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zlopas [31]
2 years ago
4

Hotel rooms in Smalltown go for $100, and 1,000 rooms are rented on atypical day.a. To raise revenue, the mayor decides to charg

e hotels a tax of $10 perrented room. After the tax is imposed, the going rate for hotel roomsrises to $108, and the number of rooms rented falls to 900. Calculate theamount of revenue this tax raises for Smalltown and the deadweightloss of the tax. (Hint: The area of a triangle is l/2Xbase X height.)h. The mayor now doubles the tax to $20. The price rises to $116, andthe number of rooms rented falls to 800. Calculate tax revenue anddeadweight loss with this larger tax. Are they double, more thandouble, or less than double? Explain.
Physics
1 answer:
Mama L [17]2 years ago
7 0

Answer:

doubling the size of the tax more than doubles the deadweight loss while less than doubles the revenue generated

Explanation:

(a)

The quantity of rooms rented before tax, Q1 = 1000 rooms.

The quantity of rooms rented after the imposition of tax Q2 = 900 rooms.

Size of the tax = $10

Price paid by buyer = $108

Price received by seller = $98  

 Deadweight loss = 1/2 x (Q2 — Q1) x (size of the tax)  

 Deadweight loss = 1/2 x (1000 — 900) x ($10) = $500  

Tax revenue generated = size of tax * (Q2) = $10 x (900) = $9000

b)

The quantity of rooms rented before tax, Q1 = 1000 rooms

The quantity of rooms rented after the imposition of tax, Q2 = 800 rooms Size of the tax = $20

Price paid by buyer = $116

Price received by seller = $96  

  Deadweight loss = 1/2 x (Q2 — Q1) x (size of the tax)  

  New Deadweight loss = 1/2 x (1000 — 800) x ($20) = $2000  

Thus, dead weight loss quadruples post doubling the size of tax. New Tax revenue generated = size of tax x (Q2) = $20 x (800) = $16000 Thus, revenue generated less than doubles post doubling the size of tax.  

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Air at 3 104 kg/s and 27 C enters a rectangular duct that is 1m long and 4mm 16 mm on a side. A uniform heat flux of 600 W/m2 is
ad-work [718]

Answer:

T_{out}=27.0000077 ºC

Explanation:

First, let's write the energy balance over the duct:

H_{out}=H_{in}+Q

It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:

m*h_{out}=m*h_{in}+Q\\m*(h_{out}-h_{in})=Q

The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.

m*Cp(T_{out}-T_{in})=Q

So, let's isolate T_{out}:

T_{out}-T_{in}=\frac{Q}{m*Cp}\\T_{out}=T_{in}+\frac{Q}{m*Cp}

The Cp of the air at 27ºC is 1007\frac{J}{kgK} (Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are T_{out} and Q.

Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.

The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:

Perimeter:

P=2*H+2*A=2*0.004m+2*0.016m=0.04m

Surface area:

A=P*L=0.04m*1m=0.04m^2

Then, the heat Q is:

600\frac{W}{m^2} *0.04m^2=24W

Finally, find the exit temperature:

T_{out}=T_{in}+\frac{Q}{m*Cp}\\T_{out}=27+\frac{24W}{3104\frac{kg}{s} *1007\frac{J}{kgK} }\\T_{out}=27.0000077

T_{out}=27.0000077 ºC

The temperature change so little because:

  • The mass flow is so big compared to the heat flux.
  • The transfer area is so little, a bigger length would be required.
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