the rate of the chemical reactions and initial gain in the temperature increased the burning rate. As, gas flow velocity also influenced to increasing the burn rate as it flowing parallel to the surface burning. Burn rate is also known as erosive burning because of the variation in flow velocity and chamber pressure.

4 0
3 years ago
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