ENGINEERING COLLEGE

Calculate the "exact" alkalinity of the water in Problem 3-2 if the pH is 9.43.Calculate the "approximate" alkalinity (in mg/L as CaCO3 ) of a water containing 120 mg/L of bicarbonate ion and 15 mg/L of carbonate ion.

Answers

Answer 1

Answer:

Answer:

A) approximate alkalinity = 123.361 mg/l

B) exact alkalinity = 124.708 mg/l

Explanation:

Given data :

A) determine approximate alkalinity first

Bicarbonate ion = 120 mg/l

carbonate ion = 15 mg/l

Approximate alkalinity = ( carbonate ion ) * 50/30 + ( bicarbonate ion ) * 50/61

= 15 * (50/30) + 120*( 50/61 ) = 123.361 mg/l as CaCO3

B) calculate the exact alkalinity of the water if the pH = 9.43

pH + pOH = 14

9.43 + pOH = 14. therefore pOH = 14 - 9.43 = 4.57

[OH^- ] = 10^-4.57 = 2.692*10^-5 moles/l

[ OH^- ] = 2.692*10^-5 * 179/mole * 10^3 mg/g = 0.458 mg/l

[ H^+ ] = 10^-9.43 * 1 * 10^3 = 3.7154 * 10^-7 mg/l

therefore the exact alkalinity can be calculated as

= ( approximate alkalinity ) + ( [ OH^- ] * 50/17 ) - ( [ H^+ ] * 50/1 )

= 123.361 + ( 0.458 * 50/17 ) - ( 3.7154 * 10^-7 * 50/1 )

= 124.708 mg/l

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Answers

Answer:

Ergr5

Explanation:

A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickness is 5 mm. Treat the wall as a slab of the insulating material Styrofoam whose area and thickness are 10 m2 and 0.15 m, respectively. Heat is lost via conduction through the wall and the window. The temperature difference between the inside and outside is the same for the wall and the window. Of the total heat lost by the wall and the window, what is the percentage lost by the window

Answers

Answer:

The percentage loss of the window is $E = 97.3$ %

Explanation:

From the question we are told that

The area of pane of glass is $A = 0.15 m^2$

The thickness is $d = 5mm = (5)/(1000) = 0.005m$

The thickness of the wall is $D = 0.15m$

The area of the wall is $a = 10m^2$

Generally the heat lost as a result of conduction of the window is

$Q_(window) = (j_(glass) * A * (\Delta T) )/(d)$

Where $j_(glass)$ is the thermal conductivity of glass which has a constant value of

$j_(glass) = 0.80 J/(s \cdot m \cdot C^o)$

Substituting values

$Q_(window) = ( 0.80 * 0.15 * (\Delta T) )/(0.005)$

$Q_(window) = 24 \Delta T$

Generally the heat lost as a result of conduction of the wall is

$Q_(wall) = (j_(styrofoam) * A * (\Delta T) )/(d)$

$j_(styrofoam)$ s the thermal conductivity of Styrofoam which has a constant value of $j_(styrofoam) = 0.010J / (s \cdot m \cdot C^o)$

Substituting values

$Q_(wall) = ( 0.010 * 10 * (\Delta T) )/(0.15)$

$Q_(wall) = 0.667 \ \Delta T$

Now the net loss of heat is

$Q_(net) = Q_(window) + Q_(wall)$

Substituting values

$Q_(net) = 24 + 0.667$

$Q_(net) = 24.667$

Now the percentage loss by the window is

$E = (Q_(window) )/(Q_(net)) * 100$

Substituting value

$E = (24)/(24 .667) * 100$

$E = 97.3$ %

A 1-m3 tank containing air at 10°C and 350 kPa is connected through a valve to another tank containing 3 kg of air at 35°C and 150 kPa. Now the valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings, which are at 19.5°C. Determine the volume of the second tank and the final equilibrium pressure of air. The gas constant of air is R = 0.287 kPa·m3/kg·K.

Answers

Answer:

V₂=1.76 m³

P=222.03 KPa

Explanation:

Given that

For tank 1

V₁=1 m³

T₁= 10°C = 283 K

P₁=350 KPa

For tank 2

m₂=3 kg

T₂=35°C = 308 K

P₂=150 KPa

We know that for air

P V = m R T

P=pressure ,V= Volume,R= gas constant ,T= temperature ,m =mass

for tank 2

P₂ V₂ = m₂ R T₂

By putting the values

150 x V₂ = 3 x 0.287 x 308

V₂=1.76 m³

Final mass = m₁+m₂

m =m₁+m₂

The final volume V= V₂+V₁

V= 1.76 + 1 m³

V= 2.76 m³

The final temperature T= 19.5°C

T= 292.5 K

$m=(PV)/(RT)$

$m_1=(P_1V_1)/(RT_1)$

$m_1=(350* 1)/(0.287* 283)$

$m_1=4.3\ kg$

m =m₁+m₂

m =4.3 + 3 = 7.3 kg

Now at final state

P V = m R T

P x 2.76 = 7.3 x 0.287 x 292.5

P=222.03 KPa

On what single factor does the efficiency of the Otto cycle depend?

Answers

Answer:

Compression ratio(r)

Explanation:

Otto cycle:

Otto cycle is an ideal cycle for all working petrol engine.It have four processes in which two are constant volume process and other two are reversible adiabatic or we can say that isentropic processes.All petrol engine works on Otto cycle.

The efficiency of Otto cycle given as follows

$\eta =1-(1)/(r^(\gamma-1))$

Where r is the compression ratio and γ is heat capacity ratio.

So from above we can say that the efficiency of Otto cycle depends onl;y on compression ratio (r).

After the load impedance has been transformed through the ideal transformer, its impedance is: + . Enter the real part in the first blank and the imaginary part in the second blank. If a value is negative, include the negative sign. Provide up to four digits of precision. If the exact value can be provided with fewer digits, merely provide the exact value. These instructions pertain to the following blanks as well. What is the total impedance seen by the source? + . What is the current phasor Ig (expressed in rectangular form)?

Answers

Answer:

Ig =7.2 +j9.599

Explanation: Check the attachment

At its current short-run level of production, a firm's average variable costs equal $20 per unit, and its average fixed costs equal $30 per unit. Its total costs at this production level equal $2,500. A. What is the firm's current output level? B. What are its total variable costs at this output level? C. What are its total fixed costs?

Answers

Answer and Explanation:

The computation is shown below

Total average cost + total variable cost = total cost

Let number of output be x

So,

Total fixed average cost = x × $30

Total variable cost = x × $15

Total cost = $2,500

Therefore,

$20 × x + $30 × x = $2,500

50 × x = $2,500

x = 50

Now the total variable cost is

= 50 × $20

= $1,000

And, the fixed cost is

= 50 × $30

= $1,500

Answer:

12 4

Explanation:

because the production average is variable

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