Answer:
.
Explanation:
Given that
L= 50 m
Pressure drop = 130 KPa
copper tube is 3/4 standard type K drawn tube.
From standard chart ,the dimension of 3/4 standard type K copper tube given as
Outside diameter=22.22 mm
Inside diameter=18.92 mm
Dynamic viscosity for kerosene

We know that

Where Q is volume flow rate
L is length of tube
is inner diameter of tube
ΔP is pressure drop
μ is dynamic viscosity
Now by putting the values



So flow rate is
.
Answer:
The rate of energy absorbed per unit time is 3500W.
Explanation:
From the question, we were given the following parameters;
Plane, opaque, gray, diffuse surface
â = 0.7
Surface area, A = 0.5m²
Incoming radiant energy, G = 10000w/m²
T = 500°C
Rate of energy absorbed is âAG;
âAG = 0.7 × 0.5 × 10000
âAG = 3500W.
The energy absorbed is measured in watts and denoted by the symbol W.
Answer:
The hydraulic will jump since the flow is subcritical ( i.e. Y2 > Yc )
Explanation:
width of channel = 3.0 m
Flow rate = 5 m^3/s
Normal depth = 0.50 m
Flow encounters a dam rise of 0.25 m
To know if the hydraulic jump will occur
we will Determine the new normal depth
Y2 = 3.77m
Yc ( critical depth )= 0.66m
Attached below is the detailed solution