Concentric Pipe Heat Exchange
Energy Balance on Cold Stream (differential)
dQ_{C} = (wC_{p})_{c} dT_{C }= C_{C}dT_{c}
Energy Balance on Hot Stream (differential)
dQ_{H} = (wC_{p })_{H} dT_{H} = C_{H}dT_{H}
Overall Energy Balance (differential)
For an adiabatic heat exchanger, the energy lost to the surroundings is zero so what is lost by one stream is gathered by the other.
dQ_{C} + dQ_{H} = 0
Heat Exchange Equation It follows that the heat exchange from the hot to the cold is expressed in terms of the temperature difference between the two streams
dQ_{H }= U (T_{H }– T_{C} )d_{A}
The proportionality constant is the “Overall” heat transfer coefficient ( discussion later)
Solution of the Energy Balances
The Energy Balance on the two streams provides a delation for the differential temperature change.
However, we should recall that we have an adiabatic heat exchanger so that
Overall Energy balances on each stream
Hot Fluid
Q_{H} = C_{H}( T_{H1} – T_{H2})
Cold fluid
Q_{C} = C_{C} (T_{C2} – T_{C1})
Overall Energy balance on the Exchanger
Q_{C} + Q_{H} = 0
The equation for DT can be modified using the overall energy balances to yield
The denominator is the energy lost by the hot stream, so
.
Application of the relation for energy transfer between the two streams yields
Integration of the relation is the basis of a design equation for a heat exchanger.
Rearrangement of the equation leads to
The Design Equation for a Heat Exchanger
Design of a Parallel Tube Heat Exchanger
The Exchanger
The Design Equation for a Heat Exchanger
Glycerinwater solution with a Pr = 50 (at 70 °C) flows through a set of parallel tubes that are plumbed between common headers. We must heat this liquid from 20 °C to 60°C with a uniform wall temperature of 100 °C. The flow rate, F, is 0.002 m^{3} /sec (31.6 gal/sec.).
Data
The heat capacity, Cp , is 4.2 kJ/kg°K
The density, r, is 1100 kg/m^{3}
The liquid has a kinematic viscosity, n = 10^{3} cm^{2} /sec.
Step 1
Calculate the heat load
Step 2
Calculate the heat transfer coefficient If the flow is laminar, likely since glycerin is quite viscous, and the Re < 2000 the Nusselt number relation for laminar flow can be expressed as
The Graetz number is
If the flow is turbulent (Re > 2000), the Nusselt numberr is given by
We do not know the flow per tube and therefore we do not know the Re. However we don’t need to know that. In Lecture 27 we observed for Heat Transfer in a Tube that
The definition of the Stanton Number is :
Given a Re and Pr, we can calculate the Nu and the Stanton Number, the latter prviding us with the temperature at length L from the previous equation. Let’s examine several configurations at L/D = 50, 100, 200. The Excel table below can be used to specify a design chart.
To obtain the numbers in the spreadsheet, we used the Nusselt number relation for laminar flow expressed as
and for turbulent flow as
Nu = 0.023Re^{0.8}Pr^{0.4}
Step 3
Calculate the Area required Base case D = 2 cm. and L = 100 D = 2 meters For this case we observe that from the calculations for θcm
We can observe that the flow rate per tube is given by
F_{nt} = F/ n_{t}
so that the Reynolds’ number is
As a consequence we can observe that the total length of tubing is not dependent on D alone but on othere considerations that might set a condition for Re, e.g. a pressure drop limitation. Wv find that for this base case, we find
We find that θcm = 0.5
Does it make sense?
Maximum Cooling Capacity of an Exchanger of Fixed Area
Water is available for use as a coolant for an oil stream in a doublepipe heat exchanger.
The flow rate of the water is 500 lb_{m}/hr.
The heat exchanger has an area of 15 ft^{2} .
The oil heat capacity, Cpo, is 0.5 BTU/lb°F
The overall heat transfer coefficient, U, is 50 BTU/hrft^{2} °F
The initial temperature of the water, T_{w0}, is 100°F
The maximum temperature of the water is 210°F
The initial temperature of the oil, T_{w0}, is 250°F
The minimum temperature of the oil, T_{w0}, is 140°F
Estimate the maximum flow rate of oil that may be cooled assuming a fixed flow rate of water at 500 lbm/hr There are two possible modes of operation Cocurrent flow Countercurrent flow Let us look at both cases
Cocurrent flow
Constraints
T_{w} < 210 ; T_{w} < T_{o} ; T_{o} ≥ 140
Energy balances Oil
Q_{o} = F_{o}C_{po} (T_{o1} – T_{o2} )= F_{o} (0.5 ) (250 – T_{o2})
Water
Q_{w} = F_{w}C_{pw} (T_{w1} – T_{w2})
F_{o}C_{p0} (T_{o1} – T_{o2} )= 500(1.0)(210 – 100) = 55,000 BTU / hr
Recall the Design equation
Now the ΔT_{lm} is given by
Using the temperatures, we obtain T_{0max} = 238.5 °F and from the heat balance for oil, we obtain
Countercurrent Flow
Constraints
T_{w} < 210 ; T_{w} < T_{o} ; T_{o} ≥ 140
Energy balances Oil
Q_{o} = F_{o}C_{po} (T_{o1} – T_{o2} )= F_{o} (0.5 ) (250 – T_{o2})
Water
Q_{w} = F_{w}C_{pw} (T_{w1} – T_{w2})
F_{o}C_{p0} (T_{o1} – T_{o2} )= 500(1.0)(210 – 100) = 55,000 BTU / hr
F_{o}C_{p0} (T_{o1} – T_{o2} )= 500(1.0)(210 – 100) = 55,000 BTU / hr
Recall the Design equation
Now the ΔT_{lm} is given by
Using the temperatures, we obtain T_{0max} = 221 °F and from the heat balance for oil, we obtain the oil flow rate as 3800 lbm/hr.
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