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Paladinen [302]
3 years ago
7

Water is stored in a tank which has vent open to the atmosphere. The water level is 1.0 m below the top the tank and the water i

s 4 m deep. A siphon with .03 m ID is used to remove water from the tank. The siphon discharge is 6.0 m below the top of the tank. Neglect the friction loss of water through the siphon. Determine the flow rate of water in unit of m^3/s.
Engineering
1 answer:
sp2606 [1]3 years ago
7 0

Answer:

6.99 x 10⁻³ m³ / s

Explanation:

Th e pressure difference at the two ends of the delivery pipe due to atmospheric pressure and water column will cause flow of water.

h = difference in the height of water column at two ends of delivery pipe

6 - 1 =  5 m

Velocity of flow of water

v = √2gh

= √ (2 x 9.8 x 5)

=  9.9 m /s

Volume of water flowing per unit time

velocity x cross sectional area

= 9.9 x 3.14 x .015²

= 6.99 x 10⁻³ m³ / s

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mote1985 [20]

Answer:

7.07%

Explanation:

Thermal efficiency can be by definition seen as the ratio of the heat utilized by a heat engine to the total heat units in the fuel consumed.

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3 years ago
An ideal vapor-compression refrigeration cycle that uses refrigerant-134a as its working fluid maintains a condenser at 800 kPa
mote1985 [20]

Answer:

COP = 3.828

W' = 39.18 Kw

Explanation:

From the table A-11 i attached, we can find the entropy for the state 1 at -20°C.

h1 = 238.43 KJ/Kg

s1 = 0.94575 KJ/Kg.K

From table A-12 attached we can do the same for states 3 and 4 but just enthalpy at 800 KPa.

h3 = h4 = hf = 95.47 KJ/Kg

For state 2, we can calculate the enthalpy from table A-13 attached using interpolation at 800 KPa and the condition s2 = s1. We have;

h2 = 275.75 KJ/Kg

The power would be determined from the energy balance in state 1-2 where the mass flow rate will be expressed through the energy balance in state 4-1.

W' = m'(h2 - h1)

W' = Q'_L((h2 - h1)/(h1 - h4))

Where Q'_L = 150 kW

Plugging in the relevant values, we have;

W' = 150((275.75 - 238.43)/(238.43 - 95.47))

W' = 39.18 Kw

Formula foe COP is;

COP = Q'_L/W'

COP = 150/39.18

COP = 3.828

4 0
3 years ago
A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 19.636
luda_lava [24]

Answer:

The original length of the specimen is found to be 76.093 mm.

Explanation:

From the conservation of mass principal, we know that the volume of the specimen must remain constant. Therefore, comparing the volumes of both initial and final state as state 1 and state 2:

Initial Volume = Final Volume

πd1²L1/4 = πd2²L2/4

d1²L1 = d2²L2

L1 = d2²L2/d1²

where,

d1 = initial diameter = 19.636 mm

d2 = final diameter = 19.661 mm

L1 = Initial Length = Original Length = ?

L2 = Final Length = 75.9 mm

Therefore, using values:

L1 = (19.661 mm)²(75.9 mm)/(19.636 mm)²

<u>L1 = 76.093 mm</u>

5 0
3 years ago
An ideal fluid flows through a pipe made of two sections with diameters of 1.0 and 3.0 inches, respectively. The speed of the fl
geniusboy [140]

Answer:

(\frac{r_1}{r_2})^2=\frac{1}{9}

Explanation:

From the question we are told that:

Diameter 1 d_1=1.0

Diameter 2 d_2=3.0

Generally the equation for Radius is mathematically given by

At Diameter 1

r_{1}=\frac{1}{2} inch

At Diameter 2

r_{2}=\frac{3}{2} inch

Generally the equation for continuity is mathematically given by

 A_1V_1=A_2V_2

Therefore

(\frac{r_1}{r_2})^2=(\frac{1/2}{3/2})^2

(\frac{r_1}{r_2})^2=\frac{1}{9}

5 0
3 years ago
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