Compute the discharge observed at a v-notch weir. The weir has an angle of 90-degrees. The height above the weir is 3 inches.

Answers

Answer 1

Answer: the discharge observed at a v-notch weir is 66.7 in³/s

Explanation:

Given that;

Notch angle ∅ = 90°

height above the weir is 3 inches { head + head correction factor) h + k = 3 in

Discharge Q = ?

To determine the discharge observed, we us the following expression

Q = 4.28Ctan(∅/2) ( h + k )^5/2

where Q is discharge, C is discharge coefficient, ∅ is notch angle, h is head and k is head correction factor

now we substitute

Q = 4.28 × 1 × tan(90/2) ( 3 )^5/2

Q = 4.28 × 1 × 1 × 15.5884

Q =  66.7 in³/s

Therefore the discharge observed at a v-notch weir is 66.7 in³/s


Related Questions

A gas stream contains 18.0 mole% hexane and the remainder nitrogen. The stream flows to a condenser, where its temperature is reduced and some of the hexane is liquefied. The hexane mole fraction in the gas stream leaving the condenser is 0.0500. Liquid hexane condensate is recovered at a rate of 1.50 L/min.

(a) What is the flow rate of the gas stream leaving the condenser in mol/min? (Hint : First calculate the molar flow rate of the condensate and note that the rates at which C6H14 and N2 enter the unit must equal the total rates at which they leave in the two exit streams.)

(b) What percentage of the hexane entering the condenser is recovered as a liquid?

Answers

Answer:

A. 72.34mol/min

B. 76.0%

Explanation:

A.

We start by converting to molar flow rate. Using density and molecular weight of hexane

= 1.59L/min x 0.659g/cm³ x 1000cm³/L x 1/86.17

= 988.5/86.17

= 11.47mol/min

n1 = n2+n3

n1 = n2 + 11.47mol/min

We have a balance on hexane

n1y1C6H14 = n2y2C6H14 + n3y3C6H14

n1(0.18) = n2(0.05) + 11.47(1.00)

To get n2

(n2+11.47mol/min)0.18 = n2(0.05) + 11.47mol/min(1.00)

0.18n2 + 2.0646 = 0.05n2 + 11.47mol/min

0.18n2-0.05n2 = 11.47-2.0646

= 0.13n2 = 9.4054

n2 = 9.4054/0.13

n2 = 72.34 mol/min

This value is the flow rate of gas that is leaving the system.

B.

n1 = n2 + 11.47mol/min

72.34mol/min + 11.47mol/min

= 83.81 mol/min

Amount of hexane entering condenser

0.18(83.81)

= 15.1 mol/min

Then the percentage condensed =

11.47/15.1

= 7.59

~7.6

7.6x100

= 76.0%

Therefore the answers are a.) 72.34mol/min b.) 76.0%

Please refer to the attachment .

An unknown impedance Z is connected across a 380 V, 60 Hz source. This causes a current of 5A to flow and 1500 W is consumed. Determine the following: a. Real Power (kW) b. Reactive Power (kvar) c. Apparent Power (kVA) d. Power Factor e. The impedance Z in polar and rectangular form

Answers

Answer:

a) Real Power (kW) = 1.5 kW

b) Reactive Power (kvar) is 1.1663 KVAR

c) Apparent Power (kVA) is 1.9 KVA

d) the Power Factor cos∅ is 0.7894

e) the impedance Z in polar and rectangular form is 76 ∠ 37.87° Ω

Explanation:

Given that;

V = 380v

i = 5A

P = 1500 W

determine;

a) Real Power (kW)

P = 1500W = 1.5 kW

therefore Real Power (kW) = 1.5 kW

b) Reactive Power (kvar)

p = V×i×cos∅

cos∅ = p / Vi

cos∅ = 1500 / ( 380 × 5 ) = 0.7894

∅ = cos⁻¹ (0.7894)

∅ = 37.87°

Q = VIsin∅

Q = 380 × 5 × sin( 37.87° )

Q = 1.1663 KVAR

Therefore Reactive Power (kvar) is 1.1663 KVAR

c) Apparent Power (kVA)

S = P + jQ

= ( 1500 + J 1166.3 ) VA

S = 1900 ∠ 37.87° VA

S = 1.9 KVA

Therefore Apparent Power (kVA) is 1.9 KVA

d) Power Factor

p = V×i×cos∅

cos∅ = p / Vi

cos∅ = 1500 / ( 380 × 5 ) = 0.7894

Therefore the Power Factor cos∅ is 0.7894

e) The impedance Z in polar and rectangular form

Z = 380 / ( S∠-37.87) = V/I

Z = ( 60 + j 46.647) Ω

Z = 76 ∠ 37.87° Ω

Therefore the impedance Z in polar and rectangular form is 76 ∠ 37.87° Ω

Un mol de gas ideal realiza un trabajo de 3000 J sobre su entorno, cuando se expande de manera isotermica a una temperatura de 58°C, cuando su volumen inicial es de 25 L. Determinar el volumen final

Answers

Answer:

74,4 litros

Explanation:

Dado que

W = nRT ln (Vf / Vi)

W = 3000J

R = 8,314 JK-1mol-1

T = 58 + 273 = 331 K

Vf = desconocido

Vi = 25 L

W / nRT = ln (Vf / Vi)

W / nRT = 2.303 log (Vf / Vi)

W / nRT * 1 / 2.303 = log (Vf / Vi)

Vf / Vi = Antilog (W / nRT * 1 / 2.303)

Vf = Antilog (W / nRT * 1 / 2.303) * Vi

Vf = Antilog (3000/1 * 8,314 * 331 * 1 / 2,303) * 25

Vf = 74,4 litros

what is an example of an innovative solution to an engineering problem? Explain briefly why you chose this answer.

Answers

Answer:

robotic technology    

Explanation:

Innovation is nothing but the use of various things such as ideas, products, people to build up a solution for the benefit of the human. It can be any product or any solution which is new and can solve people's problems.

Innovation solution makes use of technology to provide and dispatch new solutions or services which is a combination of both technology and ideas.

One such example of an innovative solution we can see is the use of "Robots" in medical science or in any military operations or rescue operation.

Sometimes it is difficult for humans to do everything or go to everywhere. Thus scientist and engineers have developed many advance robots or machines using new ideas and technology to find solutions to these problems.

Using innovations and technologies, one can find solutions to many problems which is difficult for the peoples. Robots can be used in any surveillance operation or in places of radioactive surrounding where there is a danger of humans to get exposed to such threats. They are also used in medical sciences to operate and support the patient.  

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