Question

Saturated propane vapor at 200 psia is fed to a well-insulated heat exchanger at a rate of 2600 standard cubic feet per hour. The propane leaves the exchanger as a saturated liquid (a liquid at its boiling point) at the same pressure. Cooling water enters the exchanger at 70 oF. Flowing concurrently (in the same direction) with the propane. The temperature difference between the outlet streams (liquid propane and water) is 15 oF.

Estimate the required flow rate (lbm/h) of the water to take away the heat from exchanger. (you will need to write two separate energy balances.) Assume the heat capacity of liquid water is constant at 1.0 Btu/(lbm.oF) and neglect heat losses to the outside and the effects of pressure on the heat of evaporation of propane.

Answer 1

solution:

In the given problem, water is a hot fluid and propane is a cold fluid.

Now, inlet temperature of cold fluid, Tc_i = Temperature of Propane at 200 psia = 95⁰F (from properties of Propane)

Outlet temperature of cold fluid, Tc_o= Boiling point of Propane = - 44⁰F

Inlet temperature of hot fluid, Th_i= Inlet temperature of water = 70⁰F

Now, given type of heat exchanger is parallel flow heat exchanger,

Hence, Temperature difference at outlet = Th_o - Tc_o= 15 ⁰F

theefore, Th_o = 15 + Tc_o = 15 + (-44) = -29 ^oF

Heat capacity rate of water, Cp_h= 1 Btu/(lbm.⁰F)

Flow rate of cold fluid, = 2600 ft³/hr

From references, density of cold fluid (propane) = 31.12 lb/ft³

Therefore, mass flow rate of cold fluid, m_c = 2600 * 31.12 = 80912 lb/hr

Also, from references, specific heat of cold fluid, Cp_c = 0.39 Btu/(lbm.⁰F)

Appying energy balance to cold fluid,

Q = m_c * Cp_c * ( Tc_i - Tc_o ) = 80912  * 0.39* ( 95 - (-44)) = 4386239.52 Btu / hr

Applying energy balance to hot fluid,

4386239.52 Btu / hr = m_{h *} 1 _* ( 70 - ( -29))

m_h = 44305.4497 lb/ hr

hey, if you find nay doubt please ask and please give a thumbs up. thanks