A 1-kW electric resistance heater submerged in 10-kg water is turned on and kept on for 15 min. During the process, 400 kJ of he
1 answer:
Answer:
ΔT= 11.94 °C
Explanation:
Given that
mass of water = 10 kh
Time t= 15 min
Heat lot from water = 400 KJ
Heat input to the water = 1 KW
Heat input the water= 1 x 15 x 60
=900 KJ
By heat balancing
Heat supply - heat rejected = Heat gain by water
As we know that heat capacity of water
Now by putting the values
900 - 400 = 10 x 4.187 x ΔT
So rise in temperature of water ΔT= 11.94 °C
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Answer:
Outside temperature =88.03°C
Explanation:
Conductivity of air-soil from standard table
K=0.60 W/m-k
To find temperature we need to balance energy
Heat generation=Heat dissipation
Now find the value
We know that for sphere
Given that q=500 W
so
By solving that equation we get
=88.03°C
So outside temperature =88.03°C
Answer:
≅ 111 KN
Explanation:
Given that;
A medium-sized jet has a 3.8-mm-diameter i.e diameter (d) = 3.8
mass = 85,000 kg
drag co-efficient (C) = 0.37
(velocity (v)= 230 m/s
density (ρ) = 1.0 kg/m³
To calculate the thrust; we need to determine the relation of the drag force; which is given as:
=
× CρAv²
where;
ρ = density of air wind.
C = drag co-efficient
A = Area of the jet
v = velocity of the jet
From the question, we can deduce that the jet is in motion with a constant speed; as such: the net force acting on the jet in the air = 0
SO,
We can as well say:
We can now replace in the above equation.
Therefore, =
× CρAv²
The A which stands as the area of the jet is given by the formula:
We can now have a new equation after substituting our A into the previous equation as:
=
× Cρ
Substituting our data from above; we have:
=
×
=
= 110,990N
in N (newton) to KN (kilo-newton) will be:
=
= 110.990 KN
≅ 111 KN
In conclusion, the jet engine needed to provide 111 KN thrust in order to cruise at 230 m/s at an altitude where the air density is 1.0 kg/m³.
Answer:
a) 42.08 ft/sec
b) 3366.33 ft³/sec
c) 0.235
d) 18.225 ft
e) 3.80 ft
Explanation:
Given:
b = 80ft
y1 = 1 ft
y2 = 10ft
a) Let's take the formula:
1 + 8f² = (20+1)²
= 8f² = 440
f² = 55
f = 7.416
For velocity of the faster moving flow, we have :
V1 = 42.08 ft/sec
b) the flow rate will be calculated as
Q = VA
VA = V1 * b *y1
= 42.08 * 80 * 1
= 3366.66 ft³/sec
c) The Froude number of the sub-critical flow.
V2.A2 = 3366.66
Where A2 = 80ft * 10ft
Solving for V2, we have:
= 4.208 ft/sec
Froude number, F2 =
F2 = 0.235
d)
= 18.225ft
e) for critical depth, we use :
= 3.80 ft