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3 years ago

In a steady flow device, the properties of the system remains constant with time. a)True b) False

Engineering

1 answer:

Leviafan [203]3 years ago

7 0

Answer:

True

Explanation:

By definition of steady flow we have

$\frac{\partial f(x,y,z,t) }{\partial t}=0$

where f(x,y,z,t) is any property of the system under consideration

=> f(x,y,z,t) = constant

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Two bodies have heat capacities (at constant volume) c, = a and c2 = bT and are thermally isolated from the rest of the universe

prisoha [69]

Answer:

Explanation:

The answer to the above question is given in attached files.

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3 years ago

An o ring intended for use in a hydraulic system using MIL-H-5606 (mineral base) fluid will be marked with

Alex_Xolod [135]

An o ring intended for use in a hydraulic system using MIL-H-5606 (mineral base) fluid will be marked with a blue stripe or dot.

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2 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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3 years ago

Electric Resistance Heating. A house that is losing heat at a rate of 50,000 kJ/h when the outside temperature drops to 4 0C is

ryzh [129]

Answer:

a) $\dot W = 0.978\,kW$ , b) $I = \left(50000\,\frac{kJ}{h} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}\right)\cdot \left(\frac{1}{COP_{real}} \right) - 0.978\,kW$

Explanation:

a) The ideal Coefficient of Performance for the heat pump is:

$COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}$

$COP_{HP} = \frac{298.15\,K}{298.15\,K - 277.15\,K}$

$COP_{HP} = 14.198$

The reversible work input is:

$\dot W = \frac{\dot Q_{H}}{COP_{HP}}$

$\dot W = \left(\frac{50000\,\frac{kJ}{h} }{14.198} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s} \right)$

$\dot W = 0.978\,kW$

b) The irreversibility is given by the difference between real work and ideal work inputs:

$I = \dot W_{real} - \dot W_{ideal}$

$I = \left(50000\,\frac{kJ}{h} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}\right)\cdot \left(\frac{1}{COP_{real}} \right) - 0.978\,kW$

7 0

4 years ago

The flow of a real fluid has (more —less - same ) complexity of an ideal fluid, owing to the phenomena caused by the existence o

Viefleur [7K]

Answer:

The flow of a real fluid has <u>more</u> complexity as compared to an ideal fluid owing to the phenomena caused by existence of <u>viscosity</u>

Explanation:

For a ideal fluid we know that there is no viscosity of the fluid hence the boundary condition need's not to be satisfied and the flow occur's without any head loss due to viscous nature of the fluid. The friction of the pipe has no effect on the flow of an ideal fluid. But for a real fluid the viscosity of the fluid has a non zero value, the viscosity causes boundary layer effects, causes head loss and also frictional losses due to pipe friction hugely make the analysis of the flow complex. The losses in the energy of the flow becomes complex to calculate as frictional losses depend on the roughness of the pipe and Reynolds number of the flow thus increasing the complexity of the analysis of flow.

3 0

3 years ago