TES Practice Problem Set 3 Winter 2015

Transcription

1 TES Practice Problem Set 3 Winter 2015 Solutions 1. This first question explores the basic energy conversions implied by the first law of thermodynamics. The unit of energy considered is one we can all hopefully visualize: it is the energy released through a small car s load of fuel, say burning in air 40 L of gasoline at 35 MJ/L. The TES system we will initially compare this to is a thousand cubic meter of (pure) water, with (for simplicity) an assumed density of 1000 kg per cubic meter. (In water supply work, 1000 m3 of water is a standard unit of water storage, affectionately called a megalitre, and is visualized as a cube of water 10 m on edge.) a. What speed would this cube of water have to be travelling at to have the KE equivalent of this small tank of gasoline? Ans: Total Energy released by burning the gasoline = 40 L 35 MJ/L = 1400 MJ = J cubic meter of water, m = 1000 kg/m m 3 = 10 6 kg. The Kinetic Energy is calculated by KE = 21 mv 2 (where v and m are the velocity and the mass of moving object respectively), hence: v 2kE / m = ( )/10 6 (J) = 52.9 m/s (190 km/hr; try to visualize this it is impressive!) b. How high would you have to lift this cube of water against a constant gravitational field of 9.8 m/s2 to have this amount of energy? Ans: Potential energy PE = mgh, then: h= PE/mg = (J)/ 106(kg) 9.8(m/s 2 )=143m (Hydro power requires a lot of water, or a lot of elevation, or quite a bit of both!) c. What would be the increase in temperature in this cube of water to absorb all the energy in thermal form associated with the gasoline? Ans: Consider the specific heat capacity of water we have: Q = C p m T (where Q is the heat absorbed and CP is the specific heat capacity of the water). Therefore, T = Q/Cpm = (J)/ 4200(J /kg.k) 10 6 (kg)=0.333 K =0.333! C (Temperature changes in water store an impressive amount of energy, which is why we can usually assume in hydraulic calculations that change is isothermal and talk about a loss of mechanical energy, not an increase in thermal energy. In gases, we usually have to write a thermal energy equation as well.)

2 d. How much water would have to be evaporated? Ans: Consider the latent heat of evaporation: meva = Q/Levap = (J) / 2.5(MJ / kg) = 560kg(a pretty small fraction of the original) (Note that the 2.5 MJ/kg value used here is typical of TES evaporation taking place at average earth temperatures; the standard value is a little lower than this, but is for the conversion at 100 C, and is thus too small for most of our TES purposes. We need more energy to evaporate cooler water. But the real point is that the phase chance is a power energy storage effect, and this powers many of our TES systems.) e. If the water were in the form of ice, how much could be melted with this much energy? Ans: Consider the melting of ice into water at 0 C, the mass of ice melted will be: mm = Energy / LM = (J ) / (J /kg) = 4,190 kg f. How much of the mass would have to be converted directly to energy using nuclear considerations, assuming suitable ones were available (i.e., using E = mc2)? Ans: E = mc 2, thus m = E / c 2 = (J ) /( ) 2 (m 2 / s 2 ) = (kg) (This is impressively small, and a reason a star can survive billions of years.) g. Assuming a net solar energy input with an average in specific area of 200 W/m 2, how much area would be required to absorb this much energy in one hour? Ans: A = (J) / 200(W / m2 ) 3600(s) = 1,944.4(m 2 ) (roughly a football field for an hour). h. What volume of air would be required to have the space specific heat as our megalitre of water? Water=1000m 3 x1000kg/m3 x4200j/kgk=4.2x10 9 J/K Air=Vm 3 x 1200J/m 3 K = 4.2x10 9 J/K Thus, V = 4.2 x 10 9 /1200 = 3.5 x 10 6 m 3 (which is 2500 times greater in volume). m3 2. In PPS 1 we determined the Earth continually received 174 PW from the sun. Convert that quantity to (i) an equivalent amount of combustion of gasoline assuming 35 MJ/litre; (ii) to an equivalent rate of mass extinction using Einstein s famous equation E = mc 2. Ans: i) Energy received form the sun = 174 PW = W (J/s) Energy from combustion of 1 liter of gasoline: 35 MJ = Therefore the equivalent amount of required gasoline per second to be combusted is as follows: M = (J / s) / (J /lit) = L/s

3 ii) For the equivalent rate of mass extinction we have (E = J/s, c = m/s): m = E/c 2 = / ( ) 2 = kg (a pretty impressive contrast with the first part) 3. How much energy is associated with a cubic kilometer of saturated tropical air at 35 C compared to a cubic kilometer of saturated air at 15 C? Determine the fraction of this energy that is in both sensible and latent forms. Express your answer for the total energy in MJ and in liters of gasoline equivalent. What is the significance of this vast energy amount? Note that for a preliminary or ballpark estimate like this, you can assume for simplicity that the density of air in both volumes is about 1.2 kg/m 3, the volumetric specific heat of air is 1200 J/m 3.K, and that the latent heat of evaporation is 2.5 MJ/kg. Assume also that the specific humidity of saturated air at 35 C is 35 g of water per kg of air, and that the value for air at 15 C is 10 g of water per kg of air. Ans: Moisture content of air at 35 C = kgh2o/kg air Moisture content of air at 15 C = kgh2o/kg air Air volume = 1 X 10 9 m 3 Air mass = 1 X 10 9 (m 3 ) X 1.2 (kg/m 3 ) = 1.2 X 10 9 kgair Amount of excess water content in the saturated air compared to dry air = (1.2 x 10 9 kgair) x ( ) kgh2o/kgair = 30,000,000 kgh2o The latent heat corresponding to the excess water content = 30,000,000 kgh2o x 2.5 MJ/kg = 75,000,000 MJ = 75 X 10 6 MJ or 2.1 x 10 6 (lit of gasoline, using 35 MJ per liter of gasoline). The sensible heat corresponding to the excess temperature of the saturated air compared to the dry air = mc v ΔT and m=vxρ = c v x(1/ρ).v. T=1.2 (kj/m 3 K) x 1/1.2 (m 3 /kg) x 1.2X10 9 (kg) x 20 (K) = 2.4x10 10 kj = 2.4X10 7 MJ or 6.9x 10 5 (litre of gasoline, using 35 MJ per liter of gasoline). Total excess energy = 75 x 10 6 (MJ) X10 7 (MJ) = 99 X 10 6 MJ This is naturally a huge amount of energy, and is the basic fuel of the hurricane and tropical storm systems we have introduced in class. If you gathered the gasoline equivalent together, you could have quite the fire or explosion. This vast amount of energy has the capacity to release a great amount of heat during precipitation events, typically causing warm and wet weather, but also drawing in moist air, as in the explanation above, and causing more precipitation. With this much energy in the latent heat, there is the obvious likelihood of violent storms. When Toronto experiences a warm front in the winter, we are importing this kind of energy from places like the Gulf of Mexico. 4. Consider the heat, work and energy exchanges, and the material properties, associated with the following two events. Account for as many of the observations (or expected observations) as you can. a. It is a warm, calm summer morning and you are on vacation in Ontario. You walk beside a small, flat, shallow, calm pond of water. You pick up a warm rock from the side of the pond, a toss it with a splash into the middle of the pond. Consider first the rock and then the pond as your control system. Try to be as complete as possible, and consider the signs of the various heat, work and energy terms, using the convention we presented in class (i.e., taking heat transfer TO the control

4 system, and work done BY the control system, as positive). Assume the duration of this event is a minute or two. Ans: Rock as control system: 1) Sun transfers heat to the rock (+heat input, + energy exchange) 2) You do work on the rock ( - work, + kinetic energy increase for the rock) first to lift and then to throw the rock (using stored chemical energy in your body); almost certainly some heat transfer too from the warm rock to the relative cool hand; as soon as you pick it up, its radiant energy flux terms will shift from its original location and surroundings to its new one. 3) Rock does work on the air ( + work) as it flies through the air to the pond, and dynamically changes KE to gravitational PE as it moves vertically up and down. 4) Rock does work on the water ( + work) causing change in surface tension energy and displacing some of the surface which creates a splash (which creates a 2nd order set of waves and impacts), surface waves which propagate and gradually decay into heat, and acoustic energy which spreads out from the impact location as a sequence of sound waves 5) Rock transfers heat to the pond (-heat input, - energy exchange) 6) Rock loses gravitational potential energy, as it sinks in the pond ( - PE, + KE) until it hits the bottom of the pond and the rock does work on the sand at the bottom moving it out of the way (+work). Though all its motion fluid motion and turbulent eddies are formed (translational and rotational KE) which gradually decay to thermal form. Pond as control system: 1) Sun transfers heat to pond (+ heat input, + energy exchange) some if reflected in a complex light-water interaction process that depends on the angle and frequency of the light and state of the water surface 2) Some photons will transmit through the pond water to strike the bottom of the bond, where they will be absorbed and cause warming. 3) Some photons will likely be used by plants and algae to complete photosynthesis and the creation of organic high energy molecules like sugars, starches and cellulose 4) Air probably transfers heat to pond (+ heat input, + energy exchange) Rock does work on the water by moving it through the application of force (-work, + KE) 5) Almost certainly some wind-water coupling could occur that might generate some waves, the energy of which might part build up (accumulate) and partly decay 6) There will likely be an evaporative exchange, which will transfer some heat energy from the water and (eventually) to the atmosphere (at time of condensation) 5. During class we looked at how to calculate the rate of freezing of ice on a lake on a cold night using the Stefan formula. h = λ x Accumulated FDD where h = depth of ice built in cm, λ is the thermal conductivity of a solid and FDD is the freezing degree days. The winters are long and hard in Siberia. Lake Baikal is the world s largest fresh water lake by volume, and also the world s deepest.

5 Month Nov Dec Jan Feb Mar Apr Temperature (i). Assuming that the winter in Baikal lasts from November to the end of April, calculate the freezing degree days based on the above weather data: Ans. Freezing degree day = # days x temperature below freezing = 30x x x x x x5 = 2,892 FDD (ii). How deep will the ice be at the end of December? At the end of April? Assume that there is a skim of ice on Nov 1 and that the thermal conductivity of ice = Ans. H = 2.24 x 858 (858 = 10x X31) = 65 cm End of April, = 2.24 x 2,892 = cm (iii). Name three factors that may prevent this accumulation of ice? Snow coverage. Snow is an excellent insulator with a thermal conductivity that is about 10x less than that of the ice it sits on top of River flowing underneath (well not after mid- Dec once it freezes and the snow isn t melting). This will both mix the water thereby increase the water temperature under the ice i.e. bring up water between 1 and 4C from below and replace the water that is just at 0C. AND it can simply introduce more heat to the water if the river s temperature is >4C. Wind may prevent the skim from settling into proper ice. In fact Baikal doesn t even freeze until the beginning of January, likely due to the huge volume of warm water at the end of summer. (iv). The following equation deals with the rate of heat passing through a material of known thickness (L), area (A) and thermal conductivity (λ) for a given temperature difference of ΔT. See previous questions for the thermal conductivity of ice. Q = λaδt L

6 Calculate the heat loss rate through the ice at the end of December. Why isn t this in reality not the heat loss? Ans: Assume you have 1m 2 of surface area. Assume a temperature of - 25C from the weather data (ie it will be the same as the January average which is an estimation for the end of December condition. You could use another estimate and get a slightly different answer. Q = 2.24 x 1 x 25 / 0.65 Q = 86W The rate of loss will be lower than this as the air above the ice acts as an insulator, slowing down heat transfer like a blanket. Of course if you have snow that will further slow the heat loss for reasons already discussed. 6. What is the % compression of water at the bottom of the world s deepest lake, Lake Baikal (again!). The lake is 1,600m deep, the bulk modulus of water = 2.2 x 10 9 Pa. Ans: The bulk modulus of fresh water, K = - V dp/dv The trick is that the % compression is equal to dv/v Rearranging the equation, - dv/v =dp/k dp, change in pressure = ρgh = 1,000 x 9.81 x 1,600 (though in fact there will be a very minor change in g due to the depth change and indeed to the density of water, but we don t need to go to that level of detail) dp = 15,696 kpa so dv/v = 15,696,000 / 2.2 x 10 9 = - 0.7% or a 0.7% shrink.