Answer:
\epsilon = 0.028*0.3 = 0.0084
Explanation:
\frac{P_1}{\rho} + \frac{v_1^2}{2g} +z_1 +h_p - h_l =\frac{P_2}{\rho} + \frac{v_2^2}{2g} +z_2
where P_1 = P_2 = 0
V1 AND V2 =0
Z1 =0
h_P = \frac{w_p}{\rho Q}
=\frac{40}{9.8*10^3*0.2} = 20.4 m
20.4 - (f [\frac{l}{d}] +kl) \frac{v_1^2}{2g} = 10
we know thaTV =\frac{Q}{A}
V = \frac{0.2}{\pi \frac{0.3^2}{4}} =2.82 m/sec
20.4 - (f \frac{60}{0.3} +14.5) \frac{2.82^2}{2*9.81} = 10
f = 0.0560
Re =\frac{\rho v D}{\mu}
Re =\frac{10^2*2.82*0.3}{1.12*10^{-3}} =7.53*10^5
fro Re = 7.53*10^5 and f = 0.0560
\frac{\epsilon}{D] = 0.028
\epsilon = 0.028*0.3 = 0.0084
Answer:
<em>minimum required diameter of the steel linkage is 3.57 mm</em>
<em></em>
Explanation:
original length of linkage l = 10 m
force to be transmitted f = 2 kN = 2000 N
extension e = 5 mm= 0.005 m
maximum stress σ = 200 N/mm^2 = 
maximum stress allowed on material σ = force/area
imputing values,
200 = 2000/area
area = 2000/(
) =
m^2
recall that area = 
=
= 
= 
=
m = 3.57 mm
<em>maximum diameter of the steel linkage d = 3.57 mm</em>
Answer : 9.36ohms/ temperature
Explanation:
Expression for the variation of resistance of platinum with temperature
Rt= Ro(1+*t)
Rt= resistance @ t°C
Ro= resistance @ 0°C
*= temperature coefficient of resistance
Calculate the change in resistance by putting 120ohms for Ro,
0.0039/K for *
20°C for t
Using this formula:
Rt = Ro(1+*t)
Rt- Ro = Ro*t
= (120ohms)(0.0039/K)(20°C)
= 9.36ohms/K