What type of intersection is this?

A tank with a volume of 8 m3 containing 4 m3 of 20% (by volume) NaOH solution is to be purged by adding pure water at a rate of

lawyer [7]

Answer:

The time necessary to purge 95% of the NaOH is 0.38 h

Explanation:

Given:

vfpure water(i) = 3 m³/h

vNaOH = 4 m³

xNaOH = 0.2

vfpure water(f) = 2 m³/h

pwater = 1000 kg/m³

pNaOH = 1220 kg/m³

The mass flow rate of the water is = 3 * 1000 = 3000 kg/h

The mass of NaOH in the solution is = 0.2 * 4 * 1220 = 976 kg

When the 95% of the NaOH is purged, thus the NaOH in outlet is = 0.95 * 976 = 927.2 kg

The volume of NaOH in outlet after time is = 927.2/1220 = 0.76 m³

The time required to purge the 95% of the NaOH is = 0.76/2 = 0.38 h

4 0

3 years ago

An aircraft is in a steady level turn at a flight speed of 200 ft/s and a turn rate about the local vertical of 5 deg/s. Thrust

notka56 [123]

Answer:

L= 50000 lb

D = 5000 lb

Explanation:

To maintain a level flight the lift must equal the weight in magnitude.

We know the weight is of 50000 lb, so the lift must be the same.

L = W = 50000 lb

The L/D ratio is 10 so

10 = L/D

D = L/10

D = 50000/10 = 5000 lb

To maintain steady speed the thrust must equal the drag, so

T = D = 5000 lb

5 0

3 years ago

A 150 MVA, 24 kV, 123% three-phase synchronous generator supplies a large network. The network voltage is 27 kV. The phase angle

Aleks04 [339]

Answer:

the generator induced voltage is 60.59 kV

Explanation:

Given:

S = 150 MVA

Vline = 24 kV = 24000 V

$X_{s} =1.23(\frac{V_{line}^{2} }{s} )=1.23\frac{24000^{2} }{1500} =4723.2 ohms$

the network voltage phase is

$V_{phase} =\frac{V_{nline} }{\sqrt{3} } =\frac{27}{\sqrt{3} } =15.58kV$

the power transmitted is equal to:

$|E|=\frac{P*X_{s} }{3*|V_{phase}|sinO } ;if-O=60\\|E|=\frac{300*4.723}{3*15.58*sin60} =34.98kV$

the line induced voltage is

$|E_{line} |=\sqrt{3} *|E|=\sqrt{3} *34.98=60.59kV$

7 0

3 years ago

Two previously undeformed rod-shaped specimens of copper are to be plastically deformed by reducing their cross-sectional areas.

mezya [45]

I am not sure I am stuck on this and I have been for 45 min someone please help me and this girl or boy!!

4 0

3 years ago

Two common methods of improving fuel efficiency of a vehicle are to reduce the drag coefficient and the frontal area of the vehi

qaws [65]

Answer:

$\Delta V = 209.151\,L$ , $\Delta C = 217.517\,USD$

Explanation:

The drag force is equal to:

$F_{D} = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot A$

Where $C_{D}$ is the drag coefficient and $A$ is the frontal area, respectively. The work loss due to drag forces is:

$W = F_{D}\cdot \Delta s$

The reduction on amount of fuel is associated with the reduction in work loss:

$\Delta W = (F_{D,1} - F_{D,2})\cdot \Delta s$

Where $F_{D,1}$ and $F_{D,2}$ are the original and the reduced frontal areas, respectively.

$\Delta W = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot (A_{1}-A_{2})\cdot \Delta s$

The change is work loss in a year is:

$\Delta W = (0.3)\cdot \left(\frac{1}{2}\right)\cdot (1.20\,\frac{kg}{m^{3}})\cdot (27.778\,\frac{m}{s})^{2}\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25\times 10^{6}\,m)$

$\Delta W = 2.043\times 10^{9}\,J$

$\Delta W = 2.043\times 10^{6}\,kJ$

The change in chemical energy from gasoline is:

$\Delta E = \frac{\Delta W}{\eta}$

$\Delta E = \frac{2.043\times 10^{6}\,kJ}{0.3}$

$\Delta E = 6.81\times 10^{6}\,kJ$

The changes in gasoline consumption is:

$\Delta m = \frac{\Delta E}{L_{c}}$

$\Delta m = \frac{6.81\times 10^{6}\,kJ}{44000\,\frac{kJ}{kg} }$

$\Delta m = 154.772\,kg$

$\Delta V = \frac{154.772\,kg}{0.74\,\frac{kg}{L} }$

$\Delta V = 209.151\,L$

Lastly, the money saved is:

$\Delta C = \left(\frac{154.772\,kg}{0.74\,\frac{kg}{L} }\right)\cdot (1.04\,\frac{USD}{L} )$

$\Delta C = 217.517\,USD$

4 0

3 years ago