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
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Answer:
volume of biological sludge = 28.566 m³ per day
Explanation:
given data
mass of solid = 7240 kg/day
initial moisture content = 78%
solution
here percentage of solid will be
% of solid = 100 - initial moisture content
% of solid = 100 - 78 = 22 %
so that
mass of sludge produced = kg per day
put her value
mass of sludge produced = kg
mass of sludge produced = 32909.09 kg
so
specific gravity of sludge =
and as we know that
= 1.152
so that
density of sludge = S sludge × density of water
density of sludge = 1.152 × 1000
density of sludge = 1152 kg/m³
so that
volume of biological sludge =
volume of biological sludge =
volume of biological sludge = 28.566 m³ per day
Answer:
Hello your question is incomplete below is the complete question
Electronic components are often mounted with good heat conduction paths to a finned aluminum base plate, which is exposed to a stream of cooling air from a fan. The sum of the mass times specific heat products for a base plate and components is 5000 J/K, and the effective heat transfer coefficient times surface area product is 10 W/K. The initial temperature of the plate and the cooling air temperature are 295 K when 300 W of power are switched on. 1) Find the plate temperature after 10 minutes.
answer ; 311.36 k
Explanation:
Given data :
sum of mass * specific heat products for a base plate and components ( Mcp )
= 5000 J/K
effective heat transfer coefficient * surface area ( hA ) = 10 W/K
Initial temperature of plate and cooling air temperature( Tc ) = 295 k
power ( Q = W ) = 300 W
a) Determine plate temperature after 10 minutes
10 mins = 600 secs ( t )
heat supplied = change in temp + heat loss
Q * t = mCp ( ΔT ) + hA ( ΔT ) t
300*600 = 5000 * ( T -295 ) + 10 ( T -295 ) * 600
therefore ; T - 295 = 16.363
T = 311.36 K