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umka21 [38]
3 years ago
12

Define volume flow rate Q of air flowing in a duct of area A with average velocity V

Engineering
1 answer:
Shalnov [3]3 years ago
3 0

Answer:

The volume flow rate of air is Q=A\times V

Explanation:

A random duct is shown in the below attached figure

The volume flow rate is defined as the volume of fluid that passes a section in unit amount of time

Now by definition of velocity we can see that 'v' m/s means that in 1 second the flow occupies a length of 'v' meters

From the attached figure we can see that

The volume of the prism that the flow occupies in 1 second equals

Volume=Area\times V=A\times V

Hence the volume flow rate is Q=V\times A

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b

Explanation:

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What is the net force acting on a car cruising at a constant velocity of 70 km/h (a) on a level road and (b) on an uphill road?
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a) zero b) zero

Explanation:

Newton's first law tells us that a body remains at rest or in uniform rectilinear motion, if a net force is not applied on it, that is, if there are no applied forces or If the sum of forces acting is zero. In this case there is a body that moves with uniform rectilinear motion which implies that there is no net force.

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Doubling the diameter of a solid, cylindrical wire doubles its strength in tension.
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3 years ago
An insulated piston-cylinder device contains 0.15 of saturated refrigerant-134a vapor at 0.8 MPa pressure. The refrigerant is no
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Answer:

Assumption:

1. The kinetic and potential energy changes are negligible

2. The cylinder is well insulated and thus heat transfer is negligible.

3. The thermal energy stored in the cylinder itself is negligible.

4. The process is stated to be reversible

Analysis:

a. This is reversible adiabatic(i.e isentropic) process and thus s_{1} =s_{2}

From the refrigerant table A11-A13

P_{1} =0.8MPa   \left \{ {{ {{v_{1}=v_{g}  @0.8MPa =0.025645 m^{3/}/kg } } \atop { {{u_{1}=u_{g}  @0.8MPa =246.82 kJ/kg } -   also  {{s_{1}=s_{g}  @0.8MPa =0.91853 kJ/kgK } } \right.

sat vapor

m=\frac{V}{v_{1} } =\frac{0.15}{0.025645} =5.8491 kg\\and \\\\P_{2} =0.2MPa  \left \{ {{x_{2} =\frac{s_{2} -s_{f} }{s_{fg }}=\frac{0.91853-0.15449}{0.78339}   = 0.9753 \atop {u_{2} =u_{f} +x_{2} }(u_{fg}) =  38.26+0.9753(186.25)= 38.26+181.65 =219.9kJ/kg \right. \\s_{1} = s_{2}

T_{2} =T_{sat @ 0.2MPa} = -10.09^{o}  C

b.) We take the content of the cylinder as the sysytem.

This is a closed system since no mass leaves or enters.

Hence, the energy balance for adiabatic closed system can be expressed as:

E_{in} - E_{out}  =ΔE

w_{b, out}  =ΔU

w_{b, out} =m([tex]u_{1} -u_{2)

w_{b, out}  = workdone during the isentropic process

=5.8491(246.82-219.9)

=5.8491(26.91)

=157.3993

=157.4kJ

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Determine an expression in standard form for the voltage gain VoVs. Hv(jω)=Vo(jω)Vi(jω)=R2R111+jωCR2 Hv(jω)=Vo(jω)Vi(jω)=−R2R111
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Answer:

Applying Kirchhoff's current law at inverting terminal

Explanation:

Detailed explanation is given in the attached document.

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