Answer:
(a) The maximum volume flow rate for which the flow will be laminar is 0.0190 cubic meter per second
(b) The pressure drop required to deliver the maximum flow rate is 148962.96 Pascal
(c) The corresponding wall shear stress is 7600 Pascal
Explanation:
Reynolds number = 2299, density of water = 1000kg/m^3, diameter of needle = 0.27mm = 0.00027m, Length of needle = 50mm = 0.05m, viscosity of water = 0.00089kg/ms, area = 0.05m × 0.05m = 0.0025m^2, coefficient of friction = 64 ÷ Reynolds number = 64 ÷ 2299 = 0.028
Velocity = (Reynolds number × viscosity) ÷ (density × diameter) = (2299 × 0.00089) ÷ (1000 × 0.00027) = 2.046 ÷ 0.27 = 7.58m/s
(a) Maximum volume flow rate = velocity × area of needle = 7.58 × 0.0025 = 0.0190 cubic meter per second
(b) Pressure drop = ( coefficient of friction × length × density × velocity^2) ÷ (2 × diameter) = (0.028 × 0.05 × 1000 × 7.58^2) ÷ (2 × 0.00027) = 80.44 ÷ 0.00054 = 148962.96 Pascal
(c) Wall shear stress = (density × volume flow rate) ÷ area = (1000 × 0.0190) ÷ 0.0025 = 7600 Pascal
Answer:
See attached document for complete answers and also follow the step by step explanation to get given parameters
Explanation:
Given that:
In a tempering process, glass plate, which is initially at a uniform temperature Ti, is cooled by suddenly reducing the temperature of both surfaces to Ts. The plate is 10 mm thick, and the glass has a thermal diffusivity of 6 × 10−7 m2/s. (a) How long will it take for the midplane temperature to achieve 75% of its maximum possible temperature reduction? (b) If (Ti − Ts) = 300°C, what is the maximum temperature gradient in the glass at the time calculated in part (a)? (c) Plot the temperature distribution within the glass plate at t = 10s
(
,
,
) = (3, -6, 3)
<u>Explanation:</u>
Given data,

5 packets are delivered by channel, which was randomly taken.
0(P(0)), 1(P(1)), 2(P(2)), 3(P(3)), 4(P(4))
Bob receives are
(0,3) (1,0) (2,3) (3,0) (4,3)
P(0) = 3
P(1) = 0
P(2) = 3
P(3) = 0
P(4) = 3
The given equation is

Solution:
= 3
= 3

= - 3

= 3
= 0

= 0

= 3
= 0
From P(1) and P(2)



-3 +
= 0
= 3
= -6
(
,
,
) = (3, -6, 3)