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Makovka662 [10]
3 years ago
5

A groundwater contains the following cations (expressed as the cation):

Engineering
2 answers:
Hunter-Best [27]3 years ago
6 0

Answer:

208 mg/L

Explanation:

Only Mg++ and Ca++ causes hardness not Na+

Given

Mg++ = 20 mg/L

Ca++ = 50 mg/L

Hardness\ of\ water = C^{2+}\times \frac{equivalent\ weight\ of\ CaCO3}{equivalent\ mass\ of\ C^{2+}}\ +\ mg^{2+}\times \frac{equivalent\ weight\ of\ CaCO3}{equivalent\ mass\ of\ mg^{2+}}

= 50\times\frac{50}{20}\ +\ 20\times\frac{50}{12}

= 208.33 mg/L \approx 208mg/L

Oliga [24]3 years ago
5 0

Answer:

50mg/L

Explanation:

CaCO3 -> Ca++ + CO2

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vlabodo [156]

Answer:

C

Explanation:

5 0
3 years ago
A turbojet aircraft flies with a velocity of 800 ft/s at an altitude where the air is at 10 psia and 20 F. The compressor has a
nika2105 [10]

Answer:

Pressure = 115.6 psia

Explanation:

Given:

v=800ft/s

Air temperature = 10 psia

Air pressure = 20F

Compression pressure ratio = 8

temperature at turbine inlet = 2200F

Conversion:

1 Btu =775.5 ft lbf, g_{c} = 32.2 lbm.ft/lbf.s², 1Btu/lbm=25037ft²/s²

Air standard assumptions:

c_{p}= 0.0240Btu/lbm.°R, R = 53.34ft.lbf/lbm.°R = 1717.5ft²/s².°R 0.0686Btu/lbm.°R

k= 1.4

Energy balance:

h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} + \frac{v_{a} ^{2} }{2}\\

As enthalpy exerts more influence than the kinetic energy inside the engine, kinetic energy of the fluid inside the engine is negligible

hence v_{a} ^{2} = 0

h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} \\h_{1} -h_{a} = - \frac{v_{1} ^{2} }{2} \\ c_{p} (T_{1} -T_{a})= - \frac{v_{1} ^{2} }{2} \\(T_{1} -T_{a}) = - \frac{v_{1} ^{2} }{2c_{p} }\\ T_{a}=T_{1} +  \frac{v_{1} ^{2} }{2c_{p} }

T_{1} = 20+460 = 480°R

T_{a}  =480+  \frac{(800)(800}{2(0.240)(25037}= 533.25°R

Pressure at the inlet of compressor at isentropic condition

P_{a } =P_{1}(\frac{T_{a} }{T_{1} }) ^{k/(k-1)}

P_{a} = (10)(\frac{533.25}{480}) ^{1.4/(1.4-1)}= 14.45 psia

P_{2}= 8P_{a} = 8(14.45) = 115.6 psia

4 0
3 years ago
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VikaD [51]

Answer:

option c

Explanation:

8 0
3 years ago
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A 11-cm-diameter horizontal jet of water with a velocity of 40 m/s relative to the ground strikes a flat plate that is moving in
Reika [66]

Answer:

F = 8552.7N

Explanation:

We need first our values, that are,

V_{jet} = 40m/s\\V_{Plate} = 10m/s \\D = 11cm

We start to calculate the relative velocity, that is,

V_r = V_{jet}-V_{plate}\\V_r = (40)-(10)\\V_r = 30m/s

With the relative velocity we can calculate the mass flow rate, given by,

\dot{m}_r = \rho A V_r

\dot{m}_r = (1000)(30) \frac{\pi (0.11)^2}{4}

\dot{m}_r = 285.09kg/s

We need to define the Force in the direction of the flow,

\sum\vec{F} = \sum_{out} \beta\dot{m}\vec{V} - \sum_{in} \beta\dot{m} \vec{V}\\

F = \dot{m}V_r

F = (285.09Kg/s)(30)

F = 8552.7N

8 0
3 years ago
To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105
GenaCL600 [577]

Answer:

62.14\ \text{miles}

6213727.37\ \text{miles}

Explanation:

The distance of the chain would be the product of the dislocation density and the volume of the metal.

Dislocation density = 10^5\ \text{mm}^{-2}

Volume of the metal = 1000\ \text{mm}^3

10^5\times 1000=10^8\ \text{mm}\\ =10^5\ \text{m}

1\ \text{mile}=1609.34\ \text{m}

\dfrac{10^5}{1609.34}=62.14\ \text{miles}

The chain would extend 62.14\ \text{miles}

Dislocation density = 10^{10}\ \text{mm}^{-2}

Volume of the metal = 1000\ \text{mm}^3

10^{10}\times 1000=10^{13}\ \text{mm}\\ =10^{10}\ \text{m}

\dfrac{10^{10}}{1609.34}=6213727.37\ \text{miles}

The chain would extend 6213727.37\ \text{miles}

3 0
3 years ago
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