Answer:
Work done = 125π J
Explanation:
Given:
P = P_i * ( 1 - (x/d)^2 / 25)
d = 5.0 cm
x = 5 * d cm = 25d
Pi = 12 bar
Work done = integral ( F . dx )
F (x) = P(x) * A
F (x) = (πd^2 / 4) * P_i * (1 - (x/d)^2 / 25)
Work done = integral ((πd^2 / 4) * P_i * (1 - (x/d)^2 / 25) ) . dx
For Limits 0 < x < 5d
Work done = (πd^2 / 4) * P_i integral ( (1 - (x/d)^2) / 25)) . dx
Integrate the function wrt x
Work done = (πd^2 / 4) * P_i * ( x - d*(x/d)^3 / 75 )
Evaluate Limits 0 < x < 5d :
Work done = (πd^2 / 4) * P_i * (5d - 5d / 3)
Work done = (πd^2 / 4) * P_i * (10*d / 3)
Work done = (5 π / 6)d^3 * P_i
Input the values:
Work done = (5 π / 6)(0.05)^3 * (1.2*10^6)
Work done = 125π J
Answer:
A selective medium, a differential medium, and a complex medium.
Explanation:
A selective media is a microbiological media which only support the growth of a particular specie or types of species of microorganisms,this media acts in such a way to inhibit or hinder the growth of other microorganisms.
Differential media are media that acts to Identifying particular strains of microorganisms of similar species.
Complex media are media used for the growth of microorganisms this which contains complex or a wide range of nutrients with chemical composition which may be difficult to determine.
Answer:
W=2 MW
Explanation:
Given that
COP= 2.5
Heat extracted from 85°C
Qa= 5 MW
Lets heat supplied at 150°C = Qr
The power input to heat pump = W
From first law of thermodynamics
Qr= Qa+ W
We know that COP of heat pump given as
W=2 MW
For Carnot heat pump
2.5 T₂ - 895= T₂
T₂=596.66 K
T₂=323.6 °C
