A. 1.3
B. 0.766
C. 0.87
D. None of these
Answer: Option B
This Question Belongs to Chemical Engineering >> Fluid Mechanics
A. Thermal conductivity
B. Electrical conductivity
C. Specific gravity
D. Electrical resistivity
A. $$\frac{{\text{V}}}{{{{\text{V}}_{\max }}}} = {\left( {\frac{{\text{x}}}{{\text{r}}}} \right)^{\frac{1}{7}}}$$
B. $$\frac{{\text{V}}}{{{{\text{V}}_{\max }}}} = {\left( {\frac{{\text{r}}}{{\text{x}}}} \right)^{\frac{1}{7}}}$$
C. $$\frac{{\text{V}}}{{{{\text{V}}_{\max }}}} = {\left( {{\text{x}} \times {\text{r}}} \right)^{\frac{1}{7}}}$$
D. None of these
A. d
B. $$\frac{1}{{\text{d}}}$$
C. $$\sigma $$
D. $$\frac{l}{\sigma }$$
A. $$\frac{{4\pi {\text{g}}}}{3}$$
B. $$\frac{{0.01\pi {\text{gH}}}}{4}$$
C. $$\frac{{0.01\pi {\text{gH}}}}{8}$$
D. $$\frac{{0.04\pi {\text{gH}}}}{3}$$
The coefficient of contraction (
𝐶
𝑐
C
c
) is defined as the ratio of the area of the jet at the vena-contracta to the area of the orifice.
Given:
Diameter of the orifice (
𝐷
1
D
1
) = 2 inches
Diameter of the jet at the vena-contracta (
𝐷
2
D
2
) = 1.75 inches
First, we need to calculate the areas of the orifice and the jet at the vena-contracta.
The area of a circle is given by:
𝐴
=
𝜋
(
𝐷
2
)
2
A=π(
2
D
)
2
So, the area of the orifice (
𝐴
1
A
1
) is:
𝐴
1
=
𝜋
(
𝐷
1
2
)
2
=
𝜋
(
2
2
)
2
=
𝜋
(
1
)
2
=
𝜋
A
1
=π(
2
D
1
)
2
=π(
2
2
)
2
=π(1)
2
=π
The area of the jet at the vena-contracta (
𝐴
2
A
2
) is:
𝐴
2
=
𝜋
(
𝐷
2
2
)
2
=
𝜋
(
1.75
2
)
2
=
𝜋
(
0.875
)
2
=
𝜋
(
0.765625
)
A
2
=π(
2
D
2
)
2
=π(
2
1.75
)
2
=π(0.875)
2
=π(0.765625)
Next, the coefficient of contraction (
𝐶
𝑐
C
c
) is the ratio of
𝐴
2
A
2
to
𝐴
1
A
1
:
𝐶
𝑐
=
𝐴
2
𝐴
1
=
𝜋
(
0.765625
)
𝜋
=
0.765625
C
c
=
A
1
A
2
=
π
π(0.765625)
=0.765625
Therefore, the coefficient of contraction is:
𝐶
𝑐
=
0.765625
C
c
=0.765625