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alexandr1967 [171]
3 years ago
7

A pipe leads from a storage tank on the roof of a building to the ground floor. The absolute pressure of the water in the storag

e tank where it connects to the pipe is 3.0×105Pa, the pipe has a radius of 1.0 cm where it connects to the storage tank, and the speed of flow in this pipe is 1.6 m/s. The pipe on the ground floor has a radius of 0.50 cm and is 9.0 m below the storage tank. Find the speed of flow in the pipe on the ground floor.
Engineering
1 answer:
Fudgin [204]3 years ago
3 0

Answer:

6.4 m/s

Explanation:

From the equation of continuity

A1V1=A2V2 where A1 and V1 are area and velocity of inlet respectively while A2 and V2 are the area and velocity of outlet respectively

A1=\pi (r1)^{2}

A2=\pi (r2)^{2}

where r1 and r2 are radius of inlet and outlet respectively

v1 is given as 1.6 m/s

Therefore

\pi (0.01)^{2}\times 1.6 = \pi (0.005)^{2}v2

V2=\frac {\pi (0.01)^{2}\times 1.6}{\pi (0.005)^{2}}=6.4 m/s

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Answer and Explanation:

Some of the advantages of direct water supply system are:

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Some of the advantages of indirect water supply system are:

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3 years ago
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A) Consider an air standard otto cycle that has a heat addition of 2800 kJ/kg of air, a compression ratio of 8 and a pressure an
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Answer:

a) i) The maximum pressure is approximately 122.37 bar

ii) The thermal efficiency is approximately 56.47%

iii) The mean effective pressure is approximately 20.974 bar

b) (b) Four types of internal combustion engine includes;

1) The diesel engine

2) The Otto engine

3) The Brayton engine

4) The Wankel engine

Explanation:

The parameters of the Otto cycle are;

The heat added, Q_{in} = 2,800 kJ/kg

The compression ratio, r = 8

The beginning compression pressure, P₁ = 1 bar

The beginning compression temperature, T₁ = 300 K

Cp = 1.005 kJ/kg·K

Cv = 0.718 kJ/kg·K

R = 287 kJ/kg·K

K = Cp/Cv = 1.005 kJ/kg·K/(0.718 kJ/kg·K) ≈ 1.4

T₂ = T₁×r^(k - 1)

∴ T₂ = 300 K×8^(1.4 - 1) ≈ 689.219 K

\dfrac{P_1\cdot V_1}{T_1}  = \dfrac{P_2\cdot V_2}{T_2}

P_2 = \dfrac{P_1\cdot V_1 \cdot T_2}{T_1 \cdot V_2}  = \dfrac{V_1}{V_2} \cdot  \dfrac{P_1 \cdot T_2}{T_1 } = r \cdot  \dfrac{P_1 \cdot T_2}{T_1 }

∴ P₂ = 8 × 1 bar × (689.219K)/300 K ≈ 18.379 bar

Q_{in} = m·Cv·(T₃ - T₂)

∴ Q_{in} = 2,800 ≈ 0.718 × (T₃ - 689.219)

T₃ = 2,800/0.718 + 689.219 = 4588.94 K

P₃ = P₂ × (T₃/T₂)

P₃ = 18.379 bar × 4588.94K/(689.219 K) = 122.37 bar

The maximum pressure = P₃ ≈ 122.37 bar

(ii) The thermal efficiency, \eta_{Otto}, is given as follows;

\eta_{Otto} = 1 - \dfrac{1}{r^{k - 1}}

Therefore, we have;

\eta_{Otto} = 1 - \dfrac{1}{8^{1.4 - 1}} \approx 0.5647

The thermal efficiency, \eta_{Otto} ≈ 0.5647

Therefore, the thermal efficiency ≈ 56.47%

(iii) The mean effective pressure, MEP is given as follows;

MEP = \dfrac{\left(P_3 - P_1 \cdot r^k \right) \cdot \left(1 - \dfrac{1}{r^{k-1}} \right)}{(k -1)\cdot (r - 1)}

Therefore, we get;

MEP = \dfrac{\left(122.37 - 1 \times 8^{1.4} \right) \cdot \left(1 - \dfrac{1}{8^{1.4-1}} \right)}{(1.4 -1)\cdot (8 - 1)} \approx 20.974

The mean effective pressure, MEP ≈ 20.974 bar

(b) Four types of internal combustion engine includes;

1) The diesel engine; Compression heating is the source of the ignition, with constant pressure combustion

2) The Otto engine which is the internal combustion engine found in cars that make use of gasoline as the source of fuel

The Otto engine cycle comprises of five steps; intake, compression, ignition, expansion and exhaust

3) The Brayton engine works on the principle of the steam turbine

4) The Wankel it follows the pattern of the Otto cycle but it does not have piston strokes

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The technician should start by checking the temperature rise across the indoor coil.

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