Answer:
Flow-rate = 0.0025 m^3/s
Explanation:
We need to assume that the flow-rate of pure water entering the pond is the same as the flow-rate of brine leaving the pond, in other words, the volume of liquid in the pond stays constant at 20,000 m^3. Using the previous assumption we can calculate the flow rate entering or leaving the tank (they are the same) building a separable differential equation dQ/dt, where Q is the milligrams (mg) of salt in a given time t, to find a solution to our problem we build a differential equation as follow:
dQ/dt = -(Q/20,000)*r where r is the flow rate in m^3/s
what we pose with this equation is that the variable rate at which the salt leaves the pond (salt leaving over time) is equal to the concentration (amount of salt per unit of volume of liquid at a given time) times the constant rate at which the liquid leaves the tank, the minus sign in the equation is because this is the rate at which salt leaves the pond.
Rearranging the equation we get dQ/Q = -(r/20000) dt then integrating in both sides ∫dQ/Q = -∫(r/20000) dt and solving ln(Q) = -(r/20000)*t + C where C is a constant (initial value) result of solving the integrals. Please note that the integral of dQ/Q is ln(Q) and r/20000 is a constant, therefore, the integral of dt is t.
To find the initial value (C) we evaluate the integrated equation for t = 0, therefore, ln(Q) = C, because at time zero we have a concentration of 25000 mg/L = 250000000 mg/m^3 and Q is equal to the concentration of salt (mg/m^3) by the amount of liquid (always 20000 m^3) -> Q = 250000000 mg/m^3 * 20000 m^3 = 5*10^11 mg -> C = ln(5*10^11) = 26.9378. Now the equation is ln(Q) = -(r/20000)*t + 26.9378, the only thing missing is to find the constant flow rate (r) required to reduce the salt concentration in the pond to 500 mg/L = 500000 mg/m^3 within one year (equivalent to 31536000 seconds), to do so we need to find the Q we want in one year, that is Q = 500000 mg/m^3 * 20000 m^3 = 1*10^10 mg, therefore, ln(1*10^10) = -(r/20000)*31536000 + 26.9378 solving for r -> r = 0.002481 m^3/s that is approximately 0.0025 m^3/s.
Note:
- ln() refer to natural logarithm
- The amount of liquid in the tank never changes because the flow-rate-in is the same as the flow-rate-out
- When solving the differential equation we calculated the flow-rate-out and we were asked for the flow-rate-in but because they are the same we could solve the problem
- During the solving process, we always converted units to m^3 and seconds because we were asked to give the answer in m^3/seg
Given:
Pressure, 
= 1300 kPa
Temperature, 
= 
= 100 kPa
velocity, v = 40 m/s
A = 1
Solution:
For air propertiess at
= 1300 kPa
=
= 793kJ/K
= 
and also at
= 100 kPa
= 401 KJ/K
= 
a) Mass flow rate is given by:
Now,
= 46.51 kg/s
b) for the power produced by turbine, 
= 18.231 MW
Answer:
Yes, the flow is turbulent.
Explanation:
Reynolds number gives the nature of flow. If he Reynolds number is less than 2000 then the flow is laminar else turbulent.
Given:
Diameter of pipe is 10mm.
Velocity of the pipe is 1m/s.
Temperature of water is 200°C.
The kinematic viscosity at temperature 200°C is 
m2/s.
Calculation:
Step1
Expression for Reynolds number is given as follows:
Here, v is velocity, 
is kinematic viscosity, d is diameter and Re is Reynolds number.
Substitute the values in the above equation as follows:
Re=64226.07579
Thus, the Reynolds number is 64226.07579. This is greater than 2000.
Hence, the given flow is turbulent flow.
Answer:
v₀ = 2,562 m / s = 9.2 km/h
Explanation:
To solve this problem let's use Newton's second law
F = m a = m dv / dt = m dv / dx dx / dt = m dv / dx v
F dx = m v dv
We replace and integrate
-β ∫ x³ dx = m ∫ v dv
β x⁴/ 4 = m v² / 2
We evaluate between the lower (initial) integration limits v = v₀, x = 0 and upper limit v = 0 x = x_max
-β (0- x_max⁴) / 4 = ½ m (v₀²2 - 0)
x_max⁴ = 2 m /β v₀²
Let's look for the speed that the train can have for maximum compression
x_max = 20 cm = 0.20 m
v₀ =√(β/2m) x_max²
Let's calculate
v₀ = √(640 106/2 7.8 104) 0.20²
v₀ = 64.05 0.04
v₀ = 2,562 m / s
v₀ = 2,562 m / s (1lm / 1000m) (3600s / 1h)
v₀ = 9.2 km / h
