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Eva8 [605]
4 years ago
5

The mass flow rate in a 4.0-m wide, 2.0-m deep channel is 4000 kg/s of water. If the velocity distribution in the channel is lin

ear with depth what is the surface velocity of flow in the channel?
Engineering
1 answer:
IceJOKER [234]4 years ago
4 0

Answer:

V = 0.5 m/s

Explanation:

given data:

width of channel =  4 m

depth of channel = 2 m

mass flow rate = 4000 kg/s = 4 m3/s

we know that mass flow rate is given as

\dot{m}=\rho AV

Putting all the value to get the velocity of the flow

\frac{\dot{m}}{\rho A} = V

V = \frac{4000}{1000*4*2}

V = 0.5 m/s

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Mkey [24]

Explanation:

Ohm's law is used here. V = IR, and variations. The voltage across all elements is the same in this parallel circuit. (V1 =V2 =V3)

The total supply current is the sum of the currents in each of the branches. (It = I1 +I2 +I3)

Rt = (8 V)/(8 A) = 1 Ω . . . . supply voltage divided by supply current

I3 = 8A -3A -4A = 1 A . . . . supply current not flowing through other branches

R1 = (8 V)/(3 A) = 8/3 Ω

R2 = (8 V)/(4 A) = 2 Ω

R3 = (8 V)/(I3) = (8 V)/(1 A) = 8 Ω

V1 = V2 = V3 = 8 V

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3 years ago
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The gas expanding in the combustion space of a reciprocating engine has an initial pressure of 5 MPa and an initial temperature
Anit [1.1K]

Answer:

a). Work transfer = 527.2 kJ

b). Heat Transfer = 197.7 kJ

Explanation:

Given:

P_{1} = 5 Mpa

T_{1} = 1623°C

                       = 1896 K

V_{1} = 0.05 m^{3}

Also given \frac{V_{2}}{V_{1}} = 20

Therefore, V_{2} = 1  m^{3}

R = 0.27 kJ / kg-K

C_{V} = 0.8 kJ / kg-K

Also given : P_{1}V_{1}^{1.25}=C

   Therefore, P_{1}V_{1}^{1.25} = P_{2}V_{2}^{1.25}

                     5\times 0.05^{1.25}=P_{2}\times 1^{1.25}

                     P_{2} = 0.1182 MPa

a). Work transfer, δW = \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}

                                  \left [\frac{5\times 0.05-0.1182\times 1}{1.25-1}  \right ]\times 10^{6}

                              = 527200 J

                             = 527.200 kJ

b). From 1st law of thermodynamics,

Heat transfer, δQ = ΔU+δW

   = \frac{mR(T_{2}-T_{1})}{\gamma -1}+ \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}

  =\left [ \frac{\gamma -n}{\gamma -1} \right ]\times \delta W

  =\left [ \frac{1.4 -1.25}{1.4 -1} \right ]\times 527.200

  = 197.7 kJ

6 0
3 years ago
How will you maintain the orderliness of your storage area 2pts?​
fiasKO [112]

Explanation:

Label and group products. One would think that a general cleanup would be the first step, but no, it's not. ...Clean up the area. ...Put up demarcation lines. ...Stack properly. ...Keep the aisles, paths and ramps clear. ...Have all the safety signs in place.

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What is an example of an innovation in logistics?
Scorpion4ik [409]

Answer:

geolocation technologies, drones, automated transportation vehicles

Explanation:

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