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Yuri [45]
3 years ago
10

A classroom which normally contains 26 people is to be air- conditioned. A person in the classroom typically dissipates heat at

a rate of 360 kJ/h. The room is lit with 15 light bulbs which each consume 40 W. A 200 W fan also operates in the room near a door to a computer closet. This fan pulls warm air into the classroom under the door at a rate of 200 kg/h. The room gains 10 kJ in energy (enthalpy) for each kg of warm air which enters. The air comes under the door at 10 m/s and leaves the room with negligible velocity through a large vent 3 m above the floor. Otherwise the room is well sealed. Finally, the room gains 5,000 kJ/h by heat transfer through the walls and windows. Determine the air-conditioner capacity, in kW, that is needed to keep the room at a steady temperature. Are any terms of the energy balance negligible?
Engineering
1 answer:
lutik1710 [3]3 years ago
7 0

Answer:

Explanation:

Heat dissipated by 26 people per second

= 26 x 360 x 1000 / (60 x 60)

= 2600 J

Heat dissipated by bulbs per second

= 15 x 40

= 600 J

Heat dissipated by fans per second

200 J

Heat dissipated by air coming in

= 200 x 10 x 1000 / (60 x 60)

=555.55 J

Heat gain by walls

= 5000 x 1000 / (60 x 60)

= 1388.88 J

Total heat gain per second

= 2600 + 600+200+555.55

= 3955.55 J

Capacity of air-conditioner required

= 3955.55 J/s

= 3.9 kJ/s

= 3.9kW

= 4 kW

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5 0
3 years ago
If you measure 0.7 V across a diode, the diode is probably made of
tatuchka [14]

Answer:

Made of Silicon.

Explanation:

A diode is a semiconductor device use in mostly electronic appliances. It is two terminals device consisting of a P-N junction formed either in Germanium or silicon crystal.

Diode can be forward biased or reverse biased.

When a diode is forward biased and the applied voltage is increased from zero, hardly any current flows through the device in the beginning.

It is so because the external voltage is being opposed by the internal barrier voltage whose value is 0.7v for silicon and 0.3v for germanium.

If you measure 0.7 V across a diode, the diode is probably therefore made of Silicon.

6 0
3 years ago
Refer to Figure 9-18. A #_____ electrode lead and workpiece lead should be used to carry 150 amperes of electricity 75 feet to t
Anni [7]

Answer:

AC or

Explanation:

BC

8 0
2 years ago
An insulated piston-cylinder device contains 0.15 of saturated refrigerant-134a vapor at 0.8 MPa pressure. The refrigerant is no
Stells [14]

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

4 0
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Angelina_Jolie [31]

Answer:

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4 0
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