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3 years ago

Match each titration term with its definition.

Engineering

1 answer:

Illusion [34]3 years ago

3 0

Answer:

1) titration

2) titrand

3) equivalence point

4) titrant

5) Burette

6) Indicator

Explanation:

The process in which a known volume of a standard solution is added to another solution so that the standard solution can react with the solution of unknown concentration such that its concentration is determined can be referred to as titration.

The solution which is added to another solution is called the titrant. The titrand is the solution of unknown concentration

A burette is a glassware used to slowly add a known volume of the titrant to the titrand.

The indicator used signals the point when the reaction is complete by a color change. At this point, a stoichiometric amount of titrant has been added to the titrand. This is also referred to as the equivalence point.

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A normal shock wave takes place during the flow of air at a Mach number of 1.8. The static pressure and temperature of the air u

Darina [25.2K]

Answer:

The pressure upstream and downstream of a shock wave are related as

$\frac{P_{1}}{P_{o}}=\frac{2\gamma M^{2}-(\gamma -1)}{\gamma +1}$

where,

$\gamma$ = Specific Heat ratio of air

M = Mach number upstream

We know that $\gamma _{air}=1.4$

Applying values we get

$\frac{P_{1}}{100kPa}=\frac{2\times 1.4\times 1.8^{2}-(1.4 -1)}{1.4 +1}\\\\\frac{P_{1}}{100kPa}=3.61\\\\\therefore P_{1}=361.33kPa(Absloute)$

Similarly the temperature downstream is obtained by the relation

$\frac{T_{1}}{T_{o}}=\frac{[2\gamma M^{2}-(\gamma -1)][(\gamma -1)M^{2}+2]}{(\gamma +1)^{2}M^{2}}$

Applying values we get

$\frac{T_{1}}{423}=\frac{[2\times 1.4\times 1.8^{2}-(1.4-1)][(1.4-1)1.8^{2}+2]}{(1.4+1)^{2}\times 1.8^{2}}\\\\\therefore \frac{T_{1}}{423}=1.53\\\\\therefore T_{1}=647.85K=374.85^{o}C$

The Mach number downstream is obtained by the relation

$M_{d}^{2}=\frac{(\gamma -1)M^{2}+2}{2\gamma M^{2}-(\gamma -1)}\\\\\therefore M_{d}^{2}=\frac{(1.40-1)\times 1.8^{2}+2}{2\times1.4\times 1.8^{2}-(1.4-1)}\\\\\therefore M_{d}^{2}=0.38\\\\M_{d}=0.616$

3 0

4 years ago

Determine the work done by an engine shaft rotating at 2500 rpm delivering an output torque of 4.5 N.m over a period of 30 secon

balu736 [363]

Answer:

work done= 2.12 kJ

Explanation:

Given

N=2500 rpm

T=4.5 N.m

Period ,t= 30 s

$torque =\frac{power}{2\pi N}$

$power=2\pi N\times T$

P= $2\times \pi \times2500 \times 4.5$

P=70,685W

P=70.685 KW

power= $\frac{work done}{time}$

work done = power * time

= 70.685*30=2120.55J

= 2.12 kJ

7 0

4 years ago

Consider the flow field given by V ! =xy2^i− 1 3 y3^j+xyk ^. Determine (a) the number of dimensions of the flow, (b) if it is a

Basile [38]

Answer:

a) The flow has three dimensions (3 coordinates).

b) ∇V = 0 it is a incompressible flow.

c) ap = (16/3) i + (32/3) j + (16/3) k

Explanation:

Given

V = xy² i − (1/3) y³ j + xy k

a) The flow has three dimensions (3 coordinates).

b) ∇V = 0

then

∇V = ∂(xy²)/∂x + ∂(− (1/3) y³)/∂y + ∂(xy)/∂z

⇒ ∇V = y² - y² + 0 = 0 it is a incompressible flow.

c) ap = xy²*∂(V)/∂x − (1/3) y³*∂(V)/∂y + xy*∂(V)/∂z

⇒ ap = xy²*(y² i + y k) - (1/3) y³*(2xy i − y² j + x k) + xy*(0)

⇒ ap = (xy⁴ - (2/3) xy⁴) i + (1/3) y⁵ j + (xy³ - (1/3) xy³) k

⇒ ap = (1/3) xy⁴ i + (1/3) y⁵ j + (2/3) xy³ k

At point (1, 2, 3)

⇒ ap = (1/3) (1*2⁴) i + (1/3) (2)⁵ j + (2/3) (1*2³) k

⇒ ap = (16/3) i + (32/3) j + (16/3) k

3 0

3 years ago

A compressed-air drill requires an air supply of 0.25 kg/s at gauge pressure of 650 kPa at the drill. The hose from the air comp

Klio2033 [76]

Answer:

L = 46.35 m

Explanation:

GIVEN DATA

\dot m = 0.25 kg/s

D = 40 mm

P_1 = 690 kPa

P_2 = 650 kPa

T_1 = 40° = 313 K

head loss equation

$[\frac{P_1}{\rho} +\alpha \frac{v_1^2}{2} +gz_1] -[\frac{P_2}{\rho} +\alpha \frac{v_2^2}{2} +gz_2] = h_l +h_m$

where $h_l = \frac{ flv^2}{2D}$

$h_m minor loss$

density is constant

$v_1 = v_2$

head is same so, $z_1 = z_2$

curvature is constant so $\alpha = constant$

neglecting minor losses

$\frac{P_1}{\rho} -\frac{P_2}{\rho} = \frac{ flv^2}{2D}$

we know $\dot m$ is given as $= \rho VA$

$\rho =\frac{P_1}{RT_1}$

$\rho =\frac{690 *10^3}{287*313} = 7.68 kg/m3$

therefore

$v = \frac{\dot m}{\rho A}$

$V =\frac{0.25}{7.68 \frac{\pi}{4} *(40*10^{-3})^2}$

V = 25.90 m/s

$Re = \frac{\rho VD}{\mu}$

for T = 40 Degree, $\mu = 1.91*10^{-5}$

$Re =\frac{7.68*25.90*40*10^{-3}}{1.91*10^{-5}}$

Re = 4.16*10^5 > 2300 therefore turbulent flow

for Re =4.16*10^5 , f = 0.0134

Therefore

$\frac{P_1}{\rho} -\frac{P_2}{\rho} = \frac{ flv^2}{2D}$

$L = \frac{(P_1-P_2) 2D}{\rho f v^2}$

$L =\frac{(690-650)*`10^3* 2*40*10^{-3}}{7.68*0.0134*25.90^2}$

L = 46.35 m

5 0

3 years ago

The compressor in a refrigerator compresses saturated R-134a vapor at 0°F to 200 psia. Calculate the work required by this compr

kvasek [131]

Answer:

The work required is W = 20.2 BTU per lbm

Explanation:

The value of entropy & enthalpy at initial conditions are

$h_{1}$ = 103.1

S = 0.225

Final enthalpy

$h_{2}$ = 123.3

Therefore work done

W = $h_{1}$ - $h_{2}$

W = 103.1 - 123.3

W = - 20.2 BTU per lbm

Therefore the work required is W = 20.2 BTU per lbm

6 0

3 years ago