Suppose a tank is made of glass and has the shape of a right-circular cylinder of radius 1 ft. Assume that h(0) = 2 ft corresponds to water filled to the top of the tank, a hole in the bottom is circular with radius in., g = 32 ft/s2, and c = 0.6. Use the differential equation in Problem 12 to find the height h(t) of the water.
Answer:
Height of the water = √(128)/147456 ft
Explanation:
Given
Radius, r = 1 ft
Height, h = 2 ft
Radius of hole = 1/32in
Acceleration of gravity, g = 32ft/s²
c = 0.6
Area of the hold = πr²
A = π(1/32)² ---- Convert to feet
A = π(1/32 * 1/12)²
A = π/147456 ft²
Area of water = πr²
A = π 1²
A = π
The differential equation is;
dh/dt = -A1/A2 √2gh where A1 = Area of the hole and A2 = Area of water
A1 = π/147456, A2 = π
dh/dt = (π/147456)/π √(2*32*2)
dh/dt = 1/147456 * √128
dh/dt = √128/147456 ft
Height of the water = √(128)/147456 ft
Answer:
Heat
Explanation:
Carnot cycle:
Carnot cycle is the ideal cycle for all working engine .Carnot cycle all processes are reversible.It have fore process Out of two are constant temperature process and other two are isentropic process(reversible adiabatic).
We know that area under T-s diagram represents the heat.
So 
From cycle we can say that
Answer: 16 squared + 63 squared = x squared
Explanation:
Hi, since we have a right triangle we have to apply the Pythagorean Theorem:
c^2 = a^2 + b^2
Where c is the hypotenuse of the triangle (the longest side) and a and b are the other sides.
Replacing with the values given:
x^2 = 63^2 + 16^2
So, the correct option is
16 squared + 63 squared = x squared
Feel free to ask for more if needed or if you did not understand something.
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
