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nikitadnepr [17]
3 years ago
7

Nitrogen gas is compressed at steady state from a pressure of 14.2 psi and a temperature 60o F to a pressure of 120 psi and a te

mperature of 500o F. The gas enters a compressor with a volumetric flow rate of 1200 ft3 /min. The magnitude of the heat transfer rate from the compressor to its surroundings is 5% of the compressor power input. Using the ideal gas model (with variable specific heats) and neglecting kinetic and potential energy effects, determine (a) The compressor power input (in horsepower). (b) The volumetric flow rate at the exit (in ft3 /min).
Engineering
1 answer:
brilliants [131]3 years ago
5 0

Answer:

a) 229.4281 hp.

b) 262.15 ft3/min.

Explanation:

Given data:

P1 = 14.2 psi

T1 = 60°F = 520° R

P2 = 120 psi

T2 = 500°F = 960° R

volumetric flow rate ( Av1 ) = 1200 ft^3 /min = 20 ft^3 / sec

attached below is the detailed solution

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Technician B

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4 0
3 years ago
The substance called olivine may have any composition between Mg2SiO4 and Fe2SiO4, i.e. the Mg atoms can be replaced by Fe atoms
Ipatiy [6.2K]

Answer:

The answer is "0.147 nm and  99.63 mol %"

Explanation:

In point (a):

\to nk1 = 062

\to \text{Bragg angle} \theta =37.21^{\circ}

\to \text{diffraction angle} 2 \theta = 74.42^{\circ}

\to \lambda = 0.1790 nm

find:

d(062)=?

formula:

\to nx = 2d \sin  \theta

\to  d(062) = \frac{1 \times 0.1790^{\circ}}{2 \times \sin 37.21^{\circ}}\\

               = \frac{0.1790^{\circ}}{2 \times 0.604738126}\\\\= \frac{0.29599589}{2}\\\\= 0.147 \\

In point (b):

\to Mg_2SiO_4\longleftrightarrow  Fe_2SiO_4

d= 0.14774  \ \ \ \ \ olivine = 0.147 \ \ \ \  \ 0.15153

formula:

\to d=\frac{a}{\sqrt{n^2+k^2+i^2}}\\

that's why the composition value equal to 99.63 %

3 0
3 years ago
Underground water is to be pumped by a 78% efficient 5- kW submerged pump to a pool whose free surface is 30 m above the undergr
maksim [4K]

Answer:

a) The maximum flowrate of the pump is approximately 13,305.22 cm³/s

b) The pressure difference across the pump is approximately 293.118 kPa

Explanation:

The efficiency of the pump = 78%

The power of the pump = 5 -kW

The height of the pool above the underground water, h = 30 m

The diameter of the pipe on the intake side = 7 cm

The diameter of the pipe on the discharge side = 5 cm

a) The maximum flowrate of the pump is given as follows;

P = \dfrac{Q \cdot \rho \cdot g\cdot h}{\eta_t}

Where;

P = The power of the pump

Q = The flowrate of the pump

ρ = The density of the fluid = 997 kg/m³

h = The head of the pump = 30 m

g = The acceleration due to gravity ≈ 9.8 m/s²

\eta_t = The efficiency of the pump = 78%

\therefore Q_{max} = \dfrac{P \cdot \eta_t}{\rho \cdot g\cdot h}

Q_{max} = 5,000 × 0.78/(997 × 9.8 × 30) ≈ 0.0133 m³/s

The maximum flowrate of the pump Q_{max} ≈ 0.013305 m³/s = 13,305.22 cm³/s

b) The pressure difference across the pump, ΔP = ρ·g·h

∴ ΔP = 997 kg/m³ × 9.8 m/s² × 30 m = 293.118 kPa

The pressure difference across the pump, ΔP ≈ 293.118 kPa

6 0
3 years ago
Consider the following class definitions: class smart class superSmart: public smart { { public: public: void print() const; voi
arsen [322]

Answer:

a) There is no any private member of smart which are public members of superSmart.

Explanation:

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kolezko [41]

Answer:

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Explanation:

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