Write a loop to print all elements in hourly_temperature. Separate elements with a -> surrounded by spaces. Sample output for
the given program with input: '90 92 94 95' 90 -> 92 -> 94 -> 95 Note: 95 is followed by a space, then a newline
1 answer:
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Answer:
the rate of entropy change of the air is -0.1342 kW/K
the assumptions made in solving this problem
- Air is an ideal gas.
- the process is isothermal ( internally reversible process ). the change in internal energy is 0.
- It is a steady flow process
- Potential and Kinetic energy changes are negligible.
Explanation:
Given the data in the question;
From the first law of thermodynamics;
dQ = dU + dW ------ let this be equation 1
where dQ is the heat transfer, dU is internal energy and dW is the work done.
from the question, the process is isothermal ( internally reversible process )
Thus, the change in internal energy is 0
dU = 0
given that; Air is compressed by a 40-kW compressor from P1 to P2
since it is compressed, dW = -40 kW
we substitute into equation 1
dQ = 0 + ( -40 kW )
dQ = -40 kW
Now, change in entropy of air is;
ΔS = dQ / T
given that T = 25 °C = ( 25 + 273.15 ) K = 298.15 K
so we substitute
ΔS = -40 kW / 298.15 K
ΔS = -0.13416 ≈ -0.1342 kW/K
Therefore, the rate of entropy change of the air is -0.1342 kW/K
the assumptions made in solving this problem
- Air is an ideal gas.
- the process is isothermal ( internally reversible process ). the change in internal energy is 0.
- It is a steady flow process
- Potential and Kinetic energy changes are negligible.
Answer:
The graphs are attached below
Explanation:
Ans) We know,
F(t) = K(t - to) + zΔ∅ ln[(1 + F(t)/zΔ∅]
where K = hydraulic conductivity =0.1 cm/hr
z = capillary suction =20cm
Δ∅ = ∅ (1 - Se)
∅ = effective porosity , Se = initial moisture content
Δ∅ = 0.40(1 - 0.15)
Δ∅ = 0.34
Now, F(t) = 0.1(t - to) + 20(0.34) ln[ 1 + F(t)/6.8]
F(t) = 0.1(t - to) + 6.8 ln [1+ 0.147 F(t)]
Also, infiltration rate (f),
f = K [(zΔ∅ + F)/F]
f = 0.40 [6.8 + F]/F
Condition of ponding,
Ponding time tp = K zΔ∅ i(i-k)
where, i = rainfall rate (1cm/hr)
tp = 0.40(6.8) / 0.60
tp= 4.53 hours
Now, cumulative infiltraton at ponding Fp = i tp
Fp = 1 x 4.53 or 4.53 cm
For infiltration at time less then ponding time , infiltration rate = rainfall rate
For t = 0.25 hr , f = 1cm/hr ; F = 0.25 x 1 = 0.25 cm
For t = 0.50 hr , f = 1cm/hr ; F = 0.50 cm
For t = 0.75 hr , f = 1cm/hr ; F = 0.75 cm
For t = 1 hr , f = 1cm/hr ; F = 1cm/hr
For t > tp ,
Equivalent time origin(to),
to = tp - 1/K [ Fp - z Δ∅ ln (1+ Fp/zΔ∅)
to = 4.53 - 1/0.40[ 4.53 - 6.8 ln(1 + 0.66)
to = 4.53 - 2.5 ( 4.53 - 3.44)
to = 1.82 hr
Hence,
F = 0.10( t - 1.82) + 6.8 ln[ 1+ 0.147F(t)]
Solving above equation for t by assuming F, and further solving equation for infiltration rate
Ans (a) Following is required curve :
O.S 1-5 2 time Chr)ホー Lt 2 time Chr)
Ans (c) Following is required table :
Ans (d) For time step t = 0.25 hrs
At t < tp ; i = f = 1 cm/hr
F = i x t
F = 1 x 0.25
F = 0.25 cm
Answer:
The statement regarding the mass rate of flow is mathematically represented as follows
Explanation:
A junction of 3 pipes with indicated mass rates of flow is indicated in the attached figure
As a basic sense of intuition we know that the mass of the water that is in the pipe junction at any instant of time is conserved as the junction does not accumulate any mass.
The above statement can be mathematically written as
this is known as equation of conservation of mass / Equation of continuity.
Now we know that in a time 't' the volume that enter's the Junction 'O' is
1) From pipe 1 =
1) From pipe 2 =
Mass leaving the junction 'O' in the same time equals
From pipe 3 =
From the basic relation of density, volume and mass we have
Using the above relations in our basic equation of continuity we obtain
Thus the mass flow rate equation becomes
