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patriot [66]
3 years ago
15

A venturi meter is to be installed in a 63 mm bore section of a piping system to measure the flow rate of water in it. From spac

e considerations, the maximum differential head in the mercury manometer is to be 235 mm. The maximum expected flow rate is 240 litres per minute. Design the throat diameter by assuming the discharge coefficient to be 0.8
Engineering
1 answer:
Alex Ar [27]3 years ago
4 0

Answer:

Throat diameter d_2=28.60 mm

Explanation:

 Bore diameter d_1=63mm  ⇒A_1=3.09\times 10^{-3} m^2

Manometric deflection x=235 mm

Flow rate Q=240 Lt/min⇒ Q=.004\frac{m^3}{s}

Coefficient of discharge C_d=0.8

We know that discharge through venturi meter

 Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}

h=x(\dfrac{S_m}{S_w}-1)

S_m=13.6 for Hg and S_w=1 for water.

h=0.235(\dfrac{13.6}{1}-1)

h=2.961 m

Now by putting the all value in

Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}

0.004=0.8\times \dfrac{3.09\times 10^{-3} A_2\sqrt{2\times 9.81\times 2.961}}{\sqrt{(3.09\times 10^{-3})^2-A_2^2}}

A_2=6.42\times 10^{-4} m^2

 ⇒d_2=28.60 mm

So throat diameter d_2=28.60 mm

     

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A reservoir delivers water to a horizontal pipeline 39 long The first 15 m has a diameter of 50 mm, after which it suddenly beco
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Answer:

The difference of head in the level of reservoir is 0.23 m.

Explanation:

For pipe 1

d_1=50 mm,f_1=0.0048

For pipe 2

d_2=75 mm,f_2=0.0058

Q=2.8 l/s

Q=2.8\times 10^{-3]

We know that Q=AV

Q=A_1V_1=A_2V_2

A_1=1.95\times 10^{-3}m^2

A_2=4.38\times 10^{-3} m^2

So V_2=0.63 m/s,V_1=1.43 m/s

head loss (h)

h=\dfrac{f_1L_1V_1^2}{2gd_1}+\dfrac{f_2L_2V_2^2}{2gd_2}+0.5\dfrac{V_1^2}{2g}

Now putting the all values

h=\dfrac{0.0048\times 15\times 1.43^2}{2\times 9.81\times 0.05}+\dfrac{0.0058\times 24\times 0.63^2}{2\times 9.81\times 0.075}+0.5\dfrac{1.43^2}{2\times 9.81}

So h=0.23 m

So the difference of head in the level of reservoir is 0.23 m.

8 0
3 years ago
Four of the minterms of the completely specified function f(a, b, c, d) are m0, m1, m4, and m5.
Sveta_85 [38]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) The required additional minterms  for f so that f has eight primary implicants with two literals and no other prime implicant are m_{2},m_{3},m_{7},m_{8},m_{11},m_{12},m_{13},m_{14} and m_{15}

b) The essential prime implicant are c' d',a'b',ab and cd

c) The minimum sum-of-product expression for f are

                  a'b' +ab +c'd'+cd+a'c',\\ a'b'+ab+c'd'+cd+a'd,\\a'b'+ab+c'd'+cd+bc'  and  \\ a'b'+ab+c'd' +cd+bd

Explanation:

The explanation is shown on the second third and fourth image

8 0
3 years ago
Water is the working fluid in an ideal Rankine cycle. The condenser pressure is 8 kPa, and saturated vapor enters the turbine at
sergeinik [125]

Explanation:

The obtained data from water properties tables are:

Point 1 (condenser exit) @ 8 KPa, saturated fluid

h_{f} = 173.358 \\h_{fg} = 2402.522

Point 2 (Pump exit) @ 18 MPa, saturated fluid & @ 4 MPa, saturated fluid

h_{2a} =  489.752\\h_{2b} =  313.2

Point 3 (Boiler exit) @ 18 MPa, saturated steam & @ 4 MPa, saturated steam

h_{3a} = 2701.26 \\s_{3a} = 7.1656\\h_{3b} = 2634.14\\s_{3b} = 7.6876

Point 4 (Turbine exit) @ 8 KPa, mixed fluid

x_{a} = 0.8608\\h_{4a} = 2241.448938\\x_{b} = 0.9291\\h_{4b} = 2405.54119

Calculate mass flow rates

Part a) @ 18 MPa

mass flow

\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3a}  - h_{4a}) - (h_{2a}  - h_{f})}\\\\= \frac{100*10^ 3}{(2701.26  - 2241.448938 ) - (489.752  - 173.358)}\\\\= 697.2671076 \frac{kg}{s} = 2510161.587 \frac{kg}{hr}

Heat transfer rate through boiler

Q_{in}  = mass flow * (h_{3a} -  h_{2a})\\Q_{in} = (697.2671076)*(2701.26-489.752)\\\\Q_{in} = 1542011.787 W

Heat transfer rate through condenser

Q_{out}  = mass flow * (h_{4a} -  h_{f})\\Q_{out} = (697.2671076)*(2241.448938-173.358)\\\\Q_{out} = 1442011.787 W

Thermal Efficiency

n = \frac{W_{net}  }{Q_{in} } = \frac{100*10^3}{1542011.787}  \\\\n = 0.06485

Part b) @ 4 MPa

mass flow

\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3b}  - h_{4b}) - (h_{2b}  - h_{f})}\\\\= \frac{100*10^ 3}{(2634.14  - 2405.54119 ) - (313.12  - 173.358)}\\\\= 1125 \frac{kg}{s} = 4052374.235 \frac{kg}{hr}

Heat transfer rate through boiler

Q_{in}  = mass flow * (h_{3b} -  h_{2b})\\Q_{in} = (1125.65951)*(2634.14-313.12)\\\\Q_{in} = 2612678.236 W

Heat transfer rate through condenser

Q_{out}  = mass flow * (h_{4b} -  h_{f})\\Q_{out} = (1125)*(2405.54119-173.358)\\\\Q_{out} = 2511206.089 W

Thermal Efficiency

n = \frac{W_{net}  }{Q_{in} } = \frac{100*10^3}{1542011.787}  \\\\n = 0.038275

6 0
3 years ago
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OleMash [197]

The statement that correctly describes the circuit is the circuit provided is an example of a CMOS circuit.

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This is known to be a machine that create integrated circuits and it is one that is often seen a a lot of electronic component and as such, the the circuit provided is an example of a CMOS circuit.

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Air-conditioning principles are being discussed. Technician A states that heat always
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Answer: C.

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