Solving for cup mixing temperature

Question

Consider a pipe that has a length of 150m and an inner diameter of 2.5cm as well as a wall thickness of 0.8cm. The pipe transfers hot water (at 150°C)from a boiler to a storage tank at a volumetric flow rate of $7 \times 1 0^{- 4}$ m³/min. The temperature outside of the pipe is 10°C. The thermal conductivity (k) of the pipe is 0.12 W/m-K. The heat transfer coefficient for forced convection is 180 W/m²-K for flowing water, and the heat transfer coefficient for natural convection (for the air outside the pipe) is 15 W/m²-K. The heat capacity of water can be considered 4.184 kJ/kg-K and the density of water is 1000 kg/m³. What is the cup mixing temperature of water as it enters the storage tank?

Answer

How do you know how to solve this question? From the given information, one can deduce that he/she will be dealing with heat transfer in this problem. The problem is asking you to solve for a change in temperature which is the driving force for heat transfer. The given information also includes: heat capacity, heat transfer coefficients, and thermal conductivity. Cup mixing temperature can be described as the bulk temperature of a moving fluid.The problem states that you are solving for the temperature entering the tank which is essentially the same thing as solving for the temperature leaving the pipe that connects the boiler and the tank. Therefore, the objective of this problem is to solve for T_out.

The general equation one would need to solve for heat transfer is:

$Δ_{T} \times C_{p} \times \dot{M} = \dot{Q}$

$Δ_{T}$ is the change in temperature from the inlet of the pipe to the outlet of the pipe.

$C_{p}$ is the heat capacity of water

$\dot{Q}$ is the heat transfer in Watts

Note that $\dot{M}$ is mass flow rate, and the given information includes a volumetric flow rate (not a mass flow rate). Therefore, you must multiply the volumetric flow rate by density in order to get mass flow rate.

$\dot{M}$ = $7 \times 1 0^{- 4}$ m³/min * (1min/60s) * 1000 (kg/m^3)

$\dot{M}$ = (7/600) kg/s

Heat transfer leaving or entering the water can be described as negative or positive depending on the convention you are used to. However, regardless of which convention you decide to use, the heat transfer balance should look like equation 1 below. In this problem, heat is leaving the system so either the surroundings or the system will have a negative rate of heat transfer. A common mistake for several daring chemical engineers is their neglect of that very important negative sign in equation 1.

$Q_{(} w a t e r) = \dot{M} \times C_{p} (T_{o u t}) - T_{i n})) = - U A Δ_{T L o g M e a n}$

(1)

The log mean temperature difference can be written as follows:

Δ T_{l m} = \frac{(T_{i n} - T_{i n f i n i t y}) - (T_{o u t} - T_{i n f i n i t y})}{\ln (\frac{T_{i n} - T_{i n f i n i t y}}{T_{o u t} - T_{i n f i n i t y}})}

(2)

The problem states that the ambient temperature outside the pipe is 10°C so one can refer to this temperature as T_{infinity} because we assume that the ambient temperature is going to remain a constant 10°C all around the pipe. Note that in the Delta T log mean term, the -T infinity cancels with the +T infinity in the numerator. The numerator of the Delta T log mean term then becomes ( Tout-Tin) Notice that the Delta T on the right side of equation 1 is also (Tout-Tin). As a result, one can divide both sides by (Tout-Tin) to reduce the equation to only have 1 unknown (the Tout in the ln term).

The equation then reduces to:

$\ln (\frac{T_{o u t} - T_{i n f i n i t y}}{T_{i n} - T_{i n f i n i t y}}) = \frac{- U A}{{\dot{M}}_{w a t e r} C p_{w a t e r}}$

(4)

However, before solving for Tout directly, you need to solve for the -UA term. U is the overall heat transfer coefficient.

The equation for the UA term can be defined as follows:

U A = {(\frac{1}{h_{o u t s i d e} A_{o u t s i d e}} + \frac{R_{o} - R_{i}}{k_{p w} A_{l m}} + \frac{1}{h_{i n s i d e} A_{i n s i d e}})}^{- 1}

(5)

However before you solve for the UA term, you need to define all the terms within equation 5. The Area log mean equation then becomes necessary to solve this problem because the area is changing from the inside to to the outside of the pipe. It is important to note that the area associated with heat transfer is not the cross sectional area of a cylinder, rather it is the surface area of the cylinder because heat is leaving or entering the system radially. In this case, heat is leaving the hot fluid in the radial direction because the outside temperature will cool the hot fluid that is flowing in the pipe.

This is the equation for the area log mean term in our UA equation ( equation 5). When you are told you can neglect the heat transfer across the pipe wall, you can neglect this term. Or if you assume that the pipe wall is thin you can also neglect the change in radius term in Equation 5.

$Δ A_{L} M = \frac{A r e a_{o u t s i d e} - A r e a_{i n s i d e}}{l n (\frac{A r e a_{o u t s i d e}}{A r e a_{i n s i d e}})}$

The area equations for the inside and outside of the pipe are as follows:

$A r e a = Π \times D \times L$

Simply plug in your values for the inner and outer diameter and keep in mind that the length of the pipe will remain the system for both inside and outside of the pipe.

$A r e a_{o u t} = Π \times D_{o u t} \times L$

Area out= (3.14) * (4.1/100)m * 150m Area out= 11.781 m²

but how did the outer diamete become 4.1 cm? The problem statement states that the inner diameter is 2.5cm and that the wall thickness is 0.8cm.In order to calculate the outer diameter one has to account for the wall above and below the inner diameter. Therefore, the outer diameter becomes 2.5cm+ (0.8cm*2) which equals 4.1cm.

$A r e a_{i n} = Π \times D_{i n} \times L$

Area in= (3.14) * (2.5/100)m * 150m Area in= 11.781 m²

Now that Area in and Area out are defined, one can plug both of these known values into the Area log mean equation

Area log mean= 15.244m²

Note that the third term in the UA equation includes a Δr term. This part of the UA equation is the resistance across the pipe wall, and so that Δr refers to the wall thickness (0.8/100 m) .

Inside the pipe, heat transfer occurs by forced convection so the heat transfer coefficient inside is 180 W/m²K Outside the pipe, heat transfer occurs by natural convection so the heat transfer coefficient outside is 15 W/m²K

Now that all the values for Equation 5 have been defined, one can calculate the value of -UA in Equation 4. As a result, -UA=-120.538 W/K

After Observing equation 4, one should be able to deduce that all the variables are known except the Temperature coming out of the pipe.

-UA=-120.538 W/K

$\dot{M}$ = (7/600) kg/s

The heat capacity of water is 4.184 *1000 J/kg-K

The inlet temperature of the water is 150°C

The ambient temperature (T_ infinity) 10°C

Isolate for Tout in Equation 4 and plug in all known values. To simplify the solution for Temperature out, put in the base number e to both sides to cancel the ln term on the left.

Once you have solved for T out, Tout=22°C you can solve for the bulk temperature of the water. The bulk temperature of the fluid is an arithmetic average of the temperature in and the temperature out of a fluid in a pipe. Therefore, Tbulk= (Tin+Tout)/2 Tbulk=((150+22)/2) °C