The maximum volume flow rate of water is determined as 0.029 m³/s.
<h3>Power of the pump</h3>
The power of the pump is watt is calculated as follows;
1 hp = 745.69 W
7 hp = ?
= 7 x 745.69 W
= 5,219.83 W
<h3>Mass flow rate of water</h3>
η = mgh/P
mgh = ηP
m = ηP/gh
m = (0.82 x 5,219.83)/(9.8 x 15)
m = 29.12 kg/s
<h3>Maximum volume rate</h3>
V = m/ρ
where;
- ρ is density of water = 1000 kg/m³
V = (29.12)/(1000)
V = 0.029 m³/s
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Answer:
The electrical power is 96.5 W/m^2
Explanation:
The energy balance is:
Ein-Eout=0
if:
Gsky=oTsky^4
Eb=oTs^4
qc=h(Ts-Tα)
if Gl≈El(l,5800)
lt= 2*5800=11600 um-K, at this value, F=0.941
The hemispherical emissivity is equal to:
lt=2*333=666 K, at this value, F=0
The hemispherical absorptivity is equal to:
Answer:
(iv) second law of thermodynamics
Explanation:
The Clausius inequality expresses the second law of thermodynamics it applies to the real engine cycle.It is defined as the cycle integral of change in entropy of a reversible system is zero. It is nothing but mathematical form of second law of thermodynamics . It also states that for irreversible process the cyclic integral of change in entropy is less than zero
Answer:
We can describe 15×-10 as an expression. we would describe 6×-2< 35 as an...
Explanation:
We can describe 15×-10 as an expression. we would describe 6×-2< 35 as an...
Answer:
The pressure drop across the pipe also reduces by half of its initial value if the viscosity of the fluid reduces by half of its original value.
Explanation:
For a fully developed laminar flow in a circular pipe, the flowrate (volumetric) is given by the Hagen-Poiseulle's equation.
Q = π(ΔPR⁴/8μL)
where Q = volumetric flowrate
ΔP = Pressure drop across the pipe
μ = fluid viscosity
L = pipe length
If all the other parameters are kept constant, the pressure drop across the circular pipe is directly proportional to the viscosity of the fluid flowing in the pipe
ΔP = μ(8QL/πR⁴)
ΔP = Kμ
K = (8QL/πR⁴) = constant (for this question)
ΔP = Kμ
K = (ΔP/μ)
So, if the viscosity is halved, the new viscosity (μ₁) will be half of the original viscosity (μ).
μ₁ = (μ/2)
The new pressure drop (ΔP₁) is then
ΔP₁ = Kμ₁ = K(μ/2)
Recall,
K = (ΔP/μ)
ΔP₁ = K(μ/2) = (ΔP/μ) × (μ/2) = (ΔP/2)
Hence, the pressure drop across the pipe also reduces by half of its initial value if the viscosity of the fluid reduces by half of its value.
Hope this Helps!!!