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brionna246
25.03.2020 •
Physics
A building is maintained at a temperature T by means of an ideal heat pump which uses a river at temperature T0 as a source of heat. The heat pump consumes power W, and the building loses heat to its surroundings at a rate α(T−T0), where α is a positive constant. Show that T is given by T=T0+W2α(1+1+4αT0/W‾‾‾‾‾‾‾‾‾‾‾√).
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Ответ:
We want to prove that
Show that T is given by T=To+W/2α[1+√(1+4αTo/W)]
Explanation:
The efficiency of the pump is given as
n = Q/t / P
Where Q/t is the rejected power
And P is the consumed power to cool the house
Given that,
The at rejected or loss is α(T−To)
So, Q/t = α(T−To)
Pump Consumed =W
Then, power consumed P=W/t
Then,
n = Q/t /P
n = α(T−To) / W . Equation 1
Generally,
The efficiency of a carnot engine is given as
n = TH/(TH —TC)
Where TC is cold temperature
TC=To
And TH is high temperature
TH=T
n= T/ (T—To) Equation 2
Equation the two equation we have,
α(T−To) / W = T/ (T—To)
Cross multiply
α(T−To) × (T-To) =W× T
α(T² — 2TTo + To²)=WT
Divide both sides by α
T² — 2TTo + To²=WT/α
Rearrange
T² — 2TTo + To²—WT/α=0
T² — 2TTo —WT/α + To²=0
T² — (2To + W/α)T + To²=0
This form a quadratic equation
So let use formula method.
T = [- b ± √(b²-4ac) ]/ 2a
Where a=1, b= — (2To + W/α), c= To²
Then, applying the formula
T = [- b ± √(b²-4ac) ]/ 2a
T = [- -(2To + W/α) ± √(-(2To +W/α))² - 4×1×To²)]/ 2×1
T = [(2To + W/α) ± √4To²+4ToW/α+ W²/α² - 4To²]/ 2
T = [(2To+W/α) ±√(4ToWα+W²/α²)]/2
T=[(2To+W/α)±√(W²/α²(4Toα/W+1)]/2
T=[(2To+W/α)±W/α√(4Toα/W+1)]/2
Divide through by 2
T=To+W/2α±W/2α√(4Toα/W+1)
T=To+W/2α(1 ±√(4Toα/W+1)
Rearranging to conform to what we want to proof
T=To+W/2α(1 ±√(1 + 4αTo/W)
This is the required proof..
T = To + W/2α(1 + √(1 + 4αTo/W)
OR
T=To+W/2α(1 —√(1 + 4αTo/W)
Ответ:
T = (T_o + W/2α (1 + √(4αT_o/(W) + 1)] has been proved!
Explanation:
We know that efficiency of pump is given by;
η = (Q/t)/P
Where;
P = the power consumed in the house
Q/t = the rejected power.
Now, P can also be expressed as W/t
Likewise, Q/t can also be expressed as; α(T - T_o)
From the question, pump consumed power of W. Thus, P = W in this case.
So, efficiency can be expressed as;
η = (α(T - T_o))/(W) - - - - (1)
Now, for the carnot engine of the pump, efficiency is expressed as;
η = T/(T - T_o) - - - - (2)
Equating equation 1 and 2,we have;
(α(T - T_o))/(W) = T/(T - T_o)
α(T - T_o)² = WT
Expanding, we have;
T² - (2T•T_o) + T_o² = WT/α
T² - (2T•T_o + WT/α) + T_o² = 0
T² -T(2T_o + W/α) + T_o² = 0
To find T, let's use quadratic formula which is;
x = [-b ± √(b² - 4ac)]/2a
T = -(-(2T_o + W/α)) ± √(-(2T_o + W/α)² - (4•1•T_o²)]/(2•1)
T = (2T_o + W/α)) ± √((-2T_o - W/α)² - (4T_o²)]/(2)
T = (2T_o + W/α)) ± √((4T_o² + 2T_o•W/α + (W/α)² - 4T_o²)]/(2)
T = (2T_o + W/α)) ± √(2T_o•W/α + (W/α)²)]/(2)
T = (T_o + W/2α)) ± (1/2)√(2T_o•W/α + (W/α)²)]
T = (T_o + W/2α)) ± (W/2α)√(4T_o/(W/α) + 1)]
T = (T_o + W/2α)) ± (W/2α)√(4αT_o/(W) + 1)]
Now, let's arrange to correspond with what's in the question. And it means we will make use of the positive sign where we have ±.
Thus;
T = (T_o + W/2α (1 + 1√(4αT_o/(W) + 1)]
T = (T_o + W/2α (1 + √(4αT_o/(W) + 1)]
Ответ:
she must increase the current by factor of 7
Explanation:
The magnetic field produced by a steady current flowing in a very long straight wire encircles the wire.In order to solve the question, we use this formula,
B= μo I/(2πr)
where,
'μo' represents permeability of free space i.e 4π*10-7 N/A2
B=magnetic field
I= current
r=radius
->When r= 1cm=> 0.01m
B1 = μo
/(2π x 0.01)
->when r=7cm =>0.07m
B2 = μo
/(2π x 0.07)
Now equating both of the magnetic fields, we have
B1= B2
μo
/(2π x 0.01)= μo
/(2π x 0.07)
Therefore, she must increase the current by factor of 7