Answer:
increases by a factor of 6.
Explanation:
Let us assume that the initial cross sectional area of the pipe is A m² while the initial velocity of the water is V m/s², hence the flow rate of the water is:
Initial flow rate = area * velocity = A * V = AV m³/s
The water speed doubles (2V m/s) and the cross-sectional area of the pipe triples (3A m²), hence the volume flow rate becomes:
Final flow rate = 2V * 3A = 6AV m³/s = 6 * initial flow rate
Hence, the volume flow rate of the water passing through it increases by a factor of 6.
Answer:
a. 4
b. 1 m
Explanation:
According to the question, the data is as follows
The Density of water at 20 degrees celcius is 1000 kg/m^3
Viscosity is 0.001kg/m/.s
Velocity V = 25 cm/s
V = 0.25 m/s
Now
a. The creeping motion is
As we know that
Reynold Number = (Density of water × V × d) ÷ (Viscosity)
1 = (1,000 × 0.25 × d) ÷ 0.0001
d = (1 × 0.001) ÷ (1,000 × 0.25)
= 4E - 06^m
= 4
b. Now the sphere diameter is
Reynold Number = (Density of water × V × d) ÷ (Viscosity)
250,000 = (1,000 × 0.25 × d) ÷ 0.0001
d = (250,000 × 0.001) ÷ (1,000 × 0.25)
= 1 m
Answer:
1.96 kg/s.
Explanation:
So, we are given the following data or parameters or information which we are going to use in solving this question effectively and these data are;
=> Superheated water vapor at a pressure = 20 MPa,
=> temperature = 500°C,
=> " flow rate of 10 kg/s is to be brought to a saturated vapor state at 10 MPa in an open feedwater heater."
=> "mixing this stream with a stream of liquid water at 20°C and 10 MPa."
K1 = 3241.18, k2 = 93.28 and 2725.47.
Therefore, m1 + m2= m3.
10(3241.18) + m2 (93.28) = (10 + m3) 2725.47.
=> 1.96 kg/s.
Answer: 24 pA
Explanation:
As pure silicon is a semiconductor, the resistivity value is strongly dependent of temperature, as the main responsible for conductivity, the number of charge carriers (both electrons and holes) does.
Based on these considerations, we found that at room temperature, pure silicon resistivity can be approximated as 2.1. 10⁵ Ω cm.
The resistance R of a given resistor, is expressed by the following formula:
R = ρ L / A
Replacing by the values for resistivity, L and A, we have
R = 2.1. 10⁵ Ω cm. (10⁴ μm/cm). 50 μm/ 0.5 μm2
R = 2.1. 10¹¹ Ω
Assuming that we can apply Ohm´s Law, the current that would pass through this resistor for an applied voltage of 5 V, is as follows:
I = V/R = 5 V / 2.1.10¹¹ Ω = 2.38. 10⁻¹¹ A= 24 pA