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ozzi
3 years ago
8

A fluid flows through a pipe of constant circular cross-section diameter 1.1m with a mean velocity of 50cm/s. The mass flow rate

of the flow through the pipe, given by the formula mass flow rate = (density) (mean velocity) (cross - sectional area), is measured to be 110 kg/s. Find the specific gravity, specific weight and specific volume of the fluid.
Engineering
1 answer:
allochka39001 [22]3 years ago
3 0

Answer:

Explanation:

Given

diameter(d)=1.1 m

mean velocity(u)=50 cm/s

mass flow rate(m)=110 kg/s

A=0.950 m^2

m=\rho \cdot A\cdot u

where \rho =density of fluid

110=\rho \times 0.95\times 0.5

\rho =231.46 kg/m^3

Therefore specific gravity=\frac{\rho }{\rho _{water}}

=\frac{231.46}{1000}=0.231

Specific weight=\rho \cdot g=231.46\times 9.81=2270.70 kg/m^2-s^2

specific volume=\frac{1}{density}

=\frac{1}{231.46}=0.00432 m^3/kg

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Relation between Poisson's ration, young's modulus, shear modulus for an isotropic material
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 E=2 G(1+2μ)

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3 years ago
A heat engine receives 6 kW from a 250oC source and rejects heat at 30oC. Examine each of three cases with respect to the inequa
taurus [48]

Answer:

Explanation:

Given

T_h=250^{\circ}C\approx 523\ K

T_L=30^{\circ}C\approx 303\ K

Q_1=6 kW

From Clausius inequality

\oint \frac{dQ}{T}=0  =Reversible cycle

\oint \frac{dQ}{T}  =Irreversible cycle

\oint \frac{dQ}{T}>0  =Impossible

(a)For P_{out}=3 kW

Rejected heat Q_2=6-3=3\ kW

\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}

=\frac{6}{523}-\frac{3}{303}=1.57\times 10^{-3} kW/K

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(b)P_{out}=2 kW

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\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}

=\frac{6}{523}-\frac{4}{303}=-1.73\times 10^{-3} kW/K

Possible

(c)Carnot cycle

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Q_2=3.47\ kW

\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}

=\frac{6}{523}-\frac{3.47}{303}=0

and maximum Work is obtained for reversible cycle when operate between same temperature limits

P_{out}=Q_1-Q_2=6-3.47=2.53\ kW

Thus it is possible

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