Assignment 6

Problem 1:

A tankless water heater heats water as it flows through the device. The water circulates through a copper heat exchanger. For a particular water heater the flow rate is 8.7 liter per minute. The temperature rise of the water is 25 ^oC. The water heater uses gas to heat the water in the copper coils and is 80% efficient. What is the gas energy input rate in units of kW?
conversion: 1 kcal = 4186 J

Solution:

Concepts:
Energy conservation, specific heat c
Reasoning:
Water flow rate (kg/s) * c(J/(kg^oC) * temperature change (^oC) = power input (W) * efficiency.
Details of the calculation:
Water flow rate: 8.7 kg/minute = (8.7/60) kg/s,
specific heat of water: c = 4186 J/(kg ^oC).
P = ((8.7/60) kg/s)*(4186 J/(kg^oC))*(25 ^oC)/0.8 = 18967 W ≈ 19 kW.

Problem 2:

Particles evaporating in the z-direction from a hole in a container with an ideal gas at temperature T.
(a) Compute the speed distribution, f*(v), of the evaporating particles.
(b) Compute the mean z-component of the velocity, <v_z>* of the escaping particles.
(c) Compute the mean energy of the particles leaving the container.
Assume that the equilibrium in the interior of the container is not disturbed by the evaporating particles.

Maxwell-Boltzmann speed distribution: f(v) = (m/(2πkT))^3/24πv²exp(-mv²/(2kT)

Solution:

Concepts:
The Maxwell-Boltzmann speed distribution, geometry
Reasoning:
The atoms escape through a small hole of area A. We need to find the rate at which atoms with different v or v_z escape from the hole.
Details of the calculation:

Consider an area dA and gas atoms with speeds between v and v + dv in a ring at a distance between r and r + dr from dA. Assume that the gas atoms travel in straight lines. The probability that gas atoms from the ring will reach the area dA is dA cosθ/(4πr²). This is the solid angle subtended by dA at any dV in the ring.
The number of atoms from the ring reaching dA with speeds between v and v + dv is
nf(v)dv2πr²sinθ dr dθ dA cosθ/(4πr²) = ½nf(v)dv sinθ dr dθ dA cosθ.
Here n = N/V is the density of the gas.
The total number of atoms with speeds between v and v + dv reaching dA in a time interval dt comes from rings with radii between r = 0 and dr = vdt. Per unit area per unit time it is therefore given by
∫₀^π/2 dθ nf(v)dv v sinθ cosθ/2 = ¼ n v f(v) dv.
The escape rate per unit area per unit time is
R_esc = ½n∫₀^∞dv ∫₀^π/2dθ sinθ cosθ v f(v) = ¼ n ∫₀^∞ vf(v) dv = n (kT/(2mπ))^½.
We therefore have f*(v) = ¼ n v f(v)/R_esc = ½(m²/(kT)²)v³exp(-mv²/(2kT).

(b) Atoms from a ring reaching dA with speed v have a velocity component v_z = vcosθ.
The average velocity component <v_z> for the particles escaping from the hole is given by
[½n∫₀^∞dv ∫₀^π/2dθ sinθ cos²θ v² f(v)]/R_esc
= (3/4)(2πkT/m)^½ [∫₀^π/2dθ sinθ cos²θ]/[ ∫₀^π/2dθ sinθ cosθ] = ½(2πkT/m)^½.

(c) The mean energy of the particles is given by <E> = ½m∫₀^∞v²f*(v) dv.
<E> = ¼(m³/(kT)²)∫₀^∞v⁵exp(-mv²/(2kT) dv = 2kT.

Problem 3:

A small air bubble of initial radius r = 1 cm is introduced at the bottom of a lake that is 20 m deep. The bubble expands as it rises slowly. Assume the lake has the same temperature everywhere.
(a) What is the pressure at the bottom of the lake?
(b) What is the radius of the bubble after it rises to the surface?

Solution:

Concepts:
Pressure in a fluid, the ideal gas law
Reasoning:
The pressure at the bottom of a liquid column is P_bottom = P_top + ρgh.
Details of the calculation:
P_bottom = P_surf + ρgz = 1 atm + (1000 kg/m³)(9.8 m/s²)(20 m) = 1 atm + 196000 Pa.
1 atm = 1.013 * 10⁵ Pa.
P_bottom = 2.93 atm.
Ideal gas law: PV = NkT = constant.
P_i V_i = P_f V_f.
V_f = P_i/P_f V_i = (4/3)πr_f³ = [(2.93 atm)/(1 atm)]* (4/3) π(0.01)³.
r_f = 1.43 cm.

Problem 4:

Estimate (roughly) the net rate of heat loss by your body due to radiation when you are in a room with a temperature of 10 ^oC.

Solution:

Concepts:
Stefan-Boltzmann law
Reasoning:
Heat loss = σ* T⁴_body * Area, heat gain = σ* T⁴_surrounding * Area.
Net loss = σ* (T⁴_body- T⁴_surrounding)*Area.
Details of the calculation:
T_body ~ 33 ^oC = 306 K (skin temperature).
Area ~ 1.75 m² (skin surface area).
Net loss ~ 5.67*10^-8(306⁴ - 283⁴)*1.75 W = 234 W.

Problem 5:

During phase transition, the change in the pressure P and temperature T can be expressed by the Clausius-Clapeyron relation,

dP/dT = L/(T ∆V),

where L is the latent heat and ∆V is the change in volume.
(a) How much pressure does one have to put on an ice cube to make it melt at -1^o C?
The density of ice is 917 kg/m³, and the latent heat of 1 kg of ice melting is 333000 J.
(b) Approximately how deep under a glacier does it have to be before the weight of the ice above gives the pressure you found in part (a)?

Solution:

Concepts:
The Clausius-Clapeyron relation
Reasoning:
For small changes of the melting point temperature we expand ∆P = (dP/dT)∆T.
Details of the calculation:
(a) The density of water is 1000 kg/m³. The volume of 1 kg of water is 1/1000 m³.
From the density of ice 917 kg/m³, the volume of 1 kg of ice is 1/917 m³.
For phase transition of 1 kg of ice to water ∆V =(1/1000 - 1/917) m³ = -9.051*10^-5 m³.
dP/dT = (333000 J)/(273 K * ∆V) = -1.35*10⁷ Pa/K = -135 bar.
For the melting point decrease by 1^o C we have ∆P = (dP/dT)∆T = 135 bar.
For the melting point decrease by 1^o C, the pressure should increase by 135 bar.

(b) To reduce the melting point by 1^o C, the pressure under a glacier should be ~136 bar.
P_{under_glacier} = P_air+ P_glacier = 1 bar + ρ_ice d g = 136 bar
d = (135 *10⁵ N/m²)/(917 kg/m³ * 9.8 m/s²) = 1500 m.