A = 4\pi r^2
A = 4\pi (2\mu m /2)^2 (10^{-6}m/1\mu m)^2 (1mm/10{-3})^2
A = 1.33*!0^{-5}MM^2
There are two atoms of potassium bonded to one atom of sulfur.
Answer:
Flow rate 2.34 m3/s
Diameter 0.754 m
Explanation:
Assuming steady flow, the volume flow rate along the pipe will always be constant, and equals to the product of flow speed and cross-section area.
The area at the well head is
So the volume flow rate along the pipe is
We can use the similar logic to find the cross-section area at the refinery
The radius of the pipe at the refinery is:
So the diameter is twice the radius = 0.38*2 = 0.754m
The answer is False give thanks for the answer m8 and happy Halloween
Correct temperature is 80°F
Answer:
T_f = 38.83°F
Explanation:
We are given;
Volume; V = 8 ft³
Initial Pressure; P_i = 100 lbf/in² = 100 × 12² lbf/ft²
Initial temperature; T_i = 80°F = 539.67 °R
Time for outlet flow; t_o = 90 s
Mass flow rate at outlet; m'_o = 0.03 lb/s
Final pressure; P_f = 30 lbf/in² = 30 × 12² lbf/ft²
Now, from ideal gas equation,
Pv = RT
Where v is initial specific volume
R is ideal gas constant = 53.33 ft.lbf/°R
Thus;
v = RT/P
v_i = 53.33 × 539.67/(100 × 12²)
v_i = 2 ft³/lb
Formula for initial mass is;
m_i = V/v_i
m_i = 8/2
m_i = 4 lb
Now change in mass is given as;
Δm = m'_o × t_o
Δm = 0.03 × 90
Δm = 2.7 lb
Now,
m_f = m_i - Δm
Thus; m_f = 4 - 2.7
m_f = 1.3 lb
Similarly in above;
v_f = V/m_f
v_f = 8/1.3
v_f = 6.154 ft³/lb
Again;
Pv = RT
Thus;
T_f = P_f•v_f/R
T_f = (30 × 12² × 6.154)/53.33
T_f = 498.5°R
Converting to °F gives;
T_f = 38.83°F
