Answer:
A) q = 0.33
B) Q'_L = 28.565 Kw
C) C.O.P = 2.5
Explanation:
From the table first table i attached, in state 1,at P1 = 0.1MPa, we have;
Specific enthalpy, h1 = 234.46 KJ/Kg
Specific Entropy, s1 = 0.9519 KJ/Kg.k
Now, in state 2, from the second table i attached and by interpolation under the condition s1=s2 at P2 = 0.4MPa,we have;
Specific enthalpy, h2 = 262.74 KJ/Kg
Now, in state 3, from the first table i attached, at P3 = 0.4MPa,we have;
Specific enthalpy, h3 = 255.61 KJ/Kg
Now, in state 5 and 6, from the first table i attached, at P5 = 1.4MPa,we have;
h_f = h5 = h6 = 127.25 KJ/Kg
Now, in state 7 and 8, from the first table i attached, at P7 = 0.4MPa,we have;
h_f = h7 = h8 = 63.92 KJ/Kg
A) The amount of evaporated refrigerant is simply the quality at state 6.
Thus;
q = (h6 - h7)/(h_fg,0.4MPa)
Now,from the third table i attached,
(h_evap,0.4MPa) = h_f,g = 191.62 KJ/Kg
Thus, q = (127.25 - 63.92)/(191.62)
q = 0.33
B) The enthalpy at state 9 is determined from the energy balance of the heat exchanger as;
q = (h9 - h2)/(h3 - h2)
Making h9 the subject, we have;
h9 = qh3 + (1 - q)h2
Thus, plugging in the relevant values to get ;
h9 = 0.33(255.61) + (1 - 0.33)262.74
h9 = 84.35 + 176.04
h9 = 260.39 KJ/Kg
The entropy at this state and value of enthalpy is done by interpolation from the third table i attached at P9 = 0.4 MPa
This gives; s9 = 0.9483 KJ/Kg.k
The enthalpy at state 4 and is done by interpolation from the third table i attached under the condition s4 = s9 and P4 = 1.4MPa.This gives;
h4 = 287.13 KJ/Kg
The rate of heat removal is given by the formula;
Q'_L = m'_b(h1 - h8)
This can be simplified into;
Q'_L = m'_a(1 - q)•(h1 - h8)
Plugging in the relevant values to obtain;
Q'_L = 0.25(1 - 0.33)•(234.46 - 63.92)
Q'_L = (0.25 x 0.67) x 170.54
Q'_L = 28.565 Kw
C) The formula for COP is;
COP = Q'_L/W
This gives;
C.O.P = Q'_L/[m'_a(h4 - h9) + m_b(h2 - h1)]
C.O.P = Q'_L/[m'_a((h4 - h9) + (1 - q) (h2 - h1)]
Plugging in the relevant values to obtain;
C.O.P = 28.565/[0.25((287.13 - 260.39) + (1 - 0.33)(262.74 - 234.46)]
C.O.P = 28.565/(0.25(26.74 + 18.9476)
C.O.P = 28.565/11.4219 = 2.5
Answer:
The nail exerts a force of 573.88 Pounds on the Hammer in positive j direction.
Explanation:
Since we know that the force is the rate at which the momentum of an object changes.
Mathematically
The momentum of any body is defines as
In the above problem we see that the moumentum of the hammer is reduced to zero in 0.023 seconds thus the force on the hammer is calculated using the above relations as
Answer:
capacity = 0.555 mAh
capacity = 3600 mAh
Explanation:
given data
battery = 1800 mAh
OCV = 3.9 V
solution
we get here capacity when it is in series
so here Q = 2C
capacity = 2 × ampere × second ...............1
put here value and we get
and 1 Ah = 3600 C
capacity =
capacity = 0.555 mAh
and
when it is in parallel than capacity will be
capacity = Q1 +Q2 ...............2
capacity = 1800 + 1800
capacity = 3600 mAh